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1) is revolved about the y-axis. (a) Show that sin# = . (b) Find lim V. R +h (b) Find 9, to the nearest degree, for a satellite that is 10,000 km from the Earth's surface (use R = 6378 km). 75-76 Find the average value of the function over the given interval. ? where
75. fCO =
77. Use the Mean-Value Theorem to prove that x < tan x < x (x > 0) 1 "y" A
Earth
Figure Ex-66
67. An airplane is flying at a constant height of 3000 ft above water at a speed of 400 ft/s. The pilot is to release a survival package so that it lands in the water at a sighted point P. If air resistance is neglected, then the package will follow a parabolic trajectory whose equation relative to the coordinate system in the accompanying figure is =
3000 -
78. Find a positive value of k such that the average value of /CO = l/(k2 + x2) over the interval [—k, k] is it. 79-81 Solve the initial-value problems. 3
dx
472
Chapter 6 / Exponential, Logarithmic, and Inverse Trigonometric Functions
V I I — 81 = 9t2' 3J~30 ' dt 82. Evaluate the limit by interpreting it as a Riemann sum in which the interval [0, 1] is divided into n subintervals of equal width: « M lim V^ -
83. Prove: i^a^\ sin ' —' \( x )\ — _ — sin .• —i x ,'!•. ,_ i/ \ i \o) tan \—x) = —tan x. 84. Prove: (a) cos~'(-*) = jr-cos" 1 x (b) sec"1 (—x) = 7i — sec"1 x. 85. Prove: (a) sin"1 x = tan"1
(b) cos
87. Use the result in Exercise 86 to show that (a) tan"1 \ + tan"1 | = jr/4 (b) 2 tan"1 i+tan" 1 ± = ;r/4. 88. Use identities (4) and (7) to obtain identity (11). 89. Writing Suppose that / is a nonconstant function that is twice-differentiable everywhere. Is it always possible to restrict the domain of / to an open interval so that the restricted function has an inverse? Justify your answer by appealing to appropriate theorems. 90. Writing Let 9 = tan"1 (-3/4) and explain why the triangle in the accompanying figure may be used to evaluate sin 9 and cos#. More generally, suppose that q denotes a rational number and that 9 = tan"1 q or that 9 = sin"1 q. Find right triangles, with at least two sides labeled by integers, that may be used to evaluate sin 9 and cos 9.
x =—
86. Prove:
-3
tan
x + tan
y = tan
provided — jt/2 < tan ' x + tan ' y < n/2. [Hint: Use an identity for tan (a + /6).]
•QUICK CHECK ANSWERS
•^ Figure Ex-90
6.7
1. (a) -it 12 (b) 7T/4 (c) 7T/3 (d) 7i/3 (e) 2^/3
2. (a) nil (b) 2n/l (c) jr/6 (d) 2jr/7
3.
4.
HYPERBOLIC FUNCTIONS AND HANGING CABLES In this section we will study certain combinations ofex and e x, called "hyperbolic functions." These functions, which arise in various engineering applications, have many properties in common with the trigonometric functions. This similarity is somewhat surprising, since there is little on the surface to suggest that there should be any relationship between exponential and trigonometric functions. This is because the relationship occurs within the context of complex numbers, a topic which we will leave for more advanced courses. DEFINITIONS OF HYPERBOLIC FUNCTIONS
To introduce the hyperbolic functions, observe from Exercise 61 in Section 0.2 that the function ex can be expressed in the following way as the sum of an even function and an odd function: ex ex = 2 —-v—
Even
Odd
These functions are sufficiently important that there are names and notation associated with them: the odd function is called the hyperbolic sine of x and the even function is called the hyperbolic cosine of x. They are denoted by sinhjc =
-e
and
cosh* =
6.8 Hyperbolic Functions and Hanging Cables
473
where sinh is pronounced "cinch" and cosh rhymes with "gosh." From these two building blocks we can create four more functions to produce the following set of six hyperbolic functions.
6.8.1
DEFINITION
ex - e~x
Hyperbolic sine The terms "tanh," "sech," and "csch" are pronounced "tanch," "seech," and "coseech," respectively.
ex + e-x 2 sinh x coshx coshx sinhjc
Hyperbolic cosine Hyperbolic tangent Hyperbolic cotangent Hyperbolic secant
- e~x + e~x + e~x - e~x 2 x e + e-x 1 x e — e~x ex ex ex ex
coshjc
Hyperbolic cosecant
sinhjc
TECHNOLOGY MASTERY Computer algebra systems have builtin capabilities for evaluating hyperbolic functions directly, but some calculators do not. However, if you need to evaluate a hyperbolic function on a calculator, you can do so by expressing it in terms of exponential functions, as in Example 1.
Example 1 sinh 0 = cosh 0 = sinh 2 =
e° - e~°
1-1
e° + e~°
1+ 1
e2 - e-2
=1
^ 3.6269
GRAPHS OF THE HYPERBOLIC FUNCTIONS
The graphs of the hyperbolic functions, which are shown in Figure 6.8.1, can be generated with a graphing utility, but it is worthwhile to observe that the general shape of the graph of y = cosh x can be obtained by sketching the graphs of y = \ex andy — \e~x separately and adding the corresponding ^-coordinates [see part (a) of the figure]. Similarly, the general shape of the graph of y = sinh x can be obtained by sketching the graphs of y — \ex and y = —\e~x separately and adding corresponding y -coordinates [see part (b) of the figure]. Observe that sinh x has a domain of (—oo, +00) and a range of (—00, +00), whereas cosh x hasadomainof (—oo, +00) and a range of [1, +0°). Observe also that y — ^ex andy — ^e~x are curvilinear asymptotes for y = cosh x in the sense that the graph of y = cosh x gets closer and closer to the graph of y = ^ex as x —> +00 and gets closer and closer to the graph of y = ^e~x as x -> -oo. (See Section 3.3.) Similarly, y — \e* is a curvilinear asymptote for y — sinh x as x ->• +00 and y = ~^e~x is a curvilinear asymptote as x ->• -oo. Other properties of the hyperbolic functions are explored in the exercises. Glen Allison/Stone/Getty Images The design of the Gateway Arch near St. Louis is based on an inverted hyperbolic cosine curve (Exercise 73).
HANGING CABLES AND OTHER APPLICATIONS
Hyperbolic functions arise in vibratory motions inside elastic solids and more generally in many problems where mechanical energy is gradually absorbed by a surrounding medium. They also occur when a homogeneous, flexible cable is suspended between two points, as with a telephone line hanging between two poles. Such a cable forms a curve, called a catenary (from the Latin catena, meaning "chain"). If, as in Figure 6.8.2, a coordinate
474
Chapter 6 / Exponential, Logarithmic, and Inverse Trigonometric Functions
y — tanh —I x I (c)
-\
y = coth x (d)
(e)
A Figure 6.8.1
= a cosh (x/a) + c
A Figure 6.8.2
system is introduced so that the low point of the cable lies on the y-axis, then it can be shown using principles of physics that the cable has an equation of the form y = a cosh ( — ) + c \a' where the parameters a and c are determined by the distance between the poles and the composition of the cable. HYPERBOLIC IDENTITIES
The hyperbolic functions satisfy various identities that are similar to identities for trigonometric functions. The most fundamental of these is cosh2 x — sinh2 x — I
(1)
which can be proved by writing cosh * — sinh * = (cosh* + sinh*)(cosh* — sinh*) ex + e~x =
x e
e* - e~
- e~
• e~* = 1
Other hyperbolic identities can be derived in a similar manner or, alternatively, by performing algebraic operations on known identities. For example, if we divide (1) by cosh2 *, we obtain 2 2 1 — tanh * = sech * Larry Auippy/Mira.com/Digital Railroad, Inc.
.,, cable ,, suspended , ,,between two A.aflexible poles forms a catenary.
-.„
i«
• i
^« \ i
• i 2
and if we divide (1) by J sinh *, we obtain coth * — 1 = csch *
6.8 Hyperbolic Functions and Hanging Cables 475
The following theorem summarizes some of the more useful hyperbolic identities. The proofs of those not already obtained are left as exercises.
6.8.2
THEOREM
cosh x + sinh x = ex
sinh()c + v) = sinh x cosh y + cosh x sinh y
cosh jt — sinhjc = e~
x
cosh(jc + y) = cosh x cosh y + sinh x sinh y
cosh2 x — sinh2 x = 1
sinh(jc — y) — sinh x cosh y - cosh x sinh y
2
2
1 — tanh x = sech x
cosh(jc — y) = cosh x cosh y — sinh x sinh y
com2 x — 1 = csch2 x
sinh 2x = 2 sinh x cosh x
cosh(—x) = cosh A:
cosh 2x = cosh2 x + sinh2 x
sinh(—x) = — sinhjc
cosh 2x = 2 sinh2 x + 1 = 2 cosh2 JE — 1
WHY THEY ARE CALLED HYPERBOLIC FUNCTIONS
Recall that the parametric equations (cos t, sin r)
x = cos t, 2
y — sin t
(0 < t < 2n)
2
represent the unit circle x + y = 1 (Figure 6.8.3a), as may be seen by writing x2 + y2 = cos 2 1 + sin 2 1 — 1 If 0 < t < 2n, then the parameter t can be interpreted as the angle in radians from the positive .x-axis to the point (cosf, sinf) or, alternatively, as twice the shaded area of the sector in Figure 6.8.3a (verify). Analogously, the parametric equations x = cosh?,
y = sinhf 2
(-co < t < +00)
2
represent a portion of the curve x — y = 1, as may be seen by writing x1 - y2 = cosh 2 1 — sinh2 t = 1 (cosh t, sinh ()
and observing that x = cosh t > 0. This curve, which is shown in Figure 6.8.3b, is the right half of a larger curve called the unit hyperbola', this is the reason why the functions in this section are called hyperbolic functions. It can be shown that if t > 0, then the parameter t can be interpreted as twice the shaded area in Figure 6.8.3£>. (We omit the details.) DERIVATIVE AND INTEGRAL FORMULAS
Derivative formulas for sinh x and cosh x can be obtained by expressing these functions in terms of ex and e~x: A Figure 6.8.3
— [sinh*] = — dx dx
= cosh x
e* — e
]
= sinhjt — [coshx] = — =— 2 J 2 a* ax Derivatives of the remaining hyperbolic functions can be obtained by expressing them in terms of sinh and cosh and applying appropriate identities. For example,
d — dx
d_ r s i n h ^ l
cosh x —[sinh x] - sinh * — [cosh x]
dx [coshzj
cosh2 A:
cosh x — sinh x 2
cosh x
1 cosh2 *
= sech2 x
476
Chapter 6 / Exponential, Logarithmic, and Inverse Trigonometric Functions
The following theorem provides a complete list of the generalized derivative formulas and corresponding integration formulas for the hyperbolic functions.
6.8.3
THEOREM
d du [sinh u] — cosh u ax ax
d
f cosh u du — sinh u + C
du
— [cosh u] — sinh u — ax dx d du 7 — [tanhw] = sech u — dx dx d 2 du — [cothMj — —csch u — dx dx
f sinh u du = cosh u + C I sech2 u du = tanh u + C csch2 u du = — coth u + C
d
du
d
du
— [sech u\ = —sech u tanh u — dx dx — [csch u] = —csch u coth u — dx dx
/ sech M tanh u du = —sech u + C / csch u coth udu = —csch u + C
Example 2 — [cosh(*3)] = sinh(*3) • — [*3] = 3*2 sinh(*3) dx dx d I d sech2* — [In(tanh*)] = —— - — [tanh*] = —- < dx tanh* dx tanh*
Example 3 sinh5 * cosh * dx — | sinh6 * + C
/
f sisinh* tanh * dx = I — dx cosh* J cc = ln|cosh*| + C
« = cosh x \ du = sinh x dx
— ln(cosh*) + C We were justified in dropping the absolute value signs since cosh * > 0 for all *.
> Example 4 A 100 ft wire is attached at its ends to the tops of two 50 ft poles that are positioned 90 ft apart. How high above the ground is the middle of the wire? Solution.
From above, the wire forms a catenary curve with equation y = a cosh ( — ) + c \a/
6.8 Hyperbolic Functions and Hanging Cables
477
where the origin is on the ground midway between the poles. Using Formula (4) of Section 5.4 for the length of the catenary, we have 45
100 = -45 45
By symmetry about the y-axis
dx 45
' 1 + sinh2 (-} dx \a/
C^ /x\ = 1 \ cosh I - I dx Jo
W
| By (1) and the fact , that cosh x. > 0
/*\f5 /45 = la sinh - I = la sinh — WJ0 \a Using a calculating utility's numeric solver to solve 45\
100 = la sinh —
a )
for a gives a ^ 56.01. Then -45
> = 56.01 cosh(-^-r)-25.08 \56.L)1/ A Figure 6.8.4
45 75.08 56.01 so c « -25.08. Thus, the middle of the wire is y(0) % 56.01 - 25.08 = 30.93 ft above the ground (Figure 6.8.4). •< 50 = j(45) = 56.01 cosh
INVERSES OF HYPERBOLIC FUNCTIONS
Referring to Figure 6.8.1, it is evident that the graphs of sinh x, tanh x, cothx, and csch x pass the horizontal line test, but the graphs of cosh x and sech x do not. In the latter case, restricting x to be nonnegative makes the functions invertible (Figure 6.8.5). The graphs of the six inverse hyperbolic functions in Figure 6.8.6 were obtained by reflecting the graphs of the hyperbolic functions (with the appropriate restrictions) about the line y — x. Table 6.8.1 summarizes the basic properties of the inverse hyperbolic functions. You should confirm that the domains and ranges listed in this table agree with the graphs in Figure 6.8.6. With the restriction that x > 0, , the curves y = coshjt and y = sech x pass the horizontal ; line test. A Figure 6.8.5
LOGARITHMIC FORMS OF INVERSE HYPERBOLIC FUNCTIONS
Because the hyperbolic functions are expressible in terms of ex, it should not be surprising that the inverse hyperbolic functions are expressible in terms of natural logarithms; the next theorem shows that this is so. 6.8.4 THEOREM The following relationships hold for all x in the domains of the stated inverse hyperbolic functions: sinh
x = \n(x + V x 2 + 1) l
tanh -i x = -1,In i + 2 \l-x sech ' x = In
cosh
x = ln(x
x
coth ' x = - In
JC+1
x- 1
478
Chapter 6 / Exponential, Logarithmic, and Inverse Trigonometric Functions
y = sech ' x
> Figure 6.8.6
Table 6.8.1 PROPERTIES OF INVERSE HYPERBOLIC FUNCTIONS FUNCTION
DOMAIN
(—00,
+00)
RANGE
(—00,
+00)
cosh
[l.+oo)
tanh"1 x
(-1, 1)
, _, coth ' x
(-00, -1) U (1, +<=o)
,«.,,« (-00, 0) U (0, +00)
sech lx
(0,1]
[0, +00)
(-00, 0) U (0, +00)
(-00, 0) U (0, +00)
[0, +00)
(—00,
+00)
BASIC RELATIONSHIPS
sinh '(sinh*) = x sinh(sinh~' x) = x
if -<x><x<+° if -oo < x < +°
cosh ' (cosh x) — x if x > 0 cosh(cosh~' x) = x if x > 1 tanh '(tanh x) = x tanh(tanh~' x) = x
if - o o < ^ < + if -1 < x < 1
coth '(coth*) = jt if * < 0 or* > 0 coth(coth x) = x if x < -1 or x > 1 sech (sech x) = x if * > 0 sech(sech = x if 0<x < csch '(csch jc) = * if * < 0 or x > 0 ' csch(csch ' jt) = x if x < 0 or x > 0
We will show how to derive the first formula in this theorem and leave the rest as exercises. The basic idea is to write the equation x = sinh y in terms of exponential functions and solve this equation for y as a function of x. This will produce the equation y = sinh"1 x with sinh"1 x expressed in terms of natural logarithms. Expressing x = sinh y in terms of exponentials yields ey - e~y x = sinh y =
6.8 Hyperbolic Functions and Hanging Cables
479
which can be rewritten as ey - 2x - e~y = 0
Multiplying this equation through by ey we obtain e2y - 2xey - 1 = 0 and applying the quadratic formula yields
Since ey > 0, the solution involving the minus sign is extraneous and must be discarded. Thus, , ey — x + V* 2 + I Taking natural logarithms yields
y — ln(x + V*2 + 1) or sinrT1 x = \n(x + Jx2 + 1) Example 5
sinhr1 1 = ln(l
= ln(l + V2) % 0.8814
tanrrM- = -ln
) = I In 3 % 0.5493 +
DERIVATIVES AND INTEGRALS INVOLVING INVERSE HYPERBOLIC FUNCTIONS
Formulas for the derivatives of the inverse hyperbolic functions can be obtained from show that the derivative of the function sinh x can also be obtained by letting^ = sinh"1 x and then differentiating* = sinh y implicitly.
Theorem 6.8.4. For example, ^ j -—[sinh-1 x] = —[ln(jc +
dx
1+
1 )] =
dx
C2+1
+ 1+ x
This computation leads to two integral formulas, a formula that involves sinh equivalent formula that involves logarithms:
dx
x and an
= sinh"1 x + C = \n(x + V*2 + 1) + C
2
/JC +1
The following two theorems list the generalized derivative formulas and corresponding integration formulas for the inverse hyperbolic functions. Some of the proofs appear as exercises.
6.8.5 THEOREM
d
1 U) —
dx d
dx
/^.«fU~i ,,\ dx
dx
V V+l^
ll\ (cosrT1 U)
/t n n h~i dx
du
r
,.\
—
1
1
dx
'
du
dx
du
9 2 J
IMI > 1
I
du
r^dx V
- 1— u
1 du i1 — ui2 jdx >
M
t <
J
d dx
_j
1
du
\u\Jl+u
—, 2dx
a -/- n
U f. U
480
Chapter 6 / Exponential, Logarithmic, and Inverse Trigonometric Functions 6.8.6
THEOREM
If a > 0, then
I , dU = sintT1 (-) + C or ln(w + VV + a2) + C J Va2 + u2 ^a) I : = cosh"1 (-} + C or ln(M + Ju2 - a2) + C, w > a Vfl/ •/ Ju2 - a2
r
du J a — u2
a 1 — coth a
2
/
\a
du 1 —, =— 2 2 a u^/a - u
+ C or - - l n f a +
f du 1 / — =— 2 2 •J Mx/a + w a
Example 6 Solution.
a +u or — In 2a a —u
u
,
— | + C , 0 < |u|
+ C, u £ 0
+ C or - - In a
dx
3 , x > -. 2 /4x - 9
Evaluate /
2
Let M = 2*. Thus, du = 2dx and
/•
l_ f
dx
J ^4x2 - 9
2dx
2 j ^4^2 _ 9
1. { __d 2 j ^2 _ 32
Alternatively, we can use the logarithmic equivalent of cosh"1 (2jc/3), cosh'1 —
= ln(2;c +
9 ) - In 3
(verify), and express the answer as
dx
1
2
x -9
I/QUICK CHECK EXERCISES 6.8
1.
(See page 483 for answers.)
sinh x =
represent the right half of the curve called a inating the parameter, the equation of this curve is.
2. Complete the table.
4. —[cosh*] = dx
cosh* sinh* tanh* coth* sech* csch* DOMAIN RANGE
3. The parametric equations x = coshf,
y = sinhf
(—00 < t < +00)
— [tanhx] =. dx
— [sinh*] = . dx
Elim-
6.8 Hyperbolic Functions and Hanging Cables 5
I rnsh r SJTC —
6
/ sinh r HTC —
dx d
/ tflfllh X
.
_,
_j
dx
EXERCISE SET 6.8
0 Graphing Utility
1-2 Approximate the expression to four decimal places. :S
V tanh x sech x dx
1. (a) sinh 3 (d) sinh"'(-2)
(b) cosh(-2) (e) cosh"1 3
(c) tanh(ln4) (f ) tanh"1 |
33. / tanh2xdx
2. (a) csch(-l) (d) sech"1^
(b) sech(m2) (e) coth"1 3
(c) cothl (f) csch"!(— V3)
35. /
5. In each part, a value for one of the hyperbolic functions is given at an unspecified positive number XQ. Use appropriate identities to find the exact values of the remaining five hyperbolic functions at^o(a) sinh XQ = 2 (b) cosh XQ — | (c) tanh #0 = 5 6. Obtain the derivative formulas for csch x, sech x, and coth x from the derivative formulas for sinh x, cosh x, and tanh x. 7. Find the derivatives of cosh"1 x and tanh"1 x by differentiating the formulas in Theorem 6.8.4. 8. Find the derivatives of sinh"1 x, cosh"1 x, and tanh"1 x by differentiating the equations x = sinh y, x = cosh y, and x = tanh y implicitly. m
11. y = coth (In x)
17. y = jc3 tanh2(00
18. y = sinh(cos3jc)
19. y = sinh"1 (~x)
20. y = sinh~1(l/x)
21. y = In (cosh"1 x)
22. y = cosh"1 (sinh"1 x)
23. y =
24. y = (coth-1 x)2
27. y = e
41
f I
/
/ Jo
i1 — xv2
13 r
dX
dx
(x < 0)
40
r
( sin 6 dO / J Vl+cos26» dx
f
t^. 1
44 1[^ Jo
,
-J^^
((JiT ->5/3) Jl J)
d / '
v? 2 + 1
45-48 True-False Determi ne whether the statement is true or false. Explain your answer. 45. The equation cosh x = sinh x has no solutions. 46. Exactly two of the hyperbolic functions are bounded. 47. There is exactly one hyperbolic function f ( x ) such that for all real numbers a, the equation f ( x ) = a has a unique solution x. 48. The identities in Theorem 6.8.2 may be obtained from the corresponding trigonometric identities by replacing each trigonometric function with its hyperbolic analogue.
51. Find the volume of the solid that is generated when the region enclosed by y = cosh2jc, y = sinh 2x, x = 0, and x = 5 is revolved about the x-axis.
26. y = si 28. y =
29-44 Evaluate the integrals. >M sinh x cosh x dx
/
(r ^
/
50. Find the volume of the solid that is generated when the region enclosed by y = sech x, y = 0, x = 0, and x = In 2 is revolved about the jc-axis.
16. y = sinh3(2x)
x
X
39
IS
49. Find the area enclosed by y = sinh 2x, y = 0, and x = In 3.
15. y = \/4x + cosh (5x)
tan 25. y = cosh"1 (cosh x)
dX 1f / J Vl+9x2
e* + e~x
'' Jo
12. y = In (tanh 2x) 14. y = sech(e *) 2
/•ln3gJ:_e-x
tanh x sech3 x dx
10. y = coshO4) 2
13. y = csch(l/x)
34. / coth x csch x dx
/•In 3
37
4. In each part, rewrite the expression as a ratio of polynomials, (a) cosh(lnx) (b) sinh (In x) (c) tanh (2 In x) (d) cosh(-lnjt)
9. y = sinh(4x - 8)
32. /f csch2 (3^:) dx
J\n2
3. Find the exact numerical value of each expression. (a) smh(ln3) (b) cosh(— In 2) (c) tanh (2 In 5) (d) sinh(-31n2)
9-28 Finddy/dx.
d dx
[rnsh"1 r] —
481
30. 0. / / cosh(2x - 3) dx
\io
52. Approximate the positive value of the constant a such that the area enclosed by y = cosh ax, y = 0, x = 0, and x = 1 is 2 square units. Express your answer to at least five decimal places. 53. Find the arc length of the catenary y = cosh x between x = 0 and x = In 2. 54. Find the arc length of the catenary y = a cosh(;c/a) between x = 0 and x = x\ (x\ > 0). 55. In parts (a)—(f) find the limits, and confirm that they are consistent with the graphs in Figures 6.8.1 and 6.8.6.
482
Chapter 6 / Exponential, Logarithmic, and Inverse Trigonometric Functions (a)
lim sinh x
(b)
X^+co
(c) (e)
lim sinh x X^ —cc
lim tanh x
(d) lim tanhz
lim sinh~' x
(f) lim~
*-»+ao JC—>• +co
67. The integration formulas for 1/V«2 — a2 in Theorem 6.8.6 are valid for u > a. Show that the following formula is valid for u < —a: du
= — cosh '
FOCUS ON CONCEPTS
56. Explain how to obtain the asymptotes for y = tanh x from the curvilinear asymptotes for y = cosh* and y = sinhz. 57. Prove that sinh x is an odd function of x and that cosh x is an even function of x, and check that this is consistent with the graphs in Figure 6.8.1. 58-59 Prove the identities. 58. (a) (b) (c) (d) (e) (f) (g) (h)
cosh x + sinh x = e* cosh x — sinh x = e~x sinh(x + y) = sinh x cosh y + cosh x sinh y sinh 2x — 2 sinh x cosh x cosh(;t + y) = cosh x cosh y + sinh x sinh y cosh 2x = cosh2 x + sinh2 x cosh 2x = 2 sinh2 x + 1 cosh2jc = 2 cosh2 x - 1
59. (a) 1 — tanh2 x = sech2 x tanh x + tanh y (b) tanhC* + y) = 1 + tanh x tanh y 2 tanh x (c) tanh 2x — 1 + tanh x 60. Prove: (a) cosh * = ln(* >1 1 — 1 < x < 1. (b) 61. Use Exercise 60 to obtain the derivative formulas for cosh"1 x and tanh"1 x. 62. Prove: sech"1 x = cosh"1 (1/z), com"' * = tanh-
1
<1
0<
\x\ > 1
csch"1 x = sinrT1 (1/jc), 63. Use Exercise 62 to express the integral
0
du
or
In
+C
68. Show that (sinh x + cosh x)" = sinh nx + cosh nx. 69. Show that
e'x dx =
2 sinh at
70. A cable is suspended between two poles as shown in Figure 6.8.2. Assume that the equation of the curve formed by the cable is y = a cosh(x/a), where a is a positive constant. Suppose that the x -coordinates of the points of support are x = —b and x = b, where b > 0. (a) Show that the length L of the cable is given by
b
L — la sinh a (b) Show that the sag S (the vertical distance between the highest and lowest points on the cable) is given by b S = a cosh a a 71-72 These exercises refer to the hanging cable described in Exercise 70. B 71. Assuming that the poles are 400 ft apart and the sag in the cable is 30 ft, approximate the length of the cable by approximating a. Express your final answer to the nearest tenth of a foot. [Hint: First let u = 200/a.] 72. Assuming that the cable is 120 ft long and the poles are 100 ft apart, approximate the sag in the cable by approximating a. Express your final answer to the nearest tenth of a foot. [Hint: First let u = 5Q/a.] 73. The design of the Gateway Arch in St. Louis, Missouri, by architect Eero Saarinan was implemented using equations provided by Dr. Hannskarl Badel. The equation used for the centerline of the arch was for x between -299.2239 and 299.2239. (a) Use a graphing utility to graph the centerline of the arch. (b) Find the length of the centerline to four decimal places. (c) For what values of x is the height of the arch 100 ft? Round your answers to four decimal places. (d) Approximate, to the nearest degree, the acute angle that the tangent line to the centerline makes with the ground at the ends of the arch.
64. Show that d
(
*T*
< b >rdx
x^l+x2 65. In each part, find the limit. lim (cosh"1 * — In*) (b) lim
+C
y = 693.8597 - 68.7672 cosh(0.0100333jt) ft
entirely in terms of tanh
(a)
(-3
cosh*
66. Use the first and second derivatives to show that the graph of y = tanh"1 * is always increasing and has an inflection point at the origin.
74. Suppose that a hollow tube rotates with a constant angular velocity of <w rad/s about a horizontal axis at one end of the tube, as shown in the accompanying figure (on the next page). Assume that an object is free to slide without friction in the tube while the tube is rotating. Let r be the distance
6.8 Hyperbolic Functions and Hanging Cables
(b) If the rope has a length of 15 m, how far must the person walk from the origin to bring the boat 10 m from the dock? Round your answer to two decimal places. (c) Find the distance traveled by the bow along the tractrix as it moves from its initial position to the point where it is 5 m from the dock.
from the object to the pivot point at time t > 0, and assume that the object is at rest and r = 0 when t = 0. It can be shown that if the tube is horizontal at time t = 0 and rotating as shown in the figure, then during the period that the object is in the tube. Assume that t is in seconds and r is in meters, and use g = 9.8 m/s2 and w = 2 rad/s. (a) Graph r versus t for 0 < t < 1. (b) Assuming that the tube has a length of 1 m, approximately how long does it take for the object to reach the end of the tube? (c) Use the result of part (b) to approximate dr/dt at the instant that the object reaches the end of the tube.
(a,0)
Dock
Initial position -4 Figure Ex-75
76. Writing Suppose that, by analogy with the trigonometric functions, we define cosh t and sinh t geometrically using Figure 6.S.3&: "For any real number t, define x = cosh t and y = sinh t to be the unique values of x and y such that (i) P(x, y) is on the right branch of the unit hyperbola x2-y2 = 1; (ii) t and y have the same sign (or are both 0); (iii) the area of the region bounded by the z-axis, the right branch of the unit hyperbola, and the segment from the origin to P is |f|/2." Discuss what properties would first need to be verified in order for this to be a legitimate definition. 77. Writing Investigate what properties of cosh t and sinh t can be proved directly from the geometric definition in Exercise 76. Write a short description of the results of your investigation.
•4 Figure Ex-74
75. The accompanying figure shows a person pulling a boat by holding a rope of length a attached to the bow and walking along the edge of a dock. If we assume that the rope is always tangent to the curve traced by the bow of the boat, then this curve, which is called a tractrix, has the property that the segment of the tangent line between the curve and the }>-axis has a constant length a. It can be proved that the equation of this tractrix is -- /a2 - *2 a (a) Show that to move the bow of the boat to a point ( x , y ) , the person must walk a distance _ D = a sech a from the origin. y = a sech
DICK CHECK ANSWERS 6.8
1.
ex -e-
ex
2.
cosh* DOMAIN RANGE
(—00,
+00)
[l,+oo)
sinh* (—00,
+00)
(—00,
+00)
3. unit hyperbola; x2 — y2 = 1
6.
l
tanh* (—00,
+00)
(-1,1)
coth*
(-00, 0) u (0, +00) (_oo, -1) U (1, + 00)
4. sinhjc; cosh*; sech2*
483
sech* (—00,
+00)
(0,1]
csch*
(-00, 0) u (0, +°o) (-00, 0) u (0, +00)
5. sinh* + C; cosh* + C; ln(cosh*)
484
Chapter 6 / Exponential, Logarithmic, and Inverse Trigonometric Functions
CHAPTER 6 REVIEW EXERCISES
0 Graphing Utility
1. In each part, find / ' (*) if the inverse exists. (a) /(*) = (ex)2 + 1
(b) /(*) =
4 + 7T
(e) How long does it take for the package to reach 98% of its terminal velocity? • 7T
1
(c) /(*) =
1 + 3 tan" 2. (a) State the restrictions on the domains of sin*, cos*, tan*, and sec* that are imposed to make those functions one-to-one in the definitions of sin ' x, cos"1 x, tan *, and sec *. (b) Sketch the graphs of the restricted trigonometric functions in part (a) and their inverses.
11. A breeding group of 20 bighorn sheep is released in a protected area in Colorado. It is expected that with careful management the number of sheep, N, after t years will be given by the formula
220 1 + 10(0.83') and that the sheep population will be able to maintain itself without further supervision once the population reaches a size of 80. (a) Graph N versus t. (b) How many years must the state of Colorado maintain a program to care for the sheep? (c) How many bighorn sheep can the environment in the protected area support? [Hint: Examine the graph of N versus t for large values of /.]
3. In each part, find the exact numerical value of the given expression. (a) cos[cos~'(4/5)+ sin'1 (5/13)] (b) sinfsin-' (4/5) + cos~'(5/13)] 3 4. In each part, sketch the graph, and check your work with a graphing utility. (a) /(*) = 3sin-'(*/2) (b) /(*) = COS"1 * - 7T/2 1
(c) /(*) = 2tan- (-3*)_ (d) /(*) = cos"1 * + sin"1 * 5. Suppose that the graph of y = log* is drawn with equal scales of 1 inch per unit in both the *- and y-directions. If a bug wants to walk along the graph until it reaches a height of 5 ft above the x-axis, how many miles to the right of the origin will it have to travel? 6. Find the largest value of a such that the function /(*) = xe~x has an inverse on the interval (—00, a]. 1. Express the following function as a rational function of*: 3\n(e2x(ex)3)+2e\p(\nl)
12. An oven is preheated and then remains at a constant temperature. A potato is placed in the oven to bake. Suppose that the temperature T (in ° F) of the potato t minutes later is given by T = 400 - 325(0.97'). The potato will be considered done when its temperature is anywhere between 260° F and 280°F. (a) During what interval of time would the potato be considered done? (b) How long does it take for the difference between the potato and oven temperatures to be cut in half? 13. (a) Show that the graphs of y = In * and y = x°'2 intersect, (b) Approximate the solution(s) of the equation In * = *°-2 to three decimal places. 14. (a) Show that for x > 0 and k / 0 the equations
kt
8. Suppose that y = Ce , where C and k are constants, and let Y = In y. Show that the graph of Y versus t is a line, and state its slope and F-intercept. 3 9. (a) Sketch the curves y = ±e~x/2 and y = e~x/2 sin 2* for —7T/2 < x < 3?T/2 in the same coordinate system, and check your work using a graphing utility, (b) Find all *-intercepts of the curve y = e~x'2 sin 2* in the stated interval, and find the *-coordinates of all points where this curve intersects the curves y = ±e~x/2. 10. Suppose that a package of medical supplies is dropped from a helicopter straight down by parachute into a remote area. The velocity v (in feet per second) of the package t seconds after it is released is given by v = 24.61(1 — e~L3t). (a) Graph v versus t. (b) Show that the graph has a horizontal asymptote v = c. (c) The constant c is called the terminal velocity. Explain what the terminal velocity means in practical terms. (d) Can the package actually reach its terminal velocity? Explain.
** = ex
and
1 In* = * k
have the same solutions. (b) Use the graph of y = (In*)/* to determine the values of k for which the equation xk = ex has two distinct positive solutions. (c) Estimate the positive solution(s) of* 8 = ex. 15-18 Find the limits. B 15.
lim e tanf
16. lim In (sin 29) -In (tan 9)
( -» JT/2+
,bx
17. lim I 1 + - I x-i.+aoV
X
)
18. lim (l + - V, JC->+c.A
*/
a, b > 0
19-20 Find dy/dx by first using algebraic properties of the natural logarithm function, m
19. y = In
3) 3 (*+4)<
20. y = In
sin * sec *
Chapter 6 Review Exercises
485
47. If / and g are inverse functions and / is differentiable on its domain, must g be differentiable on its domain? Give a reasonable informal argument to support your answer. 48. In each part, find (f l)'(x) using Formula (2) of Section 6.3, and check your answer by differentiating /"' directly. (a) /CO = 3/Ge + 1) (b) /(*) = ^ 49. Find a point on the graph of y = e3x at which the tangent line passes through the origin. 50. Show that the rate of change of y = 5000e'-07j: is proportional to y.
21-38 Yw&dy/dx. 21. y = In 2x 23. y =
25. _y =
29, y = g ln(j; 2 +l)
51. Show that the function y = eax sin bx satisfies
31. y = 2xe^
y" - 2ay' + (a2 + b2)y = 0
33. y = -tan" 1 2x n
for any real constants a and b. 52. Show that the function y = tan"1 x satisfies
(el)
35. y = x
y" = —2 sin y cos3 y
37. y = sec~'(2.x + 39-40 Find dy/dx using logarithmic differentiation. •
39. y =
53. Suppose that the population of deer on an island is modeled by the equation
95 40. y =
2
X +\
41. (a) Make a conjecture about the shape of the graph of y = jx — Inx, and draw a rough sketch. (b) Check your conjecture by graphing the equation over the interval 0 < x < 5 with a graphing utility. (c) Show that the slopes of the tangent lines to the curve at x = 1 and x = e have opposite signs. (d) What does part (c) imply about the existence of a horizontal tangent line to the curve? Explain. (e) Find the exact *-coordinates of all horizontal tangent lines to the curve. 42. Recall from Section 6.1 that the loudness ft of a sound in decibels (dB) is given by ft — 101og(7/7o), where 7 is the intensity of the sound in watts per square meter (W/m2) and 7o is a constant that is approximately the intensity of a sound at the threshold of human hearing. Find the rate of change of ft with respect to 7 at the point where (a) 7/70 = 10 (b) 7/7 0 = 100 (c) 7/70 = 1000. 43. A particle is moving along the curve y = x In x. Find all values of x at which the rate of change of y with respect to time is three times that of x. [Assume that dx/dt is never zero.] 44. Find the equation of the tangent line to the graph of 2 y = In(5 - x ) at x = 2. 45. Find the value of b so that the line y = x is tangent to the graph of y = logfc x. Confirm your result by graphing both y = x and y = logfc x in the same coordinate system. 46. In each part, find the value of k for which the graphs of y = /CO and y = Inx share a common tangent line at their point of intersection. Confirm your result by graphing y = /CO and y = In x in the same coordinate system. (a) /« = 4^ + k (b) /CO = ,
where P ( t ) is the number of deer t weeks after an initial observation at time t = 0. (a) Use a graphing utility to graph the function P ( t ) . (b) In words, explain what happens to the population over time. Check your conclusion by finding lim, _^ +co P (t). (c) In words, what happens to the rate of population growth over time? Check your conclusion by graphing P'(t). 54. The equilibrium constant k of a balanced chemical reaction changes with the absolute temperature T according to the law / q(T-T0y k = ko exp 2T0T where ko, q, and TQ are constants. Find the rate of change of k with respect to T. 55-56 Find the limit by interpreting the expression as an appropriate derivative. • (1+hy-l l-ln* 55. hm 56. hm h -> o h x -> e (x — e) In x 57. Suppose that lim /CO = ±00 and lim g(jc) = ±00. In each of the four possible cases, state whether lim[/CO — g(x)] is an indeterminate form, and give a reasonable informal argument to support your answer. 58. (a) Under what conditions will a limit of the form Hm[/CO/gCO] be an indeterminate form? (b) Iflirn^Q g(x) = 0,mustlirnr^ 0 [/CO/gCO]beanindeterminate form? Give some examples to support your answer. 59-62 Evaluate the given limit. •
486
Chapter 6 / Exponential, Logarithmic, and Inverse Trigonometric Functions
59. lim (e* - x2) jr->+cc
*V
61. lim —;—
60. lim 62. lim
77. Interpret the expression as a definite integral over [0, l],and then evaluate the limit by evaluating the integral.
a >0 lim
max Ajr^ -
63-64 Find: (a) the intervals on which / is increasing, (b) the intervals on which / is decreasing, (c) the open intervals on which / is concave up, (d) the open intervals on which / is concave down, and (e) the *-coordinates of all inflection points. 63. f ( x ) = 1/e*
64. /(*) = tan"
65-66 Use any method to find the relative extrema of the function /. * 65. f ( x ) = ln(l + x2) 66. f ( x ) = x2ex 67-68 In each part, find the absolute minimum m and the absolute maximum M of / on the given interval (if they exist), and state where the absolute extrema occur. H 67. f(x)=ex/x2; (0,+00) 68. f(x)=xx; (0,+00) 69. Use a graphing utility to estimate the absolute maximum and minimum values of f ( x ) = x/2 + \n(x2 + 1), if any, on the interval [—4, 0], and then use calculus methods to find the exact values. 70. Prove that x < sin"1 x for all x in [0, 1].
78. Find the limit \ln
/n i
..
lim «-»+» n by interpreting it as a limit of Riemann sums in which the interval [0, 1] is divided into n subintervals of equal length, 79-80 Find the area under the curve y = f ( x ) over the stated interval. B 80. /(*) = -; [1, e3] x 81. Solve the initial-value problems.
79. /(*) =
(a) — =cosx-5e\ y(0) = 0 dx
(b)
-= dx
\ y(0) = 0
82-84 Evaluate the integrals by making an appropriate substitution. B dx f1 dx 83. f xlnx Je Jo
'•f
2/V3
1
•dx
71-74 Evaluate the integrals. • 71. [[X'2'3-5e*]dx
72.
dx
1=1 dx ^?J
75-76 Use a calculating utility to find the left endpoint, right endpoint, and midpoint approximations to the area under the curve y = f ( x ) over the stated interval using « = 10 subintervals. B 75. y = lnjt; [1,2] 76. y =ex; [0,1]
85. Find the volume of the solid whose base is the region bounded between the curves y = ^/x and y = l/^/x for 1 < x < 4 and whose cross sections taken perpendicular to the x-axis are squares. 86. Find the average value of f ( x ) = e* + e~x over the interval [In i, In 2]. 87. In each part, prove the identity. (a) cosh 3x = 4 cosh3 x — 3 cosh x (b) cosh ijc =
+ 1)
(c) sinh ~x = ±J | (cosh x — I) 88. Show that for any constant a, the function y = sinh (a*) satisfies the equation y" = a2y.
Chapter 6 Making Connections
487
CHAPTER 6 MAKING CONNECTIONS 1. Consider a simple model of radioactive decay. We assume that given any quantity of a radioactive element, trie fraction of the quantity that decays over a period of time will be a constant that depends on only the particular element and the length of the time period. We choose a time parameter -co < t < +coandletA = A (t) denote the amount of the element remaining at time t. We also choose units of measure such that the initial amount of the element is A(0) = 1, and we let b = A(l) denote the amount at time t = 1. Prove that the function A(t) has the following properties. (a) A(—t) = ——- [Hint: For t > 0, you can interpret A(t) A(t) as the fraction of any given amount that remains after a time period of length t.] (b) A(s +t) = A(s) • A(t) [Hint: First consider positive s and t. For the other cases use the property in part (a).] (c) If n is any nonzero integer, then A Q) = (A(l))
1/n
1/n
= fe
tions agree on the set of rational numbers, then they are
equa\.] (f) If we replace the assumption that A(0) = 1 by the condition A(0) = AQ, prove that A = A^b'. 2. Refer to Figure 6.1.5. (a) Make the substitution h = l/x and conclude that (l+h)l/h <e < (l-hTllh
for/z>0
d
for h < 0
and
(b) Use the inequalities in part (a) and the Squeezing Theorem to prove that
«*-! lim - = 1 h^o h (c) Explain why the limit in part (b) confirms Figure 6.1.4. (d) Use the limit in part (b) to prove that
(d) If m and n are. integers with n /= 0, then n _ j.m/1
(e) Assuming that A(t) is a continuous function of t, then A(f) = b'. [Hint: Prove that if two continuous func-
dx 3. Give a convincing geometric argument to show that
/ In x dx + I ex dx = e
J\
Jo
PRINCIPLES OF INTEGRAL EVALUATION
O AP/Wide World Photos
The floating roof on the Stade de France sports complex is an ellipse. Finding the arc length of an ellipse involves numerical integration techniques introduced in this chapter.
In earlier chapters we obtained many basic integration formulas as an immediate consequence of the corresponding differentiation formulas. For example, knowing that the derivative of sin x is cos x enabled us to deduce that the integral of cos x is sin x. Subsequently, we expanded our integration repertoire by introducing the method of u-substitution. That method enabled us to integrate many functions by transforming the integrand of an unfamiliar integral into a familiar form. However, u-substitution alone is not adequate to handle the wide variety of integrals that arise in applications, so additional integration techniques are still needed. In this chapter we will discuss some of those techniques, and we will provide a more systematic procedure for attacking unfamiliar integrals. We will talk more about numerical approximations of definite integrals, and we will explore the idea of integrating over infinite intervals.
AN OVERVIEW OF INTEGRATION METHODS In this section we will give a brief overview of methods for evaluating integrals, and we will review the integration formulas that were discussed in earlier sections.
METHODS FOR APPROACHING INTEGRATION PROBLEMS
There are three basic approaches for evaluating unfamiliar integrals: • Technology—CAS programs such as Mathematica, Maple, and the open source program Sage are capable of evaluating extremely complicated integrals, and such programs are increasingly available for both computers and handheld calculators. • Tables—Prior to the development of CAS programs, scientists relied heavily on tables to evaluate difficult integrals arising in applications. Such tables were compiled over many years, incorporating the skills and experience of many people. One such table appears in the endpapers of this text, but more comprehensive tables appear in various reference books such as the CRC Standard Mathematical Tables and Formulae, CRC Press, Inc., 2002. • Transformation Methods—Transformation methods are methods for converting unfamiliar integrals into familiar integrals. These include w-substitution, algebraic manipulation of the integrand, and other methods that we will discuss in this chapter. 488
7.1 An Overview of Integration Methods
489
None of the three methods is perfect; for example, CAS programs often encounter integrals that they cannot evaluate and they sometimes produce answers that are unnecessarily complicated, tables are not exhaustive and may not include a particular integral of interest, and transformation methods rely on human ingenuity that may prove to be inadequate in difficult problems. In this chapter we will focus on transformation methods and tables, so it will not be necessary to have a CAS such as Mathematica, Maple, or Sage. However, if you have a CAS, then you can use it to confirm the results in the examples, and there are exercises that are designed to be solved with a CAS. If you have a CAS, keep in mind that many of the algorithms that it uses are based on the methods we will discuss here, so an understanding of these methods will help you to use your technology in a more informed way.
A REVIEW OF FAMILIAR INTEGRATION FORMULAS
The following is a list of basic integrals that we have encountered thus far: CONSTANTS, POWERS, EXPONENTIALS 1. / du = u + C
/
urdu =
2. I adu =a I du = au + C
ur+l + C, r£-\ r+l
f du 4. / — = l n | M | + C J u
5. f eudu = e"+C J
6. f b" du = — + C, b > 0, b J Infc
TRIGONOMETRIC FUNCTIONS 7. / sin u du = — cos u + C
8. / cos u du = sin u + C
9. I sec2 u du = tan u + C
10. / esc2 u du = — cot u + C
11. I sec u tan u du = sec u + C C
12. / esc u cot u du = — esc u + C
13. / tan tan u du = = — In | cos cos u \ + +C C
14. / cot u du = In | sin u \ + C
HYPERBOLIC FUNCTIONS 15. / sinh udu u du — = cosh u + C
16. / icosh udu — sinh u + C
17. / sech2 udu = tanh u + C
18. / csch2 u du = - coth u + C
19. / sech u tanh udu = —sech u + C
20. / csch u coth udu — —csch M + C
ALGEBRAIC FUNCTIONS (a > 0) _i U
C?M
21. | -—= ^— sin- —h —C a /a2 - u2 du I ,1 u 22 2 2 = - tan" - + C a +u a
(|M| < a)
•/
23.
/In
1
. w
C
(0
490
Chapter 7 / Principles of Integral Evaluation dU 24. I = ln(w + x/M2 + a 2 ) + C J Ja2 + u2 du
[
25
In
2
J Ju - a2 ~
(0 < a
C
a +u +C a —u C
28.
-
(0 < \u\
+C
-
REMARK I Formula 25 is a generalization of a result in Theorem 6.8.6. Readers who did not cover Section 6.8 I can ignore Formulas 24-28 for now, since we will develop other methods for obtaining them in this I chapter.
•QUICK CHECK EXERCISES 7.1 1.
(See page 491 for answers.)
Use algebraic manipulation and (if necessary) u -substitution to integrate the function. i 1
7
x +1 f 2x + 1
/ J
(H)
/" /
fa)
rlr —
/ V* - ' dr = 2x+t
(M
' J [?
3\nx
J
flr-
I (sin 3 v rns r + sin r rns3 r) /Vr =
(c)
2. Use trigonometric identities and (if necessary) w-substitution to integrate the function.
CM / -
1
J
X f
I fmt 2 r + 1 Wr =
J sec x + tan x 3. Integrate the function.
/ Y
(c.)
(d) /
J (ex + e~x)2
dx —
--dx-
EXERCISE SET 7.1
1-30 Evaluate the integr;ils by making appropriate M-substitutions and applying the fo rmulas reviewed in this section. c 1. l(4-2x)3dx 1. I 3V4 + 2xdx 3. / x sec2(x2)dx
4. / 4xtan(x2}dY i
C
sin 3* J 2 + cos 3x
/
*- J
aX
^
tallJC
QP^
2
V//V
r
15 1
m
^
6
dx I_X_ //r
/
/-y-
/
/7v
ax
/• cosn /7 17
r *
/• /
l
,
i •>
ex
(
/• sec (In x) tan (In x)
7. I e*sinh(ex)dx Q
n i
11. / cos5 5x sin 5x dx
/
jr /— 'l^/x
-/ »•/ ,,/ ,,/
cosx sinjtysin x + 1 1
«tan~' A
1 + j:2
dX
(x + l)cot(x 2 + 2x)dx
dx x(\nx)2
7.2 Integration by Parts
»•/. , y^M
20.
32. (a) Derive the identity
sec(sm9)tan(sm0)cos0d0
21
sech2 x
dx
1 + tanh2 x
>f
23
'/-
/
,
'•/ , /.
dx
= sech 2.x
(b) Use the result in part (a) to evaluate f sech x dx. (c) Derive the identity
<•/
29
491
sechx = :dx
(d) Use the result in part (c) to evaluate / sech x dx. (e) Explain why your answers to parts (b) and (d) are consistent. 33. (a) Derive the identity sec2x tanx
FOCUS ON CONCEPTS
31. (a) Evaluate the integral f sin * cos x dx using the substitution u = sin*. (b) Evaluate the integral / sin x cos x dx using the identity sin 2x = 2 sin * cos x . (c) Explain why your answers to parts (a) and (b) are consistent.
1 sn x cos x
(b) Use the identity sin 2x = 2 sin x cos x along with the result in part (a) to evaluate f esc x dx. (c) Use the identity cosx = sin[(?r/2) — x] along with your answer to part (a) to evaluate / sec x dx.
•QUICK CHECK ANSWERS 7.1 1. (a) x+ln\x\ (c) -cotz
(b) x + In \x + 1| + C (c) ln(x2 + 1) + tan^' x + C (d) (d) l n ( l + s i n x )
3/2
3. (a) \(x - 1)
2x+l
+ C (b) ±e
hC
2. (a) - cosx + C (b) tanx + C
(d) ^
+ C (c) isi
INTEGRATION BY PARTS In this section we will discuss an integration technique that is essentially an antiderivative formulation of the formula for differentiating a product of two functions. THE PRODUCT RULE AND INTEGRATION BY PARTS
Our primary goal in this section is to develop a general method for attacking integrals of the form r I f(x)g(x)dx As a first step, let G(x) be any antiderivative of g(x). In this case G'(x) — g(x), so the product rule for differentiating /(x)G(x) can be expressed as x)] = f ( x ) G ' ( x ) + f'(x)G(x) = f ( x ) g ( x ) + f'(x)G(x)
^
d)
This implies that /(x)G(x) is an antiderivative of the function on the right side of (1), so we can express (1) in integral form as [f(x)g(x)
+ f'(x)G(x)]dx
= f(x)G(x)
492
Chapter 7 / Principles of Integral Evaluation
or, equivalently, as f'(x)G(x)dx
(2)
This formula allows us to integrate f ( x ) g ( x ) by integrating f'(x)G(x) instead, and in many cases the net effect is to replace a difficult integration with an easier one. The application of this formula is called integration by parts. In practice, we usually rewrite (2) by letting u = f(x), v = G(x),
du = f'(x)dx dv — G'(x)dx = g(x)dx
This yields the following alternative form for (2): / udv = uv — I v du
Note that in Example 1 we omitted the constant of integration in calculating v from dv. Had we included a constant of integration, it would have eventually dropped out. This is always the case in integration by parts [Exercise 68(b)], so it is common to omit the constant at this stage of the computation. However, there are certain cases in which making a clever choice of a constant of integration to include with v can simplify the computation of fv du (Exercises 69-71).
(3)
>• Example 1 Use integration by parts to evaluate / x cos x dx. Solution. We will apply Formula (3). The first step is to make a choice for u and dv to put the given integral in the form fudv. We will let u=x
and dv = cosxdx
(Other possibilities will be considered later.) The second step is to compute du from u and i; from dv. This yields du — dx
and
v = I dv — I cosxdx = sinx
The third step is to apply Formula (3). This yields
x cos x dx — x sin x — /sir I sin x dx du
= x sin x — (— cos ;c) + C = x sin x + cos x + C
GUIDELINES FOR INTEGRATION BY PARTS
The main goal in integration by parts is to choose u and dv to obtain a new integral that is easier to evaluate than the original. In general, there are no hard and fast rules for doing this; it is mainly a matter of experience that comes from lots of practice. A strategy that often works is to choose u and dv so that u becomes "simpler" when differentiated, while leaving a dv that can be readily integrated to obtain v. Thus, for the integral fxcosxdx in Example 1, both goals were achieved by letting u — x and dv = cos x dx. In contrast, M = cos x would not have been a good first choice in that example, since du/dx = - sin x is no simpler than u. Indeed, had we chosen U = COS X
du = — sin x dx
dv — x dx
v---- I xdx = -
then we would have obtained x2 f x2 x2 1 f xcosx dx — — cos x — I —(— sinx)dx = — cos* H— I x2 sin xdx / For this choice of u and dv, the new integral is actually more complicated than the original.
7.2 Integration by Parts
The LIATE method is discussed in the article "A Technique for Integration by Parts,"/American Mathematical Monthly, Vol. 90, 1983, pp. 210-211, by Herbert Kasube.
493
There is another useful strategy for choosing u and dv that can be applied when the integrand is a product of two functions from different categories in the list Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential In this case you will often be successful if you take u to be the function whose category occurs earlier in the list and take dv to be the rest of the integrand. The acronym LIATE will help you to remember the order. The method does not work all the time, but it works often enough to be useful. Note, for example, that the integrand in Example 1 consists of the product of the algebraic function x and the trigonometric function cos x. Thus, the LIATE method suggests that we should let u = x and dv = cos x dx, which proved to be a successful choice. Example 2 Evaluate / xex dx. Solution. In this case the integrand is the product of the algebraic function x with the exponential function ex. According to LIATE we should let and dv — ex dx
u=x
so that
and u = / ex dx — ex
du = dx
Thus, from (3) x / xex dx — I udv = uv — I v du = xex — I ex dx = xe* - e + C
> Examples Evaluate / Inxdx. Solution. One choice is to let u = 1 and dv —Inxdx. But with this choice finding v is equivalent to evaluating flnxdx and we have gained nothing. Therefore, the only reasonable choice is to let u = Inx du = — dx x
dv = dx
v — I dx = x
With this choice it follows from (3) that / Inx dx = I udv = uv — I vdu = xlnx — I dx = xlnx — x + C
REPEATED INTEGRATION BY PARTS
It is sometimes necessary to use integration by parts more than once in the same problem. Example 4 Evaluate / x2e
x
dx.
Solution. Let M = X2,
= e~xdx,
du=2xdx ,
v = I e~x dx = —e
494
Chapter 7 / Principles of Integral Evaluation
so that from (3) / x2e
x
dx = I udv = uv — I vdu
= x2(-e-x)- I -e-x(2x)dx = - x2e~x + 2
xe~x dx
(4)
The last integral is similar to the original except that we have replaced x2 by x. Another integration by parts applied to f xe~x dx will complete the problem. We let u = x,
dv — e~x dx,
du = dx,
v = I e~x dx = —e~x
so that / xe~x dx = x(-e~x) - I -e~x dx = -xe~x + j e~x dx = -xe~x - e'x + C Finally, substituting this into the last line of (4) yields f x2e~xdx = -x2e~x +2 f xe~xdx = -x2e~x + 2(-xe~x - e~x) + C
The LIATE method suggests that integrals of the form / eax sin bx dx
and
/ eax cos bx dx
can be evaluated by letting u = sin bx or u = cos bx and dv = eax dx. However, this will require a technique that deserves special attention. Examples Evaluate / ex cosxdx. Solution. Let u = cosx,
dv = ex dx,
du = — sin x dx,
v — I ex dx — ex
Thus, / ex cos x dx = I udv — uv — I vdu = ex cos x + I ex sin x dx
(5)
Since the integral fex sinxdx is similar in form to the original integral fex cosxdx, it seems that nothing has been accomplished. However, let us integrate this new integral by parts. We let u = sin*,
dv = exdx,
du = cosxdx,
v = I ex dx — ex
I ex sin x dx = I udv — uv — I v du = ex sin x — I ex cos x dx
Together with Equation (5) this yields / ex cos x dx — ex cos x + ex sin x — j ex cos x dx
(6)
7.2 Integration by Parts
495
which is an equation we can solve for the unknown integral. We obtain 2 I ex cos x dx — ex cos x + ex sin x
and hence
/'
cos x dx = e* cos x + e* sin x + C
A TABULAR METHOD FOR REPEATED INTEGRATION BY PARTS
Integrals of the form f
More information on tabular integration by parts can be found in the articles "Tabular Integration by Parts," College Mathematics Journal, Vol. 21, 1990, pp. 307-311, by David Horowitz and "More on Tabular Integration by Parts," College Mathematics Journal, Vol. 22, 1991, pp. 407-410, by Leonard Gillman.
f
p(x)f(x)dx
where p ( x ) is a polynomial, can sometimes be evaluated using repeated integration by parts in which u is taken to be p(x) or one of its derivatives at each stage. Since du is computed by differentiating u, the repeated differentiation of p(x) will eventually produce 0, at which point you may be left with a simplified integration problem. A convenient method for organizing the computations into two columns is called tabular integration by parts.
Tabular Integration by Parts Step 1. Differentiate p(x) repeatedly until you obtain 0, and list the results in the first column. Step 2. Integrate f ( x ) repeatedly and list the results in the second column. Step 3. Draw an arrow from each entry in the first column to the entry that is one row down in the second column. Step 4. Label the arrows with alternating + and — signs, starting with a +. Step 5. For each arrow, form the product of the expressions at its tip and tail and then multiply that product by +1 or —1 in accordance with the sign on the arrow. Add the results to obtain the value of the integral.
This process is illustrated in Figure 7.2.1 for the integral f(x2 — x) cos x dx.
REPEATED DIFFERENTIATION
>• Figure 7.2.1
REPEATED INTEGRATION
- x) cos xdx = (x1 — x) sinx + (2x — 1) cos x — 2 sin x + C = (x2 - x - 2) sin x + (2x - l ) c o s z + C
> Example 6 In Example 9 of Section 4.3 we evaluated fx2^/x — Idx using w-substitution. Evaluate this integral using tabular integration by parts.
496
Chapter 7 / Principles of Integral Evaluation
Solution. REPEATED DIFFERENTIATION
REPEATED INTEGRATION
-105 The result obtained in Example 6 looks
' llfollowsthat /• / X2jx - 1 dx = \X2(x - 1)3/2 - ^x(x - 1)5/2 +
quite differentfrom that obtained in Example 9 of Section 4.3. Show that the two answers are equivalent.
//2
l)7 / 2 - 1
+C+
INTEGRATION BY PARTS FOR DEFINITE INTEGRALS
For definite integrals the formula corresponding to (3) is b
If udv = uv f\ — If" vdu Jo
Ja
REMARK
(7)
Ja
It is important to keep in mind that the variables u and v in this formula are functions of x and that the limits of integration in (7) are limits on the variable x. Sometimes it is helpful to emphasize this by writing (7) as
[0 I udv = uv\ J x=a
—I
lx=a
r»
vdu
(8)
J x=a
The next example illustrates how integration by parts can be used to integrate the inverse trigonometric functions. .1 *• Example 7 Evaluate / tan"1 xdx. Jo Solution. Let
u = tan x,
dv = dx,
du =
1
~dx,
1 +xz
v=x
Thus, f\an-lxdx = f'udv^uv] - f vdu Jo Jo Jo Jo — x tan ' x
]'-('Jo
Jo H-
But 1 /*'
2x
1 ^!imits°lin!!gratio?refcrto^
•^
_1 ~ 2
=-ln2 2
so /" 7o
_lln2=(^-0)-lln2^^-lnV2 ^
7.2 Integration by Parts
497
REDUCTION FORMULAS
Integration by parts can be used to derive reduction formulas for integrals. These are formulas that express an integral involving a power of a function in terms of an integral that involves a lower power of that function. For example, if n is a positive integer and n >2, then integration by parts can be used to obtain the reduction formulas
/
sin"xdx = —sin" lxcosx-\ n cos" x dx = - cos""1 x sin x H n /
n n
/ sin" 2xdx J
(9)
/ cos""2 x dx J
(10)
To illustrate how such formulas can be obtained, let us derive (10). We begin by writing cos" x as cos""1 x • cosx and letting u = cos""1 x
dv = cosxdx 2
du = (n — 1) cos"" x(— sin x) dx
v — sin x
= —(« — !) cos""2 x sin x dx so that / cos" x dx = I cos""1 x cos x dx = I udv = uv — I vdu — cos""1 x sin x + (n — 1) / sin2 x cos""2 x dx — cos""1 x sin x + (n — 1) / (1 — cos2 x) cos""2 x dx = cos""1 x sin* + (n — 1) / cos""2 xdx — (n — 1) I cos" x dx Moving the last term on the right to the left side yields n j cos" x dx — cos""1 x sinx + (n — 1) / cos""2 x dx from which (10) follows. The derivation of reduction formula (9) is similar (Exercise 63). Reduction formulas (9) and (10) reduce the exponent of sine (or cosine) by 2. Thus, if the formulas are applied repeatedly, the exponent can eventually be reduced to 0 if n is even or 1 if n is odd, at which point the integration can be completed. We will discuss this method in more detail in the next section, but for now, here is an example that illustrates how reduction formulas work. > Examples Evaluate / cos4xdx. Solution. From (10) with n = 4 /
COS4 X dx ~ j COS3 X Sin X + | I COS2 X dx
f~Now appiyTlO) with n^T-H
J
( 5 cos x sin x + \J fI dx^ = 47 cos3 x sin x +B| cos x sin x + fx + C o
498
Chapter 7 / Principles of Integral Evaluation
I/QUICK CHECK EXERCISES 7.2
(See page 500 for answers.)
1. (a) If G'O) =#(.*;), then f f(x)g(x)dx
J
/•
= f(x)G(x)-
rr/r- u —
//v —
Jt
y v^ -1
(b) If u = f ( x ) and v = G(x), then the formula in part (a) can he written in the form / u dv —
3. Use integration by parts to evaluate the integral. Jte2^*
2. Find an appropriate choice of u and dv for integration by parts of each integral. Do not evaluate the integral. r In r
(b) / ln(x
l)dx
7T/6
x sin 3.x dx
/
du —
sin3 xdx.
f
1 (r — 9) sin v dr- u —
(h)
1 sin
(c.)
dv —
EXERCISE SET 7.2
4 f 33. ^
1-38 Evaluate the integral. 1. fxe~2xdx 3. I X2exdx
fxe3xdx
2.
4. I x2e~2xdx
5. / x sin 3x dx
6. / x cos 2x dx
7. / x2 cos xdx
8. / X 2 sin ;t d*
9.
xlnxdx
11. f(lnx)2dx
10. I -Jx \nxdx 12. f -^dx
J
J
v •#
sec
0
2 f 1. ^
/•7T
/.JT
35. / xsin2xdx Jo •\H J 1*
f3
I 1
Ji
f~Z. t o r > ~ ^ l~v /-/-v Y A I till -y A W A
36. / (jc+jccos^)rfx Jo ^5J *7O«
f2
1 J
7o
In/'v^ 1 l ^ y Y v lil^A "T" 1. / C*A
39-42 True-False Determine whether the statement is true or false. Explain your answer X 39. The main goal in integration by parts is to choose u and dv to obtain a new integral that is easier to evaluate than the original.
13. fln(3x-2)dx
14.
fln(x2+4)dx
40. Applying the LIATE strategy to evaluate f x3 In x dx, we should choose u = x3 and dv = In x dx.
15. / sin"1 xdx
16.
fcos-\2x)dx
41. To evaluate / In e* dx using integration by parts, choose dv = ex dx.
17. / tan~l(3x)dx
18. / A: tan"1 ^ af jc
19. / ex sin x dx
20. Je3xcos2xdx
21. / sin(lnjc)<3(jc
22. I cos(lnx)dx
23. / x sec2 x dx
24. / x
25.
3 x
fx e *dx 2
r
27. / xe2xdx Jo 2
29. I x In xdx Ji
26. f J ( fi 28. / 30. 32.
43-44 Evaluate the integral by making a «-substitution and then integrating by parts. ;s 43. I e^d.
44.. // cos
45. Prove that tabular integration by parts gives the correct answer for P(x)f(x)dx where p(x) is any quadratic polynomial and f ( x ) is any function that can be repeatedly integrated.
V5/2
31. / ln(z J-\
42. Tabular integration by parts is useful for integrals of the form / p ( x ) f ( x ) dx, where p(x) is a polynomial and f ( x ) can be repeatedly integrated.
46. The computations of any integral evaluated by repeated integration by parts can be organized using tabular integration by parts. Use this organization to evaluate / ex cos x dx in
7.2 Integration by Parts two ways: first by repeated differentiation of cos x (compare Example 5), and then by repeated differentiation of e*. 47-52 Evaluate the integral using tabular integration by parts.
(3x2 -
(jt2 + x + 1) sin x dx
dx
499
64. In each part, use integration by parts or other methods to derive the reduction formula. f sec""2*tan* n - 2 f 2 (a) / sec x dx = 1 / sec x dx J n-l n-IJ <-orj tt ~i v r tan" x dx = / tan""2 x dx n-l J / (c) / x"e* dx = x"ex -n
x"~lex dx
49. f 4x sin 2x dx
51. fe"xsi sin bx dx
65-66 Use the reduction formulas in Exercise 64 to evaluate the integrals. X
52.
53. Consider the integral / sin x cos x dx. (a) Evaluate the integral two ways: first using integration by parts, and then using the substitution u = sin x. (b) Show that the results of part (a) are equivalent. (c) Which of the two methods do you prefer? Discuss the reasons for your preference. 54. Evaluate the integral
tan4xdx
65. (a)
(b)
sec4xdx
(c)
x3ex dx
rl (b) / xe~^ dx J Jo [Hint: First make a substitution.] 67. Let / be a function whose second derivative is continuous on[-l, 1]. Show that
r
66. (a) / x2e3x dx
*/"(*) dx = /'(I) + /'(-I) - /(I) + /(-I)
55.
56. 57.
58.
59.
using (a) integration by parts (b) the substitution u = -Jx2 + 1. (a) Find the area of the region enclosed by y = Inx, the line x = e, and the *-axis. (b) Find the volume of the solid generated when the region in part (a) is revolved about the jc-axis. Find the area of the region between y = x sin x and y = x for 0 < x < Till. Find the volume of the solid generated when the region between y = sin ;c and y = 0 for 0 < x < n is revolved about the y-axis. Find the volume of the solid generated when the region enclosed between y = cos x and y — 0 for 0 < x < jr/2 is revolved about the y-axis. A particle moving along the x -axis has velocity function v(t) = t3 sin /. How far does the particle travel from time
F O C U S ON C O N C E P T S
68. (a) In the integral / x cos x dx, let u = x, dv = cosxdx, du = dx, v = sinx + C\ Show that the constant C\ cancels out, thus giving the same solution obtained by omitting C\. (b) Show that in general
69.
70.
t = 0 to t = JT?
60. The study of sawtooth waves in electrical engineering leads to integrals of the form jr
71.
t sin(kwt) dt
/ -T where k is an integer and a> is a nonzero constant. Evaluate the integral. 61. Use reduction formula (9) to evaluate /•ir/2
(a) / sin x dx
(b)
/" Jo
sin x dx.
62. Use reduction formula (10) to evaluate
63. Derive reduction formula (9).
(b)
/Jo
with the choices
u=
r-JT/2
(a) / cos x dx
72.
thereby justifying the omission of the constant of integration when calculating v in integration by parts. Evaluate f\n(x + l)dx using integration by parts. Simplify the computation of f v du by introducing a constant of integration Ci = 1 when going from dv to v. Evaluate /m(3;c — 2)dx using integration by parts. Simplify the computation of f v du by introducing a constant of integration Ci = — | when going from dv to v. Compare your solution with your answer to Exercise 13. Evaluate / x tan~' x dx using integration by parts. Simplify the computation of f v du by introducing a constant of integration C\ = ^ when going from dv to v. What equation results if integration by parts is applied to the integral /• j / ~,— dx J xlnx
cos, x dx.
and dv = — dxl Inx x In what sense is this equation true? In what sense is it false?
500
Chapter 7 / Principles of Integral Evaluation
73. Writing Explain how the product rule for derivatives and the technique of integration by parts are related. 74. Writing For what sort of problems are the integration techniques of substitution and integration by parts "competing"
techniques? Describe situations, with examples, where each of these techniques would be preferred over the other.
UICK CHECK ANSWERS 7.2
I- (a) j f'((x)G(x)dx 3. (a) (X- - 4 ) e2
(b) uv
vdu
2. (a) hut; xdx (b) x
(x - 1)ln(* - 1) - x + C (c) i
2; sinxdx (c) sin'1*; dx (d) x;
:dX
4. -i sin2*cos* - 2 cos*
INTEGRATING TRIGONOMETRIC FUNCTIONS In the last section we derived reduction formulas for integrating positive integer powers of sine, cosine, tangent, and secant. In this section we will show how to work with those reduction formulas, and we will discuss methods for integrating other kinds of integrals that involve trigonometric functions. INTEGRATING POWERS OF SINE AND COSINE
We begin by recalling two reduction formulas from the preceding section.
/
sin" * d* = — sin" ' x cos * H n
cos" x dx = - cos""1 x sin x H n /
n
n
/ sin"" xdx
(D
/ cos"" x dx J
(2)
J
In the case where n = 2, these formulas yield / sin2 x dx — — j sin x cos x + ^ I dx = ^x — | sin * cos * + C /" cos2 * c?* = 4 cos x sin * + ^ / dx = \x. + I sin x cos ;c + C
(3) (4)
Alternative forms of these integration formulas can be derived from the trigonometric identities sin2* = ^(1 — cos 2.x) and cos2* = |(1 +cos2jc) (5-6) which follow from the double-angle formulas cos 2x = 1 — 2 sin2 x
and
cos 2x = 2 cos2 x — I
These identities yield / sin2 x dx = \ I (1 - cos 2x) dx = \x - \ sin 2x + C
(7)
j cos2 x dx = \ I (1 + cos 2x) dx = ^x + 2 sin 2x + C
(8)
7.3 Integrating Trigonometric Functions
501
Observe that the antiderivatives in Formulas (3) and (4) involve both sines and cosines, whereas those in (7) and (8) involve sines alone. However, the apparent discrepancy is easy to resolve by using the identity sin 2x — 2 sin x cos x
to rewrite (7) and (8) in forms (3) and (4), or conversely. In the case where n = 3, the reduction formulas for integrating sin3 x and cos3 x yield / sin3 x dx = — 5 sin2 x cos x + f / sin x dx = — \ sin2 x cos x — | cos x + C (9) / cos3 x dx — | cos2 x sin x + | / cos x dx = | cos2 x sin x + | sin x + C If desired, Formula (9) can be expressed in terms of cosines alone by using the identity sin2 x = 1 — cos2*, and Formula (10) can be expressed in terms of sines alone by using the identity cos2 x = 1 — sin2 x. We leave it for you to do this and confirm that TECHNOLOGY MASTERY
sin3 x dx =
The Maple CAS produces forms (11) and (12) when asked to integrate sin3* and cos3*, but Mathematica produces
sin xdx = — | cos* + ^ cos 3* + C
cos3 x — cos x + C
cos3 x dx = sin x - | sin3 x + C
(ID (12)
We leave it as an exercise to obtain the following formulas by first applying the reduction formulas, and then using appropriate trigonometric identities.
cos3xdx = | sin*
sin4 x dx = x -
+ -^ sin 3* + C Use trigonometric identities to reconcile the results of the two programs.
sin 2x +
sm
4* + C
cos4 x dx = |jc + i sin 2x + ^ sin 4x + C
(13)
(14)
> Example 1 Find the volume V of the solid that is obtained when the region under the curve y — sin2 x over the interval [0, n] is revolved about the x-axis (Figure 7.3.1). Solution. Using the method of disks, Formula (5) of Section 5.2, and Formula (13) above yields ^ V = / it sin4 xdx = n\\x — ^ sin2;c + ^ sin4jc]" = |jr2 -4 Jo INTEGRATING PRODUCTS OF SINES AND COSINES
If m and « are positive integers, then the integral A Figure 7.3.1
sinm x cos" x dx can be evaluated by one of the three procedures stated in Table 7.3.1, depending on whether m and n are odd or even.
Example 2
Evaluate (a) /"sin4 x cos5 x dx
(b) /sir / sin4 x cos4 x dx
(10)
502
Chapter 7 / Principles of Integral Evaluation Table 7.3.1 INTEGRATING PRODUCTS OF SINES AND COSINES
/ sinm x cos" x dx
PROCEDURE
RELEVANT IDENTITIES
n odd
• Split off a factor of cos x. • Apply the relevant identity. • Make the substitution u = sin x.
cos2* = 1 - sin2*
m odd
• Split off a factor of sin x. • Apply the relevant identity. • Make the substitution u = cos x.
sin 2 * = 1 - cos2*
fm even [n even
Solution (a).
• Use the relevant identities to reduce the powers on sin x and cos x.
("sin2* = jO ~ cos 2*) \cos2* = |(1 +cos2*)
Since n = 5 is odd, we will follow the first procedure in Table 7.3.1: / sin4 x cos5 x dx — I sin4 x cos4 x cos x dx = I sin 4 x(l — sin 2 jt) 2 cos;t
(u4 - 2u6 + « 8 ) du
= =
5U
~ 1U
~*~ 9U
= 5 sin5 x - | sin7 x + | sin9 * + C Solution (jb). Since m = n = 4, both exponents are even, so we will follow the third procedure in Table 7.3.1: / sin4 x cos4 x dx — I
(sin2x)2(cos2x)2dx
= I (\[l -cos2;t]) 2 (i[l+cos2jt]) 2 dx
sin 2x dx
Note that this can be obtained more directly from the original integral using the identity i sinjccos* — \ sin2*.
= i / sin u du — ^ (|w — j sin 2w + 35 sin4w) + C 128
128
1024
Sln
o* + C
Formula (i3)
7.3 Integrating Trigonometric Functions
503
Integrals of the form / sin mx cos nx dx ,
I sinmxsinnxdx,
I cos mx cos nx dx
(15)
can be found by using the trigonometric identities sin a cos ft — ±
- ft) + sm(a + ft)]
(16)
sin a sin ft = ±[cos(a - ft) - cos(a + ft)]
(17)
cos a cos ft = ±[cos(a - ft) + cos(a + ft)]
(18)
to express the integrand as a sum or difference of sines and cosines. *• Example3 Evaluate / sin7jccos3x dx. Solution.
Using (16) yields
/ sin Ix cos 3x dx = | / (sin 4x + sin lOx) dx = — ^ cos 4x — ^ cos ICbc + C < INTEGRATING POWERS OF TANGENT AND SECANT
The procedures for integrating powers of tangent and secant closely parallel those for sine and cosine. The idea is to use the following reduction formulas (which were derived in Exercise 64 of Section 7.2) to reduce the exponent in the integrand until the resulting integral can be evaluated: f tan"-'* f „, / tan" x dx = / tan""2 x dx (19) J n- 1 J
r sec" /sec" x dx —
x tan x n —2 f 1 / sec n —1 n—lj
1
x dx
(20)
In the case where n is odd, the exponent can be reduced to 1, leaving us with the problem of integrating tanx or secjc. These integrals are given by tan x dx = In I sec x I + C
(21)
sec x dx — In | sec x + tan x \ + C
(22)
Formula (21) can be obtained by writing j tan x dx = / — dx cosx J co /
= -ln|cos;c| + C = In | sec* | + C
LduM=:CO " . -saixdx ln|cos*| = -l
i |cos.r|
To obtain Formula (22) we write
[ (' sec x + tan xN sec x dx = I sec x I sec x + tan x = ln|secjc+tanx| + C
sec x + sec x tan x dx sec x + tan x
» = ^« + tan^
I d« = (sec' jc + sec x tan *) dx
504
Chapter 7 / Principles of Integral Evaluation
The following basic integrals occur frequently and are worth noting: / tan2 xdx = tanx — x + C
(23)
I sec2 x dx = tan x + C
(24)
Formula (24) is already known to us, since the derivative of tan x is sec2 jc. Formula (23) can be obtained by applying reduction formula (19) with n = 2 (verify) or, alternatively, by using the identity 1 + tan2 x = sec2 x to write
/ tan2 x dx dx = = II (sec2 x - 1) dx = tan x - x + C
The formulas I tan3 xdx = ±tan2 x -ln|sec;c|
(25)
/ sec3 xdx = 1 sec x tan x + -^In \secx + tan x\ + C
(26)
can be deduced from (21), (22), and reduction formulas (19) and (20) as follows: / tan3 xdx = ^tan2x — I tanxdx = ^tan2 x — In \secx\ + C I sec3 xdx = | sec x tan x + | / secxdx = ^ sec x tan x + ^ln \secx + tan x\ + C INTEGRATING PRODUCTS OF TANGENTS AND SECANTS
If m and n are positive integers, then the integral tanm x sec" x dx can be evaluated by one of the three procedures stated in Table 7.3.2, depending on whether m and n are odd or even.
Table 7.3.2 INTEGRATING PRODUCTS OF TANGENTS AND SECANTS
tanm x sec" x dx
PROCEDURE
RELEVANT IDENTITIES
n even
Split off a factor of sec2*. Apply the relevant identity. Make the substitution u = tan x.
sec2* = tan 2 * +
modd
Split off a factor of sec x tan x. Apply the relevant identity. Make the substitution u = sec x.
tan 2 * = sec2*- 1
Use the relevant identities to reduce the integrand to powers of sec * alone. Then use the reduction formula for powers of sec *.
tan 2 * = sec 2 *— 1
TOT even [nodd
7.3 Integrating Trigonometric Functions
505
> Example 4 Evaluate (a) / tan2 x sec4 x dx Solution (a).
(b) / tan3 x sec3 x dx
(c) / tan2 x sec x dx
Since n = 4 is even, we will follow the first procedure in Table 7.3.2:
/ tan2 x sec4 x dx = I tan2 x sec2 x sec2 x dx = / tan2 x (tan2 x + 1) sec2 ;c dx
- jM 5 + i«3 + C = i tan5 jc + 5 tan3 x + C Solution (b).
Since m = 3 is odd, we will follow the second procedure in Table 7.3.2:
/ tan3 x sec3 x dx — I tan2 x sec2 x (sec x tan x) dx = / ((sec2 x — 1) sec2 x(secx tan jc) dx = fI (u2-l)u2du = it*5 - jM 3 + C = 5 sec5 x - | sec3 x + C Solution (c). Since m = 2 is even and « = 1 is odd, we will follow the third procedure in Table 7.3.2: / tan2 x sec x dx = I (sec2jc — l)secxdx =
J
I SBC3 X dx — I Sec X dx J
See (26) and (22).
= \ sec x tan* + jln |secx +tan;c| — In |secx +
+C
= i sec x tan x — \ In | sec x + tan x | + C ^ AN ALTERNATIVE METHOD FOR INTEGRATING POWERS OF SINE, COSINE, TANGENT, AND SECANT With the aid of the identity
1 + cot? x = esc2 x the techniques in Table 7.3.2 can be adapted to evaluate integrals of the form
r
The methods in Tables 7.3.1 and 7.3.2 can sometimes be applied if m = 0 or n = 0 to integrate positive integer powers of sine, cosine, tangent, and secant without reduction formulas. For example, instead of using the reduction formula to integrate sin x, we can apply the second procedure in Table 7.3.1: / sin3 x dx = I (sin2 x) sin x dx
cot"1 x esc" x dx
= I (1 — cos2 x) sinxdx
/
It is also possible to derive reduction formulas for powers of cot and esc that are analogous to Formulas
= - I (\-u2)du
(19)
I
and (20).
u + C = | cos3 x — cos x + C which agrees with (11).
506
Chapter 7 / Principles of Integral Evaluation MERCATOR'S MAP OF THE WORLD
The integral of sec x plays an important role in the design of navigational maps for charting nautical and aeronautical courses. Sailors and pilots usually chart their courses along paths with constant compass headings; for example, the course might be 30° northeast or 135° southeast. Except for courses that are parallel to the equator or run due north or south, a course with constant compass heading spirals around the Earth toward one of the poles (as in the top part of Figure 7.3.2). In 1569 the Flemish mathematician and geographer Gerhard Kramer (1512-1594) (better known by the Latin name Mercator) devised a world map, called the Mercator projection, in which spirals of constant compass headings appear as straight lines. This was extremely important because it enabled sailors to determine compass headings between two points by connecting them with a straight line on a map (as in the bottom part of Figure 7.3.2). If the Earth is assumed to be a sphere of radius 4000 mi, then the lines of latitude at 1 ° increments are equally spaced about 70 mi apart (why?). However, in the Mercator projection, the lines of latitude become wider apart toward the poles, so that two widely spaced latitude lines near the poles may be actually the same distance apart on the Earth as two closely spaced latitude lines near the equator. It can be proved that on a Mercator map in which the equatorial line has length L, the vertical distance Dp on the map between the equator (latitude 0°) and the line of latitude f!° is
2*
sec x dx
(27)
A Figure 7.3.2 A flight path with constant compass heading from New York City to Moscow follows a spiral toward the North Pole but is a straight line segment on a Mercator projection. U1CK CHECK EXERCISES 7.3
(See page 508 for answers.)
1. Complete each trigonometric identity with an expression involving cos 2.x. (a) sin x = (b) cos2 x = 2 (c) cos x — sin x = 2. Evaluate the integral. (a) / sec2 x dx =
3. Use the indicated substitution to rewrite the integral in terms of u. Do not evaluate the integral. (a) / sin 2 *cos;tdx; M = sinjt (b) / sin 3 ;ccos 2 ;t
(b)/tan2^ =
(d) / tan3 x sec* dx; u =
(c) / sec x dx = (d) / tan x dx =
EXERCISE SET 7.3
1-52 Evaluate the integral.
5. / sin3 aQ d9
6. / cos3atdt 8. / sin3 x cos3 x dx
1. / cos3 x sin x dx
2. I sin5 3x cos 3x dx
7. I sin ax cos ax dx
3. / sin2 59 d9
4. f cos2 3x dx
9. / sin2 t cos3 t dt
10. / sin3 x cos2 x dx
7.3 Integrating Trigonometric Functions 11. / sin2 x cos2 x dx
12. / si sin2 x cos4 x dx
13. / sin 2x cos 3x dx
14. / sin 36> cos 26 dO
15. / sinxcos(x/2)dx
16. /
/* JT/2
17. / Jo
/*
cos 3 xdx
j.jr/3
-I
19. /
4
3
sin 3x cos 3x dx
fw
20. /
JO
cos 50 d'9
22. 22. /
Jo
56. The integral / tan4 x sec5 x dx is equivalent to one whose integrand is a polynomial in secx. 57. Let m, n be distinct nonnegative integers. Use Formulas (16)-(18) to prove: 2 (a) /I J(. Jo
sin si mx cos nx dx = 0
(b) /
cos mx cos nx dx = 0
2
sin kx dx
n2n
Jo 2
23. /"sec (2x-l)dx x
25.': I/ ee~* tan(e~ ) dx
(c)
24. /f tanSxdx
dx
•/
29. / tan2 x sec2 x dx
30. / tan x sec
31. f tan4xsec44xdx
32. f/ tan4 9 sec4 0 d9
33. / sec5 x tan3 x dx
34. / tan5 6 sec 9 d6 34.
35.. // tan x sec x dx
36. / tan2 x sec3 x dx
37.. / tan tan tf sec31 dt
38. / tan x sec5 x dx
39.. /[ sec 4 xdx
40. / sec5 x dx
41. /
42. / tan 4 xdx
43.. / V tan x sec
44. / tan x sec3'2 x dx a/6
fir/8
/
Jo
I
sin si mx sin nx dx = 0.
Jo
26.
27. / sec4xdx
45 . /
i
,2*
r 2n
sin 4x cos cos 2x dx
55. The trigonometric identity sin a cos ft = ^ [sin (a — ft) + sin (a + ft)] is often useful for evaluating integrals of the form
rfZn"
2
J-Jt
r-f-TT/6 7T/6
21.. /I
sin - cos — dx 2 2
18. / Jo
507
sec3 26> tan 29 d9
/•1/4
f*72 47 . / Jo
48. /
49. / cot3 jc esc3 x dx
50. / cot2 3t sec 3f dt
1
4
sec JTX tan nx dx
58. Evaluate the integrals in Exercise 57 when m and n denote the same nonnegative integer. 59. Find the arc length of the curve y = In (cos x) over the interval [0, jr/4]. 60. Find the volume of the solid generated when the region enclosed by y = tan x, y = 1, and x = 0 is revolved about the x-axis. 61. Find the volume of the solid that results when the region enclosed by y = cosx, y = sinx, x = 0, and x = n/4 is revolved about the x-axis. 62. The region bounded below by the x-axis and above by the portion of y = sin x from x = 0 to x = n is revolved about the x-axis. Find the volume of the resulting solid. 63. Use Formula (27) to show that if the length of the equatorial line on a Mercator projection is L, then the vertical distance D between the latitude lines at a° and ft° on the same side of the equator (where a < ft) is sec/3° +tan£ c D = — In 2n sec a ° + tan a c 64. Suppose that the equator has a length of 100 cm on a Mercator projection. In each part, use the result in Exercise 63 to answer the question. (a) What is the vertical distance on the map between the equator and the line at 25 ° north latitude? (b) What is the vertical distance on the map between New Orleans, Louisiana, at 30° north latitude and Winnipeg, Canada, at 50° north latitude? FOCUS ON CONCEPTS
51. / cot xdx
52. / esc x dx
53-56 True-False Determine whether the statement is true or false. Explain your answer. 53. To evaluate f sin5 x cos8 x dx, use the trigonometric identity sin x = I — cos 2 x and the substitution u = cos*. 54. To evaluate / sin8 x cos5 x dx, use the trigonometric identity sin2 x = 1 — cos2 x and the substitution u = cos x.
65. (a) Show that esc x dx = — In esc x + cot x + C
(b) Show that the result in part (a) can also be written as / cscx dx = In \cscx — cotx +C and I cscx dx = In | tan |jc| + C
508
Chapter 7 / Principles of Integral Evaluation
and use your result together with Exercise 65 to evaluate dx sin x + cos x 67. Use the method of Exercise 66 to evaluate (a, b not both zero)
a sinx + bcosx
68. (a) Use Formula (9) in Section 7.2 to show that /'ir/2 n - 1 fn/2 I sin" x dx = / sin" x dx (n > 2) Jo n Jo (b) Use this result to derive the Wallis sine formulas: nil 71 1 • 3 • 5 • • • (n - 1) ( n even sin" x dx = ^ • >2 2 2-4-6---n / /• /•JT/2
2 - 4 - 6 - - - ( « - 1 ) /nodd / sin" x dx = and > 3 Jo 3 •5 •7 •••n 69. Use the Wallis formulas in Exercise 68 to evaluate pitli
(a) / Jo
sin3 x dx
r*
it12
66. Rewrite sin x + cos x in the form
fit fi li
(b) / Jo
sin5 x dx sin x dx. (d) / . Jo 70. Use Formula (10) in Section 7.2 and the method of Exercise 68 to derive the Wallis cosine formulas: fa/2
/ cos" Jo
_ 7T
2
1 • 3 • 5 • • • (« - 1)
2-4-6---«
/ n even \ \and > 2j
f-x/2 2-4-6--.(«-!) n odd \ / cos" x dx = and > 3J Jo 3 •5 •7 •••n 71. Writing Describe the various approaches for evaluating integrals of the form r m sin x cos" x dx
Into what cases do these types of integrals fall? What procedures and identities are used in each case? 72. Writing Describe the various approaches for evaluating integrals of the form tanm x sec" x dx Into what cases do these types of integrals fall? What procedures and identities are used in each case?
sin4 x dx
I/QUICK CHECK ANSWERS 7.3 1. (a)
1 — cos 2x
(b)
1 + cos 2x
(c) cos2;t 2. (a) tanjc + C (b) tan* - x + C (c) In | secz + tan*| + C (d) In | secx| + C
3. (a) ju2du (b) /'(u 2 - l)u2 du (c) f u3 du (d)
f(u2-l)du
TRIGONOMETRIC SUBSTITUTIONS In this section we will discuss a method for evaluating integrals containing radicals by making substitutions involving trigonometric functions. We will also show how integrals containing quadratic polynomials can sometimes be evaluated by completing the square.
THE METHOD OF TRIGONOMETRIC SUBSTITUTION
To start, we will be concerned with integrals that contain expressions of the form
in which a is a positive constant. The basic idea for evaluating such integrals is to make a substitution for x that will eliminate the radical. For example, to eliminate the radical in the expression V" — x2, we can make the substitution
(1)
Jt=asin0, which yields la2 - x2 = \/V - a2 sin2 0 = v/a 2 (l - sin 2 6>) = OV COS2 9 = a\COs9\ = a COS 0
cosS > 0 since -jr/2 < 0 < 7T/2
7.4 Trigonometric Substitutions
509
The restriction on 6 in (1) serves two purposes—it enables us to replace |cos$| by cos# to simplify the calculations, and it also ensures that the substitutions can be rewritten as 0 = sin"1 (x/a), if needed. dx
> Example 1 Evaluate
Solution. To eliminate the radical we make the substitution ;t=2sin6', This yields
r
dx
dx = 2cos9d9
(
2cos9d9
(2sin0) 2 v / 4-4sin 2 0
2cos9d9
d9 sin2 9
- 1 [ csc29d9 = —cot0 4
(2)
-~4j
At this point we have completed the integration; however, because the original integral was expressed in terms of x, it is desirable to express cot 9 in terms of x as well. This can be done using trigonometric identities, but the expression can also be obtained by writing the substitution x = 2 sin 9 as sin 9 = x/2 and representing it geometrically as in Figure 7.4.1. From that figure we obtain r 2 cot 9 = A Figure 7.4.1
Substituting this in (2) yields dx
C -4
r<
Example 2 Evaluate / J\ Solution. There are two possible approaches: we can make the substitution in the indefinite integral (as in Example 1) and then evaluate the definite integral using the x -limits of integration, or we can make the substitution in the definite integral and convert the ^-limits to the corresponding 9-limits. Method 1. Using the result from Example 1 with the *-limits of integration yields •J2
/
V2
dx
V3-1
r2
Method 2. The substitution x = 2sin6> can be expressed as x/2 — sin9 or 9 = sin l(x/2), so the 9 -limits that correspond to x = 1 and x = \/2 are x = l:
9 = sin"1 (1/2) = jr/6
510
Chapter 7 / Principles of Integral Evaluation
Thus, from (2) in Example 1 we obtain
C
dx CSC 9 dO
Convert jr-limits to 0-limits.
J\
V3-1 Jjt/6
Example 3
x
Find the area of the ellipse
Solution. Because the ellipse is symmetric about both axes, its area A is four times the area in the first quadrant (Figure 7.4.2). If we solve the equation of the ellipse for y in terms of x, we obtain i y — ±— va 7 —x 2 a where the positive square root gives the equation of the upper half. Thus, the area A is given by .a b _ _ 4b .a A — 4 / - Va 2 — x2 dx = — / V a 2 — x2 dx Jo a a Jo
To evaluate this integral, we will make the substitution x — a sin 9 (so dx = a cos 9 dO) and convert the ^-limits of integration to ^-limits. Since the substitution can be expressed as 9 — sin"1 ( x / a ) , the ^-limits of integration are
A Figure 7.4.2
x=a: 9 = sii Thus, we obtain A = — / Va2 -x2dx = — I a Jo a Jo
Va2 - a2 sin2 9 -acosOdO
/2 4b r = — / a cos 9 • a cos 9 d9 a Jo
f>7T/2
/»7T/2
= 4ab
Jo
cos29d9-4ab
Jo
-1JT/2 9 + - sin 29 \ =2ab\ 2 Jo L2
[ REMARK
TECHNOLOGY MASTERY If you have a calculating utility with a numerical integration capability, use it and Formula (3) to approximate n to three decimal places.
1
-(l+cos29)d9 2 0 = nab -4 J
In the special case where a = b, the ellipse becomes a circle of radius a, and the area formula becomes A = no1, as expected. It is worth noting that
" i2 va — x2 dx = ^7 / a
(3)
since this integral represents the area of the upper semicircle (Figure 7.4.3).
Thus far, we have focused on using the substitution x = a sin 9 to evaluate integrals involving radicals of the form va 2 — x2. Table 7.4.1 summarizes this method and describes some other substitutions of this type.
7.4 Trigonometric Substitutions
511
Table 7.4.1 TRIGONOMETRIC SUBSTITUTIONS EXPRESSION IN THE INTEGRAND
SUBSTITUTION
RESTRICTION ON 0
SIMPLIFICATION
x = a sin 6
-n!2 <6
a2 - x2 = a2 - a2 sin2 0 = a 2 cos2 6
x = a tan 6
-n!2 < 9 < n!2
a2+x2 = a2 + a2 Ian2 0 = a2 sec2 8
x = a sec i
f o < 0 < 7 T / 2 (ifx>a) \n!2 <6
x2 - a2 = a2 sec2 9 - a2 = a2 tan
> Example 4 ure 7.4.4).
Find the arc length of the curve y — x2/2 from x — 0 to .x = 1 (Fig-
Solution. From Formula (4) of Section 5.4 the arc length L of the curve is 'l+x2dx
A Figure 7.4.4
The integrand involves a radical of the form Va2 + x2 with a — 1, so from Table 7.4.1 we make the substitution ,„ . x = tanO, -n/2 < 9 < n/2 dx = sec2 0 or dx = sec2 9 dO Since this substitution can be expressed as 9 = tan ' x, the 0-limits of integration that correspond to the ^-limits, x = 0 and x = 1, are x=0: e ^ t a n ^ ' O ^ O x - l : 9 = tan"1 1 = n/4 Thus, /> 1
•\/H-tan 2 0sec 2 0d0
L= / Jo /•jr/4
= /I
V/sec1/ — 0 sec 7 (9 a 6
11 + tan2 e = sec2 e j
/•7T/4
~ JO
rn/4
= I Jo
secZ 3 0 d9
' sec9 > 0 since-
r/4
1.148
Example 5 Evaluate
25 - dx, assuming that x > 5.
Formula (26) of Section 7.3
512
Chapter 7 / Principles of Integral Evaluation
Solution. The integrand involves a radical of the form «Jx2 — a2 with a = 5, so from Table 7.4.1 we make the substitution , 0<0<7r/2 — =5sec6>tan0 dB
or dx = 5sec6tan6dQ
Thus, v6c2 - 25
f ^25 sec2 9- 25/•« ~ fl tin fi"\Hfi rr 5 sec 9 5ltan0| -J -l(5sec0tan0)«/0 5 sec 6»
-1
f tan2 Odd
[^7>^)sh^eO<8 < 31/2
= 5 / (sec^0-l)rf0 = 5tan0-50 + C J
To express the solution in terms of x, we will represent the substitution x = 5 sec 6 geometrically by the triangle in Figure 7.4.5, from which we obtain tan0 = From this and the fact that the substitution can be expressed as 0 — sec ' (x/5), we obtain •dx = Vx2-25-5sec~l (-}+C * A Figure 7.4.5
INTEGRALS INVOLVING ax2 + bx + c
Integrals that involve a quadratic expression ax2 + bx + c, where a ^ 0 and b ^ 0, can often be evaluated by first completing the square, then making an appropriate substitution. The following example illustrates this idea. Example 6 Evaluate / —— 2 J
•dx.
X -'
Solution. Completing the square yields x2 - 4x + 8 = (x2 - 4x + 4) + 8 - 4 = (x - 2)2 + 4
Thus, the substitution
M — x — 2,
du = dx
yields —
uji — i
.J/ 1 2
X.
-2)2 + 4"'
[
£. M2-
I
/:
2u
2
—
si,, 1
+4d'
J
du
du M
2 /
2 _i_
i
4
du u2 + 4
1 -i /n \ ln(M 2 + 4 ) + 2 J tan ~2
= -ln[(jt-2) 2 +4]
+2 u2 + 4
u 24
ffl
7.4 Trigonometric Substitutions UICK CHECK EXERCISES 7.4
513
(See page 514 for answers.)
1. For each expression, give a trigonometric substitution that will eliminate the radical, (a)
(d)
2. If x = 2sec6> and 0 < 9 < jr/2, then (a) sin 9 = (b) cos 6 = (c) tan 0 = 3. In each part, state the trigonometric substitution that you would try first to evaluate the integral. Do not evaluate the integral.
(f)
(e)
/ / /
4. In each part, determine the substitution u. C 1 f 1 (a) / —2 dx = \ —2 du J x — 2x + 10 7 u + 32
(a) f v/9 + x2 dx (b) [V9-.
(0/VT^
EXERCISE SET 7.4
CAS
1-26 Evaluate the integral. 3 2
1. I V4-x dx 1
X
I
rlr
1C
A
9
x
f /
3x3
r
6 1
8
' J
dx
dx 2
1 x V9S^4 dx
13 f 15 f i
>/
dX r 2
dx 17 f ' J (4x2 - 9)3/2
^ dx
21. f 5.x3 /I Jo
„ f2
f 2 dX J x Jx2 - 16
10 1I X r 3 iA l\l. vJ
dx
^dx
2
fir Av ax
12. / — J t
16 /
/~
19 / exi/l
Jr
Id
1
*
dr
18
27-30 True-False Determine whether the statement is true or ? alse. Explain your answer. II 27. An integrand involving a radical of the form *Ja2 — x2 suggests the substitution x = a sin 9. 28. The trigonometric substitution x — a sin 9 is made with the restriction 0 < 0 < TT. 29. An integrand involving a radical of the form -\A2 — a2 suggests the substitution x = a cos 6. ju. me aica enuiuseu uy me ellipse -t -|- «t_y — I I S T Z / Z , .
dt
31. The integral
/•
x
dX
14 /
2> 3 2
I n
X
t
2 J x2^^^ f x2
5 f ' J (4 + x2)2
^
2. /Vl - 4x dx
J /16-.X2 dX
/
2
1
^
1 Oy^
fl/2 22. I
j-4
r/v
/
/"
j
cos6» dx
can be evaluated either by a trigonometric substitution or by the substitution u = x2 + 4. Do it both ways and show that the results are equivalent. 32. The integral ,. ^2 J x2+4 can be evaluated either by a trigonometric substitution or by algebraically rewriting the numerator of the integrand as (x2 + 4) — 4. Do it both ways and show that the results are equivalent.
22
,— /• V2^; -4
33. Find the arc length of the curve y = In x from x = 1 to jc = 2. 34. Find the arc length of the curve y = jc 2 from;t = Oto* = 1.
514
Chapter 7 / Principles of Integral Evaluation
35. Find the area of the surface generated when the curve in Exercise 34 is revolved about the ;t-axis.
50. I (xcosx + sinx)Vl + x2sin2 xdx
36. Find the volume of the solid generated when the region enclosed by x = y(l — y 2 ) 1 ^ 4 , y = 0, y = 1, and x = 0 is revolved about the y-axis.
51. (a) Use the hyperbolic substitution x — 3 sinh u, the identity cosh2 u — sinh2 u = 1, and Theorem 6.8.4 to evaluate dx
I
,,/
37-48 Evaluate the integral. S dx xi - 4X + 5
39. /
(b) Evaluate the integral in part (a) using a trigonometric substitution and show that the result agrees with that obtained in part (a). 52. Use the hyperbolic substitutionn x = cosh u, the identity sinh2 u = i(cosh 2u — 1), and the results referenced in Exercise 51 to evaluate
V3+ x2
lr
"I A 2 -6* + 10 r i. f \/3 - 2x - x2dx
f
53. Writing The trigonometric substitution x=asin9, —jr/2 < 9 < ji/2, is suggested for an integral whose integrand involves \/a2 — x2. Discuss the implications of restricting 9 to jt/2 < 9 < 3iz/2, and explain why the restriction —n/2 < 9 < 7C/2 should be preferred. 54. Writing The trigonometric substitution x = a cos 9 could also be used for an integral whose integrand involves Determine an appropriate restriction for 9 with the substitution x = a cos 9, and discuss how to apply this substitution in appropriate integrals. Illustrate your discussion by evaluating the integral in Example 1 using a substitution of this type.
45. / 47
48
'/ ./I
[c] 49-50 There is a good chance that your CAS will not be able to evaluate these integrals as stated. If this is so, make a substitution that converts the integral into one that your CAS can evaluate.
49. / cos;tsin.x:V 1 —
I/QUICK CHECK ANSWERS 7.4 1.
x = a sin 9 (b) x = a tan 9 (c) x = a sec 9
x = 3 tan 9
x (c) x = i sin6> (d) x = 3 sec6> (e) x = V3 tan6> (f) x = i tan6»
x 2 4. (a) x - 1 (b) x - 3 (c) x + 2
x = 3 sin 9
INTEGRATING RATIONAL FUNCTIONS BY PARTIAL FRACTIONS Recall that a rational function is a ratio of two polynomials. In this section we will give a general method for integrating rational functions that is based on the idea of decomposing a rational function into a sum of simple rational functions that can be integrated by the methods studied in earlier sections.
PARTIAL FRACTIONS
In algebra, one learns to combine two or more fractions into a single fraction by finding a common denominator. For example,
2
3
x-4
x+ 1
2(x + 1) + 3(* - 4)
5x- 10 -3*-4
(D
7.5 Integrating Rational Functions by Partial Fractions
515
However, for purposes of integration, the left side of (1) is preferable to the right side since each of the terms is easy to integrate: / , *~-~,dx = I --dx + I - d j c = 2 1 n | j c - 4 | + 3 1 n | j c + l | + C J x2-3x-4 J x -4 J x +l Thus, it is desirable to have some method that will enable us to obtain the left side of (1), starting with the right side. To illustrate how this can be done, we begin by noting that on the left side the numerators are constants and the denominators are the factors of the denominator on the right side. Thus, to find the left side of (1), starting from the right side, we could factor the denominator of the right side and look for constants A and B such that
A IB _5x - 10 _ __ (x - 4)(x + 1) x - 4 x+l
__
(2\
One way to find the constants A and B is to multiply (2) through by (x — 4)(x + 1) to clear fractions. This yields 5x - 10 = A(x + 1) + B(x - 4) (3) This relationship holds for all x, so it holds in particular if* = 4 or Jt = — 1. Substituting x = 4 in (3) makes the second term on the right drop out and yields the equation 10 = 5 A or A = 2; and substituting x = — 1 in (3) makes the first term on the right drop out and yields the equation — 15 = —56 or B = 3. Substituting these values in (2) we obtain 5x - 10 2 3 - =--1 -(x - 4)(x + 1) x - 4 x + l
(4)
which agrees with (1). A second method for finding the constants A and B is to multiply out the right side of (3) and collect like powers of x to obtain 5x - 10 = (A + B)x + (A- 4B)
Since the polynomials on the two sides are identical, their corresponding coefficients must be the same. Equating the corresponding coefficients on the two sides yields the following system of equations in the unknowns A and B : A+ B= 5 A-4B = -10
Solving this system yields A = 2 and B — 3 as before (verify). The terms on the right side of (4) are called partial fractions of the expression on the left side because they each constitute part of that expression. To find those partial fractions we first had to make a guess about their form, and then we had to find the unknown constants. Our next objective is to extend this idea to general rational functions. For this purpose, suppose that P(x)/Q(x) is a proper rational function, by which we mean that the degree of the numerator is less than the degree of the denominator. There is a theorem in advanced algebra which states that every proper rational function can be expressed as a sum F2(x) + ••• + Fn(x)
where F\ (x), F2(x ) , . . . , Fn(x) are rational functions of the form A (ax + b)k
or
Ax + B (ax + bx + c)k 2
in which the denominators are factors of Q(x). The sum is called the partial fraction decomposition of P(x)/Q(x), and the terms are called partialfractions. As in our opening example, there are two parts to finding a partial fraction decomposition: determining the exact form of the decomposition and finding the unknown constants.
516
Chapter 7 / Principles of Integral Evaluation FINDING THE FORM OF A PARTIAL FRACTION DECOMPOSITION
The first step in finding the form of the partial fraction decomposition of a proper rational function P(x)/Q(x) is to factor Q(x) completely into linear and irreducible quadratic factors, and then collect all repeated factors so that Q(x) is expressed as a product of distinct factors of the form and (ax2 + bx + c)m
(ax + b)m
From these factors we can determine the form of the partial fraction decomposition using two rules that we will now discuss. LINEAR FACTORS
If all of the factors of Q(x) are. linear, then the partial fraction decomposition of P(x)/Q(x) can be determined by using the following rule:
LINEAR FACTOR RULE For each factor of the form (ax + b)m, the partial fraction decomposition contains the following sum of m partial fractions: A\
r\2
**m
ax + b
(ax + b)2
(ax + b)"
where A\, A2,..., Am are constants to be determined. In the case where m = 1, only the first term in the sum appears.
Example 1 Solution,
Evaluate /
dx
-x-2
The integrand is a proper rational function that can be written as 1 _ 1 x2+x-2 ~ O - l ) O + 2)
The factors x — 1 and x + 2 are both linear and appear to the first power, so each contributes one term to the partial fraction decomposition by the linear factor rule. Thus, the decomposition has the form 1 (jt-l)O+2)
x-l
B x+2
(5)
where A and B are constants to be determined. Multiplying this expression through by (x - \)(x + 2) yields 1 = A(x + 2) + B(x - 1) (6) As discussed earlier, there are two methods for finding A and B: we can substitute values of x that are chosen to make terms on the right drop out, or we can multiply out on the right and equate corresponding coefficients on the two sides to obtain a system of equations that can be solved for A and B. We will use the first approach. Setting x — 1 makes the second term in (6) drop out and yields 1 = 3A or A = |; and setting x — — 2 makes the first term in (6) drop out and yields 1 = —35 or B = — |. Substituting these values in (5) yields the partial fraction decomposition 1 (x- l ) O + 2 )
x-l
x+2
7.5 Integrating Rational Functions by Partial Fractions
517
The integration can now be completed as follows: f dx _ 1I ff dx 11 C( dx dx dx J (x - \)(x + 2) = 3J x-l~3j x + 2 JC-1
x+2
+C
If the factors of Q(x) are linear and none are repeated, as in the last example, then the recommended method for finding the constants in the partial fraction decomposition is to substitute appropriate values of x to make terms drop out. However, if some of the linear factors are repeated, then it will not be possible to find all of the constants in this way. In this case the recommended procedure is to find as many constants as possible by substitution and then find the rest by equating coefficients. This is illustrated in the next example. Example 2
Evaluate
f 2x- __ dx. 2x2
Solution. The integrand can be rewritten as 2x + 4 2x + 4 2 x (x - 2) Although x is a quadratic factor, it is not irreducible since x = xx. Thus, by the linear factor rule, x2 introduces two terms (since m = 2) of the form A B and the factor x — 2 introduces one term (since m = 1) of the form C x-2 so the partial fraction decomposition is 2x+4 _ A B C x2(x - 2) ~ x *2 x-2
(7)
Multiplying by x2(x — 2) yields
2x + 4 = Ax(x - 2) + B(x - 2) + Cx2
(8)
which, after multiplying out and collecting like powers of x, becomes
2x + 4 = (A + C)x2 + (-2A + B)x - 2B
(9)
Setting x = 0 in (8) makes the first and third terms drop out and yields B — -2, and setting x = 2 in (8) makes the first and second terms drop out and yields C = 2 (verify). However, there is no substitution in (8) that produces A directly, so we look to Equation (9) to find this value. This can be done by equating the coefficients of x2 on the two sides to obtain
A + C = 0 or A = -C = -2 Substituting the values A — —2, B = —2, and C = 2 in (7) yields the partial fraction decomposition 2x + 4 —2 —2 2 x2(x — 2)
x
x2
x —2
Thus,
/
2x+4 [dx [dx C dx —~T; -™dx = -2J I x-- 21 2 +2 I -; x2(x-2) J x J x-:
= -2 In | | + - + 2 In \x - 2\ + C = 2 In x
2 x
518
Chapter 7 / Principles of Integral Evaluation QUADRATIC FACTORS
If some of the factors of Q(x) are irreducible quadratics, then the contribution of those factors to the partial fraction decomposition of P(x)l'Q(x) can be determined from the following rule: QUADRATIC FACTOR RULE For each factor of the form (ax2 + bx + c)m, the partial fraction decomposition contains the following sum of m partial fractions: AIX ax 2 + bx+c
A2x + (ax2 + bx + c)2
h
Amx + Bn (ax2 + bx+ c)
where A], AI, ..., Am, B\, BI, ... ,Bm are constants to be determined. In the case where m = 1, only the first term in the sum appears.
Example 3 Evaluate
x2 + x - 2 •dx. - x2 + 3x -
Solution. The denominator in the integrand can be factored by grouping: 3x3 - x2 + 3x - 1 = x2(3x - 1) + (3x - 1) = (3* - l)(x2 + 1) By the linear factor rule, the factor 3x — 1 introduces one term, namely, 3jc-l
and by the quadratic factor rule, the factor x2 + 1 introduces one term, namely,
Bx + C X2+l
Thus, the partial fraction decomposition is x2+x-2 (3x - \)(x2 + 1
Bx + C
3x-l
(10)
Multiplying by (3x - l)(x2 + 1) yields
x2 + x - 2 = A(x2 + 1) + (Bx + C)(3x - 1)
(11)
We couldfindA by substituting x = | to make the last term drop out, and thenfindthe rest of the constants by equating corresponding coefficients. However, in this case it is just as easy to find all of the constants by equating coefficients and solving the resulting system. For this purpose we multiply out the right side of (11) and collect like terms: x2 + x - 2 = (A + 3B)x2 + (-B + 3C)x + (A - C) Equating corresponding coefficients gives
A + 35 = 1 - B + 3C = 1 A - C = -2 To solve this system, subtract the third equation from the first to eliminate A. Then use the resulting equation together with the second equation to solve for B and C. Finally, determine A from the first or third equation. This yields (verify)
7.5 Integrating Rational Functions by Partial Fractions 519
Thus, (10) becomes *2+*-2
(3x-
1)
-I 3x - 1
and TECHNOLOGY MASTERY Computer algebra systems have builtin capabilities for finding partial fraction decompositions. If you have a CAS, use it to find the decompositions in Examples 1, 2, and 3.
4 f x 3 f dx f x2+x-2 ^^ _ 7 I dx dx 2 2 + < / ~7^7 + « 5J x +1 5 J x* + l J (3x- I)0t + l) 5 7 3x-l 7 2 3 = —-ln|3;c- I| + -ln(jc 2 + 1) + - tan"1 * + C <«
-dx.
Example 4 Evaluate
Solution. Observe that the integrand is a proper rational function since the numerator has degree 4 and the denominator has degree 5. Thus, the method of partial fractions is applicable. By the linear factor rule, the factor x + 2 introduces the single term
x+2 and by the quadratic factor rule, the factor (x2 + 3)2 introduces two terms (since m — 2): Bx + C Dx + E + 2 x + 3 (x2 + 3)2
Thus, the partial fraction decomposition of the integrand is 3x4 + 4x3 + I6x2 + 20x + 9
A
Bx + C
Dx + E
(12)
Multiplying by (x + 2)(x2 + 3)2 yields 3x4 + 4x3 + I6x2 + 2Qx + 9 = A(x2 + 3)2 + (Bx
C)(x2 + 3)(x + 2) + (Dx + E)(x + 2) (13)
which, after multiplying out and collecting like powers of x, becomes 3x4 + 4x3 + I6x2 + 2Qx+9 = (A + B)x4 + (2B + C)x3 + (6A + 3B + 2C + D)x2 + (6B + 3C + 2D + E)x + (9A + 6C + 2E)
(14)
Equating corresponding coefficients in (14) yields the following system of five linear equations in five unknowns: A + B= 3 2B + C= 4 6A + 3B + 2C + D= 16 (15) 6B + 3C + 2D + E = 20 9A+6C+2E = 9 Efficient methods for solving systems of linear equations such as this are studied in a branch of mathematics called linear algebra', those methods are outside the scope of this text. However, as a practical matter most linear systems of any size are solved by computer, and most computer algebra systems have commands that in many cases can solve linear systems exactly. In this particular case we can simplify the work by first substituting x = —2
520
Chapter 7 / Principles of Integral Evaluation
in (13), which yields A = 1. Substituting this known value of A in (15) yields the simpler system B= 2 3B + 2C + D=10 6B + 3C + 2D + E = 20
(16)
This system can be solved by starting at the top and working down, first substituting B = 2 in the second equation to get C = 0, then substituting the known values of B and C in the third equation to get D = 4, and so forth. This yields A = l,
B = 2,
C=0, D=4,
£=0
Thus, (12) becomes 3x4 + 4;c3 + 16jc2 + 20* + 9 (x + 2)(x2 + 3)2
1 x +2
2x x2 + 3
4x (x2 + 3)2
and so 20* + 9 -dx (x + 2)(x + 3)2 f dx f 2x 2
=
+
f
x
dX+4
]—2 J^3
JW^dX
= In \x + 2\ + ln(;c2 + 3) - -^—- + C * INTEGRATING IMPROPER RATIONAL FUNCTIONS
Although the method of partial fractions only applies to proper rational functions, an improper rational function can be integrated by performing a long division and expressing the function as the quotient plus the remainder over the divisor. The remainder over the divisor will be a proper rational function, which can then be decomposed into partial fractions. This idea is illustrated in the following example. >• Example 5 Evaluate / J
3jc4 + 3x3 - 5x2 + x dx. x2 + x-2
Solution. The integrand is an improper rational function since the numerator has degree 4 and the denominator has degree 2. Thus, we first perform the long division 3x2 +1 x - 2 I 3* + 3* - 5x2 3x4 + 3x3 - 6x2 x2 + x - 1 4
3
It follows that the integrand can be expressed as 3x4 + 3jc3 - 5*2 + x - 1 , 2 1 = (3jc2 + 1) + 2 2 * +;c-2 * +x-2 and hence x4 + 3x3-5x2+x-l J, /• f , f dx dx = = / (3* + 1) dx + x2 + x-2 J " J x2+x-2
7.5 Integrating Rational Functions by Partial Fractions
521
The second integral on the right now involves a proper rational function and can thus be evaluated by a partial fraction decomposition. Using the result of Example 1 we obtain
3x4
x3 - 5x2 + x x-l • dx = x3 + x H— In +C 2 +x-2 x+2
CONCLUDING REMARKS
There are some cases in which the method of partial fractions is inappropriate. For example, it would be inefficient to use partial fractions to perform the integration 3*24 -dx=ln | x + 2x 3
2x - 8| + C
since the substitution u = x3 + 2x — 8 is more direct. Similarly, the integration
requires only a little algebra since the integrand is already in partial fraction form. I/QUICK CHECK EXERCISES 7.5
(See page 523 for answers.)
1- A partial fraction is a rational function "f the form nr of the fr>rrn
2. (a) What is a proper rational function? (b) What condition must the degree of the numerator and the degree of the denominator of a rational function satisfy for the method of partial fractions to be applicable directly? (c) If the condition in part (b) is not satisfied, what must you do if you want to use partial fractions? 3. Suppose that the function f(x) = P(x)/Q(x) is a proper rational function, (a) For each factor of Q (x) of the form (ax + b)m , the partial fraction decomposition of / contains the following
(b) For each factor of Q(x) of the form (ax2 + bx + c)m, where ax2 + bx + c is an irreducible quadratic, the partial fraction decomposition of / contains the following sum of m partial fractions: 4. Complete the partial fraction di^composition. 2 A (a) ~3
' (x + \)(2x - 1) 2
2x -3x ' (x + l)(3x + 2) 5. Evaluate the integral. , . /• 3 J (x + 1)(1 - 2x) 2
x+ l
1 x +l
B 3x +2
2
2x2 3x (b) f ~ dx >> J (x2+l)(3x + 2)d•
sum of m partial fractions:
EXERCISE SET 7.5
HI CAS
1 -8 Write out the form of the partial fraction decomposition. (Do not find the numerical values of the coefficients.) • oix „
!
(;t-3)(*+4)
3 3 2 ' * -x v-2
1
3
2
'' x (x + 2) 4x x (x2 + 5)2 9-34 Evaluate the integral. 3 C dx ' J x2 -3x -4
" x(x2-4)
4
r
r2r*-9*-9
c 5
l
11 / 9jrUxl +/ yl1 4dx
' (x + 2)3 Qv
2
' (x-l)(x + 6) 1 3x " (x - 2)(x2 + I)2
C dx J x2 - dx - 1
r3
5X 5 [ ~ J 3x2-8x-3 C dx
1 rCv 2
9r
2
/•* -8 , J x +3 f 3;c2 -10 17 ' J x2 4x+4dX f 2x-3 10 1 2 rlr J x - 3x - 10 rx5+x2 + 2 J x3 — x
„ f 2*2+3 ,
23. / dx J x(x-l)2
2
[ X
J/
f ^0
X
11
+ l ^
xr
1
dX
x2
x2
3x + 2
dx
3x2+2x-\" V-5 _
4y3
1
1
-ix
/
[3x2-x + l ax 24. 1 J
Ji
A
522
Chapter 7 / Principles of Integral Evaluation
2.x2 - 10* + 4
(x + l)(x - 3)2 x2
26. /
— dx
•f
j
dx
'•ft
2* 2 -l
29.
f 2x2 - 2x - 1
dx
- dx
(4x - \)(x2
30.
f
dx
3
2
Y +-4- r -I- r +-t- 7 ".: dx 32. / * 2 ":..•;•'.: 2 '•/ (x + l)(x + 3) >•/ (*2 + *1)(*+2 + 2). «/*
'•/
* 3 -2* 2 + 2 * - 2 ^ dx 2 X +l
Wx2
•/
x2+6x + 10
dx
47.
48
+2x
/
33. /
[cl 47-48 Integrate by hand and check your answers using a CAS.
•f
3
x3 - lx2 + 27x - 18 dx -4x2 + 4x-l
FOCUS ON CONCEPTS
49. Show that
/' Jo
dx
50. Use partial fractions to derive the integration formula
35-38 True-False Determine whether the statement is true or false. Explain your answer, m 35. The technique of partial fractions is used for integrals whose integrands are ratios of polynomials.
1 a +x +C - dx = — In •72 -
51. Suppose that ax2 + bx + c is a quadratic polynomial and that the integration
—T2
ax + bx + c
36. The integrand in 2
(x + I) is a proper rational function.
2
dx
produces a function with no inverse tangent terms. What does this tell you about the roots of the polynomial?
dx
52. Suppose that ax2 + bx + c is a quadratic polynomial and that the integration
37. The partial fraction decomposition of
-dx ax2 + bx + c 38. If f ( x ) = P(x)l(x + 5)3 is a proper rational function, then the partial fraction decomposition of f ( x ) has terms with constant numerators and denominators (x + 5), (x + 5)2, and (x + 5)3. 39-42 Evaluate the integral by making a substitution that converts the integrand to a rational function. »• cos 6 dO 40. -=2t dt
° -4 5 + 2 In*
-5
.3*
•dx
-/
produces a function with neither logarithmic nor inverse tangent terms. What does this tell you about the roots of the polynomial? 53. Does there exist a quadratic polynomial ax2 + bx + c such that the integration
f
x dx ax2 + bx + c
produces a function with no logarithmic terms? If so, give an example; if not, explain why no such polynomial can exist.
dx
e2*+4 43. Find the volume of the solid generated when the region enclosed by y = x2/(9 — x2), y — 0, x = 0, and x = 2 is revolved about the ;t-axis. 44. Find the area of the region under the curve y = 1/(1 + ex), over the interval [— In 5, In 5]. [Hint: Make a substitution that converts the integrand to a rational function.]
54. Writing Suppose that P(x) is a cubic polynomial. State the general form of the partial fraction decomposition for
P(x) and state the implications of this decomposition for evaluating the integral f f ( x ) dx. 55. Writing Consider the functions
45-46 Use a CAS to evaluate the integral in two ways: (i) integrate directly; (ii) use the CAS to find the partial fraction decomposition and integrate the decomposition. Integrate by hand to check the results.
•I
•!
(x2
2x
5
4
3)2
dx +4x2 +4x +4
2
(x + 2)3
dx
and 2 /(*) = i—7 £(*) = x-I—7 x1 -4 -4 Each of the integrals f f ( x ) dx and f g ( x ) dx can be evaluated using partial fractions and using at least one other integration technique. Demonstrate two different techniques for evaluating each of these integrals, and then discuss the considerations that would determine which technique you would use.
7.6 Using Computer Algebra Systems and Tables of Integrals
523
•QUICK CHECK ANSWERS 7.5 1
Ax + B
2. (a) A proper rational function is a rational function in which the degree of the numerator is
less than the degree of the denominator, (b) The degree of the numerator must be less than the degree of the denominator. (c) Divide the denominator into the numerator, which results in the sum of a polynomial and a proper rational function. A, A2 Am Aix + B\ A2x + B2 Amx + Bm 2 ' ax + b (ax + b)2 + bx + c (ax2 + bx + c)2 (ax + b)m (ax2 + bx + c) 4. (a) A = 1 (b) B = 2 3 2x2 x+l dx = In C (b) dx = - In |3jc + 2| - tan"1 x + C 2 (x + l)(3x + 2) l-2x /
USING COMPUTER ALGEBRA SYSTEMS AND TABLES OF INTEGRALS In this section we will discuss how to integrate using tables, and we will see some special substitutions to try when an integral doesn't match any of the forms in an integral table. In particular, we will discuss a method for integrating rational functions of sin x and cos x. We will also address some of the issues that relate to using computer algebra systems for integration. Readers who are not using computer algebra systems can skip that material.
INTEGRAL TABLES
Tables of integrals are useful for eliminating tedious hand computation. The endpapers of this text contain a relatively brief table of integrals that we will refer to as the Endpaper Integral Table; more comprehensive tables are published in standard reference books such as the CRC Standard Mathematical Tables and Formulae, CRC Press, Inc., 2002. All integral tables have their own scheme for classifying integrals according to the form of the integrand. For example, the Endpaper Integral Table classifies the integrals into 15 categories; Basic Functions, Reciprocals of Basic Functions, Powers of Trigonometric Functions, Products of Trigonometric Functions, and so forth. The first step in working with tables is to read through the classifications so that you understand the classification scheme and know where to look in the table for integrals of different types.
PERFECT MATCHES
If you are lucky, the integral you are attempting to evaluate will match up perfectly with one of the forms in the table. However, when looking for matches you may have to make an adjustment for the variable of integration. For example, the integral x1 sin x dx
is a perfect match with Formula (46) in the Endpaper Integral Table, except for the letter used for the variable of integration. Thus, to apply Formula (46) to the given integral we need to change the variable of integration in the formula from u to x. With that minor modification we obtain I x2 sin x dx =2x sin x + (2 — x2) cos x + C
Here are some more examples of perfect matches.
524
Chapter 7 / Principles of Integral Evaluation
>• Example 1 Use the Endpaper Integral Table to evaluate (a) (c)
(b) / x2V7 + 3xdx
sinlxcos2xdx '2-x1
(d) / (x3 + 7* + 1) sin^jc dx
•dx
Solution (a). The integrand can be classified as a product of trigonometric functions. Thus, from Formula (40) with m = 7 and n — 2 we obtain
cos 9x 18
sin Ix cos 2x dx =
cos 5x hC 10
Solution (b). The integrand can be classified as a power of * multiplying ~Ja +bx. Thus, from Formula (103) with a = 1 and b = 3 we obtain
I x2 T+3xdx = J 2835
- 252* + 392)(7 + 3x)3/2 + C
Solution (c). The integrand can be classified as a power of x dividing \/a2 — x2. Thus, from Formula (79) with a = V2 we obtain _ Y2
•dx = J2-x2-
+C
Solution (d). The integrand can be classified as a polynomial multiplying a trigonometric function. Thus, we apply Formula (58) with p(x) = x3 + Ix + 1 and a = n. The successive nonzero derivatives of p(x) are p'(x) = 3x2 + 1,
=6
p"(x) = 6x,
and so
/ (x3 + 7* + 1) sin JTX dx 3x2+7 . 6x 6 cos Ttx H -- -2— sin nx -\ -- 5r cos nx -- -4 sin nx + C 7T
Tl
7T
MATCHES REQUIRING SUBSTITUTIONS
Sometimes an integral that does not match any table entry can be made to match by making an appropriate substitution.
Example 2
Use the Endpaper Integral Table to evaluate
(a) I f*&r\?*)dx
(b) I' x^fc^- 4x + 5 dx
Solution (a). The integrand does not even come close to matching any of the forms in the table. However, a little thought suggests the substitution u =e
du = jtenx dx
7.6 Using Computer Algebra Systems and Tables of Integrals
525
from which we obtain / e7lxsin~\enx)dx = -
sin"' udu
The integrand is now a basic function, and Formula (7) yields \ejlx)dx = - f a s i n - ' M + Vl - u2] + C ji L
- \enx sin'1 (enx) +
- e2™ ] + C
71 L
Solution (b). Again, the integrand does not closely match any of the forms in the table. However, a little thought suggests that it may be possible to bring the integrand closer to the form x \/x2 + a2 by completing the square to eliminate the term involving x inside the radical. Doing this yields I x^x2-4x+5dx = I x^/(x2 -4x+4)+~ldx = j x^(x - 2)2 + 1 dx
(1)
At this point we are closer to the formjcV* 2 + a2, but we are not quite there because of the (x — 2)2 rather than x2 inside the radical. However, we can resolve that problem with the substitution u = x — 2, du = dx With this substitution we have x — u + 2, so (1) can be expressed in terms of u as / x^x2 - 4x + 5dx =
(u + 2)V/M2 + 1 du = / u^/u2 + 1 du +21 ju2 1 du
The first integral on the right is now a perfect match with Formula (84) with a — 1 , and the second is a perfect match with Formula (72) with a — 1 . Thus, applying these formulas we obtain Ix
= \(u2 + 1)3
If we now replace u by x — 2 (in which case u2 + I = x2 — 4;t + 5), we obtain I x^x2-4x + 5dx = |(jt:2 - 4x + 5)3/2
(x - 2 ) x 2 - 4x + 5
+ In (x - 2 + V* 2 - 4x + 5 } + C Although correct, this form of the answer has an unnecessary mixture of radicals and fractional exponents. If desired, we can "clean up" the answer by writing -4x + 5)3/2 = (x2 -4x+
- 4x + 5
from which it follows that (verify)
In
• 2 + V*2 - 4* + 5 ) + C
MATCHES REQUIRING REDUCTION FORMULAS
In cases where the entry in an integral table is a reduction formula, that formula will have to be applied first to reduce the given integral to a form in which it can be evaluated.
526
Chapter 7 / Principles of Integral Evaluation
*• Example 3
Use the Endpaper Integral Table to evaluate /
.dx.
T+JC
Solution. The integrand can be classified as a power of x multiplying the reciprocal of V« + bx. Thus, from Formula (107) with a — 1, b — 1, and n = 3, followed by Formula (106), we obtain
/TT7_6 r
J yr
1 ~7j 2;cVl +x 6f 2
C 32
35
35
- — \ vr,/TT7 357
• SPECIAL SUBSTITUTIONS
The Endpaper Integral Table has numerous entries involving an exponent of 3/2 or involving square roots (exponent 1/2), but it has no entries with other fractional exponents. However, integrals involving fractional powers of x can often be simplified by making the substitution u = x]'n in which n is the least common multiple of the denominators of the exponents. The resulting integral will then involve integer powers of u.
> Example 4
Evaluate (a)
jj=dx
/
(b) / Vl+eM*
Solution (a). The integrand contains ;ci/2 and;c1/3, so we make the substitution u — x 1/6 , from which we obtain x = u , dx = 6u du Thus,
/r^"*=/ J
1
l
\/-^-
J
6 1/3 ^ (DM 5^ ) du — fi6
(M )
8 /
By long division
T^
1+M2
from which it follows that
r /
fv /
dx dx
^Tx 1 \ V^
= 66 I I u6 - w 4 + u2 - 1 +
1
= v (\
= fw 7 - |w5
- 6u + 6 tan~' u + C 6 t a n -l (
Solution (b). The integral does not match any of the forms in the Endpaper Integral Table. However, the table does include several integrals containing «Ja + bu. This suggests the substitution u = ex, from which we obtain x = In M ,
dx = — du u
7.6 Using Computer Algebra Systems and Tables of Integrals
527
Thus, from Formula (110) with a = 1 and b — 1, followed by Formula (108), we obtain
/ \/l +exdx = I
-du
I
f
du
= 2V1+M + /
J MV 1 + U
c
= 2 V 1 + M + In '1 + M + 1
Try finding the antiderivative in Example 4(b) using the substitution
u = -/I + e"
— 2V1 + ex + In
Vl + ex - 1
+ C
! Absolute value not needed |
VI + e* + 1
Functions that consist of finitely many sums, differences, quotients, and products of sin x and cos x are called rational functions of sin x and cos x. Some examples are sin x + 3 cos2 x sin x 3 sin5 x cos ;t + 4 sin * ' 1 + cos x — cos2 x ' l+4sin.x: The Endpaper Integral Table gives a few formulas for integrating rational functions of sin x and cos x under the heading Reciprocals of Basic Functions. For example, it follows from Formula (18) that
1 dx = tan x — sec x + C (2) 1 + sin x However, since the integrand is a rational function of sin x, it may be desirable in a particular application to express the value of the integral in terms of sin x and cos x and rewrite (2) as 1 shut — 1 - dx h C 1 + sin x cos x Many rational functions of sin x and cos x can be evaluated by an ingenious method that was discovered by the mathematician Karl Weierstrass (see p. 82 for biography). The idea is to make the substitution
I
M = tan(*/2),
-n/2 < x/2 < n/2
from which it follows that x = 2tan ' M,
dx =
1+M2
du
To implement this substitution we need to express sin x and cos x in terms of u. For this purpose we will use the identities sinjc = 2sin(;c/2) cos(jt/2)
(3)
cosx — cos 2 (x/2) - sin2(;t/2)
(4)
and the following relationships suggested by Figure 7.6.1:
u
sin (x/2) =
cos (x/2) =
and
Substituting these expressions in (3) and (4) yields A Figure 7.6.1 Mil.* —
J » W i Z I
,
I
.
2M
1+M2 1-M2
cosx = '1+M2
~ 1+M2
528
Chapter 7 / Principles of Integral Evaluation
In summary, we have shown that the substitution u — tan(jc/2) can be implemented in a rational function of sin x and cos x by letting smjc =
2u
COSJt =
l-u1
1+M2
dx =
,2'
1+M2
du
(5)
dx Example 5 Evaluate / J 1— -si sin x + cos x
Solution. The integrand is a rational function of sin x and cos x that does not match any of the formulas in the Endpaper Integral Table, so we make the substitution M = tan(x/2). Thus, from (5) we obtain 2du The substitution u = tan(jc/2)
will
1+M2
I
J 1 — sin si x + cos x
/
1
convert any rational function of sinx
2M \
/1-M2\
2
ll+M J ' (l+M2J
and cos x to an ordinary rational func-
2du
tion of M. However, the method can lead to cumbersome partial fraction J
decompositions, so it may be worthwhile to consider other methods as
f
2
(1+M )-2M
du
= jT^-u=-}n
well when hand computations are being used.
= -ln|l-tan(x/2)|
INTEGRATING WITH COMPUTER ALGEBRA SYSTEMS
Integration tables are rapidly giving way to computerized integration using computer algebra systems. However, as with many powerful tools, a knowledgeable operator is an important component of the system. Sometimes computer algebra systems do not produce the most general form of the indefinite integral. For example, the integral formula dx = In \x - 1 + C x- 1
which can be obtained by inspection or by using the substitution M = x — 1, is valid for x > 1 or for x < 1. However, not all computer algebra systems produce this form of the answer. Some typical answers produced by various implementations of Mathematica, Maple, and the CAS on a handheld calculator are ln(-l+jc),
In(z-l),
ln(|*-l|)
Observe that none of the systems include the constant of integration—the answer produced is a particular antiderivative and not the most general antiderivative (indefinite integral). Observe also that only one of these answers includes the absolute value signs; the antiderivatives produced by the other systems are valid only for x > 1. All systems, however, are able to calculate the definite integral 1/2
I
dx = -ln2 x-l
correctly. Now let us examine how these systems handle the integral f XT/x2-
+ ln(x -2 + V*2 - 4* + 5)
(6)
7.6 Using Computer Algebra Systems and Tables of Integrals
529
which we obtained in Example 2(b) (with the constant of integration included). Some CAS implementations produce this result in slightly different algebraic forms, but a version of Maple produces the result 3 5dx = \(x22 - 4x +5)3/2
f
\ (2x - 4) V* 2 - 4x + 5 + sinlT ' (x - 2)
This can be rewritten as (6) by expressing the fractional exponent in radical form and expressing sofa l(x — 2) in logarithmic form using Theorem 6.8.4 (verify). A version of Mathematica produces the result I
Expanding the expression
(x + I) 8 8 produces a constant term of §, whereas the second expression in (7) has no constant term. What is the explanation?
-4x+5dx = \(x2 -x- I)/*2 - 4x + 5 - sinh~'(2 - x)
which can be rewritten in form (6) by using Theorem 6.8.4 together with the identity sinh^ ! (— x) = — sinh"1 x (verify). Computer algebra systems can sometimes produce inconvenient or unnatural answers to integration problems. For example, various computer algebra systems produced the following results when asked to integrate (x + I) 7 :
1)8
7 , 35 7 1 -x6 + lx5 — 8 8 2 4 The first form is in keeping with the hand computation
,,7,
(* + D8
7jc 3
, 1 2
2
(7)
C
that uses the substitution u = x + 1, whereas the second form is based on expanding (x + I) 7 and integrating term by term. In Example 2(a) of Section 7.3 we showed that / sin4 x cos5 x dx — ^ sin5 x — | sin7 x + \ sin9 x + C However, a version of Mathematica integrates this as ^~ sin 9x -rjg sin x — j|j sin 3x — ^ sin 5x + -^ sin lx +
whereas other computer algebra systems essentially integrate it as — \ sin3 x cos6 x — jj- sin x cos6 x + -^ cos4 x sin x + ~ cos2 x sin x + ^ sin x Although these three results look quite different, they can be obtained from one another using appropriate trigonometric identities. COMPUTER ALGEBRA SYSTEMS HAVE LIMITATIONS
T E C H N O L O G Y MASTERY Sometimes integrals that cannot be evaluated by a CAS in their given form can be evaluated by first rewriting them in a different form or by making a substitution. If you have a CAS, make a w-substitution in (8) that will enable you to evaluate the integral with your CAS. Then evaluate the integral.
A computer algebra system combines a set of integration rules (such as substitution) with a library of functions that it can use to construct antiderivatives. Such libraries contain elementary functions, such as polynomials, rational functions, trigonometric functions, as well as various nonelementary functions that arise in engineering, physics, and other applied fields. Just as our Endpaper Integral Table has only 121 indefinite integrals, these libraries are not exhaustive of all possible integrands. If the system cannot manipulate the integrand to a form matching one in its library, the program will give some indication that it cannot evaluate the integral. For example, when asked to evaluate the integral (8)
all of the systems mentioned above respond by displaying some form of the unevaluated integral as an answer, indicating that they could not perform the integration. Sometimes computer algebra systems respond by expressing an integral in terms of another integral. For example, if you try to integrate ex using Mathematica, Maple, or
530
Chapter 7 / Principles of Integral Evaluation
Sage, you will obtain an expression involving erf (which stands for error function). The function erf(jc) is defined as -' dt
erf(jc) = —=
so all three programs essentially rewrite the given integral in terms of a closely related integral. From one point of view this is what we did in integrating l/x, since the natural logarithm function is (formally) defined as In x = I
- dt
(see Section 6.6). Example 6 time t is
A particle moves along an x-axis in such a way that its velocity v(t) at
v(t) =3Qcos7tsm4t
(t>0)
Graph the position versus time curve for the particle, given that the particle is at x = I when t = 0. Solution. by
Since dx/dt = v(t) and x = 1 when t = 0, the position function x(t) is given ^
x(t) = l+ I v(s)ds Jo Some computer algebra systems will allow this expression to be entered directly into a command for plotting functions, but it is often more efficient to perform the integration first. The authors' integration utility yields x = I 30 cos7 t sin 4 1 dt
2
= -ff sin 11 1 + lOsin 9 t - f sin7 t + 6sin51 + C
1.5 1
where we have added the required constant of integration. Using the initial condition jt(0) = 1, we substitute the values x — 1 and / = 0 into this equation to find that C — 1, so
0.5 5
A Figure 7.6.2
10
15
x(t) = -^sin u f + 10sin 9 ?- f sin7 f + 6sin 5 ? + 1 (t > 0)
20
The graph of x versus t is shown in Figure 7.6.2. -4
•Q;UICK CHECK EXERCISES 7.6
(See page 533 for answers.)
1. Find an integral formula in the Endpaper Integral Table that can be used to evaluate the integral. Do not evaluate the integral. f 2x (a) / :dx 3x+4
(b)
can be used to evaluate the integral. Do not evaluate the integral. (a)
dx; u = x2
(b) / e^dx; u = x (c) I—. x dx', u = e sin(e ) J 1+si
(d) 2. In each part, make the indicated w-substitution, and then find an integral formula in the Endpaper Integral Table that
/ T + ex
1
dx; u = 2x
3. In each part, use the Endpaper Integral Table to evaluate the integral. (If necessary, first make an appropriate substitution or complete the square.) (mm.)
7.6 Using Computer Algebra Systems and Tables of Integrals Cal
/ fir J 4-X2
N f e3* C\^) C I / 1 / s— "^Ar J Vl - e2*
Ch)
/ ens 2r POS r f i x = J
CHI l)
EXERCISE SET 7.6
1L 1 4X dx J 3*-l
X " f dx - / (4~5x)2d
3 / dx • J x(2x+5)d-
4
z
ft
AV
r
i~1//v
/ dx • J x2(l-5x)d'
J 7
1
J x^4 - 3x
1 9 •\ta
J \6-x*2 712
R
d-f
J x^3x - 4
OWy
r /
IT
v2 -/r
/ -v
VZ
*
Vx +4
J
3^2,7,,
1/;
f^^~2dx
2
1S
I
w^
Jt
^
^ JC^ - *2
19. / sin 3x sin 4j: Jx
20. / sin 2x cos 5^ dx
^
1]
1
3 r
lnr//i-
'•^
I
nJ
/
" //r
23. / e~2x sin3x dx
JY
.,
[
sin 2x J (cos2x)(3 — cos2jc) ]
it-r
u
°v 2
2i
cos 2 (e
2x
2j:
)^, M = e
2x
dx, u = -2x u — 3x + 1
[c] 37-48 (a) Make an appropriate w-substitution, and then use the Endpaper Integral Table to evaluate the integral, (b) If you have a CAS, use it to evaluate the integral (no substitution), and then confirm that the result is equivalent to that in part (a). R C S3X 37 1 ° dx ' J (sin3jt)(sin3* + I ) 2 "
lnX [ d, J x^4lnx- 1
<«
x
16JC4 - 1
/ /
. rxJ^
r i
/ J^r
'/
rlr
4rt
Ac2* dr Or 2 -ir
41 44
r e * 1
J 3 - 4e2*
rlr
r J A — Qx2 / dx J x2
c
/
i
/ —^5 ^ A:V^ 5x
//r
45. / 4^: sin 2x dx
46. / cos -^/x dx
47. I e-^dx
48. f x\n(2 + x2)dx
r2x
(4-3^)2^' "
/
rj
* e
43
4x
/7v
jZ. / - ax, u — Lx. J x V3 - 4x2 /"sin 2 (In jc)
24. / e* cos 2* rfjc
£] 25-36 (a) Make the indicated w-substitution, and then use the Endpaper Integral Table to evaluate the integral, (b) If you have a CAS, use it to evaluate the integral, and then confirm that the result is equivalent to the one that you found in part (a).
7r
^ V2-4JC 4
J Vx 1
J
u
36. / ln(3;c + 1) dx
x2jx1-2
/•
4x5
[
^1
xe
s _JLdv
/ * J x2
/7v
30. f x^/2x^+3dx, u = V2x2
dr
r /
14
2
I . /o J
/
^ V4z2-9
J
1
J /
20
M\ 1 Ay iu. Ji xi-9 ax
J-r
/
ie
J-r
\J/O L — X^
J /
X 1 dx / J2_4jc + 8rf-
E CAS
~c\ 1 -24 (a) Use the Endpaper Integral Table to evaluate the given integral, (b) If you have a CAS, use it to evaluate the integral, and then confirm that the result is equivalent to the one that you found in part (a).
I
531
"\ l^r
/ VK9* + 4) /• cos4A: 28. / ;— a x , u = sm 4x J 9 + sin24A;
[c] 49-52 (a) Complete the square, make an appropriate «substitution, and then use the Endpaper Integral Table to evaluate the integral, (b) If you have a CAS, use it to evaluate the integral (no substitution or square completion), and then confirm that the result is equivalent to that in part (a).
x2 + 6x - 7
dx
50. / ^3 J
2x
x2dx
532
Chapter 7 / Principles of Integral Evaluation
J v 5 + 4x — x2
J x + ox + \3
79. y = e~\ y = 0 , x = 0 , x = 3 80. y = lnx, y = 0, x =5
[c] 53-64 (a) Make an appropriate u -substitution of the form u = x['n or u = (x +a)' , and then evaluate the integral, (b) If you have a CAS, use it to evaluate the integral, and then confirm that the result is equivalent to the one that you found in part (a), m
f
C^
1
r
i
L/r
5 r
^ /7r
,/
3
^4
i j
i
ES
f
x
/
f
57
58
j x- yx
ft
T
f I
/
y
<1
dx
2
' J x'/ -x'/ 63
/"
x
83. y = sinx, 0 < x < n
j
dX
/
1 i_ // r
(b) Graph the position versus time curve, tii 85. v(t) = 20 cos 6 ? sin3 f, s(0)=2 86. a(t)
" J i-v*
3
64>
' 7 yi—2^
/•
/
84. y = 1/x, 1 < x < 4
[g] 85-86 Information is given about the motion of a particle moving along a coordinate line. (a) Use a CAS to find the position function of the particle for
y i + yi ^
3
82. y = 31nx, l < x < 3
83-84 Use any method to find the area of the surface generated by revolving the curve about the x-axis. IS
j yx + v*
fin /
81. y = 2x2, 0 < x < 2
sf Y
J xVx 3 — 1 dX
81-82 Use any method to find the arc length of the curve. «
x
+ sy/s'3'*
e f sin2?sin4/, v(0)
F O C U S ON C O N C E P T S
0, s(0)
10
me&UI&^$x&~&W-^'i^\^MS3&2-'ZKy
87. (a) Use the substitution u = tan(:c/2) to show that c] 65-70 (a) Make w -substitution (5) to convert the integrand to
1 - tan(x/2)
you have a CAS, use it to evaluate the integral (no substitution), and then confirm that the result is equivalent to that in part (a).
65
69
dx f J \ + sin x + cos x d9 / 1 — cos 9 dx ( J sin x + tan x
and confirm that this is consistent with Formula (22) of Section 7.3. (b) Use the result in part (a) to show that
dX f J 2 + sin x /* dx AS ' J 4 sin x — 3 cos x
66
/
88. Use the substitution u = tan(x/2) to show that
s mx 70 f ' dx J sin x + tan x
f J
71-72 Use any method to solve for x.
71
/
h '(4-0
72 f-
tit
!
Ji t^2t - I
v }
v— f i r — Or— 2>>
[l+cosxj
J 2 cosh x + sinh x without expressing the integrand in terms ofex and e~x .
x2" y — 0 x — 0 x — 4
74. y = ^9x2 - 4, y = 0, x = 2 75
2
89. Find a substitution that can be used to integrate rational functions of sinh x and cosh x. and use your substitution to evaluate ,i 1r UJir
1x>'
73-76 Use any method to find the area of the region enclosed by the curves.
73. y — v/25
i ri-cosxi
and confirm that this is consistent with the result in Exercise 65(a) of Section 7.3.
0 S ? «^ »- ^- 4
A
sec x dx = In tan ( —|— ) + C V4 2/
r^ 1
25-16x ~ 76. y = Vx~lnx, y = 0, x = 4
90-93 Some integrals that can be evaluated by hand cannot be evaluated by all computer algebra systems. Evaluate the integral ?y hand, and determine if it can be evaluated on your CAS. it, x3
/-/v
/
77-80 Use any method to find the volume of the solid generated when the region enclosed by the curves is revolved about the y-axis. »
91. / (cos32 x sin30 x - cos30 x sin32 x) dx
77. y = cosx, y = 0, x = 0, x = n/2
92. / Vx - V* 2 - 4 d x [Hint: ^(^x+2 - ~Jx - 2)2 = ?]
7.7 Numerical Integration; Simpson's Rule
1
533
(b) Check your answer in part (a) by using the CAS to find the partial fraction decomposition of /. (c) Integrate / by hand, and then check your answer by integrating with the CAS.
-dx
[Hint: Rewrite the denominator as jc lo (l 94. Let -2x5 + 26x4 + 15x3 + 6x2 + 20x + 43 /(*) = - x5 - 18jc4 - 2x3 - 39x2 -x-20 (a) Use a CAS to factor the denominator, and then write down the form of the partial fraction decomposition. You need not find the values of the constants.
•QUICK CHECK ANSWERS 7.6 1. (a) Formula (60) (b) Formula (108) (c) Formula (102) (d) Formula (50) 2. (a) Formula (25) (b) Formula (5 1 ) 1 x +2 1 1 1 e* / (c) Formula (18) (d) Formula (97) 3. (a) In 4-C (b) - sin 3x + sin* + C (c) VI 4 x-2 6 2 2 (d) - In (x2 - 4x + 8) + tan"
+C
_7
C
NUMERICAL INTEGRATION; SIMPSON'S RULE If it is necessary to evaluate a definite integral of a function for which an antiderivative cannot be found, then one must settle for some kind of numerical approximation of the integral. In Section 4.4 we considered three such approximations in the context of areas—left endpoint approximation, right endpoint approximation, and midpoint approximation. In this section we will extend those methods to general definite integrals, and we will develop some new methods that often provide more accuracy with less computation. We will also discuss the errors that arise in integral approximations. A REVIEW OF RIEMANN SUM APPROXIMATIONS
Recall from Section 4.5 that the definite integral of a continuous function / over an interval [a, b} may be computed as
/'
f(x) dx =
Ja
lim
max Ajct
,o£.
where the sum that appears on the right side is called a Riemann sum. In this formula, is the width of the fcth subinterval of a partition a = x0 < x\ < JC2 < • • • < xn = bof [a, b] into n subintervals, and x*k denotes an arbitrary point in the fcth subinterval. If we take all subintervals of the same width, so that Ajtt — (b — a)/n, then as n increases the Riemann sum will eventually be a good approximation to the definite integral. We denote this by writing
r"
(b-a\
\ f(x)dx*t(Ja
\
n
-)[/(*?) + /(*£) + • • • + /(*:)]
d)
/
If we denote the values of / at the endpoints of the subintervals by yo = f(a),
J\=f(x\),
)>2
and the values of / at the midpoints of the subintervals by
ym,,ym2, ••-,ymn then it follows from (1) that the left endpoint, right endpoint, and midpoint approximations discussed in Section 4.4 can be expressed as shown in Table 7.7. 1 . Although we originally
Hk
534
Chapter 7 / Principles of Integral Evaluation Table 7.7.1 LEFT ENDPOINT APPROXIMATION
RIGHT ENDPOINT APPROXIMATION
MIDPOINT APPROXIMATION
ly
\ .
obtained these results for nonnegative functions in the context of approximating areas, they are applicable to any function that is continuous on [a,b]. TRAPEZOIDAL APPROXIMATION
It will be convenient in this section to denote the left endpoint, right endpoint, and midpoint approximations with n subintervals by Ln, Rn, and Mn, respectively. Of the three approximations, the midpoint approximation is most widely used in applications. If we take the average of Ln and Rn , then we obtain another important approximation denoted by called the trapezoidal approximation: Trapezoidal Approximation 'b-a In
/' Ja Trapezoidal approximation A Figure 7.7.1
\-1yn-\ + yn]
(2)
The name "trapezoidal approximation" results from the fact that in the case where / is nonnegative on the interval of integration, the approximation Tn is the sum of the trapezoidal areas shown in Figure 7.7.1 (see Exercise 51). > Example 1 In Table 7.7.2 we have approximated I n 2 = I -dx Ji x using the midpoint approximation and the trapezoidal approximation.* In each case we used n = 10 subdivisions of the interval [1, 2], so that b-a 2-1 b-a 2-1 = 0.1 and = = 0.05 ^ 10 20 In Midpoint
Trapezoidal
Throughout this section we will show numerical values to nine places to the right of the decimal point. If your calculating utility does not show this many places, then you will need to make the appropriate adjustments. What is important here is that you understand the principles being discussed.
7.7 Numerical Integration; Simpson's Rule
535
Table 7.7.2 MIDPOINT AND TRAPEZOIDAL APPROXIMATIONS FOR /
MIDPOINT APPROXIMATION
TRAPEZOIDAL APPROXIMATION
MIDPOINT
i
m
i
1
1.05
2 3 4 5 6 7 8 9 10
1.15 1.25 1.35 1.45 1.55 1.65 1.75 1.85 1.95
^ dx
MULTIPLIER
ENDPOINT
ymi=f(mi) = l/mi
i
0.952380952 0.869565217 0.800000000 0.740740741 0.689655172 0.645161290 0.606060606 0.571428571 0.540540541 0.512820513 6.928353603
0 1
1.2 1.3 1.4 1.5 1.6
2
3 4
5 6 7 8 9 10
1.7
1.8 1.9 2.0
:
/
«W
w
i
1.000000000 0.909090909 0.833333333 0.769230769 0.714285714 0.666666667 0.625000000 0.588235294 0.555555556 0.526315789 0.500000000
1.0
1.1
i dx = (0.1)(6.928353603) = 0.692835360
^—^^""-—•--••-•^^^"^^^ By rewriting (3) and (4) in the form
yi=f(xi) = \lxt
*,'
1.000000000 1.818181818 1.666666667 1.538461538 1.428571429 1.333333333 1.250000000 1.176470588 1.111111111 1.052631579 0.500000000 13.875428063
1
2
2 2 2 2 2 2 2 2 1
±
COMPARISON OF THE MIDPOINT AND TRAPEZOIDAL APPROXIMATIONS We define the errors in the midpoint and trapezoidal approximations to be , b
rb
EM = I
I f(x) dx = approximation + error
Ja
f(x) dx — Mn
and
ET — I
Ja
f(x) dx - Tn
Ja
we see that positive values of EM and ET correspond to underestimates and negative values to overestimates.
respectively, and we define I EM \ and | ET \ to be the absolute errors in these approximations. The absolute errors are nonnegative and do not distinguish between underestimates and overestimates. Example 2
The value of In 2 to nine decimal places is 2 i -dx «s 0.693 147181 / x
(5)
so we see from Tables 7.7.2 and 7.7.3 that the absolute errors in approximating In 2 by M\Q andr, 0 are
| m 2 _ M l o | ^0.000311821 \ET\ = I In 2 - r,0| « 0.000624222
Thus, the midpoint approximation is more accurate than the trapezoidal approximation in this case. -^ Table 7.7.3 In 2 (NINE DECIMAL PLACES)
0.693147181 0.693147181
APPROXIMATION
ERROR
M| 0 » 0.692835360 T, n«0.69377 1403
E M = l n 2 - M 1 0 = 0.000311821 ET = In 2 - r,0 = -0.000624222
536
Chapter 7 / Principles of Integral Evaluation
The shaded triangles have equal areas. A Figure 7.7.2
It is not accidental in Example 2 that the midpoint approximation of In 2 was more accurate than the trapezoidal approximation. To see why this is so, we first need to look at the midpoint approximation from another point of view. To simplify our explanation, we will assume that / is nonnegative on [ a , b ] , though the conclusions we reach will be true without this assumption. If / is a differentiable function, then the midpoint approximation is sometimes called the tangent line approximation because for each subinterval of [a, b] the area of the rectangle used in the midpoint approximation is equal to the area of the trapezoid whose upper boundary is the tangent line to y = f ( x ) at the midpoint of the subinterval (Figure 7.7.2). The equality of these areas follows from the fact that the shaded areas in the figure are congruent. We will now show how this point of view about midpoint approximations can be used to establish useful criteria for determining which of Mn or Tn produces the better approximation of a given integral. In Figure 7.7.3a we have isolated a subinterval of [a, b] on which the graph of a function / is concave down, and we have shaded the areas that represent the errors in the midpoint and trapezoidal approximations over the subinterval. In Figure 1.1.3b we show a succession of four illustrations which make it evident that the error from the midpoint approximation is less than that from the trapezoidal approximation. If the graph of / were concave up, analogous figures would lead to the same conclusion. (This argument, due to Frank Buck, appeared in The College Mathematics Journal, Vol. 16, No. 1, 1985.) - Midpoint error Trapezoidal error
Justify the conclusions in each step of
Blue area
Blue area
Figure 7.7.36.
(a)
=
Blue area
< Yellow area
(b)
A Figure 7.7.3
Figure 7.7.3a also suggests that on a subinterval where the graph is concave down, the midpoint approximation is larger than the value of the integral and the trapezoidal approximation is smaller. On an interval where the graph is concave up it is the other way around. In summary, we have the following result, which we state without formal proof: 7.7.1 THEOREM Let f be continuous on [a, b], and let \ EM \ and \Ej\be the absolute errors that result from the midpoint and trapezoidal approximations offa f ( x ) dx using n subintervals. (a)
If the graph off is either concave up or concave down on(a,b), then \ EM I < \Ej\, that is, the absolute error from the midpoint approximation is less than that from the trapezoidal approximation.
(b) If the graph of f is concave down on (a,b), then fb Tn < I
f(x)dx<Mn
Ja
(c) If the graph of f is concave up on (a, b), then fb Mn < I f(x)dx
7.7 Numerical Integration; Simpson's Rule
537
> Example 3 Since the graph of f ( x ) = l/x is continuous on the interval [1,2] and concave up on the interval (1, 2), it follows from part (a) of Theorem 7.7.1 that Mn will always provide a better approximation than Tn for - dx — In 2 x WARNING
Moreover, if follows from part (c) of Theorem 7.7.1 that Mn < In 2 < Tn for every positive integer n. Note that this is consistent with our computations in Example 2. -4
Do not conclude that the midpoint approximation is always better than the trapezoidal approximation; the trapezoidal approximation may be better if the function changes concavity on the interval of integration.
> Example 4 The midpoint and trapezoidal approximations can be used to approximate sin 1 by using the integral l sin 1 = / cos x dx
Jo
Since f ( x ) = cosx is continuous on [0,1] and concave down on (0, 1), it follows from parts (a) and (b) of Theorem 7.7.1 that the absolute error in Mn will be less than that in Tn, and that Tn < sin 1 < Mn for every positive integer n. This is consistent with the results in Table 7.7.4 for n = 5 (intermediate computations are omitted). •« Table 7.7.4 sin 1 (NINE DECIMAL PLACES)
APPROXIMATION
ERROR
0.841470985 0.841470985
M5 = 0.842875074 = 0.838664210
EM = sin 1 - M5 « - 0.001404089 ET = sin 1 - T5 « 0.002806775
> Example 5 Table 7.7.5 shows approximations for sin 3 = /0 cos x dx using the midpoint and trapezoidal approximations with n = 10 subdivisions of the interval [0, 3]. Note that |E M < \ET\ and Tw < sin 3 < M\o, although these results are not guaranteed by Theorem 7.7.1 since f ( x ) = cosx changes concavity on the interval [0, 3]. -4 Table 7.7.5
sin 3 (NINE DECIMAL PLACES)
0.141120008 0.141120008
APPROXIMATION
i - 0.141650601 , = 0.140060017
ERROR
EM = sin 3 - M10 = -0.000530592 ET =sin3-r 1 0 « 0.001059991
SIMPSON'S RULE
When the left and right endpoint approximations are averaged to produce the trapezoidal approximation, a better approximation often results. We now see how a weighted average of the midpoint and trapezoidal approximations can yield an even better approximation. The numerical evidence in Tables 7.7.3, 7.7.4, and 7.7.5 reveals that Ej ~ —2EM, so that 2EM + ET ^ 0 in these instances. This suggests that f(x)dx =2
f(x)dx Ja
+ v/' ci
f(x)dx
= 2(Mk + EM) I-(7* +\-ET) = (2Mk + Tk) (2EM ET) * 2M, + 7*
538
Chapter 7 / Principles of Integral Evaluation
This gives (6)
/' Ja WARNING Note that in (7) the subscript n in Sn is
always even since it is twice the value
of the subscripts for the corresponding midpoint and trapezoidal approximations. For example,
The midpoint approximation Mk in (6) requires the evaluation of / at k points in the interval [a, b], and the trapezoidal approximation Tk in (6) requires the evaluation of / at k + 1 points in [a, b]. Thus, \(2Mk + Tk) uses 2k + 1 values of /, taken at equally spaced . . . . f These omts • , • i< • • • r . 1 ~, omts ln the mterval
P [«' H P ^ obtained by partitioning [a, 6] into 2k equal subintervals indicated by the left endpoints, right endpoints, and midpoints used in Tk and Mk, respectively. Setting n = 2k, we use Sn to denote the weighted average of Mk and Tk in (6). That is,
and
or
Sn = S2k = \ (2Mk
Sn = \ (2Mn/2 + Tn/2)
(7)
Table 7.7.6 displays the approximations Sn corresponding to the data in Tables 7.7.3 to 7.7.5. Table 7.7.6 FUNCTION VALUE (NINE DECIMAL PLACES)
APPROXIMATION
ERROR
In 2 = 0.693147181
fi\\/x) dx = S2Q 20 = } (2M IO + r lo ) = 0.693147375
-0.000000194
sin 1=0.841470985
focosxdx» " --S SwW= =±(2M j (2MS5 + T5) =0.841471453
-U.UUUUUUTOO -0.000000468
sin 3 = 0.141120008
tfcosxdx- = S20 = } (2M10 + 710) = 0.141120406
-0.000000398
Using the midpoint approximation formula in Table 7.7.1 and Formula (2) for the trapezoidal approximation, we can derive a similar formula for Sn. We start by partitioning the interval [a, b] into an even number of equal subintervals. If n is the number of subintervals, then each subinterval has length (b — a)In. Label the endpoints of these subintervals successively by a = jco, x\, x2,..., xn = b. Then jtn, x2, X4, ..., xn define a partition of [a, b] into n/2 equal intervals, each of length 2(b — a)In, and the midpoints of these subintervals are x\, x$, x$,..., xn^\, respectively, as illustrated in Figure 7.7.4. Using y,- — /(*;), we ^
9 J2(b-a)\ 2Mn/2 = 2 [yi + y3 H
r- y«-i]
b-a
Noting that (b — a)/\2(n/2)} — (b — a)/n, we can express Tn/2 as (b~a\ Tn/2 = I —— I [yo + 2y2 + 2y4 -\
\- 2;
Thus, Sn = | (2Mn/2 + Tn/2) can be expressed as
3\
«
'
+2y2 + 4y 3 +2y4 + - - - + 2yn-2
(8)
The approximation (9)
/' Ja
with Sn as given in (8) is known as Simpson's rule. We denote the error in this approximation by ,& Es= / f(x)dx-Sn (10) Ja
As before, the absolute error in the approximation (9) is given by | Es \.
7.7 Numerical Integration; Simpson's Rule b —a
b —a X
*2
6
midpoint
> Figure 7.7.4
539
midpoint
midpoint
midpoint
2 ~/b-a\ --
> Example 6 In Table 7.7.7 we have used Simpson's rule with n = 10 subintervals to obtain the approximation In2= /
-dx ^ S{0 =0.693150231
For this approximation,
1 fb-a\ 3\ n
1 /2-1
3 V 10
1 30
Table 7.7.7 AN APPROXIMATION TO In 2 USING SIMPSON'S RULE MULTIPLIER
ENDPOINT
i
x
y,- =/(*,-) =l/x,-
0 1
1.0
2
5 6
1.2 1.3 1.4 1.5 1.6
7
1.7
8 9 10
1.8 1.9 2.0
1.000000000 0.909090909 0.833333333 0.769230769 0.714285714 0.666666667 0.625000000 0.588235294 0.555555556 0.526315789 0.500000000
3 4
i
1.1
wt 1
4 2 4
2 4
2 4 2 4 1
wlyi 1.000000000 3.636363636 1.666666667 3.076923077 1.428571429 2.666666667 1.250000000 2.352941176 1.111111111 2.105263158 0.500000000 20.794506921
/ \dx = (rU(20.794506921) = 0.693150231 \M/
_ / ] Ji
Thomas Simpson (1710-1761) English mathematician. Simpson was the son of a weaver. He was trained to follow in his father's footsteps and had little formal education in his early life. His interest in science and mathematics was aroused in 1724, when he witnessed an eclipse of the Sun and received two books from a peddler, one on astrology and the other on arithmetic. Simpson quickly absorbed their contents and soon became a successful local fortune teller. His improved financial situation enabled him to give up weaving and marry his landlady. Then in 1733 some mysterious "unfortunate incident" forced him to move. He settled in Derby, where he taught in an evening school and worked at weaving during the day. In 1736 he
moved to London and published his first mathematical work in a periodical called the Ladies' Diary (of which he later became the editor). In 1737 he published a successful calculus textbook that enabled him to give up weaving completely and concentrate on textbook writing and teaching. His fortunes improved further in 1740 when one Robert Heath accused him of plagiarism. The publicity was marvelous, and Simpson proceeded to dash off a succession of best-selling textbooks: Algebra (ten editions plus translations), Geometry (twelve editions plus translations), Trigonometry (five editions plus translations), and numerous others. It is interesting to note that Simpson did not discover the rule that bears his name—it was a well-known result by Simpson's time.
540
Chapter 7 / Principles of Integral Evaluation
Although SIQ is a weighted average of MS and T$, it makes sense to compare SIQ to M\Q and TIQ, since the sums for these three approximations involve the same number of terms. Using the values for M\o and TIQ from Example 2 and the value for S\o in Table 7.7.7, we have \EU\ = I n 2 - M 1 0
10.693147181 - 0.6928353601 = 0.000311821
\ET\= I n 2 - r , 0 |
|0.693147181-0.693771403| =0.000624222 [0.693147181 - 0.693150231 = 0.000003050
Comparing these absolute errors, it is clear that of In 2 than either M10 or TIQ. -4
is a much more accurate approximation
GEOMETRIC INTERPRETATION OF SIMPSON'S RULE
y =/« y = Ax2 + Bx + C
The midpoint (or tangent line) approximation and the trapezoidal approximation for a definite integral are based on approximating a segment of the curve y — f ( x ) by line segments. Intuition suggests that we might improve on these approximations using parabolic arcs rather than line segments, thereby accounting for concavity of the curve y — f ( x ) more closely. At the heart of this idea is a formula, sometimes called the one-third rule. The one-third rule expresses a definite integral of a quadratic function g (x) = Ax2 + Bx + C in terms of the values Yo,Y\, and ¥2 of g at the left endpoint, midpoint, and right endpoint, respectively, of the interval of integration [m — A*, m + Ax] (see Figure 7.7.5): ,
(Ax2
AJC
= — [Y0
47, + Y2]
(11)
Jm — Ax
Verification of the one-third rule is left for the reader (Exercise 53). By applying the one-third rule to subintervals [x2k-2, x2k], & = ! > • • • > n/2, one arrives at Formula (8) for Simpson's rule (Exercise 54). Thus, Simpson's rule corresponds to the integral of a piecewise-quadratic approximation to f ( x ) . ERROR BOUNDS
With all the methods studied in this section, there are two sources of error: the intrinsic or truncation error due to the approximation formula, and the roundoff error introduced in the calculations. In general, increasing n reduces the truncation error but increases the roundoff error, since more computations are required for larger n. In practical applications, it is important to know how large n must be taken to ensure that a specified degree of accuracy is obtained. The analysis of roundoff error is complicated and will not be considered here. However, the following theorems, which are proved in books on numerical analysis, provide upper bounds on the truncation errors in the midpoint, trapezoidal, and Simpson's rule approximations.
7.7.2 i HECIREM (Midpoint and Trapezoidal Error Bounds) If f" is continuous on [a,b] and if K i is the maximum value of\ f"(x)\ on[a,b], then (a) \EM
11 Ja
f" f(r~\dr J\x)ax
M M
(b) \ET
f" f(r}Hr x ax 11 J\ / Ja
T in
n
. (b-a)3K2 24n2 ^ (b~a)3K2 I2n2
(12)
(13)
7.7 Numerical Integration; Simpson's Rule
541
7.7.3 THEOREM (Simpson Error Bound} /f/ (4) is continuous on [a, b] and if K4 is f/ze maximum value of\ / (4) (x)| on [a, b], then tb
1 Z7
/
)a
Example 7
ff,,\
J_
^
C
V
(b-a)' 5K4 4
(14)
180«4
"
Find an upper bound on the absolute error that results from approximating In 2
•r
J\ x
dx
using (a) the midpoint approximation MW, (b) the trapezoidal approximation T\Q, and (c) Simpson's rule S\o, each with n — 10 subintervals. Solution,
We will apply Formulas (12), (13), and (14) with = -, x
a = l,
b = 2,
and n = 10
We have 2
,
Thus,
24 /"(*)! =
Note that the upper bounds calculated in Example 7 are consistent with the values | EM \, |£>|,and |£$| calculated in Example 6 but are considerably greater than those values. It is quite common that the upper bounds on the absolute errors given in Theorems 7.7.2 and 7.7.3 substantially exceed the actual absolute errors. However, that does not diminish the utility of these bounds.
24 V5
where we have dropped the absolute values because f"(x) and /(4) (x) have positive values for 1 < x < 2. Since \f"(x)\ and |/4(j:)| are continuous and decreasing on [1, 2], both functions have their maximum values at jc = l;for f"(x)\ this maximum value is 2 and for |/4(jc)| the maximum value is 24. Thus we can take K2 = 2 in (12) and (13), and K4 = 24 in (14). This yields
\EM\< \ET\<
\ES\< > Example 8
(b-a)3K2 24n2
I3-2 24 • 102
(b-a)3K2 _ 2
I3 - 2
" 1 2 - 102
12n
(b-a)5K4
I 5 - 24
180«4
180- 104
• 0.000833333
; 0.001666667 0.000013333
How many subintervals should be used in approximating
r
In 2 = /
-dx x
by Simpson's rule for five decimal-place accuracy? Solution. To obtain five decimal-place accuracy, we must choose the number of subintervals so that 0.000005 = 5 x l O - 6 From (14), this can be achieved by taking n in Simpson's rule to satisfy
542
Chapter 7 / Principles of Integral Evaluation Taking a — 1, b = 2, and K$ •= 24 (found in Example 7) in this inequality yields
I 5 • 24 fi < 5 x 10~6 180-n 4 ~ which, on taking reciprocals, can be rewritten as 2 x 106 75 Thus,
8 x 104 3
20 n > -— % 12.779
Since n must be an even integer, the smallest value of n that satisfies this requirement is n — 14. Thus, the approximation Su using 14 subintervals will produce five decimal-place accuracy. -^
REMARK
In cases where it is difficult to find the values of K2 and Ka, in Formulas (12), (13), and (14), these constants may be replaced by any larger constants. For example, suppose that a constant K can be easily found with the certainty that | f"(x)\ < K on the interval. Then KI < K and \t-T\ <
(15)
I2n2
so the right side of (15) is also an upper bound on the value of \ET\. Using K, however, will likely increase the computed value of n needed for a given error tolerance. Many applications involve the resolution of competing practical issues, illustrated here through the trade-off between the convenience of finding a crude bound for | f"(x)\ versus the efficiency of using the smallest possible n for a desired accuracy.
Example 9
How many subintervals should be used in approximating
i
Jo
cos(;t2) dx
by the midpoint approximation for three decimal-place accuracy? Solution.
To obtain three decimal-place accuracy, we must choose n so that \EM\ < 0.0005 = 5 x 1(T4
(16)
2
From (12) with f ( x ) = cos(jc ), a = 0, and b = 1, an upper bound on \EM\ is given by
\EM\<
(17)
24n2
where \K2\ is the maximum value of | f"(x)\ on the interval [0, 1]. However, /'(*) = -2x sin(jt2) f"(x) = -4x2cos(x2) - 2sin(x2) = -[4jc2cos(*2) + 2sin(z2)] f"(x)\ = |4* 2 cos(jc 2 )
01
(18)
It would be tedious to look for the maximum value of this function on the interval [0, 1]. For x in [0, 1], it is easy to see that each of the expressions x2, cos(;t2), and sin(;c2) is bounded in absolute value by 1, so \4x2 cos(x2) + 2sin(%2)| < 4 + 2 = 6 on [0, 1]. We can improve on this by using a graphing utility to sketch | f"(x)\, as shown in Figure 7.7.6. It is evident from the graph that A Figure 7.7.6
I f"(x)\ < 4
for
0< x <1
7.7 Numerical Integration; Simpson's Rule
543
Thus, it follows from (17) that \E
24n2 ~ 24«2 and hence we can satisfy (16) by choosing n so that 1 < 5 x 10~4 6«2 which, on taking reciprocals, can be written as If/ "30"
or n >
102
18.257
The smallest integer value of n satisfying this inequality is n = 19. Thus, the midpoint approximation M\<) using 19 subintervals will produce three decimal-place accuracy. -4 A COMPARISON OF THE THREE METHODS
Of the three methods studied in this section, Simpson's rule generally produces more accurate results than the midpoint or trapezoidal approximations for the same amount of work. To make this plausible, let us express (12), (13), and (14) in terms of the subinterval width A* =
b-a
We obtain
\EU\ <
(19)
\ET\ < — K2(b
(20) (21)
15U
(verify). For Simpson's rule, the upper bound on the absolute error is proportional to (Ax) 4 , whereas the upper bound on the absolute error for the midpoint and trapezoidal approximations is proportional to (Ax)2. Thus, reducing the interval width by a factor of 10, for example, reduces the error bound by a factor of 100 for the midpoint and trapezoidal approximations but reduces the error bound by a factor of 10,000 for Simpson's rule. This suggests that, as n increases, the accuracy of Simpson's rule improves much more rapidly than that of the other approximations. As a final note, observe that if f ( x ) is a polynomial of degree 3 or less, then we have f(4)(x) = 0 for all x, so K4 — 0 in (14) and consequently \ES\ = 0. Thus, Simpson's rule gives exact results for polynomials of degree 3 or less. Similarly, the midpoint and trapezoidal approximations give exact results for polynomials of degree 1 or less. (You should also be able to see that this is so geometrically.)
•QUICK CHECK EXERCISES 7.7
(See page 547 for answers.)
I. Let Tn be the trapezoidal approximation for the definite integral of f(x) over an interval [a, b] using n subintervals. (a) Expressed in terms of Ln and Rn (the left and right endpoint approximations), Tn = (b) Expressed in terms of the function values yo,y\, ... ,yn at the endpoints of the subintervals, Tn =
2. Let / denote the definite integral of / over an interval [a, b] with Tn and Mn the respective trapezoidal and midpoint approximations of / for a given n. Assume that the graph of / is concave up on the interval [a, b] and order the quantities Tn, Mn, and 7 from smallest to largest:
544
Chapter 7 / Principles of Integral Evaluation
3. Let Se be the Simpson's rule approximation for fa f(x) dx using n = 6 subintervals. (a) Expressed in terms of M$ and T$ (the midpoint and trapezoidal approximations), S^ = (b) Using the function values yo, y\, y 2 , . . . , y$ at the endpoints of the subintervals, 5g = 4. Assume that /(4) is continuous on [0,1 ] and that / (t) (x) satisfies \f(ky(x)\ < 1 on [0, 1], k = 1, 2, 3,4. Find an upper
EXERCISE SET 7.7
bound on the absolute error that results from approximating the integral of / over [0, 1] using (a) the midpoint approximation MIO; (b) the trapezoidal approximation T^o; and (c) Simpson's rule SIQ.
r3 i
5. Approximate / —2 dx using the indicated method.
h x
(a) M, = (c) S2 =
(b) r, = -
CAS
1-6 Approximate the integral using (a) the midpoint approximation MIO, (b) the trapezoidal approximation T\o, and (c) Simpson's rule approximation $20 using Formula (7). In each case, find the exact value of the integral and approximate the absolute error. Express your answers to at least four decimal places. 2. I -^dx
1 dx
Ji, V*
,2
3. f cos x dx Jo
23-24 Find a function g(x) of the form g(x) = Ax2 + Bx + C whose graph contains the points (m — AJC, f(m — Ax)), (m, f(m)), and (m + Ax, f(m + Ax)), for the given function f(x) and the given values of m and Ax. Then verify Formula (11): i-m+&x ^x I g(x)dx = -—[Y0 + 4Y1 + Y2] Jm—A.x
3
where Y0 = f(m - Ax), Yl = f ( m ) , and F2 = f(m + Ax).
•
,3
4. / sinxdx Jo
5. I Ji
• dx
-dx 23. f(x) = -• x
7-12 Use inequalities (12), (13), and (14) to find upper bounds on the errors in parts (a), (b), or (c) of the indicated exercise, m 7. Exercise 1 8. Exercise 2 9. Exercise 3 10. Exercise 4
11. Exercise 5
12. Exercise 6
13-18 Use inequalities (12), (13), and (14) to find a number n of subintervals for (a) the midpoint approximation Mn, (b) the trapezoidal approximation Tn, and (c) Simpson's rule approximation Sn to ensure that the absolute error will be less than the given value. 13. Exercise 1; 5 x 1CT4 14. Exercise 2; 5 x 10~4 3
15. Exercise 3; 10~
4
17. Exercise 5; 1(T
3
m
= 3,Ax = l
24. f(x) = sin2(jrjc); m = \, Ax = \ 25-30 Approximate the integral using Simpson's rule S\Q and compare your answer to that produced by a calculating utility with a numerical integration capability. Express your answers to at least four decimal places. 25. 27.
cos(xz)dx
(lnx)3/2dx
16. Exercise 4; 1(T
18. Exercise 6; 1(T4
19-22 True-False Determine whether the statement is true or false. Explain your answer. 19. The midpoint approximation, Mn, is the average of the left and right endpoint approximations, Ln and Rn, respectively. 20. If f(x) is concave down on the interval (a, b), then the trapezoidal approximation Tn underestimates fa f(x) dx. 21. The Simpson's rule approximation S$o for fa f(x) dx is a weighted average of the approximations MSQ and TSO, where MSO is given twice the weight of TSQ in the average. 22. Simpson's rule approximation £50 for fa f ( x ) d x corresponds to fa q(x)dx, where the graph of q is composed of 25 parabolic segments joined at points on the graph of /.
31-32 The exact value of the given integral is n (verify). Approximate the integral using (a) the midpoint approximation MIO, (b) the trapezoidal approximation T\Q, and (c) Simpson's rule approximation £20 using Formula (7). Approximate the absolute error and express your answers to at least four decimal places.
•dx x2+4 33. In Example 8 we showed that taking n = 14 subdivisions ensures that the approximation of dx x by Simpson's rule is accurate to five decimal places. Confirm this by comparing the approximation of In 2 produced by Simpson's rule with n = 14 to the value produced directly by your calculating utility. In 2
-IJi
(a) I cos(x ) dx
(b)
Jo
/'
2
cos(x ) dx
h/2
35-36 Find a value of n to ensure that the absolute error in approximating the integral by the midpoint approximation will be less than 10^4. Estimate the absolute error, and express your answers to at least four decimal places. £
f2 35. / xsinxdx
36. / eco" dx
37-38 Show that the inequalities (12) and (13) are of no value in finding an upper bound on the absolute error that results from approximating the integral using either the midpoint approximation or the trapezoidal approximation, i: I
/•!
x^fxdx
/
38.
39-40 Use Simpson's rule approximation SIQ to approximate the length of the curve over the stated interval. Express your answers to at least four decimal places. & 39. y = sin x from x = 0 to x = n 40. y = x~2 from x = 1 to x = 2 FOCUS ON CONCEPTS
41. A graph of the speed v versus time t curve for a test run of a Mitsubishi Galant ES is shown in the accompanying figure. Estimate the speeds at times t = 0, 2.5, 5, 7.5, 10, 12.5, 15 s from the graph, convert to ft/s using 1 mi/h = f| ft/s, and use these speeds and Simpson's rule to approximate the number of feet traveled during the first 15 s. Round your answer to the nearest foot. [Hint: Distance traveled = / 0 v(t)dt.] Source: Data from Car and Driver, November 2003.
O
i
/
.... X _J—^ 1
5. 20 0
2.5
5
7.5
10 12.5 15 17.5
Tlmer(s)
-4 Figure Ex-41
42. A graph of the acceleration a versus time t for an object moving on a straight line is shown in the accompanying figure. Estimate the accelerations at t = 0, 1, 2 , . . . , 8 seconds (s) from the graph and use Simpson's rule to approximate the change in velocity from t = 0 to t = 8 s. Round your answer to the nearest tenth cm/s. [Hint: Change in velocity = fQ a(t) dt.]
2
/
A zq
3
4
ZJ ZJ J
o AJ IVJ _H
5
6
TimeJ(s)
H8
7
•< Figure Ex-42
43-46 Numerical integration methods can be used in problems where only measured or experimentally determined values of the integrand are available. Use Simpson's rule to estimate the value of the relevant integral in these exercises. • 43. The accompanying table gives the speeds, in miles per second, at various times for a test rocket that was fired upward from the surface of the Earth. Use these values to approximate the number of miles traveled during the first 180 s. Round your answer to the nearest tenth of a mile. [Hint: Distance traveled = /0180 v(t) dt.] TIME t (s) SPEED v (mi/s)
0 30 60 90 120 150 180
0.00 0.03 0.08 0.16 0.27 0.42 0.65
<4 Table Ex-43
44. The accompanying table gives the speeds of a bullet at various distances from the muzzle of a rifle. Use these values to approximate the number of seconds for the bullet to travel 1800 ft. Express your answer to the nearest hundredth of a second. [Hint: If v is the speed of the bullet and x is the distance traveled, then v = dx/dt so that dt/dx = l/v and
DISTANCE X (ft)
C/3
_
_ ,
(
545
— H)
!
L/i
34. In each part, determine whether a trapezoidal approximation would be an underestimate or an overestimate for the definite integral.
Acceleration a (cm/s2) p *-*
7.7 Numerical Integration; Simpson's Rule
SPEED V (ft/s)
0
3100
300 600 900 1200 1500 1800
2908 2725 2549 2379
2216 2059
« Table Ex-44
45. Measurements of a pottery shard recovered from an archaeological dig reveal that the shard came from a pot with a flat bottom and circular cross sections (see the accompanying
546
Chapter 7 / Principles of Integral Evaluation figure below). The figure shows interior radius measurements of the shard made every 4 cm from the bottom of the pot to the top. Use those values to approximate the interior volume of the pot to the nearest tenth of a liter (1 L = 1000 cm3). [Hint: Use 5.2.3 (volume by cross sections) to set up an appropriate integral for the volume.] A y (cm)
16
16.8cm-
12
— 15.4cm—:%
8
—13.8cm-
4
-11.5 cm•* Figure Ex-45
46. Engineers want to construct a straight and level road 600 ft long and 75 ft wide by making a vertical cut through an intervening hill (see the accompanying figure). Heights of the hill above the centerline of the proposed road, as obtained at various points from a contour map of the region, are shown in the accompanying figure. To estimate the construction costs, the engineers need to know the volume of earth that must be removed. Approximate this volume, rounded to the nearest cubic foot. [Hint: First set up an integral for the cross-sectional area of the cut along the centerline of the road, then assume that the height of the hill does not vary between the centerline and edges of the road.] HORIZONTAL DISTANCE X (ft) 0 100 200
300 400 500 600
HEIGHT fc(ft)
0 7 16 24 25 16 0
Centerline "75ft
A Figure Ex-46
47. Let f ( x ) = cos(x2). (a) Use a CAS to approximate the maximum value of |/"(*) | on the interval [0, 1]. (b) How large must n be in the midpoint approximation of fa f ( x ) dx to ensure that the absolute error is less than 5 x 10~4? Compare your result with that obtained in Example 9. (c) Estimate the integral using the midpoint approximation with the value of n obtained in part (b). 48. Let f ( x ) = V l + * 3 . (a) Use a CAS to approximate the maximum value of "(x)| on the interval [0, 1].
(b) How large must n be in the trapezoidal approximation of fQ f ( x ) dx to ensure that the absolute error is less than 10~3? (c) Estimate the integral using the trapezoidal approximation with the value of n obtained in part (b). [£] 49. Let f(x) = cos(x - x2). (a) Use a CAS to approximate the maximum value of f ( 4 ) ( x ) \ on the interval [0, 1]. (b) How large must the value of n be in the approximation Sn of f 0 f ( x ) dx by Simpson's rule to ensure that the absolute error is less than 10~4? (c) Estimate the integral using Simpson's rule approximation Sn with the value of n obtained in part (b).
@ 50. Let f ( x ) = V2 + JC3. (a) Use a CAS to approximate the maximum value of | /(4)(*)| on the interval [0, 1]. (b) How large must the value of n be in the approximation Sn of f0f(x)dx by Simpson's rule to ensure that the absolute error is less than 10~6? (c) Estimate the integral using Simpson's rule approximation Sn with the value of n obtained in part (b). FOCUS ON CONCEPTS
51. (a) Verify that the average of the left and right endpoint approximations as given in Table 7.7.1 gives Formula (2) for the trapezoidal approximation. (b) Suppose that / is a continuous nonnegative function on the interval [a, b] and partition [a, b] with equally spaced points, a = XQ < x\ < • • • < xn = b. Find the area of the trapezoid under the line segment joining points (xk, /(**)) and (xk+i, /ta+i)) and above the interval [xk, xk+{\. Show that the right side of Formula (2) is the sum of these trapezoidal areas (Figure 7.7.1). 52. Let / be a function that is positive, continuous, decreasing, and concave down on the interval [a,b]. Assuming that [a, b] is subdivided into n equal subintervals, arrange the following approximations of fa f ( x ) dx in order of increasing value: left endpoint, right endpoint, midpoint, and trapezoidal. 53. Suppose that Ax > 0 and g ( x ) = Ax2 + Bx + C. Let m be a number and set YQ — g(m — Ax), Y\ = g(m), and Y2 = g(m + Ax). Verify Formula (11):
_ Ax
/" t/m54. Suppose that / is a continuous nonnegative function on the interval [a, ft], n is even, and [a, b] is partitioned using n + 1 equally spaced points, a = XQ < x\ < • • • < xn = b. Set y0 = f(x0), y\ = /(*i), ..., yn = f(xn). Let gi, g2, • • • , gn/2 be the quadratic functions of the form gi(x) = Ax2 + Bx + C so that (com.)
7.8 Improper Integrals 547 • the graph of g\ passes through the points (XQ, yo), (xi,yi),and(x2,yz)', • the graph of #2 passes through the points fe , >2 ) . (xi,y3), and (x4, y*); •
55. Writing Discuss two different circumstances under which numerical integration is necessary.
• the graph of gn/2 passes through the points (xn-2, yn-2), (xn-i,yn-\), and (xn, yn). Verify that Formula (8) computes the area under a piecewise quadratic function by showing that
56. Writing For the numerical integration methods of this section, better accuracy of an approximation was obtained by increasing the number of subdivisions of the interval. Another strategy is to use the same number of subintervals, but to select subintervals of differing lengths. Discuss a scheme for doing this to approximate fQ */x dx using a trapezoidal approximation with 4 subintervals. Comment on the advantages and disadvantages of your scheme.
«/2
E
I {b-a\ =T 3 V n )
4}>3
•QUICK CHECK ANSWERS 7.7 1. (a) i(L B
«(^
(b)
[yo + 2yi + • • • + 1yn-i + yn] 2y4 + 4y5 + y6)
4. (a)
2. Mn < I < Tn
2400
(b)
1200
(c)
3. (a) §M3 1 1,800,000
5. (a) Mi = i (b) T, = f (c) 52 = |f
IMPROPER INTEGRALS Up to now we have focused on definite integrals with continuous integrands and finite intervals of integration. In this section we will extend the concept of a definite integral to include infinite intervals of integration and integrands that become infinite within the interval of integration.
IMPROPER INTEGRALS It is assumed in the definition of the definite integral
/'
f(x)dx
Ja
that [a, b] is a finite interval and that the limit that defines the integral exists; that is, the function / is integrable. We observed in Theorems 4.5.2 and 4.5.8 that continuous functions are integrable, as are bounded functions with finitely many points of discontinuity. We also observed in Theorem 4.5.8 that functions that are not bounded on the interval of integration are not integrable. Thus, for example, a function with a vertical asymptote within the interval of integration would not be integrable. Our main objective in this section is to extend the concept of a definite integral to allow for infinite intervals of integration and integrands with vertical asymptotes within the interval of integration. We will call the vertical asymptotes infinite discontinuities, and we will call
548
Chapter 7 / Principles of Integral Evaluation
integrals with infinite intervals of integration or infinite discontinuities within the interval of integration improper integrals. Here are some examples: • Improper integrals with infinite intervals of integration:
r* r e dx, r>+co
x
l
!\ h
xx
j_ J_ aaa
1 + X*-
j_x
Improper integrals with infinite discontinuities in the interval of integration: r3 —, dx /f2 dx -, / r tan x d x 2 J-3 X
Ji
X-l
Jo
Improper integrals with infinite discontinuities and infinite intervals of integration:
/•+" dx_ Jo
/•+'
V*'
dx 2
/-oo x -9'
r Ji
sec x dx
INTEGRALS OVER INFINITE INTERVALS
To motivate a reasonable definition for improper integrals of the form
/
f ( x ) dx
Ja
let us begin with the case where / is continuous and nonnegative on [a, +0°), so we can think of the integral as the area under the curve y = f ( x ) over the interval [a, +°o) (Figure 7.8.1). At first, you might be inclined to argue that this area is infinite because the region has infinite extent. However, such an argument would be based on vague intuition rather than precise mathematical logic, since the concept of area has only been defined over intervals of finite extent. Thus, before we can make any reasonable statements about the area of the region in Figure 7.8.1, we need to begin by defining what we mean by the area of this region. For that purpose, it will help to focus on a specific example. Suppose we are interested in the area A of the region that lies below the curve y = l/x2 and above the interval [1, +00) on the x-axis. Instead of trying to find the entire area at once, let us begin by calculating the portion of the area that lies above a finite interval [1, b], where b > 1 is arbitrary. That area is
A Figure 7.8.1
fh dx _ J \
X
1
1
JC
(Figure 7.8.2). If we now allow b to increase so that b—> +00, then the portion of the area over the interval [ l , b ] will begin to fill out the area over the entire interval [1, +co) (Figure 7.8.3), and hence we can reasonably define the area A under y = \/x2 over the interval [ 1, +00) to be
r+rx' dx
rb dx
-; =1
— = lim / — lim h x2 fc^Wi Thus, the area has a finite value of 1 and is not infinite as we first conjectured. A=
Area =
(1)
Area = 1
> Figure 7.8.3
With the preceding discussion as our guide, we make the following definition (which is applicable to functions with both positive and negative values).
7.8 Improper Integrals
If / is nonnegative over the interval [a, +00), then the improper integral in Definition 7.8.1 can be interpreted to be the area under the graph of / over theinterval [a, +00). If the integral converges, then the area is finite and equal to the value of the integral, and if the integral diverges, then the area is regarded to be infinite.
:
7.8.1 DEFINITION to be
549
The improper integral off over the interval [a, +00) is denned
f ( x ) dx — lim
Ja f
rb
1 f ( x ) dx
In the case where the limit exists, the improper integral is said to converge, and the limit is defined to be the value of the integral. In the case where the limit does not exist, the improper integral is said to diverge, and it is not assigned a value.
Example 1
Evaluate 3
(a) /
i
dx x
Solution (a). Following the definition, we replace the infinite upper limit by a finite upper limit 6, and (hen take (he limi( of (he resulting integral. This yields
dx
cb dx
m
m
r i ib
/i
i \
T = h' / ~~r = li ~^^ = n'm -x3 fe^+oo J, x3 fe^+oo [_ 2jc 2 J, fc^+» \2
i
;r~r2 — ~ 2b J 2
Since the limit is finite, the integral converges and its value is 1 /2. Solution (b). — = Hm / — = lim L[in*], = lim lnb = +oo x b^+nJi x b^+* *->+«, In this case the integral diverges and hence has no value. ^
A Figure 7.8.4
Because the functions l/x3, l/x2, and l/x are nonnegative over the interval [1, +00), it follows from (1) and the last example that over this interval the area under y = l/x3 is |, the area under y = l/x2 is 1, and the area under y — l/x is infinite. However, on the surface the graphs of the three functions seem very much alike (Figure 7.8.4), and there is nothing to suggest why one of the areas should be infinite and the other two finite. One explanation is that 1 /x3 and l/x2 approach zero more rapidly than 1 /x as x -> +co, so that the area over the interval [ l , b ] accumulates less rapidly under the curves y = l/x3 and y = l/x2 than under y = l/x as b^ +00, and the difference is just enough that the first two areas are finite and the third is infinite.
/
+°° dx — converge?
Solution. We know from the preceding example that the integral diverges if p = 1 , so let us assume that p ^ 1 . In this case we have
r+-dx J|
r
— = lim / x XP
b^+fx>Jl
p
= lim
dx = lim
b1-"
---
l —p p
i \—p\
If p > 1, then the exponent 1 — p is negative and b' —> 0 as b -> +00; and if /? < 1, then the exponent I — pis positive and bl~p -^~ +°° as b^ +00. Thus, the integral converges if p > 1 and diverges otherwise. In the convergent case the value of the integral is
r+DC dx ,+00
I, iH'-
p-1
550
Chapter 7 / Principles of Integral Evaluation The following theorem summarizes this result. 7.8.2
THEOREM
1 p-1 diverges
+c
° dx XP
/
if
P > 1
if
/>< 1
/•+OC
Example 3
((1 - x)e~x dx.
Evaluate / Jo
Solution. We begin by evaluating the indefinite integral using integration by parts. Setting u = 1 — x and dv = e~x dx yields f (I- x)e~x dx = -
C
C = xe
Thus, /•+<*>
/
J0
= (\-x)e~
A Figure 7.8.5
I
b^+oo L
JO
b^+tx e
f+OO
(1 -x)e~xdx = lim — = 0
We can interpret this to mean that the net signed area between the graph ofy = ( l — x)e~ and the interval [0, +00) is 0 (Figure 7.8.5). •« DEFINITION
The improper integral off over the interval (— <x>, b] is defined fb
fb
/
If / is nonnegative over the interval (—cxi, +c»), then the improper integral
can be interpreted to be the area under the graph of / over the interval (—00, +00). The area is finite and equal to the value of the integral if the integral converges and is infinite if it diverges.
b
The limit is an indeterminate form of type oo/oo, so we will apply L'Hopital's rule by differentiating the numerator and denominator with respect to b. This yields
7.8.3 to be
f(x)dx
b
/ (1 - x)e~x dx = lim \xe~*\ = lim —
b^+txjQ
Jo
The net signed area between the 1 graph and the interval [0, +00) is I zero.
fb
(1 - x)e~x dx = lim
f(x)dx = a lim / f(x)dx
7-o=
(2)
^-">Ja
The integral is said to converge if the limit exists and diverge if it does not. '• The improper integral of f over the interval (—00, +00) is defined as
I
f(x)dx= f
J—co
f(x)dx+ f
J— co
f(x)dx
(3)
Jc
! where c is any real number. The improper integral is said to converge if both terms converge and diverge if either term diverges.
+OQ
/ Solution, we obtain Although we usually choose c = 0 in (3), the choice does not matter because it can be proved that neither the convergence nor the value of the integral is affected by the choice of c.
dx -
OT t
We will evaluate the integral by choosing c = 0 in (3). With this value for c
00 /•+ •+ dx rh dx I / -- — lim 2 - = lim Jo l+x b^+*J0 l+x2 b^+^l
f°
I
dx
-- — lim
J-vl+.X2
C°
<*-+-<» Ja
dx
r
/ r = lim 2 l+X
a^-» L
, f n tan x\ — lim (tan b) = — Jo (>-*+= 2 _, f
tan Ja
_
ji
x\ = lim (— tan a) = a-+-«
2
7.8 Improper Integrals
551
Thus, the integral converges and its value is +
°°
Area = n
A Figure 7.8.6
dx
/ .„ l + *
r° 2=
f+co
dx
.Ll+*
2+
Jo
2
I+x
2
n 2
Since the integrand is nonnegative on the interval (—00, +00), the integral represents the area of the region shown in Figure 7.8.6. •*
INTEGRALS WHOSE INTEGRANDS HAVE INFINITE DISCONTINUITIES
Next we will consider improper integrals whose integrands have infinite discontinuities. We will start with the case where the interval of integration is a finite interval [a, b] and the infinite discontinuity occurs at the right-hand endpoint. To motivate an appropriate definition for such an integral let us consider the case where / is nonnegative on [a,b], so we can interpret the improper integral fa f ( x ) dx as the area of the region in Figure 7.8.7'a. The problem of finding the area of this region is complicated by the fact that it extends indefinitely in the positive ^-direction. However, instead of trying to find the entire area at once, we can proceed indirectly by calculating the portion of the area over the interval [a, k], where a < k < b, and then letting k approach b to fill out the area of the entire region (Figure 7.8.7fo). Motivated by this idea, we make the following definition.
7.8.4 DEFINITION If / is continuous on the interval [a, b}, except for an infinite discontinuity at b, then the improper integral of f over the interval [a, b] is defined as
/ f(x) dx = lim / f(x) dx
Ja
"
k-^b-Ja
(4)
In the case where the indicated limit exists, the improper integral is said to converge, and the limit is defined to be the value of the integral. In the case where the limit does not exist, the improper integral is said to diverge, and it is not assigned a value. A Figure 7.8.7
Example 5
2 -
/•' dx Evaluate / . Jo V1 — x
Solution. The integral is improper because the integrand approaches +00 as x approaches the upper limit 1 from the left (Figure 7.8.8). From (4), /•' dx fk dx ,I —= = lim / —:= = lim I - 2^T
= lim [-2V1 - k + 2\ = 2 -4 t-> i" L
A Figure 7.8.8
Improper integrals with an infinite discontinuity at the left-hand endpoint or inside the interval of integration are defined as follows.
552
Chapter 7 / Principles of Integral Evaluation
7.8.5 DEFINITION If / is continuous on the interval [a, b], except for an infinite discontinuity at a, then the improper integral off over the interval [a, b] is defined as f ( x ) d x = lim
f(x)dx
(5)
The integral is said to converge if the indicated limit exists and diverge if it does not. If / is continuous on the interval [a, b], except for an infinite discontinuity at a point c in (a,b), then the improper integral off over the interval [a, b] is defined as I f(x)dx=
/' f ( x ) d x + I
Ja
f(x)dx
i
f(x)dx is improper.
Ja
f(x)dx
(6)
Jc
where the two integrals on the right side are themselves improper. The improper integral on the left side is said to converge if both terms on the right side converge and diverge if either term on the right side diverges (Figure 7.8.9).
Example 6 Evaluate
A Figure 7.8.9
dx
(a)
(x - 2)2/3
Solution (a). The integral is improper because the integrand approaches —oo as x approaches the lower limit 1 from the right (Figure 7.8.10). From Definition 7.8.5 we obtain = lim [-ln|l -x\\ fc-*l+
I —x
l—X
L
J
= lim
= lim L[-ln|-l
-
= -oo
k^l+
so the integral diverges. l-x
Solution (b). The integral is improper because the integrand approaches +00 at x = 2, which is inside the interval of integration. From Definition 7.8.5 we obtain A Figure 7.8.10
f4 J,
dx
dx
r
(X~ 2)2/3
J,
(X- 2)2/3 (JC
dx J2
(x_
2)2/3
and we must investigate the convergence of both improper integrals on the right. Since lim lim 7-^-3= [3*-^ -3d -2)^1 = 3 (.x-2) 2 / 3 k-+2-Jf1 7-^2/3= (x-2)2/3 t->2- L
(x-
we have from (7) that
f
dx -
2)2/3
= 3 + 3-2/2
(7)
7.8 Improper Integrals
WARNING
553
It is sometimes tempting to apply the Fundamental Theorem of Calculus directly to an improper integral without taking the appropriate limits. To illustrate what can go wrong with this procedure, suppose we fail to recognize that the integral
dx
f
(8)
(x - I)2
Jo
is improper and mistakenly evaluate this integral as
This result is clearly incorrect because the integrand is never negative and hence the integral cannot be negative! To evaluate (8) correctly we should first write
r2
_ /•'
dx
Jo (x ~ 1)
dx
dx
I)2 /I (* -1)2
Jo (•*• ~ 1
and then treat each term as an improper integral. For the first term, /•'
/„ c
dx
fk
dx
so (8) diverges.
ARC LENGTH AND SURFACE AREA USING IMPROPER INTEGRALS
In Definitions 5.4.2 and 5.5.2 for arc length and surface area we required the function / to be smooth (continuous first derivative) to ensure the integrability in the resulting formula. However, smoothness is overly restrictive since some of the most basic formulas in geometry involve functions that are not smooth but lead to convergent improper integrals. Accordingly, let us agree to extend the definitions of arc length and surface area to allow functions that are not smooth, but for which the resulting integral in the formula converges. > Example 7
Derive the formula for the circumference of a circle of radius r.
Solution. For convenience, let us assume that the circle is centered at the origin, in which case its equation is x2 + y2 = r2. We will find the arc length of the portion of the circle that lies in the first quadrant and then multiply by 4 to obtain the total circumference (Figure 7.8.11). Since the equation of the upper semicircle is y = vV2 — x2, it follows from Formula (4) of Section 5.4 that the circumference C is
A Figure 7.8.11
This integral is improper because of the infinite discontinuity at x = r, and hence we evaluate it by writing = 4,lim f k^r-J0
^L_ _ X2
r2
r ) J0
Formula (77) in the Endpaper Integral Table
= 4r lim sin ' I - | — sin 10 k^r- [_ \rj
= 4r[sin~1 1 - sin"1 0] = 4r (- - o) = 2nr
554
Chapter 7 / Principles of Integral Evaluation
I/QUICK CH ECK EXERCISES 7.8
(See page 557 for answers.)
1. In each part, determine whether the integral is improper, and if so, explain why. Do not evaluate the integrals. <-3jT/4
converges tn
Jn/4
/•+•»
1 xV2
JO
[+"
_i_ 1 + l
1 ~-2 X -
J\
1
2. Express each improper integral in Quick Check Exercise 1 in terms of one or more appropriate limits. Do not evaluate the limits.
EXERCISE SET 7.8
B Graphing Utility
/
(P)
-
Ji x-3
h x +3
/•
4. Evaluate t he integrals that converge. /•+« x (b) / e^ rfjc (a) I .? dx Jo Jo /•' 1 - dx (d) / -^—=2 dx Jo *• Jo V*
(0 /" 4
/17T/2
23 / L/3 vT - 2cosx
r -«^
r8
J-l /»+00
10
2
'J3
7.
*
2
ft r -^ 6. / Jo
d* -l
/•+" 1
1 3 ^
/"
dx
ze
,.+00
8
—/
Y
d fv{ V
r 32. / " Jo
dx
dx*•)
fj Y —— -Jx(x + 1)
+CC
ax
3
x-4/
.
1
x
33-36 True- False Determine whether the statement is tr ueor false. Explain your answer.
10
r
30 r
1
/
\
1 - tanjc
f1 Jo (%
27. / x-'/ 3
Jo v-*-v^ + 1)
3-32 Evaluate the integral;> that converge. Bs /•+00 2l /•+a> jc 3. / e ax J-i 1+^ 2 Jo
Jo
if.
31 /"
5
prnvirlpH
"Jo '
/•+" 3 dx rn/4 (d) / e~zdx (e) I V/r^ ri /„ Ji J-oo ^0 2. In each part, determine i ill values of p for which the integral is improper. 2 dx 1 /•' dx (ft\ 1 Chi W / W Ji x- p Jo Jo XP
5
P A v
0 CAS
1. In each part, determine whether the integral is improper, and if so, explain why. C5 dx f5 dx f1 !„ „ ,/„ C \ I (\*\ 1 t^\ l n
Y
h
cot x dx
(b) /
Jji/4
(."•)
/•+00
/>7T
cot x dx
(a) /
3. The imprciper integral
rlv
dx converges to 3.
/_ 34. If / is coiitinuous on [a +00] and lim _»_)_ f(x) — 1 then dx converges.
dx
10 f
-1ST 3<
/ /
f
Jl
r\
-H* -
1
r i
36. /
— d),; = 0
+00
15
37-40 Make the w-substitution and evaluate the resulting definite integral.
-co 1
'• /^
17
-/v
x dx
/
(Z2 + 3) 2
f4 'Jo
dx (--4) 2
/>JT/2
19. / tan x dx Jo /*' dx ''I / 7 J*o v ! -x2
d
/7r
1fi
L d i+ e - 2 ' /
is r *
'/^
/•+•» e-v
17 / '• /
x / Jo \3/7
-o f
4
^
Jo V4^c
T>
/
^^
/-
0.]
Jo V^ /•+»
'' Jl2 /•+oo
39
V^'(x+4)' (
,—x
/
' Jo vT [Note: u-->• 1 as x^>+<x>.]
+00.]
7.8 Improper Integrals
40. / Jo
:
dx; u = e
555
I 53-56 Use the results in Exercise 52. I 53. (a) Confirm graphically and algebraically that
fc] 4T-42 Express the improper integral as a limit, and then evaluate that limit with a CAS. Confirm the answer by evaluating the integral directly with the CAS.
(b) Evaluate the integral
,,+=0
+»
42. / xe~3* dx / Jo [c] 43. In each part, try to evaluate the integral exactly with a CAS. If your result is not a simple numerical answer, then use the CAS to find a numerical approximation of the integral. e~"cosxdx
/•
(a) / dx /_ '_„ x*+x + l l
(c) / -^dx J\ ex J\ x[c] 44. In each part, confirm the result with a CAS. f+c° sin* sin* (a) / dx = — Jo Jx f 1 In* n2 (c) / - dx = -Jo l+x 12
(b)
-V5 £
dx '-' / (c) What does the result obtained in part (b) tell you about the integral /•+<» x , e~ dxl 54. (a) Confirm graphically and algebraically that 1 ex (*>0) 2x + l ~ 2x (b) Evaluate the integral
dx 2x + (c) What does the result obtained in part (b) tell you about the integral
fJo
45. Find the length of the curve y = (4 — jc 2 ' 3 ) 3 ' 2 over the interval [0, 8]. 2
46. Find the length of the curve y = V4 — x over the interval [0, 2]. 47-48 Use L'Hopital's rule to help evaluate the improper integral. 47.
48. fin
[nxdx
J]
X2
dx
49. Find the area of the region between the x-axis and the curve y = e~3x for x > 0. 50. Find the area of the region between the jc-axis and the curve 3; = 8/(;t 2 -4)for;t > 4.
+CO
f10
Jo
2x+\
-dx?
55. Let R be the region to the right of x = 1 that is bounded by the .r-axis and the curve y = l/x. When this region is revolved about the .r-axis it generates a solid whose surface is known as Gabriel's Horn (for reasons that should be clear from the accompanying figure). Show that the solid has a finite volume but its surface has an infinite area. [Note: It has been suggested that if one could saturate the interior of the solid with paint and allow it to seep through to the surface, then one could paint an infinite surface with a finite amount of paint! What do you think?]
51. Suppose that the region between the z-axis and the curve y = e~* for x > 0 is revolved about the x-axis. (a) Find the volume of the solid that is generated. (b) Find the surface area of the solid. •^ Figure Ex-55 F O C U S ON CONCEPTS
52. Suppose that / and g are continuous functions and that
0 < f(x) < g(x) if x > a. Give a reasonable informal argument using areas to explain why the following results are true. (a) If/Q+CC/(JC) dx diverges, then /a+°°g(x) dx diverges. (b) If fa <xg(x)dx converges, then fa ™f(x)dx converges and f^"f(x) dx < f^"g(x) dx. [Note: The results in this exercise are sometimes called comparison tests for improper integrals.]
56. In each part, use Exercise 52 to determine whether the integral converges or diverges. If it converges, then use part (b) of that exercise to find an upper bound on the value of the integral.
(a) (c)
f f
JI Jo
y. xe*
•dx
dx
(b)
fJ2
•dx
1
556
Chapter 7 / Principles of Integral Evaluation
FOCUS ON CONCEPTS
62-63 A transform is a formula that converts or "transforms" one function into another. Transforms are used in applications to convert a difficult problem into an easier problem whose solution can then be used to solve the original difficult problem. The Laplace transform of a function /(?), which plays an important role in the study of differential equations, is denoted by ^£{f(t)} and is defined by
57. Sketch the region whose area is r+°° dx
Jo 1 + x2 and use your sketch to show that
Tif^
Jo '+x 58. (a) Give a reasonable informal argument, based on areas, that explains why the integrals ,+co +co
/•+»
sin x dx
/
and
Jo
/
cos x dx
Jo
diverge. (b) Show that
f
cos^/x
Jo
>0
(a)
dx diverges.
1
r, s >2
-, s > 0
-, s > 0.
63. In each part, find the Laplace transform. (a) f ( t ) = t, s > 0 (b) f ( t ) = t2, s > 0 (r\
4 / M V T" tf dv ^ \2RTJ JQ Jo where v is the molecular speed, T is the gas temperature, M is the molecular weight of the gas, and R is the gas constant, (a) Use a CAS to show that 2 2
(b)
f ^i
(c) /(?) =
[c"| 60. The average speed, v, of the molecules of an ideal gas is given by '• /•+» M / ,VM"Jo and the root-mean-square speed, i^ms, by
:
(d) ££{cosf} =
L
where A', /, r, k, and a are constants. Find u.
/o
In this formula s is treated as a constant in the integration process; thus, the Laplace transform has the effect of transforming f ( t ) into a function of.. Use this formula in these exercises. K 62. Show that
59. In electromagnetic theory, the magnetic potential at a point on the axis of a circular coil is given by IjiNlr Mr ff+* dx
r+«
Jo
(c)
k
e-s'f(t)dt
= I
|"c] 64. Later in the text, we will show that r.+oo
dx = \ Jn Jo Confirm that this is reasonable by using a CAS or a calculator with a numerical integration capability. 65. Use the result in Exercise 64 to show that
5F
(a)/
e-x2/2a2 dx= 1, o- > 0 .
(b)
1 ^«
and use this result to show that ij = (b) Use a CAS to show that
a>0
nM).
A+00
a >0 8«5 and use this result to show that tirms = \/3RT/M.
fJo
Jo
61. In Exercise 25 of Section 5.6, we determined the work required to lift a 6000 Ib satellite to an orbital position that is 1000 mi above the Earth's surface. The ideas discussed in that exercise will be needed here. (a) Find a definite integral that represents the work required to lift a 6000 Ib satellite to a position b miles above the Earth's surface. (b) Find a definite integral that represents the work required to lift a 6000 Ib satellite an "infinite distance" above the Earth's surface. Evaluate the integral. [Note: The result obtained here is sometimes called the work required to "escape" the Earth's gravity.]
66-67 A convergent improper integral over an infinite interval can be approximated by first replacing the infinite limit(s) of integration by finite limit(s), then using a numerical integration technique, such as Simpson's rule, to approximate the integral with finite limit(s). This technique is illustrated in these exercises. * 66. Suppose that the integral in Exercise 64 is approximated by first writing it as />+cc
/ Jo
vK
e~x2 dx= I
/•+
e~xl dx + /
Jo
dx
JK
then dropping the second term, and then applying Simpson's rule to the integral f K
fJo
e~
Jo
The resulting approximation has two sources of error: the error from Simpson's rule and the error
-fJK
'dx
Chapter 7 Review Exercises
557
that results from discarding the second term. We call E the [c] 70. It is sometimes possible to convert an improper integral truncation error. into a "proper" integral having the same value by making (a) Approximate the integral in Exercise 64 by applying an appropriate substitution. Evaluate the following integral Simpson's rule with n = 10 subdivisions to the integral by making the indicated substitution, and investigate what happens if you evaluate the integral directly using a CAS. dx Jo Round your answer to four decimal places and compare it to 5 ^/JT rounded to four decimal places, (b) Use the result that you obtained in Exercise 52 and the 71-72 Transform the given improper integral into a proper infact that e~x < \xe~x for* > 3 to show that the truntegral by making the stated w-substitution; then approximate cation error for the approximation in part (a) satisfies the proper integral by Simpson's rule with n = 10 subdivisions. 0 < E <2.1 x IQ-5. Round your answer to three decimal places. 67. (a) It can be shown that cos* • dx; u = 1
/
j
1
Jo Approximate this integral by applying Simpson's rule with n = 20 subdivisions to the integral 1
r
-dx
Round your answer to three decimal places and compare it to n/3 rounded to three decimal places, (b) Use the result that you obtained in Exercise 52 and the fact that l/(x6 + 1) < l/x6 for x > 4 to show that the truncation error for the approximation in part (a) satisfies 0 < E < 2 x 1Q-4. X.+00
epx dx converge?
68. For what values of p does /
/"' sin* 72. / —= Jo \l\-
= V1 — x
73. Writing What is "improper" about an integral over an infinite interval? Explain why Definition 4.5.1 for fa f(x) dx fails for /a+c° f ( x ) d x . Discuss a strategy for assigning a value to /o+°° f(x) dx. 74. Writing What is "improper" about a definite integral over an interval on which the integrand has an infinite discontinuity? Explain why Definition 4.5.1 for fa f(x) dx fails if the graph of / has a vertical asymptote at x = a. Discuss a strategy for assigning a value to fa f(x) dx in this circumstance.
69. Show that / dx/xp converges if p < I and diverges if
i
I/QUICK CHECK ANSWERS 7.8
1. (a) proper (b) improper, since cot x has an infinite discontinuity at x = n (c) improper, since there is an infinite interval of integration (d) improper, since there is an infinite interval of integration and the integrand has an infinite discontinuity at x = 1 r r i dx (d) lim /f2 —, 1 -dx+ lim /fb —. 1 dx 3. 1 -; p>l 2. (b) lim / cotxdx (c) lim / — b^n" Jx/4 b^+^J0 X + 1 p-l 4. (a) 1 (b) diverges (c) diverges (d) 3
CHAPTER 7 REVIEW EXERCISES
\ -6 Evaluate the given integral with the aid of an appropriate ^-substitution. • 1 dx 1. / V4 + 9x dx 2.
7. (a) Evaluate the integral
sec TO:
3. / Vc08 x sin x dx 5.
xtan2(x2)sec2(x2)dx
4. 6.
dx xlnx ' ^ •dx o *+ 9
three ways: using the substitution M = *Jx, using the substitution u = V2 — x, and completing the square. (b) Show that the answers in part (a) are equivalent.
558
Chapter 7 / Principles of Integral Evaluation 35-40 Use the Endpaper Integral Table to evaluate the integral, m
8. Evaluate the integral (a) using integration by parts (b) using the substitution u = \lx2 + 1.
:
dx
sin Ix cos 9x dx
9-T2 Use integration by parts to evaluate the integral. 9. / xe~* dx
10. / x sin 2x dx
, 11. / ln(2x + 3) dx
2 + x2
,1 • 1/2
12.
tan \2x)dx
4
13. Evaluate f 8x cos 2x dx using tabular integration by parts. 14. A particle moving along the x -axis has velocity function v(t) = t2e~'. How far does the particle travel from time t = 0 to t = 5?
41-42 Approximate the integral using (a) the midpoint approximation MIQ, (b) the trapezoidal approximation T\Q, and (c) Simpson's rule approximation 820- In each case, find the exact value of the integral and approximate the absolute error. Express your answers to at least four decimal places. •
42.
f3
15-20 Evaluate the integral. 15.;.// si sin2 59 dO
16. / sin3 2x cos2 2x dx
17. /si sin x cos 2x dx
18. /
f JT/6
sin 2x cos 4x dx
Jo
20 . I xcos5(x2)dx
19. /si sin2xdx
21-26 Evaluate the integral by making an appropriate trigonometric substitution. » dx
27-32 Evaluate the integral using the method of partial fractions. B '
•h
dX
»•/
X
_+X
30>. I
29.
-dx
3. Consider the ii
I
-dx
43-44 Use inequalities (12), (13), and (14) of Section 7.7 to find upper bounds on the errors in parts (a), (b), or (c) of the indicated exercise. « 43. Exercise 41 44. Exercise 42 45-46 Use inequalities (12), (13), and (14) of Section 7.7 to find a number n of subintervals for (a) the midpoint approximation M n , (b) the trapezoidal approximation 7 K , and (c) Simpson's rule approximation Sn to ensure the absolute error will be less thanl(T 4 . » 45. Exercise 41 46. Exercise 42 47-50 Evaluate the integral if it converges. • dx 47. 48.
dx
27,
l
32
f
_*
-dx
dx
• h^~x J
A
49.
so./'
f
/9-x Jo Jo 2x - 1 51. Find the area that is enclosed between the jc-axis and the curve y = (In x — \')/x2 for x > e. 52. Find the volume of the solid that is generated when the region between the .r-axis and the curve y = e~x for x > 0 is revolved about the y-axis. 53. Find a positive value of a that satisfies the equation Jo
T A.
f
Jo
/^
(a) Evaluate the integral using the substitution x = sec6>. For what values of x is your result valid? (b) Evaluate the integral using the substitution x = smO. For what values of x is your result valid? (c) Evaluate the integral using the method of partial fractions. For what values of x is your result valid? 34. Find the area of the region that is enclosed by the curves y = (x- 3)/(jc3 + x2), y = 0,x = l,andx = 2.
x2 + a2
-dx = 1
54. Consider the following methods for evaluating integrals: w-substitution, integration by parts, partial fractions, reduction formulas, and trigonometric substitutions. In each part, state the approach that you would try first to evaluate the integral. If none of them seems appropriate, then say so. You need not evaluate the integral. (a) / x sin x dx
(b) / cos x sin x dx (cont.)
Chapter 7 Making Connections (c) / tan7 x dx
(d) / tan7 x sec2 x dx
1
(.t.
64
j (x _ \)(x + 2)(x — 3) 3jc
/
J/
]
(g)
/* 1
(f\
/7v
_|_ 11 A 3 -j"
ttm~ xdx
3jc
v
J4-x dx
,-1/2
69. /
57 f /4 tan 7 ^
/ • e 2 *cos3.xrfjt
°
J sin 0 — 6 sin 0 + 12
cin 2 9v pr>c3 9v /7v
/ /
C S0
53 f
jo
1
fid
A
/•4 /
Jfl
i 2
,Vv
Jo (*-3) / 5 62. / (1 - 2x2)3/2 dx J-lA/2
CHAPTER 7 MAKING CONNECTIONS
f f(x) dx = bf(b) - af(a) - 1 xf'(x) dx Ja
Ja
(b) Use the result in part (a) to show that if y = f ( x ) , then r-b
ff(b)
I f(x) dx = bf(b) - af(a) - I
f-\y) dy
Jf(a)
(c) Show that if we let a = /(a) and ft = /(£), then the result in part (b) can be written as
1 f-l(x)dx=pf~l(p)-af~l(a)-
1
Ja
f(x)dx Jf~*(a)
2. In each part, use the result in Exercise 1 to obtain the equation, and then confirm that the equation is correct by performing the integrations. /*l/2
(a) / Jo (b)
/*7r/6
sin~' x dx = ^ sin"1 (|) — / Jo
r/ 2 In x dx
•=. (2e
2 f — e) — 1 e
x
dx
73
r*
-
~r i)
dX
70. / tan5 4x sec4 4x dx r «pp2 ft 72 / rffl 3 J tan 9 - tan2 6
dx
dX
71 1r JQ
1 -if
/•
sin~lxdx
\X
Jrx
63 {
dx
^ flr 7^2 + 2x + 2 Ja
/
aL? +, b1,9zx27 '
ab~V '
E CAS
1. Recall from Theorem 6.3.1 and the discussion preceding it that if /'(*) > 0, then the function / is increasing and has an inverse function. Parts (a), (b), and (c) of this problem show that if this condition is satisfied and if /' is continuous, then a definite integral of /"' can be expressed in terms of a definite integral of /. (a) Use integration by parts to show that
Ja
/• In 2
66
Y 4X+
/
(4 ~" 9-* )
J0
1
67 [
55-74 Evaluate the integral. • C dx r 55. / . 56. / jc cos 3x dx J (3 -|_ j(2\3/2 J
~A
-——fit
x
/
2
(h)
j~^
8
fJY
r ~(~ 4- 1^3 i,A I)
X
/
Jo
559
sinxdx
3. The Gamma function, V(x), is defined as
r(x)= I
tx-le-'dt
Jo It can be shown that this improper integral converges if and only if x > 0. (a) FindT(l). (b) Prove: r(x + 1) = xY(x} for all x > 0. [Hint: Use integration by parts.] (c) Use the results in parts (a) and (b) to find T (2) , T (3) , and F(4); and then make a conjecture about r(n) for positive integer values of n. (d) Show that T (i) = ^/JT. [Hint: See Exercise 64 of Section 7.8.] (e) Use the results obtained in parts (b) and (d) to show that
4. Refer to the Gamma function defined in Exercise 3 to show that 1
r
(a) / (In*) dx = (— 1) F(w + 1), n > 0 Jo [Hint: Let f = -ln;c.] /•+» x /„ _i_ i\ (h} i1 r~ HY rl 1i, n ^ ~~ 0 \v) K ux 11 u.
Jo \ n / [Hint: Let t = x". Use the result in Exercise 3(b).] [c] 5. A simple pendulum consists of a mass that swings in a vertical plane at the end of a massless rod of length L, as shown in the accompanying figure. Suppose that a simple pendulum is displaced through an angle do and released from rest. It can be
560
Chapter 7 / Principles of Integral Evaluation
shown that in the absence of friction, the time T required for the pendulum to make one complete back-and-forth swing, called the period, is given by ,
1
Vcos 6 - cos 00
dO
(1)
where 0 = G(t) is the angle the pendulum makes with the vertical at time t. The improper integral in (1) is difficult to evaluate numerically. By a substitution outlined below it can be shown that the period can be expressed as
and then making the change of variable sin(f> =
sin(0/2) sin(00/2)
sin(0/2)
(b) Use (2) and the numerical integration capability of your CAS to estimate the period of a simple pendulum for which L = 1.5 ft, 00 = 20°, and g = 32 ft/s2.
— d (2) Jo VI - k2 sin2 <j> where k = sin(0o/2). The integral in (2) is called a complete elliptic integral of the first kind and is more easily evaluated by numerical methods, (a) Obtain (2) from (1) by substituting
•4 Figure Ex-5
cos0 = l-2sin 2 (0/2) cos 00 = 1 -2 sin2' k = sin(00/2)
EXPANDING THE CALCULUS HORIZON To learn how numerical integration can be applied to the cost analysis of an engineering project, see the module entitled Railroad Design at: www.wiley.com/college/anton
Photo by Milton Bell, Texas Archeological Research Laboratory, The University of Texas at Austin.
In the 1920's, excavation of an archeological site in Folsom, N.M. uncovered a collection of prehistoric stone spearheads now known as "Folsom points." In 1950, carbon dating of charred bison bones found nearby confirmed that human hunters lived in the area between 9000 B.C. and 8000 B.C. We will study carbon dating in this chapter.
MATHEMATICAL MODELING WITH DIFFERENTIAL EQUATIONS
Many of the principles in science and engineering concern relationships between changing quantities. Since rates of change are represented mathematically by derivatives, it should not be surprising that such principles are often expressed in terms of differential equations. We introduced the concept of a differential equation in Section 4.2, but in this chapter we will go into more detail. We will discuss some important mathematical models that involve differential equations, and we will discuss some methods for solving and approximating solutions of some of the basic types of differential equations. However, we will only be able to touch the surface of this topic, leaving many important topics in differential equations to courses that are devoted completely to the subject.
MODELING WITH DIFFERENTIAL EQUATIONS In this section we will introduce some basic terminology and concepts concerning differential equations. We will also discuss the general idea of modeling with differential equations, and we will encounter important models that can be applied to demography, medicine, ecology, and physics. In later sections of this chapter we will investigate methods that may be used to solve these differential equations.
TERMINOLOGY
Table 8.1.1 DIFFERENTIAL EQUATION
ORDER
dy
dx2 3
dy
dx dy
2
dt
t
3
y'-y = e2*
1
y"+y'
2
= cos t
Recall from Section 4.2 that a differential equation is an equation involving one or more derivatives of an unknown function. In this section we will denote the unknown function by y = y(x) unless the differential equation arises from an applied problem involving time, in which case we will denote it by y = y(t). The order of a differential equation is the order of the highest derivative that it contains. Some examples are given in Table 8.1.1. The last two equations in that table are expressed in "prime" notation, which does not specify the independent variable explicitly. However, you will usually be able to tell from the equation itself or from the context in which it arises whether to interpret y' as dy/dx or dy/dt. • SOLUTIONS OF DIFFERENTIAL EQUATIONS
A function y = y(x) is a solution of a differential equation on an open interval if the equation is satisfied identically on the interval when y and its derivatives are substituted 561
562
Chapter 8 / Mathematical Modeling with Differential Equations
into the equation. For example, y — e2x is a solution of the differential equation
2,
(1)
dx on the interval (-00, +00), since substituting y and its derivative into the left side of this equation yields d d ¥--y= — [e2x] - e2* = 2e2x - e2x = e2x dx dx for all real values of x. However, this is not the only solution on (—00, +00); for example, 2x the function (2) y = ezx + Cex
is also a solution for every real value of the constant C, since
dy__ dx
The first-order equation (1) has a single arbitrary constant in its general solution (2). Usually, the general solution of an nth-order differential equation will contain n arbitrary constants. This is plausible, since « integrations are needed to recover a function from its nth derivative.
_ d dx
,2*
Cex] - (e2x + Cex) = (2e2x + Cex) - (e2x + Cex)=e,,2*
After developing some techniques for solving equations such as (1), we will be able to show that all solutions of (1) on (—00, +00) can be obtained by substituting values for the constant C in (2). On a given interval, a solution of a differential equation from which all solutions on that interval can be derived by substituting values for arbitrary constants is called a general solution of the equation on the interval. Thus (2) is a general solution of (1) on the interval (-00, +00). The graph of a solution of a differential equation is called an Integral curve for the equation, so the general solution of a differential equation produces a family of integral curves corresponding to the different possible choices for the arbitrary constants. For example, Figure 8.1.1 shows some integral curves for (1), which were obtained by assigning values to the arbitrary constant in (2). C=3
C=2
c=4A/,c=i
> Figure 8.1.1
INITIAL-VALUE PROBLEMS
When an applied problem leads to a differential equation, there are usually conditions in the problem that determine specific values for the arbitrary constants. As a rule of thumb, it requires n conditions to determine values for all n arbitrary constants in the general solution of an nth-order differential equation (one condition for each constant). For a firstorder equation, the single arbitrary constant can be determined by specifying the value of the unknown function y(x) at an arbitrary x-value XQ, say y(x0) = y®. This is called an initial condition, and the problem of solving a first-order equation subject to an initial condition is called a first-order initial-value problem. Geometrically, the initial condition y(x0) — y0 has the effect of isolating the integral curve that passes through the point (XQ, y0) from the complete family of integral curves.
8.1 Modeling with Differential Equations
563
^ Example 1 The solution of the initial-value problem d
f-y ax
= e^,
y(0) = 3
can be obtained by substituting the initial condition x = 0, y = 3 in the general solution (2) to find C. We obtain 3 = e° + Ce° = 1 + C Thus, C = 2, and the solution of the initial-value problem, which is obtained by substituting this value of C in (2), is y = e2" + 2ex Geometrically, this solution is realized as the integral curve in Figure 8.1.1 that passes through the point (0, 3). •< Since many of the fundamental laws of the physical and social sciences involve rates of change, it should not be surprising that such laws are modeled by differential equations. Here are some examples of the modeling process. UNINHIBITED POPULATION GROWTH
One of the simplest models of population growth is based on the observation that when populations (people, plants, bacteria, and fruit flies, for example) are not constrained by environmental limitations, they tend to grow at a rate that is proportional to the size of the population—the larger the population, the more rapidly it grows. To translate this principle into a mathematical model, suppose that y — y (t) denotes the population at time t. At each point in time, the rate of increase of the population with respect to time is dy/dt, so the assumption that the rate of growth is proportional to the population is described by the differential equation Hank Morgan—Rainbow/Getty Images When the number of bacteria is small, an uninhibited population growth model can be used to model the growth of bacteria in a petri dish.
= ky
(3)
where k is a positive constant of proportionality that can usually be determined experimentally. Thus, if the population is known at some point in time, say y = yo at time t = 0, then a formula for the population y(t) can be obtained by solving the initial-value problem dy — = ky,
y(0) = y0
INHIBITED POPULATION GROWTH; LOGISTIC MODELS
The uninhibited population growth model was predicated on the assumption that the population y = y(t) was not constrained by the environment. While this assumption is reasonable as long as the size of the population is relatively small, environmental effects become increasingly important as the population grows. In general, populations grow within ecological systems that can only support a certain number of individuals; the number L of such individuals is called the carrying capacity of the system. When y > L, the population exceeds the capacity of the ecological system and tends to decrease toward L; when y < L, the population is below the capacity of the ecological system and tends to increase toward L; when y = L, the population is in balance with the capacity of the ecological system and tends to remain stable. To translate this into a mathematical model, we must look for a differential equation in which y > 0, L > 0, and
T dt
T dt
if
L
dt
564
Chapter 8 / Mathematical Modeling with Differential Equations
Moreover, when the population is far below the carrying capacity (i.e., y/L » 0), then the environmental constraints should have little effect, and the growth rate should behave like the uninhibited population model. Thus, we want •j-^ky if - % 0 at L A simple differential equation that meets all of these requirements is dy_ dt where k is a positive constant of proportionality. Thus if k and L can be determined experimentally, and if the population is known at some point, say y(G) — yo, then a formula for the population y ( t ) can be determined by solving the initial-value problem
-*('-£)>
) ?• y(0) = yo (4) dt This theory of population growth is due to the Belgian mathematician P. F. Verhulst (1804-1849), who introduced it in 1838 and described it as "logistic growth."* Thus, the differential equation in (4) is called the logistic differential equation, and the growth model described by (4) is called the logistic model. PHARMACOLOGY
When a drug (say, penicillin or aspirin) is administered to an individual, it enters the bloodstream and then is absorbed by the body over time. Medical research has shown that the amount of a drug that is present in the bloodstream tends to decrease at a rate that is proportional to the amount of the drug present—the more of the drug that is present in the bloodstream, the more rapidly it is absorbed by the body. To translate this principle into a mathematical model, suppose that y = y (t) is the amount of the drug present in the bloodstream at time t. At each point in time, the rate of change in y with respect to t is dy/dt, so the assumption that the rate of decrease is proportional to the amount y in the bloodstream translates into the differential equation dy
i = ~ky
(5)
where k is a positive constant of proportionality that depends on the drug and can be determined experimentally. The negative sign is required because y decreases with time. Thus, if the initial dosage of the drug is known, say y = yo at time t = 0, then a formula for y ( t ) can be obtained by solving the initial- value problem ^ = -ky,
y(0) = y0
SPREAD OF DISEASE
Suppose that a disease begins to spread in a population of L individuals. Logic suggests that at each point in time the rate at which the disease spreads will depend on how many individuals are already affected and how many are not— as more individuals are affected, the opportunity to spread the disease tends to increase, but at the same time there are fewer individuals who are not affected, so the opportunity to spread the disease tends to decrease. Thus, there are two conflicting influences on the rate at which the disease spreads.
Verhulst's model fell into obscurity for nearly a hundred years because he did not have sufficient census data to test its validity. However, interest in the model was revived during the 1930s when biologists used it successfully to describe the growth of fruit fly and flour beetle populations. Verhulst himself used the model to predict that an upper limit of Belgium's population would be approximately 9,400,000. In 2006 the population was about 10,379,000.
8.1 Modeling with Differential Equations
565
To translate this into a mathematical model, suppose that y — y ( t ) is the number of individuals who have the disease at time t, so of necessity the number of individuals who do not have the disease at time t is L — y. As the value of y increases, the value of L — y decreases, so the conflicting influences of the two factors on the rate of spread dy/dt are taken into account by the differential equation
dy_ = ky(L - y) dt Show that the model for the spread of disease can be viewed as a logistic model with constant of proportionality kL by rewriting (6) appropriately.
where k is a positive constant of proportionality that depends on the nature of the disease and the behavior patterns of the individuals and can be determined experimentally. Thus, if the number of affected individuals is known at some point in time, say y = yo at time t = 0, then a formula for y(t) can be obtained by solving the initial-value problem
dy_ = ky(L - y), dt
y(0) = y0
(6)
NEWTON'S LAW OF COOLING
If a hot object is placed into a cool environment, the object will cool at a rate proportional to the difference in temperature between the object and the environment. Similarly, if a cold object is placed into a warm environment, the object will warm at a rate that is again proportional to the difference in temperature between the object and the environment. Together, these observations comprise a result known as Newton's Law of Cooling. (Newton's Law of Cooling appeared previously in the exercises of Section 2.2 and was mentioned briefly in Section 6.4.) To translate this into a mathematical model, suppose that T = T(t) is the temperature of the object at time t and that Te is the temperature of the environment, which is assumed to be constant. Since the rate of change dT/dt is proportional to T — Te, we have dT
_=* ( r - r < ,)
Natural position
where k is a constant of proportionality. Moreover, since dT /dt is positive when T < Te, and is negative when T > Te, the sign of fe must be negative. Thus if the temperature of the object is known at some time, say T — T0 at time t = 0, then a formula for the temperature T(t) can be obtained by solving the initial-value problem i T""
— = k(T- Te), dt Stretched
Released
A Figure 8.1.2
Natural position
Y
H
A Figure 8.1.3
7X0) = To
(7)
VIBRATIONS OF SPRINGS
We conclude this section with an engineering model that leads to a second-order differential equation. As shown in Figure 8.1.2, consider a block of mass m attached to the end of a horizontal spring. Assume that the block is then set into vibratory motion by pulling the spring beyond its natural position and releasing it at time t = 0. We will be interested in finding a mathematical model that describes the vibratory motion of the block over time. To translate this problem into mathematical form, we introduce a horizontal *-axis whose positive direction is to the right and whose origin is at the right end of the spring when the spring is in its natural position (Figure 8.1.3). Our goal is to find a model for the coordinate x = x(t) of the point of attachment of the block to the spring as a function of time. In developing this model, we will assume that the only force on the mass m is the restoring force of the spring, and we will ignore the influence of other forces such as friction, air resistance, and so forth. Recall from Hooke's Law (Section 5.6) that when the connection point has coordinate x(t), the restoring force is — kx(t), where k is the spring constant. [The negative sign is due to the fact that the restoring force is to the left when x(t) is positive, and the restoring force is to the right when x(t) is negative.] It follows from Newton's
566
Chapter 8 / Mathematical Modeling with Differential Equations Second Law of Motion [Equation (5) of Section 5.6] that this restoring force is equal to the product of the mass m and the acceleration d2x/dt2 of the mass. In other words, we have
d2x m—= —kx dt2 which is a second-order differential equation for x. If at time t = 0 the mass is released from rest at position x(Q) = x$, then a formula for x(t) can be found by solving the initial-value problem ,2
m—^ = -kx, jt(0) = x0, at
x'(0) = 0
(8)
[If at time t = 0 the mass is given an initial velocity VQ /= 0, then the condition x'(0) — 0 must be replaced by x'(0) = i>o-]
QUICK CHECK EXERCISES 8.1
(See page 568 for answers.)
1. Match each differential equation with its family of solutions. (a)x-y-=y
(i)y = x2 + C
dx (b) y" = 4y
(ii) y = Ci sin 2x + C2 cos 2x
(0^=2* dx
(d)
(iv) y =
dx2
2. If y = C\ e2* + C2xe2* is the general solution of a differential equation, then the order of the equation is , and a solution to the differential equation that satisfies the initial conditions y(Q) = 1, y'(0) = 4 is given by y =
3. The graph of a differentiable function y = y(x) passes through the point (0, 1) and at every point P(x, y) on the graph the tangent line is perpendicular to the line through P and the origin. Find an initial-value problem whose solution is y(x). 4. A glass of ice water with a temperature of 36° F is placed in a room with a constant temperature of 68 °F. Assuming that Newton's Law of Cooling applies, find an initial-value problem whose solution is the temperature of water t minutes after it is placed in the room. [Note: The differential equation will involve a constant of proportionality.]
EXERCISE SET 8.1 1. Confirm that y = 3ex~ is a solution of the initial- value problem y' = 3x2y, y(0) = 3. 2. Confirm that y = |*4 + 2 cos x + 1 is a solution of the initial-value problem y' = x3 — 2 sin*, y(0) = 3. 3-4 State the order of the differential equation, and confirm that the functions in the given family are solutions.
dy
3. (a) (l+x)-/-=y; y = c(] + x)
dx (b) y" + y = 0; y = c\ sin t + c2 cos t x-3
4. (a) 2—+y = x-l; y =
dx (b) y"-y = Q; y = c,e<
5-8 True-False Determine whether the statement is true or false. Explain your answer. 5. The equation
dy
dx
^r dx
is an example of a second-order differential equation.
6. The differential equation
dy -^ = 2 dx has a solution that is constant. 7. We expect the general solution of the differential equation d3y , ^d2y
dy_ dx to involve three arbitrary constants.
- — +4y = 0
8. If every solution to a differential equation can be expressed in the form y = Aex+b for some choice of constants A and b, then the differential equation must be of second order. 9-14 In each part, verify that the functions are solutions of the differential equation by substituting the functions into the equation.
9. y" + y' - 2y = 0 (a) e~2x and e* (b) c\e~2x + c2e* (c\, c2 constants)
8.1 Modeling with Differential Equations 10. y" -y' -6y=0
(a) e-2x and e3x (b) cie^2x + C2e3x (ci, 02 constants) 11. y" - 4y' + 4y = 0 (a) e2x and xe2x (b) Cie2x + C2xe2x (c\,C2 constants) 12. y" - 8y' + 16y = 0 (a) e4x and xe4x (b) C]64x + cixe4x (c\,C2 constants) 13. y" + 4y = 0 (a) sin 2x and cos 2x (b) C] sin 2x + ci cos 2x (c\, c^ constants) 14. y" + 4y' + 13y = 0 (a) e~2x sin 3>x and e~2x cos 3x (b) e~2j:(ci sinSjc + caCosSz) (cj , C2 constants) 1 5 20 Use the results of Exercises 9-14 to find a solution to the initial- value problem. 15. y" + y' - 2y = 0, y(0) = -1, y'(0) = -4 16. y" - / - 6y = 0, y(0) = 1, y'(0) = 8 17. y" - 4y' + 4y = 0, y(0) = 2, y'(0) = 2 18. y" - 8y' + 16y = 0, y(0) = 1, y'(0) = 1 19. y" + 4y = 0, y(0) = 1, y'(0) = 2 20. y" + 4y' + 13y = 0, y(0) = -1, y'(0) = -1 21-26 Find a solution to the initial- value problem. 21. y' + 4x = 2, y(0) = 3 22. y" + 6x=0, y(0) = 1, y'(0) = 2 23. y' — y 2 = 0, y(l) = 2 [//mf: Assume the solution has an inverse function x = x ( y ) . Find, and solve, a differential equation that involves x'(y).] 24. / = 1 + y 2 , y(0) = 0 (See Exercise 23.) 25. jc 2 y' + 2xy = 0, y(l) = 2 [Hint: Interpret the left-hand side of the equation as the derivative of a product of two functions.] 26. xy' + y = ex , y(l) = 1 + e (See Exercise 25.)
(b) Suppose that a quantity y = y ( t ) changes in such a way that dy/dt = —ky3, where k > 0. Describe how y changes in words. 29. (a) Suppose that a particle moves along an s -axis in such a way that its velocity v(t) is always half of s(t). Find a differential equation whose solution is s(t). (b) Suppose that an object moves along an s-axis in such a way that its acceleration a (t) is always twice the velocity. Find a differential equation whose solution is s(t). 30. Suppose that a body moves along an s-axis through a resistive medium in such a way that the velocity v = v(t) decreases at a rate that is twice the square of the velocity. (a) Find a differential equation whose solution is the velocity v(t). (b) Find a differential equation whose solution is the position s(t).
31. Consider a solution y = y ( t ) to the uninhibited population growth model. (a) Use Equation (3) to explain why y will be an increasing function of t. (b) Use Equation (3) to explain why the graph y = y(f) will be concave up. 32. Consider the logistic model for population growth. (a) Explain why there are two constant solutions to this model. (b) For what size of the population will the population be growing most rapidly? 33. Consider the model for the spread of disease. (a) Explain why there are two constant solutions to this model. (b) For what size of the infected population is the disease spreading most rapidly? 34. Explain why there is exactly one constant solution to the Newton's Law of Cooling model. 35. Show that if ci and C2 are any constants, the function x = x(t) = i cos
27. (a) Suppose that a quantity y = y (t) increases at a rate that is proportional to the square of the amount present, and suppose that at time t = 0, the amount present is yo. Find an initial-value problem whose solution is y(t). (b) Suppose that a quantity y = y(t) decreases at a rate that is proportional to the square of the amount present, and suppose that at a time t = 0, the amount present is yo. Find an initial-value problem whose solution is y(t). 28. (a) Suppose that a quantity y = y(t) changes in such a way that dy/dt = k«/y, where k > 0. Describe how y changes in words.
567
k
— t I + c2 sin ,/ — t m I \Vm
is a solution to the differential equation for the vibrating spring. (The corresponding motion of the spring is referred to as simple harmonic motion.) 36. (a) Use the result of Exercise 35 to solve the initial-value problem in (8). (b) Find the amplitude, period, and frequency of your answer to part (a), and interpret each of these in terms of the motion of the spring. 37. Writing Select one of the models in this section and write a paragraph that discusses conditions under which the model would not be appropriate. How might you modify the model to take those conditions into account?
568
Chapter 8 / Mathematical Modeling with Differential Equations
I/QUICK CHECK ANSWERS 8.1 1. (a) (iv) (b) (iii) (c) (i) (d) (ii)
2. 2; e2x +2xe2x
dy 3. -f- = —, y(0) = 1 dx y
dT 4. —- = k(T - 68), T(0) = 36
dt
SEPARATION OF VARIABLES In this section we will discuss a method, called "separation of variables," that can be used to solve a large class of first-order differential equations of a particular form. We will use this method to investigate mathematical models for exponential growth and decay, including population models and carbon dating.
FIRST-ORDER SEPARABLE EQUATIONS
Some writers define a separable equation to be one that can be written in the form dy/dx = G(x)H(y). Explain why this is equivalent to our definition.
We will now consider a method of solution that can often be applied to first-order equations that are expressible in the form (1) dx Such first-order equations are said to be separable. Some examples of separable equations are given in Table 8.2.1. The name "separable" arises from the fact that Equation (1) can be rewritten in the differential form h(y)dy = g(x)dx
(2)
in which the expressions involving x and y appear on opposite sides. The process of rewriting (1) in form (2) is called separating variables. Table 8.2.1 EQUATION
ydy_ = x
dy _ x dx y dy
2 3
Tx=xy
y
?£=*2
g(x) x
1
2
ldy
dy_ dx~y
*y-=y-y dx x y
h(y}
FORM (1)
y dx 1 dy y dx
\ x
J
'
1
j_ 1
y
x
To motivate a method for solving separable equations, assume that h(y) and g(x) are continuous functions of their respective variables, and let H(y) and G(x) denote antiderivatives of h(y) and g(x), respectively. Consider the equation that results if we integrate both sides of (2), the left side with respect to y and the right side with respect to x. We then have j h(y)dy =
g(x)dx
(3)
or, equivalently,
H(y) = G(x) + C
(4)
where C denotes a constant. We claim that a differentiable function y = y(x) is a solution to (1) if and only if y satisfies Equation (4) for some choice of the constant C.
8.2 Separation of Variables
569
Suppose that y = y(x) is a solution to (1). It then follows from the chain rule that dH dy dy dG d -f- = g(x) = — (5) dy dx ax ax Since the functions H(y) and G(x) have the same derivative with respect to x, they must differ by a constant (Theorem 3.8.3). It then follows that y satisfies (4) for an appropriate choice of C. Conversely, if y = y(x) is defined implicitly by Equation (4), then implicit differentiation shows that (5) is satisfied, and thus y(x) is a solution to (1) (Exercise 67). Because of this, it is common practice to refer to Equation (4) as the "solution" to (1). In summary, we have the following procedure for solving (1), called separation of variables: Separation of Variables Step 1. Separate the variables in (1) by rewriting the equation in the differential form h(y)dy = g(x)dx Step 2. Integrate both sides of the equation in Step 1 (the left side with respect to y and the right side with respect to x): fh(y)dy=
I g(x)dx
Step 3. If H(y) is any antiderivative of h(y) and G(x) is any antiderivative of g(x), then the equation H(y) = G(x) + C will generally define a family of solutions implicitly. In some cases it may be possible to solve this equation explicitly for y.
Example 1 Solve the differential equation
T ax For an initial-value problem in which the differential equation is separable, you can either use the initial condition
and then solve the initial-value problem dy = -4xy2, dx
y(0) = 1
to solve for C, as in Example 1, or replace the indefinite integrals in Step 2
Solution. For y ^ 0 we can write the differential equation in form (1) as
by definite integrals (Exercise 68).
1 dy _
y2 dx Separating variables and integrating yields — dy — —4x dx
— dy = I —4x dx or 1
y Solving for y as a function of x, we obtain 2x2-C
570 Chapter 8 / Mathematical Modeling with Differential Equations The initial condition y(0) = 1 requires that y = 1 when x = 0. Substituting these values into our solution yields C — — 1 (verify). Thus, a solution to the initial-value problem is
1
(6)
2x2
Some integral curves and our solution of the initial-value problem are graphed in Figure 8.2.1. •*
dy
Integral curves for — = -4xy dx
,
A Figure 8.2.1
One aspect of our solution to Example 1 deserves special comment. Had the initial condition been y(0) = 0 instead of y(0) = 1, the method we used would have failed to yield a solution to the resulting initial-value problem (Exercise 25). This is due to the fact that we assumed y ^ 0 in order to rewrite the equation dy/dx — —4xy2 in the form
1 dy — —4r y*dx ~ It is important to be aware of such assumptions when manipulating a differential equation algebraically.
Example 2
Solve the initial-value problem (4y-cosy) — -3xz = 0,
dx
Solution. The solution of an initial-value problem in x and y can sometimes be expressed explicitly as a function of x [as in Formula (6) of Example 1], or explicitly as a function of y [as in Formula (8) of Example 2]. However, sometimes the solution cannot be expressed in either such form, so the only option is to express it implicitly as an equation in x and y.
y(0) = 0
We can write the differential equation in form (1) as
(4y — cosy)— = 3x2 dx Separating variables and integrating yields
(4y — cos y) dy = 3x2 dx
I (4y - cos y) dy - I 3x2 dx
or 2y2 - sin y = x3 + C
(7)
For the initial-value problem, the initial condition y (0) = 0 requires that y = 0 if x = 0. Substituting these values into (7) to determine the constant of integration yields C = 0 (verify). Thus, the solution of the initial-value problem is 2y2 — siny = x3
or - sin y
Integral curves for
(8)
Some integral curves and the solution of the initial-value problem in Example 2 are graphed in Figure 8.2.2. Initial-value problems often result from geometrical questions, as in the following example.
(4y - cos y) — - 3x2 = 0 A Figure 8.2.2
> Example 3 Find a curve in the jcy-plane that passes through (0, 3) and whose tangent line at a point (x, y) has slope 2x/y2.
8.2 Separation of Variables TECHNOLOGY MASTERY Some computer algebra systems can graph implicit equations. Figure 8.2.2 shows the graphs of (7) for C = 0, ± 1, ±2, and ±3. If you have a CAS that can graph implicit equations, try to duplicate this figure.
Solution.
571
Since the slope of the tangent line is dy/dx, we have
dy
2x
(9)
and, since the curve passes through (0, 3), we have the initial condition
Equation (9) is separable and can be written as y2 dy = 2x dx
so
\ y1dy =
2xdx
or iy 3 = x2 + C
It follows from the initial condition that y = 3 if x = 0. Substituting these values into the last equation yields C — 9 (verify), so the equation of the desired curve is Iy3=A.2+9
Qr
y = (3X2 + 27)1/3
4
EXPONENTIAL GROWTH AND DECAY MODELS
The population growth and pharmacology models developed in Section 8.1 are examples of a general class of models called exponential models. In general, exponential models arise in situations where a quantity increases or decreases at a rate that is proportional to the amount of the quantity present. More precisely, we make the following definition.
8.2.1 DEFINITION A quantity y = y(t) is said to have an exponential growth model if it increases at a rate that is proportional to the amount of the quantity present, and it is said to have an exponential decay model if it decreases at a rate that is proportional to the amount of the quantity present. Thus, for an exponential growth model, the quantity y(t) satisfies an equation of the form
(10) at and for an exponential decay model, the quantity y(t) satisfies an equation of the form
dy -j- = ~ky
(k > 0)
(11)
The constant k is called the growth constant or the decay constant, as appropriate.
Equations (10) and (11) are separable since they have the form of (1), but with t rather than x as the independent variable. To illustrate how these equations can be solved, suppose that a positive quantity y = y(f) has an exponential growth model and that we know the amount of the quantity at some point in time, say y = yo when t — 0. Thus, a formula for y(t) can be obtained by solving the initial-value problem
~ = ky,
y(0) = y0
Separating variables and integrating yields
flydy or (since y > 0)
=
Jkdt
In y = kt + C
(12)
572
Chapter 8 / Mathematical Modeling with Differential Equations
The initial condition implies that y = yo when / — 0. Substituting these values in (12) yields C = In yo (verify). Thus, In y = kt + In y0
from which it follows that or, equivalently,
(13)
We leave it for you to show that if y = y ( t ) has an exponential decay model, and if y(0) = 3*0, then ~ - ^r*< (14) INTERPRETING THE GROWTH AND DECAY CONSTANTS
The significance of the constant k in Formulas (13) and (14) can be understood by reexamining the differential equations that gave rise to these formulas. For example, in the case of the exponential growth model, Equation (10) can be rewritten as
k= It is standard practice in applications to call (15) the growth rate, even though it is misleading (the growth rate is dy/dt). However, the practice is so common that we will follow it here.
dy/dt
y
(15)
which states that the growth rate as a fraction of the entire population remains constant over time, and this constant is k. For this reason, k is called the relative growth rate of the population. It is usual to express the relative growth rate as a percentage. Thus, a relative growth rate of 3% per unit of time in an exponential growth model means that k = 0.03. Similarly, the constant k in an exponential decay model is called the relative decay rate. >• Example 4 According to United Nations data, the world population in 1998 was approximately 5.9 billion and growing at a rate of about 1.33% per year. Assuming an exponential growth model, estimate the world population at the beginning of the year 2023. Solution. We assume that the population at the beginning of 1998 was 5.9 billion and let t — time elapsed from the beginning of 1998 (in years) y = world population (in billions) Since the beginning of 1998 corresponds to t = 0, it follows from the given data that yo
= y(Q) = 5.9 (billion)
Since the growth rate is 1.33% (k = 0.0133), it follows from (13) that the world population at time t will be ,., n.om/ kt = 5.9eu (16) y(t) = yoe In Example 4 the growth rate was given, so there was no need to calculate it. If the growth rate or decay rate is unknown, then it can be calculated using the initial condition and the value of y at another point in time (Exercise 44).
Since the beginning of the year 2023 corresponds to an elapsed time of t = 25 years (2023 - 1998 = 25), it follows from (16) that the world population by the year 2023 will be
y(25) = 5.9eaoi33(25) « 8.2
which is a population of approximately 8.2 billion. •< DOUBLING TIME AND HALF-LIFE
If a quantity y has an exponential growth model, then the time required for the original size to double is called the doubling time, and if y has an exponential decay model, then the time required for the original size to reduce by half is called the half-life. As it turns out, doubling time and half-life depend only on the growth or decay rate and not on the amount present initially. To see why this is so, suppose that y = y(t) has an exponential growth model (17) y = kt
8.2 Separation of Variables
573
and let T denote the amount of time required for y to double in size. Thus, at time t = T the value of y will be 2yo, and hence from (17) 2yo = y0ekl
KI or e,kT =2
Taking the natural logarithm of both sides yields kT = In 2, which implies that the doubling time is . T = -In2 (18) k We leave it as an exercise to show that Formula (18) also gives the half-life of an exponential decay model. Observe that this formula does not involve the initial amount yo, so that in an exponential growth or decay model, the quantity y doubles (or reduces by half) every T units (Figure 8.2.3). > Example 5 It follows from (18) that with a continued growth rate of 1.33% per year, the doubling time for the world population will be 1 T = In 2 ss 52.116 0.0133 or approximately 52 years. Thus, with a continued 1.33% annual growth rate the population of 5.9 billion in 1998 will double to 11.8 billion by the year 2050 and will double again to 23.6 billion by 2102. + RADIOACTIVE DECAY
Exponential decay model with half-life T A Figure 8.2.3
It is a fact of physics that radioactive elements disintegrate spontaneously in a process called radioactive decay. Experimentation has shown that the rate of disintegration is proportional to the amount of the element present, which implies that the amount y = y(t) of a radioactive element present as a function of time has an exponential decay model. Every radioactive element has a specific half-life; for example, the half-life of radioactive carbon-14 is about 5730 years. Thus, from (18), the decay constant for this element is In 2 k = - In 2 = 0.000121 T '
5730
and this implies that if there are yo units of carbon-14 present at time t — 0, then the number of units present after t years will be approximately y(t) = y
(19)
> Example 6 If 100 grams of radioactive carbon-14 are stored in a cave for 1000 years, how many grams will be left at that time? Solution. From (19) with y0 = 100 and t — 1000, we obtain y(1000) = iooe-aoo°121(1000) = 100e-°-121 « 88.6 Thus, about 88.6 grams will be left. -^ CARBON DATING
When the nitrogen in the Earth's upper atmosphere is bombarded by cosmic radiation, the radioactive element carbon-14 is produced. This carbon-14 combines with oxygen to form carbon dioxide, which is ingested by plants, which in turn are eaten by animals. In this way all living plants and animals absorb quantities of radioactive carbon-14. In 1947 the American nuclear scientist W. F. Libby* proposed the theory that the percentage of
*W. F. Libby, "Radiocarbon Dating," American Scientist, Vol. 44, 1956, pp. 98-112.
574
Chapter 8 / Mathematical Modeling with Differential Equations
carbon-14 in the atmosphere and in living tissues of plants is the same. When a plant or animal dies, the carbon-14 in the tissue begins to decay. Thus, the age of an artifact that contains plant or animal material can be estimated by determining what percentage of its original carbon-14 content remains. Various procedures, called carbon dating or carbon-14 dating, have been developed for measuring this percentage.
> Example 7 In 1988 the Vatican authorized the British Museum to date a cloth relic known as the Shroud of Turin, possibly the burial shroud of Jesus of Nazareth. This cloth, which first surfaced in 1356, contains the negative image of a human body that was widely believed to be that of Jesus. The report of the British Museum showed that the fibers in the cloth contained between 92% and 93% of their original carbon-14. Use this information to estimate the age of the shroud. Solution.
From (19), the fraction of the original carbon-14 that remains after t years is = e -0.000121f
Taking the natural logarithm of both sides and solving for t, we obtain
0.000121 \ yo Thus, taking y(r)/y 0 to be 0.93 and 0.92, we obtain 1 ln(0.93) ^ 600 0.000121
Patrick Mesner/Liaison Agency, Inc./Getty Images The Shroud of Turin
1 0.000121 This means that when the test was done in 1988, the shroud was between 600 and 689 years old, thereby placing its origin between 1299 A.D. and 1388 A.D. Thus, if one accepts the validity of carbon-14 dating, the Shroud of Turin cannot be the burial shroud of Jesus of Nazareth. -^
•QUICK CHECK EXERCISES 8.2
(See page 579 for answers.)
1. Solve the first-order separable equation
dyL -/ dx by completing the following steps: Step 1. Separate the variables by writing the equation in the differential form Step 2. Integrate both sides of the equation in Step 1: Step 3. If H(y) is any antiderivative of h(y), G(x) is any antiderivative of g(x), and C is an unspecified constant, then, as suggested by Step 2, the equation will generally define a family of solutions to h(y) dy/dx = g(x) implicitly.
2. Suppose that a quantity y = y (t) has an exponential growth model with growth constant k > 0. (a) y(t) satisfies a first-order differential equation of the form dy/dt = (b) In terms of k, the doubling time of the quantity is (c) If yo = y (0) is the initial amount of the quantity, then an explicit formula for y(t) is given by y(t) = 3. Suppose that a quantity y = y(?) has an exponential decay model with decay constant k > 0. (a) y(t) satisfies a first-order differential equation of the form dy/dt = (b) In terms of k, the half-life of the quantity is (c) If yo = y (0) is the initial amount of the quantity, then an explicit formula for y(t) is given by y(t) =
8.2 Separation of Variables 575 4. The initial-value problem
has solution y(x) = EXERCISE SET 8.2
0 Graphing Utility
HI CAS
1-10 Solve the differential equation by separation of variables. Where reasonable, express the family of solutions as explicit functions of x.
y
dy
2.
2
} -x^ dx 6. y' = -xy
1. e~ smx-y'cos x = Q
dx
23. If a radioactive element has a half-life of 1 minute, and if a container holds 32 g of the element at 1 :00 P.M., then the amount remaining at 1 :05 P.M. will be 1 g.
y 2
8. y' - (1 + x)(\ + y ) = 0 10. y
sn x
is not separable.
=
4.
1 + y dx 5. (2 + 2y2)y' = exy y
dx
dy
dx
secjc = 0
11-14 Solve the initial-value problem by separation of variables.
11. y ' =
2y + cos y 12. y' -xey = 2ey,
dy_
2r
,
y(0) = . y(0) = 0
y(0) = -i
' ~dt
2y-2' 14. y' cosh2 x - y cosh 2x = 0, y (0) = 3 15. (a) Sketch some typical integral curves of the differential equation y' = y/2x. (b) Find an equation for the integral curve that passes through the point (2, 1). 16. (a) Sketch some typical integral curves of the differential equation y' = —x/y. (b) Find an equation for the integral curve that passes through the point (3,4). 17-18 Solve the differential equation and then use a graphing utility to generate five integral curves for the equation.
17.
dx
18. (cosy)y' = cos*
[c] 19-20 Solve the differential equation. If you have a CAS with implicit plotting capability, use the CAS to generate five integral curves for the equation. v-2
19. y' =
1-v
2
20. y' =
22. A differential equation of the form
y
21-24 True-False Determine whether the statement is true or false. Explain your answer. 21. Every differential equation of the form y' = /(y) is separable.
24. If a population is growing exponentially, then the time it takes the population to quadruple is independent of the size of the population. 25. Suppose that the initial condition in Example 1 had been y(0) = 0. Show that none of the solutions generated in Example 1 satisfy this initial condition, and then solve the initial-value problem dy_ = -4xy2, y(0) = 0 dx Why does the method of Example 1 fail to produce this particular solution? 26. Find all ordered pairs (XQ, yo) such that if the initial condition in Example 1 is replaced by y(xo) = yo, the solution of the resulting initial-value problem is defined for all real numbers. 27. Find an equation of a curve with x -intercept 2 whose tangent line at any point (x, y) has slope xe~y. 28. Use a graphing utility to generate a curve that passes through the point (1,1) and whose tangent line at (x, y) is perpendicular to the line through ( x , y) with slope — 2y/(3x2). 29. Suppose that an initial population of 10,000 bacteria grows exponentially at a rate of 2% per hour and that y = y (t) is the number of bacteria present t hours later. (a) Find an initial-value problem whose solution is y ( t ) . (b) Find a formula for y(f). (c) How long does it take for the initial population of bacteria to double? (d) How long does it take for the population of bacteria to reach 45,000? 30. A cell of the bacterium E. coli divides into two cells every 20 minutes when placed in a nutrient culture. Let y = y ( t ) be the number of cells that are present t minutes after a single cell is placed in the culture. Assume that the growth of the bacteria is approximated by an exponential growth model. (a) Find an initial-value problem whose solution is y(t). (b) Find a formula for y(t). (com.)
576
Chapter 8 / Mathematical Modeling with Differential Equations (c) How many cells are present after 2 hours? (d) How long does it take for the number of cells to reach 1,000,000?
31. Radon-222 is a radioactive gas with a half-life of 3.83 days. This gas is a health hazard because it tends to get trapped in the basements of houses, and many health officials suggest that homeowners seal their basements to prevent entry of the gas. Assume that 5.0 x 107 radon atoms are trapped in a basement at the time it is sealed and that y (t) is the number of atoms present t days later. (a) Find an initial-value problem whose solution is y(t). (b) Find a formula for y ( t ) . (c) How many atoms will be present after 30 days? (d) How long will it take for 90% of the original quantity of gas to decay? 32. Polonium-210 is a radioactive element with a half-life of 140 days. Assume that 10 milligrams of the element are placed in a lead container and that y(t) is the number of milligrams present t days later. (a) Find an initial-value problem whose solution is y(t). (b) Find a formula for y ( t ) . (c) How many milligrams will be present after 10 weeks? (d) How long will it take for 70% of the original sample to decay? 33. Suppose that 100 fruit flies are placed in a breeding container that can support at most 10,000 flies. Assuming that the population grows exponentially at a rate of 2% per day, how long will it take for the container to reach capacity? 34. Suppose that the town of Grayrock had a population of 10,000 in 1998 and a population of 12,000 in 2003. Assuming an exponential growth model, in what year will the population reach 20,000? 35. A scientist wants to determine the half-life of a certain radioactive substance. She determines that in exactly 5 days a 10.0-milligram sample of the substance decays to 3.5 milligrams. Based on these data, what is the half-life? 36. Suppose that 30% of a certain radioactive substance decays in 5 years. (a) What is the half-life of the substance in years? (b) Suppose that a certain quantity of this substance is stored in a cave. What percentage of it will remain after t years?
37. (a) Make a conjecture about the effect on the graphs of y = y0ekt and y = yoe~kl of varying k and keeping yo fixed. Confirm your conjecture with a graphing utility. (b) Make a conjecture about the effect on the graphs of y = yoekt and y = yoe~kt of varying y0 and keeping k fixed. Confirm your conjecture with a graphing utility.
38. (a) What effect does increasing yo and keeping k fixed have on the doubling time or half-life of an exponential model? Justify your answer, (b) What effect does increasing k and keeping yo fixed have on the doubling time and half-life of an exponential model? Justify your answer. 39. (a) There is a trick, called the Rule of 70, that can be used to get a quick estimate of the doubling time or half-life of an exponential model. According to this rule, the doubling time or half-life is roughly 70 divided by the percentage growth or decay rate. For example, we showed in Example 5 that with a continued growth rate of 1.33% per year the world population would double every 52 years. This result agrees with the Rule of 70, since 70/1.33 « 52.6. Explain why this rule works. (b) Use the Rule of 70 to estimate the doubling time of a population that grows exponentially at a rate of 1 % per year. (c) Use the Rule of 70 to estimate the half-life of a population that decreases exponentially at a rate of 3.5% per hour. (d) Use the Rule of 70 to estimate the growth rate that would be required for a population growing exponentially to double every 10 years. 40. Find a formula for the tripling time of an exponential growth model. 41. In 1950, a research team digging near Folsom, New Mexico, found charred bison bones along with some leaf-shaped projectile points (called the "Folsom points") that had been made by a Paleo-Indian hunting culture. It was clear from the evidence that the bison had been cooked and eaten by the makers of the points, so that carbon-14 dating of the bones made it possible for the researchers to determine when the hunters roamed North America. Tests showed that the bones contained between 27% and 30% of their original carbon14. Use this information to show that the hunters lived roughly between 9000 B.C. and 8000 B.C. 42. (a) Use a graphing utility to make a graph of prem versus t, where pKm is the percentage of carbon-14 that remains in an artifact after t years. (b) Use the graph to estimate the percentage of carbon-14 that would have to have been present in the 1988 test of the Shroud of Turin for it to have been the burial shroud of Jesus of Nazareth (see Example 7). 43. (a) It is currently accepted that the half-life of carbon-14 might vary ±40 years from its nominal value of 5730 years. Does this variation make it possible that the Shroud of Turin dates to the time of Jesus of Nazareth (see Example 7)? (b) Review the subsection of Section 2.9 entitled Error Propagation, and then estimate the percentage error that
8.2 Separation of Variables
results in the computed age of an artifact from an r% error in the half-life of carbon-14. 44. Suppose that a quantity y has an exponential growth model y — y0ekt or an exponential decay model y = yoe~kt, and it is known that y = y\ if t = 11. In each case find a formula for k in terms of yo, y\, and t\, assuming that t\ / 0. 45. (a) Show that if a quantity y = y(t) has an exponential model, and if y ( t i ) = yi and yfe) = yz, then the doubling time or the half-life T is
r= (b) In a certain 1-hour period the number of bacteria in a colony increases by 25%. Assuming an exponential growth model, what is the doubling time for the colony? 46. Suppose that P dollars is invested at an annual interest rate of r x 100%. If the accumulated interest is credited to the account at the end of the year, then the interest is said to be compounded annually, if it is credited at the end of each 6-month period, then it is said to be compounded semiannually; and if it is credited at the end of each 3-month period, then it is said to be compounded quarterly. The more frequently the interest is compounded, the better it is for the investor since more of the interest is itself earning interest. (a) Show that if interest is compounded n times a year at equally spaced intervals, then the value A of the investment after t years is
48. What is the effective annual interest rate for an interest rate of r% per year compounded continuously? 49. Assume that y = y(t) satisfies the logistic equation with y0 = y(Q) the initial value of y. (a) Use separation of variables to derive the solution >o + (L - yo)e~kt (b) Use part (a) to show that lim y(t) = L. 50. Use your answer to Exercise 49 to derive a solution to the model for the spread of disease [Equation (6) of Section 8.1]. 51. The graph of a solution to the logistic equation is known as a logistic curve, and if yo > 0, it has one of four general shapes, depending on the relationship between yo and L. In each part, assume that k = 1 and use a graphing utility to plot a logistic curve satisfying the given condition. (a) y0 > L (b) y0 = L (c) L/2 < y0 < L (d) 0 < y0 < L/2 52-53 The graph of a logistic model
y= —
n -» +00
(c)
47. (a)
(b)
(c)
\
n'
Use the fact that lim^o (1 + x)l/* — e to prove that A = Pert. Use the result in part (b) to show that money invested at continuous compound interest increases at a rate proportional to the amount present. If $1000 is invested at 8% per year compounded continuously (Exercise 46), what will the investment be worth after 5 years? If it is desired that an investment at 8% per year compounded continuously should have a value of $10,000 after 10 years, how much should be invested now? How long does it take for an investment at 8% per year compounded continuously to double in value?
- yo)e-kt
is shown. Estimate yo, L, and k. 52. 1000 .y
A\y\ 600 / / i' / < 200
(b) One can imagine interest to be compounded each day, each hour, each minute, and so forth. Carried to the limit one can conceive of interest compounded at each instant of time; this is called continuous compounding. Thus, from part (a), the value A of P dollars after t years when invested at an annual rate of r x 100%, compounded continuously, is / r\n A= lim P ( 1 + - )
577
200
}
]
10
***
8 6 4 2
>y \ i }
'•Jr?
i ! 'y i j/i j/r i
zn ^ !
600
j I |
i
j !
1000
1
1
\ \ \ \ \ -
2 4 6
10
54. Plot a solution to the initial- value problem
55. Suppose that the growth of a population y = y(t) is given by the logistic equation 60 ~ 5 + 7e~> (a) What is the population at time t = 0? (b) What is the carrying capacity LI (c) What is the constant kl (d) When does the population reach half of the carrying capacity? (e) Find an initial-value problem whose solution is y(t). 56. Suppose that the growth of a population y = y(t) is given by the logistic equation 1000 ~ (a) What is the population at time t = 0? (b) What is the carrying capacity L? (c) What is the constant kl
(com.)
578 Chapter 8 / Mathematical Modeling with Differential Equations (d) When does the population reach 75% of the carrying capacity? (e) Find an initial-value problem whose solution is y(t). ] 57. Suppose that a university residence hall houses 1000 students. Following the semester break, 20 students in the hall return with the flu, and 5 days later 35 students have the flu. (a) Use the result of Exercise 50 to find the number of students who will have the flu t days after returning to school. (b) Make a table that illustrates how the flu spreads day to day over a 2-week period. (c) Use a graphing utility to generate a graph that illustrates how the flu spreads over a 2-week period. 58. Suppose that at time t = 0 an object with temperature TQ is placed in a room with constant temperature Ta. If TO < Ta, then the temperature of the object will increase, and if TO > Ta, then the temperature will decrease. Assuming that Newton's Law of Cooling applies, show that in both cases the temperature T(t) at time t is given by
62. A bullet of mass m, fired straight up with an initial velocity of VQ, is slowed by the force of gravity and a drag force of air resistance kv2, where k is a positive constant. As the bullet moves upward, its velocity v satisfies the equation dv m— = —(kv + mg) dt where g is the constant acceleration due to gravity. (a) Show that if x = x(t) is the height of the bullet above the barrel opening at time t, then dv 2 mv — = -(kv + mg) dx (b) Express x in terms of v given that x = 0 when t; = VQ. (c) Assuming that v0 = 988 m/s, g = 9.8 m/s2
m = 3.56 x 10~3 kg, k = 7.3 x 10~6 kg/m use the result in part (b) to find out how high the bullet rises. [Hint: Find the velocity of the bullet at its highest point.]
7X0 = Ta + (T0 - Ta)e-kl where A: is a positive constant. 59. A cup of water with a temperature of 95 °C is placed in a room with a constant temperature of 21 °C. (a) Assuming that Newton's Law of Cooling applies, use the result of Exercise 58 to find the temperature of the water t minutes after it is placed in the room. [Note: The solution will involve a constant of proportionality.] (b) How many minutes will it take for the water to reach a temperature of 51 °C if it cools to 85°C in 1 minute? 60. Aglass of lemonade with a temperature of 40°F is placed in a room with a constant temperature of 70° F, and 1 hour later its temperature is 52°F. Show that t hours after the lemonade is placed in the room its temperature is approximated b y r = 70-30e-°-5'. 61. A rocket, fired upward from rest at time t = 0, has an initial mass of m.Q (including its fuel). Assuming that the fuel is consumed at a constant rate k, the mass m of the rocket, while fuel is being burned, will be given by m = mo — kt. It can be shown that if air resistance is neglected and the fuel gases are expelled at a constant speed c relative to the rocket, then the velocity v of the rocket will satisfy the equation
63-64 Suppose that a tank containing a liquid is vented to the air at the top and has an outlet at the bottom through which the liquid can drain. It follows from Torricelli's law in physics that if the outlet is opened at time t = 0, then at each instant the depth of the liquid h(t) and the area A(h) of the liquid's surface are related by dh A(h)— = -kVh dt where k is a positive constant that depends on such factors as the viscosity of the liquid and the cross-sectional area of the outlet. Use this result in these exercises, assuming that h is in feet, A(h) is in square feet, and t is in seconds. 63. Suppose that the cylindrical tank in the accompanying figure is filled to a depth of 4 feet at time t = 0 and that the constant in Torricelli's law is k = 0.025. (a) Findfc(0. (b) How many minutes will it take for the tank to drain completely? 64. Follow the directions of Exercise 63 for the cylindrical tank in the accompanying figure, assuming that the tank is filled to a depth of 4 feet at time t = 0 and that the constant in Torricelli's law is k = 0.025.
dv m— = ck — mg at
where g is the acceleration due to gravity. (a) Find v(t) keeping in mind that the mass m is a function off. (b) Suppose that the fuel accounts for 80% of the initial mass of the rocket and that all of the fuel is consumed in 100 s. Find the velocity of the rocket in meters per second at the instant the fuel is exhausted. [Note: Take g = 9.8 m/s2 and c = 2500 m/s.]
4ft
A Figure Ex-63
A Figure Ex-64
8.3 Slope Fields; Euler's Method
65. Suppose that a particle moving along the jt-axis encounters a resisting force that results in an acceleration of a = dv/dt = — 551>2. If x = 0 cm and v — 128 cm/s at time t = 0, find the velocity v and position x as a function of t for t > 0. 66. Suppose that a particle moving along the ;c-axis encounters a resisting force that results in an acceleration of a = dv/dt = -O.Q2^/v. Given that x = 0 cm and v = 9 cm/s at time t = 0, find the velocity D and position x as a function of t for t > 0. FOCUS ON CONCEPTS
67. Use implicit differentiation to prove that any differentiable function defined implicitly by Equation (4) will be a solution to (1). 68. Prove that a solution to the initial-value problem dy h(y)—=g(x), y(x0) = y0
579
is defined implicitly by the equation I h(r)dr= I Jyo
g(s)ds
Jxo
69. Let L denote a tangent line at (x, y) to a solution of Equation (1), and let (x\, y\), (xi, ^2) denote any two points on L. Prove that Equation (2) is satisfied by dy = Ay = yi — y] and dx = Ax = X2 — x\. 70. Writing A student objects to the method of separation of variables because it often produces an equation in x and y instead of an explicit function y = f ( x ) . Discuss the pros and cons of this student's position. 71. Writing A student objects to Step 2 in the method of separation of variables because one side of the equation is integrated with respect to x while the other side is integrated with respect to y. Answer this student's objection. [Hint: Recall the method of integration by substitution.]
•QUICK CHECK ANSWERS 8.2 /*
/
lr» O
h(y)dy = I g(x)dx; Step 3: H(y) = G(x) + C
2. (a) ky (b) — (c) y0ekt
SLOPE FIELDS; EULER'S METHOD In this section we will reexamine the concept of a slope field and we will discuss a method for approximating solutions of first-order equations numerically. Numerical approximations are important in cases where the differential equation cannot be solved exactly.
FUNCTIONS OF TWO VARIABLES
We will be concerned here with first-order equations that are expressed with the derivative by itself on one side of the equation. For example,
y In applied problems involving time, it is usual to use t as the independent variable, in which case one would be concerned with equations of the form y' = f(t, y), where y' = dy/dt.
and
y' =
The first of these equations involves only x on the right side, so it has the form y' = f ( x ) . However, the second equation involves both x and y on the right side, so it has the form y' = f ( x , y), where the symbol f(x, y) stands for a function of the two variables x and y. Later in the text we will study functions of two variables in more depth, but for now it will suffice to think of f(x, y) as a formula that produces a unique output when values of x and y are given as inputs. For example, if f ( x , y ) —x+3y
580
Chapter 8 / Mathematical Modeling with Differential Equations and if the inputs are x = 2 and y = —4, then the output is
/(2, -4) = 22 + 3(-4) = 4 - 12 = -i
SLOPE FIELDS
In Section 4.2 we introduced the concept of a slope field in the context of differential equations of the form y' = f ( x ) ; the same principles apply to differential equations of the
Slope =f(x,y)
form
J
ft y)^ y ' — f(x,
To see why this is so, let us review the basic idea. If we interpret y' as the slope of a tangent line, then the differential equation states that at each point ( x , y) on an integral curve, the slope of the tangent line is equal to the value of / at that point (Figure 8.3.1). For example, suppose that f(x, y) = y — x, in which case we have the differential equation
y = y -x
At each point (x, y) on an integral curve of / =/(*, y), the tangent line has slope f(x,y).
(1)
A geometric description of the set of integral curves can be obtained by choosing a rectangular grid of points in the xy-plane, calculating the slopes of the tangent lines to the integral curves at the gridpoints, and drawing small segments of the tangent lines through those points. The resulting picture is called a slope field or a direction field for the differential equation because it shows the "slope" or "direction" of the integral curves at the gridpoints. The more gridpoints that are used, the better the description of the integral curves. For example, Figure 8.3.2 shows two slope fields for (1)—the first was obtained by hand calculation using the 49 gridpoints shown in the accompanying table, and the second, which gives a clearer picture of the integral curves, was obtained using 625 gridpoints and a CAS.
A Figure 8.3.1
VALUES OF fix, y) = y - X
y = -3 y = -2 y = -\ y = 0
y=i
y =2
y =3
x = -3
0
1
1
3
4
5
6
x = -2
-1
0
1
2
3
4
5
x = -l
-2
-1
0
1
2
3
4
x =Q
-3
-2
-1
0
1
2
3
-3
-2
/
i
-I
x= I
-4
-3
-2
-1
0
1
2
/
x =2
-5
-4
-3
_2
-1
0
1
/
— \ -2
x =3
-6
-5
-4
-3
-2
-1
0
—
\
- -1
\ -3
A Figure 8.3.2
It so happens that Equation (1) can be solved exactly using a method we will introduce in Section 8.4. We leave it for you to confirm that the general solution of this equation is
y = x + 1 + Cex Confirm that the first slope field in Figure 8.3.2 is consistent with the accompanying table in that figure.
(2)
Figure 8.3.3 shows some of the integral curves superimposed on the slope field. Note that it was not necessary to have the general solution to construct the slope field. Indeed, slope fields are important precisely because they can be constructed in cases where the differential equation cannot be solved exactly.
8.3 Slope Fields; Euler's Method
581
EULER'S METHOD
Consider an initial-value problem of the form y(xo) = yo
y' = f(x, y),
A Figure 8.3.3
The slope field for the differential equation y' = f ( x , y) gives us a way to visualize the solution of the initial-value problem, since the graph of the solution is the integral curve that passes through the point (XQ, yo)- The slope field will also help us to develop a method for approximating the solution to the initial-value problem numerically. We will not attempt to approximate y(x) for all values of x; rather, we will choose some small increment Ax and focus on approximating the values of y(x) at a succession of x-values spaced Ax units apart, starting from XQ. We will denote these jt-values by X\ = XQ + Ax,
X2 = X\ + Ax,
XT, = X^ +
X4 = XT, + Ax ,
and we will denote the approximations of y(jc) at these points by
yfe),
Ax A Figure 8.3.4
y(x4),
The technique that we will describe for obtaining these approximations is called Euler's Method. Although there are better approximation methods available, many of them use Euler's Method as a starting point, so the underlying concepts are important to understand. The basic idea behind Euler's Method is to start at the known initial point (XQ, yo) and draw a line segment in the direction determined by the slope field until we reach the point (x\, yi) with x -coordinate x\ = XQ + Ax (Figure 8.3.4). If Ax is small, then it is reasonable to expect that this line segment will not deviate much from the integral curve y = y(x), and thus y\ should closely approximate y(x\). To obtain the subsequent approximations, we repeat the process using the slope field as a guide at each step. Starting at the endpoint (x\, y\), we draw a line segment determined by the slope field until we reach the point 0*2- J2) with jc-coordinate xi = x\ + Ax, and from that point we draw a line segment determined by the slope field to the point (jt3, y^) with x -coordinate x^ = X2 + Ax, and so forth. As indicated in Figure 8.3.4, this procedure produces a polygonal path that tends to follow the integral curve closely, so it is reasonable to expect that the y-values y2, ys, y4, - - will closely approximate yfe), y(x^), y(x$), To explain how the approximations y\, y%, y j , , . . . can be computed, let us focus on a typical line segment. As indicated in Figure 8.3.5, assume that we have found the point (xn, y«), and we are trying to determine the next point (xn+\, y«+i), where xn+\ = xn + Ax. Since the slope of the line segment joining the points is determined by the slope field at the starting point, the slope is f(xn, yn), and hence
Slope =f(xn,ya)
yn+\ - yn xn+\ A Figure 8.3.5
xn
Ax
= f(xn, Jn)
which we can rewrite as
y«+i = y« + f(xn, y This formula, which is the heart of Euler's Method, tells us how to use each approximation to compute the next approximation.
582
Chapter 8 / Mathematical Modeling with Differential Equations
Euler's Method To approximate the solution of the initial-value problem
/ = f ( x , y),
yOo) = Jo
proceed as follows: Step 1. Choose a nonzero number Ax to serve as an increment or step size along the jt-axis, and let X\ — XQ + Ax,
X2 = X\ + Ax,
XT, — X2 + Ax , . . .
Step 2. Compute successively
y\ = f(xi,y\)Ax J2 = yi = yi + yn+\ =yn + f(xn,yn)Ax The numbers y\, y(x2),
a, . . . in these equations are the approximations of y(x\),
> Example 1 Use Euler's Method with a step size of 0.1 to make a table of approximate values of the solution of the initial-value problem
y' = y-x,
y(0) = 2
(3)
over the interval 0 < x < 1. Solution. In this problem we have f ( x , y) = y — x, XQ = 0, and yo = 2. Moreover, since the step size is 0.1, the jc-values at which the approximate values will be obtained are jd=0.1,
x2=0.2,
X3=0.3,...,
jc 9 =0.9,
Jtio = 1
The first three approximations are yi = yo + f ( x o , yo)Ax = 2 + (2 - 0)(0.1) = 2.2 J2 = yi + f(xltyi)Ax = 2.2 + (2.2 - 0.1)(0.1) = 2.41 y3 = y2 + f ( X 2 , y2)Ax = 2.41 + (2.41 - 0.2)(0.1) = 2.631 Here is a way of organizing all 10 approximations rounded to five decimal places: EULER'S METHOD FOR y' = y-x, y(Q) = 2 WITH A* = 0.1
n
*
0 1 2 3 4 5 6
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
7
8 9 10
yn
f(xn,yn)Ax
2.00000 2.20000 2.41000 2.63100 2.86410 3.11051 3.37156 3.64872 3.94359 4.25795 4.59374
0.20000 0.21000 0.22100 0.23310 0.24641 0.26105 0.27716 0.29487 0.31436 0.33579 —
;vn+l=yn+f(Xn,yn)*x 2.20000 2.41000 2.63100 2.86410 3.11051 3.37156 3.64872 3.94359 4.25795 4.59374 —
8.3 Slope Fields; Euler's Method 583 Observe that each entry in the last column becomes the next entry in the third column. This is reminiscent of Newton's Method in which each successive approximation is used to find the next. •* ACCURACY OF EULER'S METHOD As a rule of thumb, the absolute error in an approximation produced by Euler's Method is proportional to the step size. Thus, reducing the step size by half reduces the absolute and percentage errors by roughly half. However, reducing the step size increases the amount of computation, thereby increasing the potential for more roundoff error. Such matters are discussed in courses on differential equations or numerical analysis.
It follows from (3) and the initial condition y(0) = 2 that the exact solution of the initialvalue problem in Example 1 is y = x + 1+ e
Thus, in this case we can compare the approximate values of y(x) produced by Euler's Method with decimal approximations of the exact values (Table 8.3.1). In Table 8.3.1 the absolute error is calculated as | exact value — approximation | and the percentage error as |exact value — approximation)
| exact value |
— ^ i uvr/v
Table 8.3.1 X
Notice that the absolute error tends to increase as x moves away from xo.
^ QUICK CHECK EXERCISES 8.3
EXACT SOLUTION
EULER APPROXIMATION
ABSOLUTE ERROR
PERCENTAGE ERROR
0.00
0
2.00000
2.00000
0.00000
0.1 0.2 0.3 0.4 0.5
2.20517
2.20000
2.42140 2.64986 2.89182 3.14872
2.41000 2.63100 2.86410 3.11051
0.00517 0.01140 0.01886 0.02772 0.03821
0.6 0.7 0.8 0.9 1.0
3.42212 3.71375 4.02554 4.35960 4.71828
3.37156 3.64872 3.94359 4.25795 4.59374
0.05056 0.06503 0.08195 0.10165 0.12454
0.23 0.47 0.71
0.96
1.21 1.48 1.75 2.04 2.33 2.64
(See page 586 for answers.)
1. Match each differential equation with its slope field. (a) y' = 2xy2 (b) y' = e~y _ (c) / = y (d) y' = 2xy _
I I I I I / I /
i \ i i
i i i i
i i i i
i i i i
i i i i in
\ \ \ \
\ \ \ \
\ \ \ \
\ \ \ \
IV
A Figure Ex-1
2. The slope field for y' = y/x at the 16 gridpoints (x, y), where x — -2, -1, 1, 2 and y = -2, -1, 1, 2 is shown in
584
Chapter 8 / Mathematical Modeling with Differential Equations the accompanying figure. Use this slope field and geometric reasoning to find the integral curve that passes through the point (1,2). N
-2
\
2
N
1
-1
1
-2
3. When using Euler's Method on the initial-value problem y' = f(x, y), y(x0) = y0, we obtain y n+1 from yn, xn, and Ax by means of the formula yn+\ = 4. Consider the initial-value problem y' = y, y(O) = 1. (a) Use Euler's Method with two steps to approximate y(l). (b) What is the exact value of y(l)?
2
•4 Figure Ex-2
EXERCISE SET 8.3
S Graphing Utility
1. Sketch the sloj>e field for y' = xy/4 at the 25 gridpoints (x, y), where x = -2,-l,...,2mdy = -2, -1, . . . , 2.
(e) y' = * x-y
y \ \ > \ - \
2. Sketch the slof >e field for y' + y = 2 at the 25 gridpoints (x, y), where x = 0 , 1 , . . . , 4 and? = 0 , 1 , . . . , 4. 3. A slope field i or the differential equation y' = 1 — y is shown in the a ccompanying figure. In each part, sketch the graph of tht 5 solution that satisfies the initial condition, (a) y(0) - -1 (b) y(0) = 1 (c) y(0) = 2 i
\.y
\ \ \ \ \ \3
r \ \ \ \ \ \ \ \ X \ \ \ 1r \ \ \ \ \ \ \ \ \ \ \ N2" r \ \ \ \ \ \
u(f) / yV
— ("in (>>iii .c)(siny)
y
\ \ \ \
\ \ i i
\ \ i /
\ i i /
i i / /
/ / / / / / //
/ / / /
/ / / /
i / / /
/ 1 1 1 1 1 1 / / / / / / / / / / / / / / /. ////// , X X X X X X ^
/ / / ; / i
/ / / i i i
/ / / i \ \
/ / i i \ \
/ i \ \ \ \
i \ \ \ \ \
• •' • • • •
^ / /
xxxxxx / / / / / /
/ / / / / /
/ / / / / / / / / / / / f 1 1 1 1 1
II
]
1
y y y y
L. \,
^' - - - - -
x x
-2 /-? /-' / ,- / 1 / ? / 3 / / / / / -i
x x
1 1 1 1 I 2 <- / / / / / / / / / / / / - / / / / / / 1
.y / / / ^^^ 1 / / ' - -• >>._x x \ \
kj
x
xx
~
x x
- - V. X S \
"
*• •" —
xx
^ - ~ . ^ x \ \
-jj.
> x-x N \_ \
>
- . ,
/ / X ^ -- —
1 / / ' ^ ^
' ' '
^ Figure Ex-3
/
/
x
*
S ^ ^ ^
/ / / ^ "~^ ^ X X \
4. Solve the initial- value problems in Exercise 3, and use a graphing utility to confirm that the integral curves for these solutions are consistent with the sketches you obtained from the slope field.
-- /
/
— X N -
. x x - x x ^
F O C U S ON C O N C E P T S E^^^^^^^^^^^^^^^^^S
5. Use the slope field in Exercise 3 to make a conjecture about the behavior of the solutions of y' = 1 — y as x —> +00, and confirm your conjecture by examining the general solution of the equation. 6. In parts (a)-(f ), match the differential equation with the slope field, and explain your reasoning. (a) y' = 1/x (b) y' = l/y (c) y' = e-*2 (d) v' = v2 - 1
/ X X ^ -- / / X x- -
I [I
x---
'' ' '' / "'' s'. • / -* '
V A Figure Ex-6
-
IV > ,y
.y '
1 / / ** *~ --
\
-X'XXN \ ~- x X X \ \
-^ ^ ^ ^ ^ ^ *
^ ^ ^ ^ ^ ^
/
/ / / / / /
/ / / /
/ / / / / /
\ s \ s.\ \ ' ;
\ \ \ \ \ \ \ \ \ \ \ \
. X X X X X X
VI
x
8.3 Slope Fields; Euler's Method 7-10 Use Euler's Method with the given step size Az or Ar to approximate the solution of the initial-value problem over the stated interval. Present your answer as a table and as a graph. 7. dy/dx = 1/y, y(0) = 1, 0 < x < 4, Ax = 0.5 9. dy/dt •= cosy, y(0) - I , 0 < t < 2, At = 0.5 10. dy/dt = e~y, y(0) =0, 0 < t < 1, At = 0.1 11. Consider the initial-value problem y(0) = 0
Use Euler's Method with five steps to approximate y(l). 12-15 True-False Determine whether the statement is true or false. Explain your answer. 12. If the graph of y = f ( x ) is an integral curve for a slope field, then so is any vertical translation of this graph. xy
13. Every integral curve for the slope field dy/dx = e graph of an increasing function of x.
is the
14. Every integral curve for the slope field dy/dx = ey is concave up. 15. If p(y) is a cubic polynomial in y, then the slope field dy/dx = p(y) has an integral curve that is a horizontal line. FOCUS ON CONCEPTS
16. (a) Show that the solution of the initial-value problem
y(x)
-I"
dt
(b) Use Euler's Method with Ax = 0.05 to approximate the value of
XI)
-/
dt
Jo
and compare the answer to that produced by a calculating utility with a numerical integration capability. 17. The accompanying figure shows a slope field for the differential equation y' = —x/y. (a) Use the slope field to estimate y(|) for the solution that satisfies the given initial condition y(0) = 1. (b) Compare your estimate to the exact value of y(|).
.,y / , S s .
2
X X X X V
1--
/
/ /--N \
•-/ I I \ \ \ \1
-2
.._2
•4 Figure Ex-17
18. Refer to slope field II in Quick Check Exercise 1. (a) Does the slope field appear to have a horizontal line as an integral curve? (b) Use the differential equation for the slope field to verify your answer to part (a). 19. Refer to the slope field in Exercise 3 and consider the integral curve through (0, —1). (a) Use the slope field to estimate where the integral curve intersects the z-axis. (b) Compare your estimate in part (a) with the exact value of the .r-intercept for the integral curve.
8. dy/dx = x -y2, y(Q) = I , 0 < x < 2, Ax = 0.25
/ = sinjrt,
585
20. Consider the initial-value problem
dx
2 '
(a) Use Euler's Method with step sizes of Ax = 0.2, 0 . 1 , and 0 .05 to obtain three approximations of y ( 1 ) . (b) Find y(l) exactly. 21. A slope field of the form y' = /(y) is said to be autonomous. (a) Explain why the tangent segments along any horizontal line will be parallel for an autonomous slope field. (b) The word autonomous means "independent." In what sense is an autonomous slope field independent? (c) Suppose that G(y) is an antiderivative of l/[/(y)] and that C is a constant. Explain why any differentiable function defined implicitly by G(y) — x = C will be a solution to the equation y' = /(y). 22. (a) Solve the equation y' = ^/y and show that every nonconstant solution has a graph that is everywhere concave up. (b) Explain how the conclusion in part (a) may be obtained directly from the equation y' = ^/y without solving. 23. (a) Find a slope field whose integral curve through (1, 1) satisfies Jty3 — x2y = 0 by differentiating this equation implicitly. (b) Prove that if y (x) is any integral curve of the slope field in part (a), then x[y(jc)]3 — x2y(x) will be a constant function. (c) Find an equation that implicitly defines the integral curve through (— 1 , — 1 ) of the slope field in part (a). 24. (a) Find a slope field whose integral curve through (0, 0) satisfies xey + yex = 0 by differentiating this equation implicitly. (b) Prove that if y (x) is any integral curve of the slope field in part (a), then xeyM + y(x)e* will be a constant function. (c) Find an equation that implicitly defines the integral curve through (1, 1) of the slope field in part (a).
586
Chapter 8 / Mathematical Modeling with Differential Equations
25. Consider the initial-value problem y' = y, v(0) = 1, and let yn denote the approximation of y(l) using Euler's Method with n steps. (a) What would you conjecture is the exact value of lira,, _> +00 yn 1 Explain your reasoning. (b) Find an explicit formula for yn and use it to verify your conjecture in part (a).
26. Writing Explain the connection between Ruler's Method and the local linear approximation discussed in Section 2.9. 27. Writing Given a slope field, what features of an integral curve might be discussed from the slope field? Apply your ideas to the slope field in Exercise 3.
•QUICK CHECK ANSWERS 8.3 1. (a) IV (b) III (c) I (d) II
2. y = 2x,x > 0
3. yn + f(xn, y n )A*
4. (a) 2.25 (b) e
FIRST-ORDER DIFFERENTIAL EQUATIONS AND APPLICATIONS In this section we will discuss a general method that can be used to solve a large class of first-order differential equations. We will use this method to solve differential equations related to the problems of mixing liquids and free fall retarded by air resistance.
FIRST-ORDER LINEAR EQUATIONS The simplest first-order equations are those that can be written in the form
dy
(1)
= «(*)
dx Such equations can often be solved by integration. For example, if dy T dx
=x
3
(2)
then r = —+ C
is the general solution of (2) on the interval (—00, +00). More generally, a first-order differential equation is called linear if it is expressible in the form dy
(3)
Equation (1) is the special case of (3) that results when the function p(x) is identically 0. Some other examples of first-order linear differential equations are
dx
= 0,
i* o
—- + 5y = 2 dx
We will assume that the functions p(x) and q(x) in (3) are continuous on a common interval, and we will look for a general solution that is valid on that interval. One method for doing this is based on the observation that if we define IJL — n(x) by (4)
8.4 First-Order Differential Equations and Applications
587
then = efp(X)dx
I
p(x)dx==
dx
Thus,
d dy d[L dy (My) = M— + — y = //,— + np(x)y dx dx dx dx If (3) is multiplied through by /a, it becomes
(5)
»£-» dx Combining this with (5) we have
_£.
(6) dx This equation can be solved for y by integrating both sides with respect to x and then dividing through by /z to obtain -If ~' »•}
(7)
which is a general solution of (3) on the interval. The function /z in (4) is called an integrating factor for (3), and this method for finding a general solution of (3) is called the method of integrating factors. Although one could simply memorize Formula (7), we recommend solving first-order linear equations by actually carrying out the steps used to derive this formula:
The Method of Integrating Factors Step 1. Calculate the integrating factor
Since any /z will suffice, we can take the constant of integration to be zero in this step. Step 2. Multiply both sides of (3) by /x and express the result as — dx Step 3. Integrate both sides of the equation obtained in Step 2 and then solve for y. Be sure to include a constant of integration in this step.
Example 1
Solve the differential equation dx
- y — e"
Solution. Comparing the given equation to (3), we see that we have a first-order linear equation with p(x) = -1 and q (x) — e2x. These coefficients are continuous on the interval (—00, +00), so the method of integrating factors will produce a general solution on this interval. The first step is to compute the integrating factor. This yields
588
Chapter 8 / Mathematical Modeling with Differential Equations
Next we multiply both sides of the given equation by /z to obtain -x dy
-x
-K 2x
~dx~ which we can rewrite as
__^^^^^_^_^^^^^^_
dx Integrating both sides of this equation with respect to x we obtain
Confirm that the solution obtained in Example 1 agrees with that obtained by
—x
x , f~*
A differential equation of the form P(x)^-+ Q(x)y = R(x) dx can be solved by dividing through by P(x) to put the equation in the form of (3) and then applying the method of integrating factors. However, the resulting solution will only be valid on intervals where p(x) = Q(x)/P(x) and q(x) = R(x)/ P(x) are both continuous.
> Example 2
Solve the initial-value problem x-
y = x,
y(l) =2
Solution. This differential equation can be written in the form of (3) by dividing through by x. This yields
/ - -y =i dx x
(8)
where q(x) = 1 is continuous on (—00, +00) and p(x) = — l / x is continuous on (—00, 0) and (0, +00). Since we need p(x) and q(x) to be continuous on a common interval, and since our initial condition requires a solution for x — 1, we will find a general solution of (8) on the interval (0, +00). On this interval we have \x\ = x, so that fp(x)dx J '
=-
f -dx = -ln\X\ = -\nX J X
[™^c^**rf] integration to be 0
Thus, an integrating factor that will produce a general solution on the interval (0, +00) is x
Multiplying both sides of Equation (8) by this integrating factor yields
Idy x dx
1
1
or d fl dx I x
x
8.4 First-Order Differential Equations and Applications
589
Therefore, on the interval (0, +00), It is not accidental that the initial-value problem in Example 2 has a unique solution. If the coefficients of (3) are continuous on an open interval that contains the point XQ, then for any yo there will be a unique solution of (3) on that interval that satisfies the initial condition y(xo) = yo [Exercise 29(b)].
1 f 1 -y= / -dx =lnx + C x J x
from which it follows that
y = x In x + Cx
(9)
The initial condition y(\) = 2 requires that y — 2 if x = 1. Substituting these values into (9) and solving for C yields C — 2 (verify), so the solution of the initial-value problem is
y = x In x + 2x •* We conclude this section with some applications of first-order differential equations. MIXING PROBLEMS
In a typical mixing problem, a tank is filled to a specified level with a solution that contains a known amount of some soluble substance (say salt). The thoroughly stirred solution is allowed to drain from the tank at a known rate, and at the same time a solution with a known concentration of the soluble substance is added to the tank at a known rate that may or may not differ from the draining rate. As time progresses, the amount of the soluble substance in the tank will generally change, and the usual mixing problem seeks to determine the amount of the substance in the tank at a specified time. This type of problem serves as a model for many kinds of problems: discharge and filtration of pollutants in a river, injection and absorption of medication in the bloodstream, and migrations of species into and out of an ecological system, for example. 5 gal/min
> Example 3 At time t = 0, a tank contains 4 Ib of salt dissolved in 100 gal of water. Suppose that brine containing 2 Ib of salt per gallon of brine is allowed to enter the tank at a rate of 5 gal/min and that the mixed solution is drained from the tank at the same rate (Figure 8.4.1). Find the amount of salt in the tank after 10 minutes.
100 gal
Solution. Let y(t) be the amount of salt (in pounds) after t minutes. We are given that y(Q) — 4, and we want to find y(10). We will begin by finding a differential equation that is satisfied by y(t). To do this, observe that dy/dt, which is the rate at which the amount of salt in the tank changes with time, can be expressed as 5 gal/min
dy_ A Figure 8.4.1
dt
(10)
= rate in — rate out
where rate in is the rate at which salt enters the tank and rate out is the rate at which salt leaves the tank. But the rate at which salt enters the tank is rate in = (2 Ib/gal) • (5 gal/min) = 10 Ib/min Since brine enters and drains from the tank at the same rate, the volume of brine in the tank stays constant at 100 gal. Thus, after t minutes have elapsed, the tank contains y(t) Ib of salt per 100 gal of brine, and hence the rate at which salt leaves the tank at that instant is rate out = ( — Ib/gal) - (5 gal/min) = ~ Ib/min y 100
/
20
Therefore, (10) can be written as dy
= 10 - f-
or
dy -f-
590
Chapter 8 / Mathematical Modeling with Differential Equations
which is a first-order linear differential equation satisfied by y(t). Since we are given that y(Q) = 4, the function y(t) can be obtained by solving the initial- value problem | + £ = 10, X0)=4 The integrating factor for the differential equation is
If we multiply the differential equation through by fj,, then we obtain -(et/20y) = Wel/2° at e"20y = I 10et/2Qdt = 200et/2 «-'/20
(11)
The initial condition states that y = 4 when t = 0. Substituting these values into (11) and solving for C yields C = —196 (verify), so y ( t ) = 200 - 196e-'/2°
(12)
The graph of (12) is shown in Figure 8.4.2. At time t = 10 the amount of salt in the tank is
£ 200
y ( l O ) = 200 - 196e~05 ^ 81.1 Ib -4
£ 150 ro
2O 100 50
|
Notice that it follows from (11) that
O
£
0
0
10 20 30 40 50 60 70 80
Time /(min)
lim y ( t ) = 200
r->+o>
A Figure 8.4.2
The graph shown in Figure 8.4.2 suggests that y(t) -> 200 as t ->• +». This means that over an extended period of time the amount of salt in the tank tends toward 200 Ib. Give an informal physical argument to explain why this result is to be expected.
for all values of C, so regardless of the amount of salt that is present in the tank initially, the amount of salt in the tank will eventually stabilize at 200 Ib. This can also be seen geometrically from the slope field for the differential equation shown in Figure 8.4.3. This slope field suggests the following: If the amount of salt present in the tank is greater than 200 Ib initially, then the amount of salt will decrease steadily over time toward a limiting value of 200 Ib; and if the amount of salt is less than 200 Ib initially, then it will increase steadily toward a limiting value of 200 Ib. The slope field also suggests that if the amount present initially is exactly 200 Ib, then the amount of salt in the tank will stay constant at 200 Ib. This can also be seen from (11), since C = 0 in this case (verify). A MODEL OF FREE-FALL MOTION RETARDED BY AIR RESISTANCE
In Section 4.7 we considered the free-fall model of an object moving along a vertical axis near the surface of the Earth. It was assumed in that model that there is no air resistance and that the only force acting on the object is the Earth's gravity. Our goal here is to find a model that takes air resistance into account. For this purpose we make the following assumptions:
300 250 200 -
150 100
0
50
A Figure 8.4.3
100
150
• The object moves along a vertical s-axis whose origin is at the surface of the Earth and whose positive direction is up (Figure 4.7.7). • At time t = 0 the height of the object is s^ and the velocity is VQ. • The only forces on the object are the force FG = —mg of the Earth's gravity acting down and the force FR of air resistance acting opposite to the direction of motion. The force FR is called the drag force.
8.4 First-Order Differential Equations and Applications 591
In the case of free-fall motion retarded by air resistance, the net force acting on the object FG + PR = ~mg + FR
and the acceleration is d2s/dt2, so Newton's Second Law of Motion [Equation (5) of Section 5.6] implies that ^2S -mg + FR=m — (13) Experimentation has shown that the force FR of air resistance depends on the shape of the object and its speed—the greater the speed, the greater the drag force. There are many possible models for air resistance, but one of the most basic assumes that the drag force FR is proportional to the velocity of the object, that is, FR = -CV
where c is a positive constant that depends on the object's shape and properties of the air.* (The minus sign ensures that the drag force is opposite to the direction of motion.) Substituting this in (13) and writing d2s/dt2 as dv/dt, we obtain
dv
—mg — cv = m — dt Dividing by m and rearranging we obtain
dv dt
— H
c v — -g m
which is a first-order linear differential equation in the unknown function v — v(t) with p(t) = c/m and q ( t ) = —g [see (3)]. For a specific object, the coefficient c can be determined experimentally, so we will assume that m, g, and c are known constants. Thus, the velocity function v = v(t) can be obtained by solving the initial-value problem dv c (14) — H -- u = -g, v(0) — v0 dt m Once the velocity function is found, the position function s = s(t) can be obtained by solving the initial-value problem
dt
(r),
5(0) = 50
(15)
In Exercise 25 we will ask you to solve (14) and show that
(16) Note that
lim v(t) =
mg
(17)
(verify). Thus, the speed \v(t)\ does not increase indefinitely, as in free fall; rather, because of the air resistance, it approaches a finite limiting speed VT given by mg mg (18) c c This is called the terminal speed of the object, and (17) is called its terminal velocity.
REMARK
Intuition suggests that near the limiting velocity, the velocity v(t) changes very slowly; that is, dv/dt ~ 0. Thus, it should not be surprising that the limiting velocity can be obtained informally from (14) by setting dv/dt = 0 in the differential equation and solving for v. This yields
mg which agrees with (17).
Other common models assume that FR = —cv or, more generally, FR = —cvp for some value of p.
597
Chapter 8 / Mathematical Modeling with Differential Equations
I/QUICK CH ECK EXERCISES 8.4
(See page 594 for answers.)
1. Solve the first-order linear differential equation
2. An integrating factor for
dy — + p(x)y = q(x) dx
dy y — + - = q(x)
x
dx
by completing the following steps:
is. Step 1. Calculate the integrating factor [i = Step 2. Multiply both sides of the equation by the integrating factor and express the result as d -]=•
dx
3. At time t = 0, a tank contains 30 oz of salt dissolved in 60 gal of water. Then brine containing 5 oz of salt per gallon of brine is allowed to enter the tank at a rate of 3 gal/min and the mixed solution is drained from the tank at the same rate. Give an initial-value problem satisfied by the amount of salt y ( t ) in the tank at time t. Do not solve the problem.
Step 3. Integrate both sides of the equation obtained in Step 2 and solve for y =
EXERCISE SET 8.4
Q Graphing Utility
1-6 Solve the differential equation by the method of integrating factors.
.
2. -/-+2xy=x dx
dx
42
3. y' + y = cos(e*)
5.
/
dx 6
- dx T"
dx
1 — ex
=0
7-10 Solve the initial- value problem.
7. x— +y = x, dx
y(l) = 2
14. In our model for free-fall motion retarded by air resistance, the terminal velocity is proportional to the weight of the falling object. 15. A slope field for the differential equation y' = 2y — x is shown in the accompanying figure. In each part, sketch the graph of the solution that satisfies the initial condition. (a) y(l) = 1 (b) y(0) = -1 (c) y(-l) = 0 y i i i i i i ; 2 •i i i i i i i i -i i i i i i i i •i i i i i i i i •i i i i i i ii / 1 i ii i ii i i i ill i i i / / I L
.
dx
9. —- 2xy = 2x, y(0) = 3 dx 10.
IIL
-2 / x / s*
,
y(0) =
11-14 True-False Determine whether the statement is true or false. Explain your answer. 11. If y\ and yi are two solutions to a first-order linear differential equation, then y = y\ + yi is also a solution. 12. If the first-order linear differential equation
dy_ dx
P(x)y = q(x)
has a solution that is a constant function, then q(x) is a constant multiple of p(x). 13. In a mixing problem, we expect the concentration of the dissolved substance within the tank to approach a finite limit over time.
\ \\ \ \ \ \ \ \
,-1 X \
\
\ \ \ \
\ \ \ \ \ \ \ \ \-2
\ \ \ \
\ \ \ \
1
\ \ \ \
\ \ \ \
\ \ \ \ \
\ \ \ \ \
2 \ \ \ \
\ \ \ \ \\ \ \ i t \ \ \ \ \ \ \\
•4 Figure Ex-15
16. Solve the initial-value problems in Exercise 15, and use a graphing utility to confirm that the integral curves for these solutions are consistent with the sketches you obtained from the slope field. FOCUS ON CONCEPTS
17. Use the slope field in Exercise 15 to make a conjecture about the effect of yo on the behavior of the solution of the initial-value problem y' = 2y — x, y(0) = yo as x -» +t», and check your conjecture by examining the solution of the initial-value problem. 18. Consider the slope field in Exercise 15. (a) Use Euler's Method with Ax = 0.1 to estimate y (5) for the solution that satisfies the initial condition y(0) = 1.
8.4 First-Order Differential Equations and Applications (b) Would you conjecture your answer in part (a) to be greater than or less than the actual value of y(|)? Explain. (c) Check your conjecture in part (b) by finding the exact value of y(j).
19. (a) Use Euler's Method with a step size of AJC = 0.2 to approximate the solution of the initial-value problem
/ = x + y,
y(0) = 1
over the interval 0 < x < 1. (b) Solve the initial-value problem exactly, and calculate the error and the percentage error in each of the approximations in part (a). (c) Sketch the exact solution and the approximate solution together. 20. It was stated at the end of Section 8.3 that reducing the step size in Euler's Method by half reduces the error in each approximation by about half. Confirm that the error in v(l) is reduced by about half if a step size of Ax =0.1 is used in Exercise 19. 21. At time t = 0, a tank contains 25 oz of salt dissolved in 50 gal of water. Then brine containing 4 oz of salt per gallon of brine is allowed to enter the tank at a rate of 2 gal/min and the mixed solution is drained from the tank at the same rate. (a) How much salt is in the tank at an arbitrary time tl (b) How much salt is in the tank after 25 min? 22. A tank initially contains 200 gal of pure water. Then at time t =0 brine containing 5 Ib of salt per gallon of brine is allowed to enter the tank at a rate of 20 gal/min and the mixed solution is drained from the tank at the same rate. (a) How much salt is in the tank at an arbitrary time t? (b) How much salt is in the tank after 30 min? 23. A tank with a 1000 gal capacity initially contains 500 gal of water that is polluted with 50 Ib of particulate matter. At time t = 0, pure water is added at a rate of 20 gal/min and the mixed solution is drained off at a rate of 10 gal/min. How much particulate matter is in the tank when it reaches the point of overflowing? 24. The water in a polluted lake initially contains 1 Ib of mercury salts per 100,000 gal of water. The lake is circular with diameter 30 m and uniform depth 3 m. Polluted water is pumped from the lake at a rate of 1000 gal/h and is replaced with fresh water at the same rate. Construct a table that shows the amount of mercury in the lake (in Ib) at the end of each hour over a 12-hour period. Discuss any assumptions you made. [Note: Use 1 m3 = 264 gal.] 25. (a) Use the method of integrating factors to derive solution (16) to the initial-value problem (14). [Note: Keep in mind that c,m, and g are constants.] (b) Show that (16) can be expressed in terms of the terminal speed (18) as
593
(c) Show that if .s (0) = SQ, then the position function of the object can be expressed as s(t) = SQ — vTt -\—-(VQ + vr)(\ — e~g g
r
)
26. Suppose a fully equipped skydiver weighing 240 Ib has a terminal speed of 120 ft/s with a closed parachute and 24 ft/s with an open parachute. Suppose further that this skydiver is dropped from an airplane at an altitude of 10,000 ft, falls for 25 s with a closed parachute, and then falls the rest of the way with an open parachute. (a) Assuming that the skydiver's initial vertical velocity is zero, use Exercise 25 to find the skydiver's vertical velocity and height at the time the parachute opens. [Note: Take g = 32 ft/s 2 .] (b) Use a calculating utility to find a numerical solution for the total time that the skydiver is in the air. 27. The accompanying figure is a schematic diagram of a basic RL series electrical circuit that contains a power source with a time-dependent voltage of V(t) volts (V), a resistor with a constant resistance of R ohms (£2), and an inductor with a constant inductance of L henrys (H). If you don't know anything about electrical circuits, don't worry; all you need to know is that electrical theory states that a current of 7(f) amperes (A) flows through the circuit where 7(f) satisfies the differential equation
L— + 7?7 = V(t) at (a) Find I(t) if R = 10 fi, L = 5 H, V is a constant 20 V, and 7(0) = OA. (b) What happens to the current over a long period of time?
•4 Figure Ex-27
28. Find 7 (t) for the electrical circuit in Exercise 27 if R = 6 £2, L = 3 H, V(t) = 3 sin t V, and 7(0) = 15 A. FOCUS ON C O N C E P T S
29. (a) Prove that any function y = y(x) defined by Equation (7) will be a solution to (3). (b) Consider the initial-value problem
dy dx where the functions p(x) and q(x) are both continuous on some open interval. Using the general solution for a first-order linear equation, prove that this initial-value problem has a unique solution on the interval.
594
Chapter 8 / Mathematical Modeling with Differential Equations
30. (a) Prove that solutions need not be unique for nonlinear initial-value problems by finding two solutions to
y(0) = 0 dx (b) Prove that solutions need not exist for nonlinear initial-value problems by showing that there is no solution for
dy
y— = -x, dx
31. Writing Explain why the quantity n in the Method of Integrating Factors is called an "integrating factor" and explain its role in this method. 32. Writing Suppose that a given first-order differential equation can be solved both by the method of integrating factors and by separation of variables. Discuss the advantages and disadvantages of each method.
y(0) = 0
•QUICK CHECK ANSWERS 8.4 1. Step 1:e!P(x)dx\ Step 2: /zy, nq(x); Step 3: - f fj,q(x) dx ;LI J
CHAPTER 8 REVIEW EXERCISES
2. Which of the given differential equations are separable?
dx
=
f(x)g(y)
(c) ^ = f ( x ) + g(y) dx
(b)
dy
f(x)
+
14. dy/dx = siny, y(0) = 1, 0 < x < 2, Ax = 0.5
y(0) = 1
Use Euler's Method with five steps to approximate y(l).
6-8 Solve the initial-value problem by the method of separation of variables. 9 y
13. dy/dx = Jy, y(0) = 1, 0 < x < 4, Ax = 0.5
y' = cos27rf,
5. (1 + y 2 )y' = e*y
7. y' =
13-14 Use Euler's Method with the given step size Ax to approximate the solution of the initial-value problem over the stated interval. Present your answer as a table and as a graph.
15. Consider the initial-value problem
dx g(y) dy (d) / = dx
3-5 Solve the differential equation by the method of separation of variables. • dy 4. 3 tan y sec x = 0 • i — v i i j /-^ dx
6. y' = 1 + y2, y(0) = 1
3. — + — = 15, y(0) = 30 dt 20
CAS
1. Give an informal explanation of why one might expect the number of arbitrary constants in the general solution of a differential equation to be equal to the order of the equation.
(a)
2. x
y(l) = 1
8. y' = 4y 2 sec 2 2x, y(;r/8) = 1 9. Sketch the integral curve of y' = — 2xy2 that passes through the point (0, 1). 10. Sketch the integral curve of 2yy' = 1 that passes through the point (0, 1) and the integral curve that passes through the point (0, -1). 11. Sketch the slope field for y' = xy/8 at the 25 gridpoints (x, y), where x = 0, 1 , . . . , 4 and y = 0, 1 , . . . , 4. 12. Solve the differential equation y' = xy/8, and find a family of integral curves for the slope field in Exercise 11.
16. Use Euler's Method with a step size of A? =0.1 to approximate the solution of the initial-value problem y'
= l+5t-y,
y(l) = 5
over the interval [1,2]. 17. Cloth found in an Egyptian pyramid contains 78.5% of its original carbon- 14. Estimate the age of the cloth. 18. Suppose that an initial population of 5000 bacteria grows exponentially at a rate of 1% per hour and that y = y(t) is the number of bacteria present after t hours. (a) Find an initial-value problem whose solution is y(t). (b) Find a formula for y(t). (c) What is the doubling time for the population? (d) How long does it take for the population of bacteria to reach 30,000? 1 9-20 Solve the differential equation by the method of integrating factors, B
dx
20. dx
1 + e*
21-23 Solve the initial-value problem by the method of integrating factors. •
Chapter 8 Making Connections
21. y-xy = x, y(0) = 3 2
22. xy' + 2y = 4x , y(l)=2 23. /cosher + ysinhx = cosh 2 x, y(0) ~2 [c] 24. (a) Solve the initial-value problem
y' — v- = x sin 3*,
y(0) = 1
by the method of integrating factors, using a CAS to perform any difficult integrations. (b) Use the CAS to solve the initial-value problem directly, and confirm that the answer is consistent with that obtained in part (a). (c) Graph the solution. 25. Classify the following first-order differential equations as separable, linear, both, or neither. dy . dy (a) 3y = sin x (b) -± dx dx (d) — + xy2 ~ sin(xy) (0 , £ - * - ! dx dx 26. Determine whether the methods of integrating factors and separation of variables produce the same solutions of the differential equation dy 4xy = x dx
595
27. A tank contains 1000 gal of fresh water. At time? = Omin, brine containing 5 oz of salt per gallon of brine is poured into the tank at a rate of 10 gal/min, and the mixed solution is drained from the tank at the same rate. After 15 min that process is stopped and fresh water is poured into the tank at the rate of 5 gal/min, and the mixed solution is drained from the tank at the same rate. Find the amount of salt in the tank at time t = 30 min. 28. Suppose that a room containing 1200 ft3 of air is free of carbon monoxide. At time t = 0 cigarette smoke containing 4% carbon monoxide is introduced at the rate of 0.1 ft3 /min, and the well-circulated mixture is vented from the room at the same rate. (a) Find a formula for the percentage of carbon monoxide in the room at time t. (b) Extended exposure to air containing 0.012% carbon monoxide is considered dangerous. How long will it take to reach this level? Source: This is based on a problem from William E. Boyce and Richard C. DiPrima, Elementary Differential Equations, 7th ed., John Wiley & Sons, New York, 2001.
CHAPTER 8 MAKING CONNECTIONS 1. Consider the first-order differential equation
dy , dx+Py
3. A first-order differential equation is homogeneous if it can be written in the form
=q
where p and q are constants. If y = y(x) is a solution to this equation, define u = u(x) = q — py(x). (a) Without solving the differential equation, show that u grows exponentially as a function of x if p < 0, and decays exponentially as a function of x if 0 < p. (b) Use the result of part (a) and Equations (1 3-14) of Section 8.2 to solve the initial-value problem
dx where / is a function of a single variable. If y = y(x) is a solution to a first-order homogeneous differential equation, define u = u(x) = y(x)/x. (a) Find a separable differential equation that is satisfied by the function u. (b) Use your answer to part (a) to solve
dy_ _ x-y dx x +y
y(0) = -l
dx
2. Consider a differential equation of the form
dy ~ = f(ax + by + c) dx where / is a function of a single variable. If y = y (x) is a solution to this equation, define u = u(x) = ax + by(x) + c. (a) Find a separable differential equation that is satisfied by the function u. (b) Use your answer to part (a) to solve
dy_ dx
1
x +y
4. A first-order differential equation is called a Bernoulli equation if it can be written in the form — + p(x)y = q(x)y" for n ^ 0, 1 dx If y = y(x) is a solution to a Bernoulli equation, define (a) Find a first-order linear differential equation that is satisfied by u. (b) Use your answer to part (a) to solve the initial-value problem dy , 1
'dx
yd) = -
INFINITE SERIES
iStockphoto
Perspective creates the illusion that the sequence of railroad ties continues indefinitely but converges toward a single point infinitely far away.
In this chapter we will be concerned with infinite series, which are sums that involve infinitely many terms. Infinite series play a fundamental role in both mathematics and science—they are used, for example, to approximate trigonometric functions and logarithms, to solve differential equations, to evaluate difficult integrals, to create new functions, and to construct mathematical models of physical laws. Since it is impossible to add up infinitely many numbers directly, one goal will be to define exactly what we mean by the sum of an infinite series. However, unlike finite sums, it turns out that not all infinite series actually have a sum, so we will need to develop tools for determining which infinite series have sums and which do not. Once the basic ideas have been developed we will begin to apply our work; we will show how infinite series are used to evaluate such quantities as In 2, e, sin 3°, and n, how they are used to create functions, and finally, how they are used to model physical laws.
SEQUENCES In everyday language, the term "sequence" means a succession of things in a definite order—chronological order, size order, or logical order, for example. In mathematics, the term "sequence" is commonly used to denote a succession of numbers whose order is determined by a rule or a function. In this section, we will develop some of the basic ideas concerning sequences of numbers.
DEFINITION OF A SEQUENCE
Stated informally, an infinite sequence, or more simply a sequence, is an unending succession of numbers, called terms. It is understood that the terms have a definite order; that is, there is a first term a\, a second term a2, a third term «3, a fourth term a4, and so forth. Such a sequence would typically be written as a\, 02, «3, 04, ...
where the dots are used to indicate that the sequence continues indefinitely. Some specific examples are 1234 i i i i 2' 3' 4
2,4,6,8,..., Each of these sequences has a definite pattern that makes it easy to generate additional terms if we assume that those terms follow the same pattern as the displayed terms. However, 596
9.1 Sequences
597
such patterns can be deceiving, so it is better to have a rule or formula for generating the terms. One way of doing this is to look for a function that relates each term in the sequence to its term number. For example, in the sequence 2,4,6,8,... each term is twice the term number; that is, the nth term in the sequence is given by the formula In. We denote this by writing the sequence as 2, 4, 6, 8 , . . . , In, ... We call the function f(ri) — 2n the general term of this sequence. Now, if we want to know a specific term in the sequence, we need only substitute its term number in the formula for the general term. For example, the 37th term in the sequence is 2 • 37 = 74.
Example 1 In each part, find the general term of the sequence. \
2 3 4
2' 3' 4' 5'
TERM NUMBER
1 2 3 4
n
TERM
1 2 3 4 2 3 4 5
n+l
•••
n
Table 9.1.2 TERM NUMBER TERM
1 2 3 4 ...
n
...
1 1 1 1 1 2 22 23 24 " 2" "
-
1I ! J 2> 4' 8' 16'
(d) 1,3,5,7,.
3' 4
Table 9.1.1
(fK» D)
Solution (a). In Table 9.1.1, the four known terms have been placed below their term numbers, from which we see that the numerator is the same as the term number and the denominator is one greater than the term number. This suggests that the nth term has numerator n and denominator n + 1, as indicated in the table. Thus, the sequence can be expressed as 1 2' n +l Solution (b\ In Table 9.1.2, the denominators of the four known terms have been expressed as powers of 2 and the first four terms have been placed below their term numbers, from which we see that the exponent in the denominator is the same as the term number. This suggests that the denominator of the «th term is 2", as indicated in the table. Thus, the sequence can be expressed as
1 1 1 1 1 2' 4' 8' 16'"" 2"'"" Solution (c). This sequence is identical to that in part (a), except for the alternating signs. Thus, the nth term in the sequence can be obtained by multiplying the nth term in part (a) by (—!)"+'. This factor produces the correct alternating signs, since its successive values, starting with n = 1, are 1, — 1, 1, —1, Thus, the sequence can be written as
I 2' Table 9.1.3 TERM NUMBER TERM
1 2 3 4
n
2 3 3' 4'
(-D
n+l
n+l
Solution (d). In Table 9.1.3, the four known terms have been placed below their term numbers, from which we see that each term is one less than twice its term number. This suggests that the «th term in the sequence is 2n — 1, as indicated in the table. Thus, the sequence can be expressed as
1 3 5 7 — 2n-l —
1,3,5,7,. . , 2 n - l , . . < When the general term of a sequence GI, 02, a-j, ...,an, ...
(D
598
Chapter 9 / Infinite Series
is known, there is no need to write out the initial terms, and it is common to write only the general term enclosed in braces. Thus, (1) might be written as
MH-I °ras {««ir=i
For example, here are the four sequences in Example 1 expressed in brace notation. A sequence cannot be uniquely determined from a few initial terms. For example, the sequence whose general term is
BRACE NOTATION
SEQUENCE
1 T 1 2' 1
/(«) = ! (3-5n + 6 n 2 - n 3 ) has 1, 3, and 5 as its first three terms, but its fourth term is also 5.
2 3' 1 4' 2
3 4 n 4' 5" " n + 1 " 1 1 1 8' 16" ' 2" " 3 4 - ,}n+\ n
2 ' 3' 4' 5 " 1, 3, 5, 7, ...,2n-l, ...
n+1"
f
n
|+°°
1/1+ 1 J n = l
1 1 j+~
1 2" ]„=!
f I
n+1
n ]+~ n + 1 J«=i
"
The letter n in (1) is called the index for the sequence. It is not essential to use n for the index; any letter not reserved for another purpose can be used. For example, we might view the general term of the sequence a\, 02, 03, ... to be the A:th term, in which case we would denote this sequence as {fl*}^. Moreover, it is not essential to start the index at 1; sometimes it is more convenient to start it at 0 (or some other integer). For example, consider t h e sequence 1 1 1 2' 2 2 '
One way to write this sequence is
1
However, the general term will be simpler if we think of the initial term in the sequence as the zeroth term, in which case we can write the sequence as
We began this section by describing a sequence as an unending succession of numbers. Although this conveys the general idea, it is not a satisfactory mathematical definition because it relies on the term "succession," which is itself an undefined term. To motivate a precise definition, consider the sequence 2,4, 6, 8, . . . , 2 n , . . . If we denote the general term by /(n) = 2n, then we can write this sequence as /(I), /(2),/(3),..., /(«),... which is a "list" of values of the function f ( n ) = 2n,
n = 1,2,3,...
whose domain is the set of positive integers. This suggests the following definition. 9.1.1
DEFINITION A sequence is a function whose domain is a set of integers.
Typically, the domain of a sequence is the set of positive integers or the set of nonnegative integers. We will regard the expression (an}+™, to be an alternative notation for the function f ( n ) = an,n = 1, 2, 3 , . . . , and we will regard {an}^T0 to be an alternative notation for the function /(n) = an, n — 0, 1, 2, 3,
9.1 Sequences
599
GRAPHS OF SEQUENCES
Since sequences are functions, it makes sense to talk about the graph of a sequence. For example, the graph of the sequence {l/nJ^T, is the graph of the equation When the starting value for the index of a sequence is not relevant to the discussion, it is common to use a notation such as {an} in which there is no reference to the starting value of n. We can distinguish between different sequences by using different letters for their general terms; thus, {«„}, {bn}, and (c,,| denote three different sequences.
Because the right side of this equation is denned only for positive integer values of n, the graph consists of a succession of isolated points (Figure 9. 1 . la). This is different from the graph of
>l
y = X-, which is a continuous curve (Figure 9.\.\b).
.y 1- • 1
> Figure 9.1.1
2
T
T
3
4
J 5
n
(a)
LIMIT OF A SEQUENCE
Since sequences are functions, we can inquire about their limits. However, because a sequence {an} is only denned for integer values of n, the only limit that makes sense is the limit of an as n -> +°°. In Figure 9.1.2 we have shown the graphs of four sequences, each of which behaves differently as n —>• +0°: • The terms in the sequence {n + 1} increase without bound. • The terms in the sequence {(—1)" +1 } oscillate between —1 and 1. • The terms in the sequence {«/(« + 1)} increase toward a "limiting value" of 1. • The terms in the sequence {l + (—5)"} also tend toward a "limiting value" of 1, but do so in an oscillatory fashion.
1 2 3 4 5 6 7 8 9
-i 1 2 3 4 5
A Figure 9.1.2
Informally speaking, the limit of a sequence {«„} is intended to describe how an behaves as n -> +00. To be more specific, we will say that a sequence {an} approaches a limit L if the terms in the sequence eventually become arbitrarily close to L. Geometrically, this
600 Chapter 9 / Infinite Series y y —L + e
L
1 in the sequence areall within e units of L.
• •
^/ n
> Figure 9.1.3
1
2
3
4
• • •
N
means that for any positive number € there is a point in the sequence after which all terms lie between the lines y = L — e and y — L + € (Figure 9.1.3). The following definition makes these ideas precise.
How would you define these limits?
lim an = +co
n~++
lim an = -co
9.1.2 DEFINITION A sequence {an} is said to converge to the limit L if given any e > 0, there is a positive integer N such that \an — L\ < € for n > N. In this case we write lim an = L n—> +00
A sequence that does not converge to some finite limit is said to diverge.
> Example 2 The first two sequences in Figure 9.1.2 diverge, and the second two converge to 1; that is, lim •
lim Lfl + V(-!)"! 2 J = 1
= 1 and
n-»+oo j
n^+oo
'
The following theorem, which we state without proof, shows that the familiar properties of limits apply to sequences. This theorem ensures that the algebraic techniques used to find limits of the form lim can also be used for limits of the form lim .
9.1.3 THEOREM Suppose that the sequences {an} and {bn} converge to limits L \ and LI, respectively, and c is a constant. Then: (a) (b) (c) (d) Additional limit properties follow from those in Theorem 9.1.3. For example,
(e)
lim c = c
n-+ +00
lim can = c lim an = cL\
n —>- +00
n —» +00
lim (an + bn) — lim an + lim bn=L\-\- L2
n —» +00
n —> +GO
lim (an — bn) = lim an — lim bn = L\ - L2
n —> +00
lim (flnTw = Lj m
n —> +00
«—> +00
lim (anbn) = lim an • lim bn = L\L-i
n —> +00
n —> +00
use part (e) to show that if an —> L and m is a positive integer, then
n —> +00
(
n -^ +00
lim \ inn aunn j " — 1 = "^+°° = —
(ifL
0)
If the general term of a sequence is f(ri), where f(x) is a function defined on the entire interval [1, +00), then the values of f ( n ) can be viewed as "sample values" of f(x) taken
9.1 Sequences
601
at the positive integers. Thus, if f ( x ) —> L as x —> +00,
then
/(n) —>• L as n —>• +00
(Figure 9.1.4a). However, the converse is not true; that is, one cannot infer that f ( x ) ->• L as x -> +00 from the fact that f(n) -»• L as n -> +00 (Figure 9.1.4&).
If f(x) -» L as x -> +co; then /(«) -> L as n —» +0°.
> Example 3 In each part, determine whether the sequence converges or diverges by examining the limit as n —»• +00. f
+00
(a)
1 +00
(-1) n+l
(b)
(a) +co
L
T—i—i—
Solution (a). lim
Dividing numerator and denominator by n and using Theorem 9.1.3 yields
= lim 1 2+ 0
/(n) -» L as n -» +00, but /(jc) i diverges by oscillation as *—>+oo.
lim 1
lim 1
1 2+1/n
n~> +co
lim (2+1/n)
n—> +00
lim 2 + lim 1/n
« —> +co
n —> +00
2
Thus, the sequence converges to |. A Figure 9.1.4
Solution (b). This sequence is the same as that in part (a), except for the factor of (—1)"+1, which oscillates between +1 and — 1. Thus, the terms in this sequence oscillate between positive and negative values, with the odd-numbered terms being identical to those in part (a) and the even-numbered terms being the negatives of those in part (a). Since the sequence in part (a) has a limit of \, it follows that the odd-numbered terms in this sequence approach ^, and the even-numbered terms approach — i. Therefore, this sequence has no limit—it diverges. Solution (c). Since l/n-^-0, the product (—l)" +1 (l/«) oscillates between positive and negative values, with the odd-numbered terms approaching 0 through positive values and the even-numbered terms approaching 0 through negative values. Thus, 1 lim (—1)"+ — = 0 n-»+c°
n
so the sequence converges to 0. Solution (d).
lim (8 — In) = — oo, so the sequence {8 — 2n}+™, diverges. •*
> Example 4 its limit.
In each part, determine whether the sequence converges, and if so, find l
1
(b) 1,2, 2 2 , 2 3 , . . . , 2 " , . . .
Solution. Replacing n by x in the first sequence produces the power function (1 /2)*, and replacing n by x in the second sequence produces the power function 2*. Now recall that if 0 < b < 1, then bx —> 0 as x -> +00, and if b > 1, then V ->• +00 as x ->• +00 (Figure 6.1.1).
602 Chapter 9 / Infinite Series Thus,
lim — = 0
and
-> +os 2"
n
lim 2" = +00 —*• +°°
So, the sequence {1/2"} converges to 0, but the sequence {2"} diverges. co
( —n |i ~'~ Solution.
The expression
lim —
is an indeterminate form of type oo/oo, so L'Hopital's rule is indicated. However, we cannot apply this rule directly to n/e" because the functions n and e" have been defined here only at the positive integers, and hence are not differentiable functions. To circumvent this problem we extend the domains of these functions to all real numbers, here implied by replacing n by x, and apply L'Hopital's rule to the limit of the quotient x/ex. This yields lim — — lim — = 0
X —> +00 g1
X -» +CO £*
from which we can conclude that lim — = 0
> Examples Show that lim tyn = I. Solution. lim !ft=
n —> +GO
lim nlln = lim
n —> +00
n —> +co
(1/ )lnn e
"
= e° = I
By L'Hopital's rule j applied to (1 /*) In x I
Sometimes the even-numbered and odd-numbered terms of a sequence behave sufficiently differently that it is desirable to investigate their convergence separately. The following theorem, whose proof is omitted, is helpful for that purpose. 9.1.4 THEOREM A sequence converges to a limit L if and only if the sequences of even-numbered terms and odd-numbered terms both converge to L.
> Example 7 The sequence 1 1 1 1 1 1 2' 3' 22' 31' 21' 33"" converges to 0, since the even-numbered terms and the odd-numbered terms both converge to 0, and the sequence ii, 1, i, i I, i,1 I, . . .
\
2
3
4
diverges, since the odd-numbered terms converge to 1 and the even-numbered terms converge to 0. -^
THE SQUEEZING THEOREM FOR SEQUENCES
If
a,, -^ L and CB -•> L, then
A Figure 9.1.5
The following theorem, illustrated in Figure 9.1.5, is an adaptation of the Squeezing Theorem (1.6.2) to sequences. This theorem will be useful for finding limits of sequences that cannot be obtained directly. The proof is omitted.
9.1 Sequences 9.1.5 THEOREM (The Squeezing Theorem for Sequences) quences such that
an < bn < cn
603
Let {«„}, [bn], and {cn} be se-
(for all values ofn beyond some index N)
If the sequences [an] and {<;„} have a common limit L as n—>• +00, then [bn] also has the limit L as n —> +co.
Recall that if n is a positive integer, then n\ (read "n factorial") is the product of the first n positive integers. In addition, it is convenient to define 0! = 1.
> Example 8 quence
Use numerical evidence to make a conjecture about the limit of the sef
and then confirm that your conjecture is correct. Table 9.1.4
Solution. Table 9.1.4, which was obtained with a calculating utility, suggests that the limit of the sequence may be 0. To confirm this we need to examine the limit of
"1 n" 1 2 3 4 5 6 7 8 9 10 11 12
1.0000000000 0.5000000000 0.2222222222 0.0937500000 0.0384000000 0.0154320988 0.0061198990 0.0024032593 0.0009366567 0.0003628800 0.0001399059 0.0000537232
n\ an = — n"
as n -» +00. Although this is an indeterminate form of type oo/oo, L'Hopital's rule is not helpful because we have no definition of x\ for values of x that are not integers. However, let us write out some of the initial terms and the general term in the sequence: _1-2_1 _l-2-3_2 1 _l-2-3-4_3 a3 < 4 ~ 2~^2 ~ 2' ~3-3-3~9 3' ° ~ 4 • 4 - 4 - 4 ~ 3 2 < If n > 1, the general term of the sequence can be rewritten as
a\ = 1,
an =
1 -2-3---n 1 /2-3---n — n • n •n • • •n n \n • n • • • n /
from which it follows that an < l/n (why?). It is now evident that 1
However, the two outside expressions have a limit of 0 as n -> +00; thus, the Squeezing Theorem for Sequences implies that an —> 0 as n -» +00, which confirms our conjecture.
The following theorem is often useful for finding the limit of a sequence with both positive and negative terms—it states that if the sequence {\an\} that is obtained by taking the absolute value of each term in the sequence {an} converges to 0, then {an} also converges toO.
9.1.6
THEOREM If lim \an — 0, then lim an = 0.
PROOF Depending on the sign of an, either an = \an\ or an = —\an\. Thus, in all cases we have -\an
604
Chapter 9 / Infinite Series
^ Example 9
Consider the sequence I
1
_L
«J_
' 2' 22 ' 2 3 ' ' ' ' 2" ' ' ' ' If we take the absolute value of each term, we obtain the sequence I
J_
J_
J_
' 2' 22' 2 3 ' ' " ' 2""" which, as shown in Example 4, converges to 0. Thus, from Theorem 9.1.6 we have lim
(-1)"— = 0 + 2"
SEQUENCES DEFINED RECURSIVELY
Some sequences do not arise from a formula for the general term, but rather from a formula or set of formulas that specify how to generate each term in the sequence from terms that precede it; such sequences are said to be defined recursively, and the defining formulas are called recursion formulas. A good example is the mechanic's rule for approximating square roots. In Exercise 23 of Section 3.7 you were asked to show that 1 /
x
n+\ — - \Xn -\ 2V
a
(2)
Xn
describes the sequence produced by Newton's Method to approximate ~Ja as a zero of the function /(jc) — x2 — a. Table 9.1.5 shows the first five terms in an application of the mechanic's rule to approximate \/2. Table 9.1.5 xn+1= x\= I
1
DECIMAL APPROXIMATION
(Starting value)
1.00000000000
*-!D+fH _ 1[3 2 L2
1.50000000000
2 ] _ 17 3/2 J 12
2
3
3
r 4
4
5
1 [ 577 2 L 408
5
6
l l " 665,857 21470,832
1.41666666667
_ 1 [ 17 , 2 ] _ 577 2 L 12 17/12 J 408 2 1 577/408 J
1.41421568627 665,857 470,832
2 1 665,857/470,832 J
1.41421356237 886,731,088,897 627,013,566,048
1.41421356237
It would take us too far afield to investigate the convergence of sequences defined recursively, but we will conclude this section with a useful technique that can sometimes be used to compute limits of such sequences. > Example 10 Assuming that the sequence in Table 9.1.5 converges, show that the limit is V2.
9.1 Sequences
605
Solution. Assume that xn —>• L, where L is to be determined. Since n + 1 —>• +00 as n -> +00, it is also true that xv+\ —>• L as n —» +00. Thus, if we take the limit of the expression
1 / 2 — r I *« H 2 V xn as n —>• +00, we obtain
which can be rewritten as L 2 = 2. The negative solution of this equation is extraneous because xn > 0 for all n, so L = \/2. •<
•QUICK CHECK EXERCISES 9.1
(See page 607 for answers.)
1. Consider the sequence 4, 6, 8, 10, 12, ____ (a) If {an}^™i denotes this sequence, then a\ = _ , an = _ , and a~i = __ The general term is an = __ +™ 0 denotes this sequence, then bo = (b) If {bn }+™ The general term , andfcg= is bn =
2. What does it mean to say that a sequence {an} converges? 3. Consider sequences [an} and \bn}, where an —> 2 as n -*• +00 and bn = (—1)". Determine which of the following se-
EXERCISE SET 9.1
quences converge and which diverge. If a sequence converges, indicate its limit. (a) {bn} (b) (So, - 1} (c) {b\\
4. Suppose that {an}, {bn}, and {<:„} are sequences such that an < bn < cn for all n > 10, and that {an} and {cn} both converge to 12. Then the _ Theorem for Sequences implies that {£>„} converges to __
0 Graphing Utility
1. In each part, find a formula for the general term of the sequence, starting with n = 1.
1
(a) 1,1, 1,1,. (b)l,-lj, 3 9 27 3 9 27 1 16 1 3 5 (d) 2' 4' 6' 2. In each part, find two formulas for the general term of the sequence, one starting with n = 1 and the other with n = 0. (a) 1, -r, r2, -r3,... (b) r, -r2, r3, -r4,... 3. (a) Write out the first four terms of the sequence {1 + (-1)"}, starting with « = 0. (b) Write out the first four terms of the sequence {cos nn], starting with n = 0. (c) Use the results in parts (a) and (b) to express the general term of the sequence 4, 0, 4, 0, ... in two different ways, starting with n = 0. 4. In each part, find a formula for the general term using factorials and starting with n = 1. (a) 1 • 2, 1 • 2 • 3 • 4, 1 • 2 • 3 • 4 • 5 • 6, 1-2-3-4-5-6-7-8,... (b) 1, 1 • 2 • 3, 1 • 2 • 3 • 4 • 5, 1 • 2 • 3 • 4 • 5 • 6 • 7 , . . .
5. (a) Does lim* _,.+,„ f ( x ) exist? Explain. (b) Evaluate a\, 0.2, 03, 04, and 05. (c) Does [an} converge? If so, find its limit. 6. (a) Evaluate b\, bj,, b^, b$, and b$. (b) Does {bn} converge? If so, find its limit. (c) Does {/(«)} converge? If so, find its limit. 7-22 Write out the first five terms of the sequence, determine whether the sequence converges, and if so find its limit. 9. {2}+;,
17. 5-6 Let / be the function f(x~) — cos ( — x\ and define sequences {«„) and [bn] by an = f(2n) and bn — f(2n + 1).
9.1 Sequences
605
Solution. Assume that xn —>• L, where L is to be determined. Since n + 1 -> +<*> as n -> +00, it is also true that xv+\ ->• L as n ->• +00. Thus, if we take the limit of the expression
1 *"+1 ~ 2 as «—»• +00, we obtain L
=2
£+ -
which can be rewritten as L 2 = 2. The negative solution of this equation is extraneous because xn > 0 for all n, so L = \/2. -4
•QDICK CHECK EXERCISES 9.1
(See page 607 for answers.)
1. Consider the sequence 4, 6, 8, 10, 12, (a) If {an}^Ti denotes this sequence, then a\ = , ci4 = , and a-i = The general term is a, =. (b) If [bn}^"0 denotes this sequence, then bo = , Z>4 = , and b$ = The general term is bn = 2. What does it mean to say that a sequence {an} converges? 3. Consider sequences {an} and [bn}, where an —>• 2 as n -> +00 and bn = (—1)". Determine which of the following se-
EXERCISE SET 9.1
4. Suppose that {«„}, {&„}, and {cn} are sequences such that o-n < bn < cn for all n > 10, and that {an} and [cn] both converge to 12. Then the Theorem for Sequences implies that {bn} converges to
Graphing Utility
1. In each part, find a formula for the general term of the sequence, starting with n = 1. (a) 1,1, 1,1. 3 9 27
quences converge and which diverge. If a sequence converges, indicate its limit. (a) {*,„} (b) [3an - 1} (c) 02}
(b)l,-l,l, 3 9 I
1 27
16 /7T
2. In each part, find two formulas for the general term of the sequence, one starting with n = 1 and the other with n = 0. (a) 1, -r, r 2 , -r3, . . . (b) r, -r2, r 3 , -r 4 ,... 3. (a) Write out the first four terms of the sequence {1 + (-1)"}, starting with n = 0. (b) Write out the first four terms of the sequence {cos nn}, starting with n = 0. (c) Use the results in parts (a) and (b) to express the general term of the sequence 4, 0, 4, 0, ... in two different ways, starting with « = 0. 4. In each part, find a formula for the general term using factorials and starting with n = 1. (a) 1 • 2, 1 • 2 • 3 • 4, 1 • 2 • 3 • 4 • 5 • 6, 1-2-3-4-5-6-7-8,... (b) 1, 1 • 2 • 3, 1 • 2 • 3 • 4 • 5, 1 • 2 • 3 • 4 • 5 • 6 • 7 , . . .
5. (a) Does linij:^ + oo/U) exist? Explain. (b) Evaluate a\, 02, 03, 04, and a$. (c) Does {an} converge? If so, find its limit. 6. (a) Evaluate b\,b2,bj,,b4, and b$. (b) Does {bn} converge? If so, find its limit. (c) Does {/(«)} converge? If so, find its limit. 7-22 Write out the first five terms of the sequence, determine whether the sequence converges, and if so find its limit. +00
8.
7.
n=l
12.
11.
14.
13.
tt
16. 1-
l2"Jn=l
(n + !)(„ + 2)
5-6 Let / be the function f ( x ) = cos l—x\ and define sequences [an] and {&„} by an = /(2n) and bn = f(2n + 1). :
9.
19. [n2
+00
n \n=\
606
Chapter 9 / Infinite Series
21. 23-30 Find the general term of the sequence, starting with n = 1, determine whether the sequence converges, and if so find its limit.
73
2 4 6 8 I I I 25. 3' 9' 27'
I . 8T"
7
(d) Determine whether the sequences in parts (a), (b), and (c) converge. For those that do, find the limit.
24. 0, —2 , 2 2" ' 2 3 4 26. -1,2, _3 4, -5,...
I ( 2J ' I 3
1
40. For what positive values of b does the sequence b, 0, b2, 0, £>3, 0, b*, ... converge? Justify your answer.
l\ <:\) ' '"
2 J ' V 3 4 / ' \5 ^ 28 3 28 3 ' ' 2' * ' # " " 29. (V2-V3), (V3-V4), ( V 4 - V 5 ) , . . . 30 1 ' 35'
11 36' 37'
(c) Starting with « = 1, and considering the even and odd terms separately, find a formula for the general term of the sequence 1 1 1 1 1 1 1 1 ' 3' 3' 5' 5' 7' 7' 9' 9""
41. Assuming that the sequence given in Formula (2) of this section converges, use the method of Example 10 to show that the limit of this sequence is ^/a. 42. Consider the sequence
a\ = V6
18 3 "
31-34 True-False Determine whether the statement is true or false. Explain your answer. 31. Sequences are functions. 32. If {an} and {bn} are sequences such that {«„ + bn} converges, then {an} and {bn} converge. 33. If {an} diverges, then an —>• +00 or an —>• —oo. 34. If the graph of y = f ( x ) has a horizontal asymptote as x —> +00, then the sequence {/(«)} converges.
(a) Find a recursion formula for an+\. (b) Assuming that the sequence converges, use the method of Example 10 to find the limit.
35-36 Use numerical evidence to make a conjecture about the limit of the sequence, and then use the Squeezing Theorem for Sequences (Theorem 9.1.5) to confirm that your conjecture is correct. sin 2 n 35. lim 36. lim
43. (a) A bored student enters the number 0.5 in a calculator display and then repeatedly computes the square of the number in the display. Taking OQ = 0.5, find a formula for the general term of the sequence {an} of numbers that appear in the display. (b) Try this with a calculator and make a conjecture about the limit of an. (c) Confirm your conjecture by finding the limit of an. (d) For what values of GO will this procedure produce a convergent sequence?
H — » +CO
FOCUS ON CONCEPTS
37. Give two examples of sequences, all of whose terms are between —10 and 10, that do not converge. Use graphs of your sequences to explain their properties. s
38. (a) Suppose that / satisfies lim^^o+ /OO = +°°- I it possible that the sequence {/(!/«)} converges? Explain. (b) Find a function / such that \imx^o+ /OO does not exist but the sequence {/(!/«)} converges. 39. (a) Starting with n = 1, write out the first six terms of the sequence {an}, where
an =
1, n,
if n is odd if n is even
(b) Starting with n = 1, and considering the even and odd terms separately, find a formula for the general term of the sequence
1 1 31 5 16 ' 22' 24 2 '"
_ I2x, 0 < x < 0.5 " \2x -I, 0.5 < x < 1 Does the sequence /(0.2), /(/(0.2)), /(/(/(0.2))), . . . converge? Justify your reasoning.
44. Let
45. (a) Use a graphing utility to generate the graph of the equation y = (2X + y)llx, and then use the graph to make a conjecture about the limit of the sequence (b) Confirm your conjecture by calculating the limit. 46. Consider the sequence {an }^j whose nth term is !_ A a
n
1_
" ~ f^ 1 + ( fc /«) Show that lim,, _,. +00 an = In 2 by interpreting an as the Riemann sum of a definite integral.
9.2 Monotone Sequences 47. The sequence whose terms are 1, 1, 2, 3, 5, 8, 13, 21, ... is called the Fibonacci sequence in honor of the Italian mathematician Leonardo ("Fibonacci") da Pisa (c. 1170-1250). This sequence has the property that after starting with two 1's, each term is the sum of the preceding two. (a) Denoting the sequence by {an} and starting with a \ = 1 and ai = 1, show that = 1H
an+\
., if n > 1
(b) Give a reasonable informal argument to show that if the sequence {an+i lan} converges to some limit L, then the sequence {an+2/an+\} must also converge to L. (c) Assuming that the sequence {an+\ lan} converges, show that its limit is (1 + \/5 )/2. 48. If we accept the fact that the sequence {l/n}^ converges to the limit L = 0, then according to Definition 9.1.2, for every e > 0, there exists a positive integer N such that an — L\ = |(l/«) — 0| < 6 whenw > N. In each part, find the smallest possible value of N for the given value of e. (a) e = 0.5 (b) e = 0.1 (c) e = 0.001
607
49. If we accept the fact that the sequence
converges to the limit L = 1 , then according to Definition 9.1.2, for every e > 0 there exists an integer N such that n \an-L\ = -1 n +l when n > N. In each part, find the smallest value of N for the given value of e. (a) 6 = 0.25 (b) e = 0.1 (c) e = 0.001 50. Use Definition 9.1.2 to prove that (a) the sequence {l/n}^ converges to 0 t
(b) the sequence \
i ^~°°
\
converges to 1.
(n + l\n=l
51. Writing Discuss, with examples, various ways that a sequence could diverge. 52. Writing Discuss the convergence of the sequence {r"} considering the cases \r\ < 1, |r| > 1, r = 1, and r = —1 separately.
I/QUICK CHECK ANSWERS 9.1 1. (a) 4; 10; 16; 2w + 2 (b) 4; 12; 20; 2n + 4
2. lim an exists rt-»+oo
(d) diverges (e) converges to | (f) diverges
3. (a) diverges (b) converges to 5 (c) converges to 1
4. Squeezing; 12
MONOTONE SEQUENCES There are many situations in which it is important to know whether a sequence converges, but the value of the limit is not relevant to the problem at hand. In this section we will study several techniques that can be used to determine whether a sequence converges.
TERMINOLOGY
We begin with some terminology.
9.2.1
Note that an increasing sequence need not be strictly increasing, and a decreasing sequence need not be strictly decreasing.
DEFINITION A sequence {an}£?i is called strictly increasing if
a\ < ai < a 3 < • • • < an < • • •
increasing if
a\ < a2 < a^ < • • • < an < • • •
strictly decreasing if
a\ > a 2 > aj, > • • • > an > • • •
decreasing if
a\ > 02 > UT, > • • • > an > • • •
A sequence that is either increasing or decreasing is said to be monotone, and a sequence that is either strictly increasing or strictly decreasing is said to be strictly monotone.
Some examples are given in Table 9.2.1 and their corresponding graphs are shown in Figure 9.2.1. The first and second sequences in Table 9.2.1 are strictly monotone; the third
608
Chapter 9 / Infinite Series
and fourth sequences are monotone but not strictly monotone; and the fifth sequence is neither strictly monotone nor monotone. Table 9.2.1 SEQUENCE
DESCRIPTION
Strictly increasing
2' 3' 4" " n + 1 J_
ill
Strictly decreasing
' 2' 3 ' ' n' 1, 1, 2, 2, 3, 3 , . . . , , 1 1 1 1 1,1, 2- 2 ' r r " 1
1
Increasing; not strictly increasing Decreasing; not strictly decreasing
1
)«+i_L
Neither increasing nor decreasing
w
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2 2
4
6
8 10 12
2
4
6
2
8 10 12
4
6
8
10 12
_ n+ ljn=l
1
1 0.8
0.5
0.6 0.4
1
0.2
3 « 5 • 7 • 9 *11*
-0.5 2
4
6
8 10 12
Can a sequence be both increasing and decreasing? Explain.
n»=i A Figure 9.2.1 TESTING FOR MONOTONICITY
Frequently, one can guess whether a sequence is monotone or strictly monotone by writing out some of the initial terms. However, to be certain that the guess is correct, one must give a precise mathematical argument. Table 9.2.2 provides two ways of doing this, one based Table 9.2.2 DIFFERENCE BETWEEN
RATIO OF
SUCCESSIVE TERMS
SUCCESSIVE TERMS
an+\ -an>0 an+] -an<0
an+l/an an+l/an a n+l/an an+l/an
n+\ -an>0
a
««+! -an<0
>1 <1 2. 1 <1
CONCLUSION
Strictly increasing Strictly decreasing Increasing Decreasing
9.2 Monotone Sequences
609
on differences of successive terms and the other on ratios of successive terms. It is assumed in the latter case that the terms are positive. One must show that the specified conditions hold for all pairs of successive terms. > Example 1 Use differences of successive terms to show that
i
!_ 2 3 n 2' 3' 4'"" n + 1"" (Figure 9.2.2) is a strictly increasing sequence.
0.8 0.6 0.4 0.2 2
4
A Figure 9.2.2
6
8 10 12
Solution. The pattern of the initial terms suggests that the sequence is strictly increasing. To prove that this is so, let n We can obtain an+\ by replacing n by n + I in this formula. This yields
n+l (n + 1) + 1
n+l n +2
Thus, for n > 1
n2 + 2n + I - n2 - In n +2 n+l (n + l)(n + 2) which proves that the sequence is strictly increasing. •« an+i — an =
n+l
1 >0 ( n + !)(« +2)
> Example 2 Use ratios of successive terms to show that the sequence in Example 1 is strictly increasing. Solution. As shown in the solution of Example 1,
n n+l and an+] = n+l n +2 Forming the ratio of successive terms we obtain an =
a n+ i an
(n + 1)1 (n + 2) n/(n + 1)
n+l n n+2 n
n2 + 2n
(1)
from which we see that an+\ lan > 1 for n > 1. This proves that the sequence is strictly increasing. •* The following example illustrates still a third technique for determining whether a sequence is strictly monotone. > Example 3
In Examples 1 and 2 we proved that the sequence
n I - I 2' 3' 4'"" ^+~l"" is strictly increasing by considering the difference and ratio of successive terms. Alternatively, we can proceed as follows. Let x /(*) = \ -r i so that the nth term in the given sequence is an = f ( n ) . The function / is increasing for x > 1 since ^x _|_ l)(l) — x(l) l
f'(x) =
:
—
(x + I)2
=
— >0
(X + I)2
610
Chapter 9 / Infinite Series
Table 9.2.3
Thus,
CONCLUSION FOR DERIVATIVE OF THE SEQUENCE /FOR X > 1 WITH an = f(ri)
an = /(«) < /(« + 1) — a,,+i . . . ls strlctl
which proves that the given
.
y mcreasmg.
Strictly increasing In general, if f(ri) = an is the nth term of a sequence, and if / is differentiable for x > 1, Strictly decreasing then the results in Table 9.2.3 can be used to investigate the monotonicity of the sequence. Increasing Decreasing B PROPERTIES THAT HOLD EVENTUALLY Sometimes a sequence will behave erratically at first and then settle down into a definite pattern. For example, the sequence
f'(x) > 0 f'(x) < 0 f'(x) > 0 /'(*) < 0
9, -8, -17, 12, 1,2, 3,4,
(2)
is strictly increasing from the fifth term on, but the sequence as a whole cannot be classified as strictly increasing because of the erratic behavior of the first four terms. To describe such sequences, we introduce the following terminology.
9.2.2 DEFINITION If discarding finitely many terms from the beginning of a sequence produces a sequence with a certain property, then the original sequence is said to have that property eventually.
For example, although we cannot say that sequence (2) is strictly increasing, we can say that it is eventually strictly increasing. 10" Example 4 Show that the sequence \ - [
3000
is eventually strictly decreasing.
J«=i 2000
Solution. We have 10" an = —-
1000 i
5
A Figure 9.2.3
10
15
i
i
20
25
and an+\ —
SO
10"+1/(n +1)! 10"/n!
10"+1n! 10"(n+l)!
10 n\ («+!)«! n+1 From (3), an+\/an < 1 for all n > 10, so the sequence is eventually strictly decreasing, as confirmed by the graph in Figure 9.2.3. •+
an+l an
AN INTUITIVE VIEW OF CONVERGENCE
Informally stated, the convergence or divergence of a sequence does not depend on the behavior of its initial terms, but rather on how the terms behave eventually. For example, t h e sequence 1 1 1 3, -9, -13, 17, 1, -,, -,, - ,, . . . 2 3 4 eventually behaves like the sequence
and hence has a limit of 0. CONVERGENCE OF MONOTONE SEQUENCES
The following two theorems, whose proofs are discussed at the end of this section, show that a monotone sequence either converges or becomes infinite — divergence by oscillation cannot occur.
9.2 Monotone Sequences
611
9.2.3 THEOREM If a sequence {an} is eventually increasing, then there are two possibilities: (a)
There is a constant M, called an upper bound for the sequence, such that an < M for all n, in which case the sequence converges to a limit L satisfying L < M.
(b) No upper bound exists, in which case lim an = +00. n—> +co
9.2.4 THEOREM If a sequence {an} is eventually decreasing, then there are two possibilities: Theorems 9.2.3 and 9.2.4 are examples of existence theorems', they tell us whether a limit exists, but they do not provide a method for finding it.
(a)
There is a constant M, called a lower bound for the sequence, such that an > M for all n, in which case the sequence converges to a limit L satisfying L > M.
I (b) No lower bound exists, in which case lim an = — oo.
Example 5
10" Show that the sequence -
converges and find its limit.
Solution. We showed in Example 4 that the sequence is eventually strictly decreasing. Since all terms in the sequence are positive, it is bounded below by M = 0, and hence Theorem 9.2.4 guarantees that it converges to a nonnegative limit L. However, the limit is not evident directly from the formula 10"/n! for the nth term, so we will need some ingenuity to obtain it. It follows from Formula (3) of Example 4 that successive terms in the given sequence are related by the recursion formula **,-^U
(4,
« + 1
where an = 10"/«L We will take the limit as n -> +00 of both sides of (4) and use the fact that lim a \ = lim a = L n —> +00
n+
n —> +GC
n
We obtain 10
L = lim an+\ = lim n^+co
so that
-an
n^+ca \ n -\- I
L-
\ J
lim
n -* +OD
= hm
10
lim an = 0 • L = 0
n-^+<x n + \ n —> +co
10" fi I
= 0 -4
In the exercises we will show that the technique illustrated in the last example can be adapted to obtain lim — = 0 (5) n-»+oo n\
for any real value of x (Exercise 29). This result will be useful in our later work. THE COMPLETENESS AXIOM In this text we have accepted the familiar properties of real numbers without proof, and indeed, we have not even attempted to define the term real number. Although this is sufficient for many purposes, it was recognized by the late nineteenth century that the study of limits
612
Chapter 9 / Infinite Series
and functions in calculus requires a precise axiomatic formulation of the real numbers analogous to the axiomatic development of Euclidean geometry. Although we will not attempt to pursue this development, we will need to discuss one of the axioms about real numbers in order to prove Theorems 9.2.3 and 9.2.4. But first we will introduce some terminology. If S is a nonempty set of real numbers, then we call u an upper bound for S if u is greater than or equal to every number in S, and we call / a lower bound for S if / is smaller than or equal to every number in S. For example, if S is the set of numbers in the interval (1,3), then u = 10, 4, 3.2, and 3 are upper bounds for S and / = —10, 0, 0.5, and 1 are lower bounds for S. Observe also that u — 3 is the smallest of all upper bounds and / — 1 is the largest of all lower bounds. The existence of a smallest upper bound and a largest lower bound for S is not accidental; it is a consequence of the following axiom.
9.2.5 AXIOM (The Completeness Axiom) If a nonempty set S of real numbers has an upper bound, then it has a smallest upper bound (called the least upper bound), and if a nonempty set S of real numbers has a lower bound, then it has a largest lower bound (called the greatest lower bound).
PROOF OF THEOREM 9.2-3
(a) We will prove the result for increasing sequences, and leave it for the reader to adapt the argument to sequences that are eventually increasing. Assume there exists a number M such that an < M for n = 1, 2, Then M is an upper bound for the set of terms in the sequence. By the Completeness Axiom there is a least upper bound for the terms; call it L. Now let e be any positive number. Since L is the least upper bound for the terms, L - e is not an upper bound for the terms, which means that there is at least one term a# sucn that ON > TL — 6 Moreover, since {«„} is an increasing sequence, we must have an > aN > L - e
(6)
when n > N. But an cannot exceed L since L is an upper bound for the terms. This observation together with (6) tells us that L > an > L — e for n > N, so all terms from the Mh on are within e units of L. This is exactly the requirement to have lim an = L
n —> +GO
Finally, L < M since M is an upper bound for the terms and L is the least upper bound. This proves part (a). (b) If there is no number M such that an < M for n — 1, 2, . . . , then no matter how large we choose M, there is a term a^ such that ON > M
and, since the sequence is increasing, an > a/y > M
when n > N. Thus, the terms in the sequence become arbitrarily large as n increases. Thatis
'
limfln = +ooB
n—> +00
We omit the proof of Theorem 9.2.4 since it is similar to that of 9.2.3.
9.2 Monotone Sequences I/QUICK CHECK EXERCISES 9.2
(See page 614 for answers.)
1. Classify each sequence as (I) increasing, (D) decreasing, or (N) neither increasing nor decreasing. {2«} {2-"} [5-nl
613
f-1
[2n + (-1)")
{n + (-1)"} (3n + (-!)"} 3
-
Since
n/[2(n + l)] (n -
the sequence {(n — l)/(2n)} is strictly
(-1)"
4. Since 2. Classify each sequence as (M) monotonic, (S) strictly monotonic, or (N) not monotonic.
d — [ ( J C - 8 ) 2 ] >OfOTX
>
dx
the sequence {(« — 8)2} is.
. strictly
EXERCISE SET 9.2
1-6 Use the difference an+{ — an to show that the given sequence {«„} is strictly increasing or strictly decreasing. .
j.i-i «
4.
4n- 1
7-12 Use the ratio an+\/an to show that the given sequence [an] is strictly increasing or strictly decreasing. ; n
7.
8.
10"
l+2»Jn= +co
9.
-
n
,
5"
13-16 True-False Determine whether the statement is true or false. Explain your answer. •••'13. If an+\ — an > 0 for all n > 1, then the sequence {an} is strictly increasing. 14. A sequence {an} is monotone if an+\ — an /Oforalln > 1. 15. Any bounded sequence converges. 16. If [an] is eventually increasing, then aioo < ^20017-20 Use differentiation to show that the given sequence is strictly increasing or strictly decreasing. IS n=l
19. {tan
20.
21-24 Show that the given sequence is eventually strictly increasing or eventually strictly decreasing. •** 21. \2n2-ln}t-,
22. {n3 - 4«2}+",
24. n=\
FOCUS ON CONCEPTS
25. Suppose that {«„} is a monotone sequence such that 1 < o-n < 2 for all n. Must the sequence converge? If so, what can you say about the limit? 26. Suppose that {an\ is a monotone sequence such that an < 2 for all n. Must the sequence converge? If so, what can you say about the limit? 27. Let {an} be the sequence defined recursively by a\ = V2 and an+i = ^/2 + an for n > 1. (a) List the first three terms of the sequence. (b) Show that an < 2 for n > 1. (c) Show that a2+1 - a 2 = (2 - «„)(! + a n ) for « > 1. (d) Use the results in parts (b) and (c) to show that [an } is a strictly increasing sequence. [Hint: If x and y are positive real numbers such that x2 — y2 > 0, then it follows by factoring that x — y > 0.] (e) Show that {an\ converges and find its limit L. 28. Let {«„} be the sequence defined recursively by a\ = 1 and «„+! = \\an + (3/an)] f o r n > 1. (a) Show that an > \/3 for n > 2. [Hint: What is the minimum value of ~ [x + (3/x)] for x > 0?] (b) Show that {an} is eventually decreasing. [Hint: Examine an+i — an or an+\/an and use the result in part (a).] (c) Show that {an} converges and find its limit L. 29. The goal of this exercise is to establish Formula (5), namely,
lim
=0
Let an = \x\"/n\ and observe that the case where x = 0 is obvious, so we will focus on the case where x ^ 0. (a) Show that —/i^i
.
- —11
n+\ (b) Show that the sequence {an} is eventually strictly decreasing, (com.)
614
Chapter 9 / Infinite Series
(c) Show that the sequence [an] converges. 30. (a) Compare appropriate areas in the accompanying figure to deduce the following inequalities for n > 2: n
/
/-n+1
In x dx < ln«! < / Jt
Inxdx
2 3 ••• A Figure Ex-30
(b) Use the result in part (a) to show that
n—1
< n\ <
(n + 1)"+' e"
31. Use the left inequality in Exercise 30(b) to show that lim vn! = +co n >1
32. Writing Give an example of an increasing sequence that is not eventually strictly increasing. What can you conclude about the terms of any such sequence? Explain. 33. Writing Discuss the appropriate use of "eventually" for various properties of sequences. For example, which is a useful expression: "eventually bounded" or "eventually monotone"?
(c) Use the Squeezing Theorem for Sequences (Theorem 9.1.5) and the result in part (b) to show that /n! _ 1 n e
lim
UICK CHECK ANSWERS 9.2
1. I; D; N; I; N
2. N; M; S
3. 1; increasing
4. 8; eventually; increasing
INFINITE SERIES The purpose of this section is to discuss sums that contain infinitely many terms. The most familiar examples of such sums occur in the decimal representations of real numbers. For example, when we write | in the decimal form i — 0.3333 . . . , we mean - = 0.3 + 0.03 + 0.003 + 0.0003 + • • • which suggests that the decimal representation of | can be viewed as a sum of infinitely many real numbers. SUMS OF INFINITE SERIES
Our first objective is to define what is meant by the "sum" of infinitely many real numbers. We begin with some terminology.
93.1 DEFINITION An infinite series is an expression that can be written in the form * = «i +«2 + «3 H
\-uk H
k=\
The numbers MI, u^, u^,... are called the terms of the series.
Since it is impossible to add infinitely many numbers together directly, sums of infinite series are defined and computed by an indirect limiting process. To motivate the basic idea, consider the decimal „ oooo fi^
9.3 Infinite Series
615
This can be viewed as the infinite series 0.3 + 0.03 + 0.003 + 0.0003 + • • or, equivalently, 3
+
3 2 +
3 3+
To To " To
3 44
(2)
To
Since (1) is the decimal expansion of |, any reasonable definition for the sum of an infinite series should yield | for the sum of (2). To obtain such a definition, consider the following sequence of (finite) sums: ^1=0-3
3
_i
3
3
_i
3
><2
o
"To 3 "
3
3
3
T o ^~Tr? H"To 3
+
To4
= 0.3333
The sequence of numbers s\, s2, s j , 54,... (Figure 9.3.1) can be viewed as a succession of approximations to the "sum" of the infinite series, which we want to be |. As we progress through the sequence, more and more terms of the infinite series are used, and the approximations get better and better, suggesting that the desired sum of | might be the limit of this sequence of approximations. To see that this is so, we must calculate the limit of the general term in the sequence of approximations, namely,
0.4
1/3 0.3
3
To2"
o
(3)
10"
The problem of calculating 1
2
3
4
1 0.3,0.33,0.333,... A Figure 9.3.1
lim sn = lim
. n —> +co
ll
n ^ +cc
I
\
f r\
1
'
3
To"
-i f^t^l
is complicated by the fact that both the last term and the number of terms in the sum change with n. It is best to rewrite such limits in a closed form in which the number of terms does not vary, if possible. (See the discussion of closed form and open form following Example 2 in Section 4.4.) To do this, we multiply both sides of (3) by -^ to obtain
1
3
(4)
TCP and then subtract (4) from (3) to obtain I
" ~ To5" ~ To ~
s
i d ^ i o l 1 io» s
"=
"
Since 1/10" -> 0 as n ->• +00, it follows that lim sn — lim - I 1 ^+o=
n^+x, 3 V
10"
616
Chapter 9 / Infinite Series
which we denote by writing I A 3 ~ TO 10" Motivated by the preceding example, we are now ready to define the general concept of the "sum" of an infinite series MI +u2
\-uk H
We begin with some terminology: Let sn denote the sum of the initial terms of the series, up to and including the term with index n. Thus, = M I + «2
= U\ +U2 + M3 + • •
k=l
The number 5n is called the nf A partial sum of the series and the sequence {sn }^, is called the sequence of partial sums. As n increases, the partial sum sn = u\ + U2 + • • • + «„ includes more and more terms of the series. Thus, if sn tends toward a limit as n —»• +00, it is reasonable to view this limit as the sum of all the terms in the series. This suggests the following definition.
9.3.2 WARNING In everyday language the words "sequence" and "series" are often used interchangeably. However, in mathematics there is a distinction between these two words—a sequence is a success/on whereas a series is a sum. It is essential that you keep this distinction in mind.
DEFINITION Let [sn} be the sequence of partial sums of the series MI +u2 + M 3 H
\-uk H
If the sequence {sn} converges to a limit S, then the series is said to converge to S, and S is called the sum of the series. We denote this by writing
If the sequence of partial sums diverges, then the series is said to diverge. A divergent series has no sum.
> Example 1 Determine whether the series 1- 1 + 1-1 + 1- ! + •••
converges or diverges. If it converges, find the sum. Solution, It is tempting to conclude that the sum of the series is zero by arguing that the positive and negative terms cancel one another. However, this is not correct; the problem is that algebraic operations that hold for finite sums do not carry over to infinite series in all cases. Later, we will discuss conditions under which familiar algebraic operations can be applied to infinite series, but for this example we turn directly to Definition 9.3.2. The partial sums are 1 51 = 1
52 = 1 - 1 = 0 53 = 1 - 1 + 1 = 1
54 = 1 - 1 + 1 - 1 = 0
9.3 Infinite Series 617
and so forth. Thus, the sequence of partial sums is 1.5
1,0,1,0,1,0,...
1
(Figure 9.3.2). Since this is a divergent sequence, the given series diverges and consequently has no sum. -^
0.5
4
6
8
10
GEOMETRIC SERIES
A Figure 9.3.2
In many important series, each term is obtained by multiplying the preceding term by some fixed constant. Thus, if the initial term of the series is a and each term is obtained by multiplying the preceding term by r, then the series has the form ark = a + ar + ar2 + ar3 -\
h ark +
(5)
k=0
Such series are called geometric series, and the number r is called the ratio for the series. Here are some examples:
_L
J_
lQ3 + ' '+ TO*
.. . + 1 + .
1 + x + x2 + x3 H ----- h xk H
The following theorem is the fundamental result on convergence of geometric series. Sometimes it is desirable to start the index of summation of an infinite series at k = 0 rather than k = 1, in which case we would call «o the zeroth term
9.3.3
THEOREM A geometric series
and sg = «o the zeroth partial sum.
ark - a + ar + ar2 H ----- h ark +
One can prove that changing the starting value for the index of summation of an infinite series has no effect on the convergence, the divergence, or the sum.
4=0
converges if\r\ < 1 and diverges if\r\ > 1. If the series converges, then the sum is
If we had started the index at 00
it = 1 in (5), then the series would be
E
expressed as
arkk =
k=0
Since this expression is more compli-
PROOF
a
1 -r
Let us treat the case \r \ = 1 first. If r = 1, then the series is
cated than (5), we started the index at
a +a +a + a-\
so the nth partial sum is sn = (n + l)a and
lim sn = lim (n + l)a = ±co
n —> +co
n —> +co
(the sign depending on whether a is positive or negative). This proves divergence. If r = — I , the series is a —a + a — a + •• • so the sequence of partial sums is a,0,a,0,a,0,... which diverges.
618 Chapter 9 / Infinite Series Now let us consider the case where \r\ ^ 1. The nth partial sum of the series is sn = a + ar + ar2 +
h ar"
(6)
Multiplying both sides of (6) by r yields rsn — ar + ar2 H
1- ar" + arn+l
(7)
and subtracting (7) from (6) gives sn — rsn = a — arn+}
or (1 - r)sn =a- arn+l
(8)
Since r ^ 1 in the case we are considering, this can be rewritten as Sn =
a — ar n 1 -r
(9)
1 -r
If \r\ < 1, then rn+ goes to 0 as n —>• +00 (can you see why?), so {sn} converges. From (9) Note that (6) is an open form for sn, while (9) is a closed form for sn. In general, one needs a closed form to calculate the limit.
lim sn =
n —> +cc
If \r\ > 1 , then either r > 1 orr < — 1. In the case r > l,r" +1 increases without bound as n —> +00, and in the case r < — 1, r"+1 oscillates between positive and negative values that grow in magnitude, so {sn} diverges in both cases. • *• Example 2 sum.
A?
20/3 6
In each part, determine whether the series converges, and if so find its ro 5
So-
k=0
1 2 3 4 5 6 7 Partial sums for
""
E^
A Figure 9.3.3
i — r
k=l
Solution (a). This is a geometric series with a = 5 and r = \. Since r| = | < 1, the series converges and the sum is a 5 _ 20 1-r ~ 1 -I " T (Figure 9.3.3). Solution (b). This is a geometric series in concealed form, since we can rewrite it as t-i
Since r = | > 1, the series diverges. Example 3
Find the rational number represented by the repeating decimal 0.784784784..
TECHNOLOGY MASTERY Computer algebra systems have commands for finding sums of convergent series. If you have a CAS, use it to compute the sums in Examples 2 and 3
Solution. We can write 0.784784784... = 0.784 + 0.000784 + 0.000000784 + so me
given decimal is the sum of a geometric series with a = 0.784 and r = 0.001. Thus, 0.784784784
l-r
0.784 0.784 1 - 0.001 0.999
784 999
9.3 Infinite Series
619
>• Example 4 In each part, find all values of x for which the series converges, and find the sum of the series for those values of x. (a) y^ xk
^-^
(b) 3
k=0
Solution (a).
1
2
4
1
8
1
; —x
k
2
H
The expanded form of the series is
The series is a geometric series with a = 1 and r = x, so it converges if |x \ < 1 and diverges otherwise. When the series converges its sum is — x
k=0
Solution (b). This is a geometric series with a — 3 and r — —x/2. It converges if | —x/2\ < 1, or equivalently, when x\ < 2. When the series converges its sum is -^
/
X\*
^(-2)
=
3
7T7IfT
2+x
TELESCOPING SUMS
> Example 5
Determine whether the series A 1 £-;'fc(jfc + 1) X
1 1 •2
_
I
1 2-3
_
_
I
1 3-4
_
_
I
1 4-5 _
_
I
converges or diverges. If it converges, find the sum. Solution. The nth partial sum of the series is 1 k=l
1
k(k
1-2
1
1
2-3
1 r-
3-4
n(n + 1)
We will begin by rewriting sn in closed form. This can be accomplished by using the method of partial fractions to obtain (verify) 1 1 1 k(k+l) k k+l from which we obtain the sum 1 k=\ The sum in (10) is an example of a telescoping sum. The name is derived from the fact that in simplifying the sum, one term in each parenthetical expression cancels one term in the next parenthetical expression, until the entire sum collapses (like a folding telescope) into just two terms.
k+l
2
3/
\3
4
n
n+l
1
(10)
620
Chapter 9 / Infinite Series
Thus,
E^TT) = ^ s » = « 1 i%( 1 -^TT) = 1 HARMONIC SERIES
One of the most important of all diverging series is the harmonic series, 1 1
1
which arises in connection with the overtones produced by a vibrating musical string. It is not immediately evident that this series diverges. However, the divergence will become apparent when we examine the partial sums in detail. Because the terms in the series are all positive, the partial sums
6 (a)
8
10
i 3 2' form a strictly increasing sequence
12
I 1 2 3 '
I I I 2 3 4 ' "
Sl < S2 < ST, < • • • < Sn < • • •
(Figure 9.3.4a). Thus, by Theorem 9.2.3 we can prove divergence by demonstrating that there is no constant M that is greater than or equal to every partial sum. To this end, we will consider some selected partial sums, namely, $2, $4, s%, s^, 532, . . . . Note that the subscripts are successive powers of 2, so that these are the partial sums of the form sy> (Figure 9.3.4i>). These partial sums satisfy the inequalities
2" 20 2'
22
23 24
25 26 27
(*)
Partial sums for the harmonic series
J ____ L _L J_ 14 T 15 -f , 6
A Figure 9.3.4
135 X V L Sfmmt fniii afiian tuTnumu fr«flfen*l:*n, , 1 + A -f i-t-i + je.1 ,f,fml-. Id primus depnhendit Fritcr : invent! tumquc per prxccd. lumma fuiei i + i + T ^ + ^ + jlj.&c. vifiuwjwrA.tpUcaKi. ffict ex ifl» fcrie, i+ i+ Ji+£ + ± , &e. fi refolvemur thodo Prap. XIV. colkgit p-opoliliomi vertuton ex abfurdi , li fumma fenei hannonicK finiu llatuc-
52
n+1
"
If M is any constant, we can find a positive integer n such that (n + l)/2 > M. Butforthisn s-,, > "-^- > M +tH-,j+,V.««-:»E-A:»i r«,m. &e.cp &cjcitr,ft. i n c m CX^.ICfllmpiit., (ifumntlftnlu diet.
Courtesy Lilly Library, Indiana University This is a proof of the divergence of the harmonic series, as it appeared in an appendix of Jakob Bernoulli's posthumous publication, Ars Conjectandi, which appeared in 1713.
so that no constant M is greater than or equal to every partial sum of the harmonic series. This proves divergence. This divergence proof, which predates the discovery of calculus, is due to a French bishop and teacher, Nicole Oresme (1323-1382). This series eventually attracted the interest of Johann and Jakob Bernoulli (p. 700) and led them to begin thinking about the general concept of convergence, which was a new idea at that time.
9.3 Infinite Series I/QUICK CHECK EXERCISES 9.3
(See page 623 for answers.)
1. In mathematics, the terms "sequence" and "series" have different meanings: a is a succession whereas a
4. A geometric series is a series of the form CO
V z_, fc=0
is a aim
2. Consider the series 50
This series converges tn Hivp.rgss if
J
2^2* k=\
if
This series
5. The harmonic series has the form
If {«„ } is the sequence of partial sums for this series, then
V
.9/1 —
k=\
CO
and xn —
Does the harmonic series converge or diverge?
3. What does it mean to say that a series £] u^ converges!
EXERCISE SET 9.3
621
@ CAS
1-2 In each part, find exact values for the first four partial sums, find a closed form for the nth partial sum, and determine whether the series converges by calculating the limit of the nth partial sum. If the series converges, then state its sum.
2 2 2 5+ 52 H ' 5*-i ' 2 22 2*-' b 1 ( ) 47 + 47 + T + • • • + ~r 4 4 +••• 1 1 1 1 (c] 1 1 1 •i 1 1 ••• ' 2 - 3 3 - 4 4 - 5 ' (k+l)(k + 2) ' W 2+
15. Match a series from one of Exercise s 3, 5, 7, or 9 with the graph of its sequence of partial sums (a) |y Cb) ' y 8 1 -• • ' •. 0.6 . •• 4 ~ 0.4 2 i 2 (C)
i 4
i 6
i i 8 10
4>
- feW
^
1
1
°- '
\
^U + 3-* + 4 J
•
-
n
2
4
6
8 10
i
i
i
i
2
4
b
»
IU
- y
(d)
3
2-> rvi (a)vVM (b) 4 1 (M'Vf c) CM V4*-
ur\ 9 z
^n
02
0.2 -
,•
• •••' 0.!
-•
0.1 -.
3-14 Determine whether the series converges, and if so find its sum. a
3
=
/
*£(D3 +1 6 f%(>) r V
!
fc
2 fe
1
3
+ )( + )
9 V
•£-9t» + 3t-2 [
fe^-2
11 V
=° Ak+2
n V / > 'ijt— i k=l
4
6
8
10
n fc
* /9\*+2
o\*-l
-£H) 7 s £f(( i)*-} ^
T•fY(V
n 2
s"ft^^r * Mv 2
in V
2t+
1 2
'fe^ -l
12 r VTf" 1
-^u =° 3
1
t4 ^y^j *^ -* k=\
16. Match a series from one of Exercises 4, 6, 8, or 10 with the graph of its sequence of partial sums. (a) *y (b) f > 60 0.8 -
•
30
0.6 -
•
AA - .* * 0.4
* * *
1 "3 *5 » 7
0.2 i 2
i 4
i 6
i i 8 10
B
•
-60
(d)
1 0.8 0.6 -
>
-30
^
(c) *y
n '
9
.y 0.6
. °'4
•'
• •
0
0.2-/ 2
1 4
6
8 10
«
2
4
6
8 10
622
Chapter 9 / Infinite Series
17-20 True-False Determine whether the statement is true or false. Explain your answer, -i 17. An infinite series converges if its sequence of terms converges. 18. The geometric series a + ar + ar2 + • • • + arn + • • • converges provided \r\ < 1. 19. The harmonic series diverges.
In -
(a) V(-1)V =
1 21-24 Express the repeating decimal as a fraction. 21. 0.9999..
22. 0.4444..
23. 5.373737..
24. 0.451141414..
25. Recall that a terminating decimal is a decimal whose digits are all 0 from some point on (0.5 = 0.50000..., for example). Show that a decimal of the form O.ai 02 • • -an9999..., where a,, / 9, can be expressed as a terminating decimal.
1
(*+i:
30. Use geometric series to show that CO
20. An infinite series converges if its sequence of partial sums is bounded and monotone.
1-
4-,
k=0
1
if
-1 < x < 1
if
2 < x <4
if
-1<JC<1.
31. In each part, find all values of x for which the series converges, and find the sum of the series for those values of x . (a) x — y? + x5 x1 + x9 — • 1 2 4 8 16 -6 X°
FOCUS ON CONCEPTS
(c)
26. The great Swiss mathematician Leonhard Euler (biography on p. 3) sometimes reached incorrect conclusions in his pioneering work on infinite series. For example, Euler deduced that 1=1-1+1-1 and
by substituting x = — 1 and x = 2 in the formula , = 1 + x + x2 + 1 -x What was the problem with his reasoning? 27. A ball is dropped from a height of 10 m. Each time it strikes the ground it bounces vertically to a height that is | of the preceding height. Find the total distance the ball will travel if it is assumed to bounce infinitely often. 28. The accompanying figure shows an "infinite staircase" constructed from cubes. Find the total volume of the staircase, given that the largest cube has a side of length 1 and each successive cube has a side whose length is half that of the preceding cube.
32. Show that for all real values of x 1., 1 i 14 2 sin* sin x -- sin x -\— sin x -- sin x + • • • = 2 4 8 2 + sinx 33. Let a\ be any real number, and let {an} be the sequence defined recursively by an+i = ^(an + 1) Make a conjecture about the limit of the sequence, and confirm your conjecture by expressing an in terms of a\ and taking the limit. 34. Show: ^ k=\
35. Show: 3 4' 1-3 2-4 3-5 1 1 + 1 + 1 4 37. Show: 2' 38. In his Treatise on the Configurations of Qualities and Motions (written in the 1350s), the French Bishop of Lisieux, Nicole Oresme, used a geometric method to find the sum of the series
36. Show:
rn? 3^5 5^7
. . . •* Figure Ex-28
29. In each part, find a closed form for the nth partial sum of the series, and determine whether the series converges. If so, find its sum.
*=i
2*
1 2 3 4 2 + 4+ 8+ 16
In part (a) of the accompanying figure, each term in the series is represented by the area of a rectangle, and in
9.4 Convergence Tests
part (b) the configuration in part (a) has been divided into rectangles with areas A], AI, A 3 , . . . . Find the sum A, + A 2 + A 3 + - - - .
6Z3
Source.' This problem is based on "Trisection of an Angle in an Infinite Number of Steps" by Eric Kincannon, which appeared in The College Mathematics Journal, Vol. 21, No. 5, November 1990.
3/
Ui
1 A3I 1
A
2\
A, 1(a)
-
1
Initial side -4 Figure Ex-39
1
»
|c| 40. In each part, use a CAS to find the sum of the series if it converges, and then confirm the result by hand calculation.
(b) Not to scale
A Figure Ex-38
k=\
39. As shown in the accompanying figure, suppose that an angle 9 is bisected using a straightedge and compass to produce ray R\, then the angle between R\ and the initial side is bisected to produce ray ^2- Thereafter, rays ^3, R$, R$,... are constructed in succession by bisecting the angle between the preceding two rays. Show that the sequence of angles that these rays make with the initial side has a limit of 0/3.
4k2 - 1
41. Writing Discuss the similarities and differences between what it means for a sequence to converge and what it means for a series to converge. 42. Writing Read about Zeno's dichotomy paradox in an appropriate reference work and relate the paradox in a setting that is familiar to you. Discuss a connection between the paradox and geometric series.
I/QUICK CHECK ANSWERS 9.3 1. sequence; series
2. - ; - ; - ; — ; 1
3. The sequence of partial sums converges.
4. ark (a / 0); ——; r\<\;\r > 1 5. -; diverge 1 —r k
CONVERGENCE TESTS In the last section we showed how to find the sum of a series by finding a closed form for the nth partial sum and taking its limit. However, it is relatively rare that one can find a closed form for the nth partial sum of a series, so alternative methods are needed for finding the sum of a series. One possibility is to prove that the series converges, and then to approximate the sum by a partial sum with sufficiently many terms to achieve the desired degree of accuracy. In this section we will develop various tests that can be used to determine whether a given series converges or diverges.
THE DIVERGENCE TEST
In stating general results about convergence or divergence of series, it is convenient to use the notation ^ u^ as a generic notation for a series, thus avoiding the issue of whether the sum begins with k — 0 or k — 1 or some other value. Indeed, we will see shortly that the starting index value is irrelevant to the issue of convergence. The Ath term in an infinite series J^ MJ. is called the general term of the series. The following theorem establishes
624
Chapter 9 / Infinite Series a relationship between the limit of the general term and the convergence properties of a series.
9.4.1
(a)
THEOREM (The Divergence Test)
If lim UK j^ 0, then the series ^ u^ diverges. k->+ca
(b)
If lim u/f = 0, then the series ^ u^ may either converge or diverge.
PROOF (a) To prove this result, it suffices to show that if the series converges, then lim^+co Uk — 0 (why?). We will prove this alternative form of (a). Let us assume that the series converges. The general term u^ can be written as
where Sk is the sum of the terms through u^ and Sk-i is the sum of the terms through u^-\ • If S denotes the sum of the series, then lim^ _,. +oc Sk = S, and since (k — 1) —>• +00 as k —> +00, we also have lim^+co Sk-i — S. Thus, from (1)
lim Uk = lim (:
PROOF (b) To prove this result, it suffices to produce both a convergent series and a divergent series for which lim,t^+00 u^ — 0. The following series both have this property:
1 1 1 - + -i + - - - + -I + - - Z*
£
1 1 1 and 1 + - + - + ... + - + ...
A.
2.
3
K.
The first is a convergent geometric series and the second is the divergent harmonic series.
The alternative form of part (a) given in the preceding proof is sufficiently important that we state it separately for future reference.
WARNING The converse of Theorem 9.4.2 is false; i.e., showing that
lim ui =0 does not prove that ]P u^ converges, since this property may hold for divergent as well as convergent series. This is illustrated in the proof of part (b) of Theorem 9.4.1.
9.4.2
THEOREM
Example 1
If the series ^ MJ- converges, then lim u^ — 0.
The series
E — =2 ^k+l diverges since
lim
fe
2
= lim
3
1
;- = 1 /= 0 •«
ALGEBRAIC PROPERTIES OF INFINITE SERIES
For brevity, the proof of the following result is omitted.
9.4 Convergence Tests 9.4.3 See Exercises 27 and 28 for an exploration of what happens when £) u/f or J^ m diverge.
625
THEOREM
(a) If^Uk and £] vk are convergent series, then XXM* + vk) ana convergent series and the sums of these series are related by
~~ vk)
are
£< k=l
(b) Ifc is a nonzero constant, then the series ^2 uk ana 5Z cuk both converge or both diverge. In the case of convergence, the sums are related by
k=\
k=\
(c)
Convergence or divergence is unaffected by deleting a finite number of terms from a series; in particular, for any positive integer K, the series
WARNING Do not read too much into part (c) of Theorem 9.4.3. Although convergence is not affected when finitely many terms are deleted from the beginning of a convergent series, the sum of the series is changed by the removal of those terms.
= M! + M 2 + M3
k=\
uk = UK
uK+2
k=K
both converge or both diverge.
Example 2
Find the sum of the series 3
Solution.
2
The series 3 4*
_3 i 3 I 3 | 4
42
43
is a convergent geometric series (a = |, r — |), and the series
A
2
2
2
2
5*-! is also a convergent geometric series (a = 2, r = ^). Thus, from Theorems 9.4.3(a) and 9.3.3 the given series converges and L-, Ak k=\
L-i k=\
626
Chapter 9 / Infinite Series
Example 3
Determine whether the following series converge or diverge.
+
f
k=\0
Solution. The first series is a constant times the divergent harmonic series, and hence diverges by part (b) of Theorem 9.4.3. The second series results by deleting the first nine terms from the divergent harmonic series, and hence diverges by part (c) of Theorem 9.4.3.
THE INTEGRAL TEST
The expressions T^T \
•\l
^k y
1 2 3 4
••• (a)
«+ 1
=f(x)
and
r
are related in that the integrand in the improper integral results when the index k in the general term of the series is replaced by x and the limits of summation in the series are replaced by the corresponding limits of integration. The following theorem shows that there is a relationship between the convergence of the series and the integral.
9.4.4 THEOREM (The Integral Test) Let J] u^ be a series with positive terms. If f is a function that is decreasing and continuous on an interval [a, +00) and such that u k = f(k)forallk > a, then and
f ( x ) dx
both converge or both diverge.
The proof of the integral test is deferred to the end of this section. However, the gist of the proof is captured in Figure 9.4.1: if the integral diverges, then so does the series (Figure 9.4. la), and if the integral converges, then so does the series (Figure 9.4. Ib). > Example 4 Show that the integral test applies, and use the integral test to determine whether the following series converge or diverge. 1
Solution (a). We already know that this is the divergent harmonic series, so the integral test will simply illustrate another way of establishing the divergence. Note first that the series has positive terms, so the integral test is applicable. If we replace k by x in the general term l/k, we obtain the function f ( x ) = 1/x, which is decreasing and continuous for x > 1 (as required to apply the integral test with a — 1). Since /•H-oo J
/
Jl
rb J
- dx = lim
X
/
b^+^J]
- dx — lim [In b — In 1] = +00
x
b^+cc
the integral diverges and consequently so does the series.
9.4 Convergence Tests WARNING In part (b) of Example 4, do not erroneously conclude that the sum of the series is 1 because the value of the corresponding integral is 1. You can see that this is not so since the sum of the first two terms alone exceeds 1. Later, we will see that the sum of the series is actually jr/6.
627
Solution (b). Note first that the series has positive terms, so the integral test is applicable. If we replace k by x in the general term \/k2, we obtain the function f(x) = \/x2, which is decreasing and continuous for x > 1. Since
/
+™ ±dx= i x2
b f dx lim / ^ =
Z>-*+°°,/i x2
r 11*
11
r
lim -- = lim l-± = b^+^l j c j , />^+» |_ b\
the integral converges and consequently the series converges by the integral test with a = 1 .
p-SERIES
The series in Example 4 are special cases of a class of series called p-series or hyperharmonic series. A p-series is an infinite series of the form
A i 1 1 i V — = i + — + — + ••• + — + ••• *—' kP 2P 3P kP where p > 0. Examples of p-series are
ti-+H+-
1 +
k=1
k
k=\ J_
V— = Vk
V2
J_ +
p=,
+ —- + 4k
V1
The following theorem tells when a p-series converges.
9.4.5
THEOREM (Convergence ofp-Series) A 1 \
_
_1
- P
1
_
1
2P
1
1 ___,
_j_
.
._(_
_
1
.
IP
converges if p > 1 and diverges ifO
PROOF
To establish this result when p ^ 1, we will use the integral test.
/.+oo j
/ J]
/.fe
p
xi-p
— d x = lim / x~ dx = lim xP fc^+»Ji ft^+»l-pj,
-ib
r
— lim fe-*+oo|_l-P ]
p
fri-p
i
-i
1-pJ
Assume first that ^ > 1. Then 1 - p < 0, so b ~ -^Q as &^ +00. Thus, the integral converges [its value is —1/(1 — p)] and consequently the series also converges. Now assume that 0 < p < 1. It follows that 1 — p > 0 and b[~p -> +00 as b—> +00, so the integral and the series diverge. The case p — 1 is the harmonic series, which was previously shown to diverge. •
Example 5
1+ diverges since it is a p-series with p = | < 1.
628
Chapter 9 / Infinite Series PROOF OF THE INTEGRAL TEST
Before we can prove the integral test, we need a basic result about convergence of series with nonnegative terms. If MI + HI + UT, + • • • + u^ + • • • is such a series, then its sequence of partial sums is increasing, that is,
Thus, from Theorem 9.2.3 the sequence of partial sums converges to a limit S if and only if it has some upper bound M, in which case S < M. If no upper bound exists, then the sequence of partial sums diverges. Since convergence of the sequence of partial sums corresponds to convergence of the series, we have the following theorem.
9.4.6 THEOREM If Yl uk is a series with nonnegative terms, and if there is a constant M such that sn = u i + M 2 + h un < M for every n, then the series converges and the sum S satisfies S < M. If no such M exists, then the series diverges.
In words, this theorem implies that a series with nonnegative terms converges if and only if its sequence of partial sums is bounded above. PROOF OF THEOREM 9.4.4 We need only show that the series converges when the integral converges and that the series diverges when the integral diverges. For simplicity, we will limit the proof to the case where a — I . Assume that f ( x ) satisfies the hypotheses of the theorem for x > 1 . Since
/(I) = m,/(2) = «2, ...,/(«) = « „ , . . . the values of u\, U2, . . . ,un, • • • can be interpreted as the areas of the rectangles shown in Figure 9.4.2. The following inequalities result by comparing the areas under the curve y = f ( x ) to the areas of the rectangles in Figure 9.4.2 for n > 1 : «
y =/(*)
Sn — MI = M 2 + M 3 H
1 2 3 4
\-un=sn
U\
/ +
un <
r
f(x)dx
n+l
These inequalities can be combined as
(a)
«+!
/
pn
f ( x ) dx < sn < M! + / f ( x ) dx
(2)
If the integral f} °° f ( x ) dx converges to a finite value L, then from the right-hand inequality in (2) sn < MI + / f ( x ) d x < MI + / f ( x ) d x — MI + L /.+00
y=f(x)
/" Ji
Thus, each partial sum is less than the finite constant u\ + L, and the series converges by Theorem 9.4.6. On the other hand, if the integral /j+c° f ( x ) dx diverges, then A Figure 9.4.2
lim / ^+=°J\ so that from the left-hand inequality in (2), sn -> +00 as n -> +00. This implies that the series also diverges. • n
9.4 Convergence Tests
•QUICK CHECK EXERCISES 9.4
(See page 631 for answers.)
1. The divergence test says that if £ uk diverges. 2. Given that a\ — 3,
0, then the series
= 1,
629
and
= +00, the 3. Since /f test applied to . shows that this series. the series 4. A p-series is a series of the form
=5
Ek=l
it follows that
_. This series diverges if
This series converges if.
bk) =
and k=l
EXERCISE SET 9.4
Graphing Utility
CAS
1. Use Theorem 9.4.3 to find the sum of each series.
8-W
9-24 Determine whether the series converges. 2. Use Theorem 9.4.3 to find the sum of each series.
10. -r^
14
13.
"
3-4 For each given p-series, identify p and determine whether the series converges. ,-2/3
-
5-6 Apply the divergence test and state what it tells you about the series. s>
'
/F + i
22. EfcV*3 23. E7*~'-°' k=S
(c)
20 CO
k=l
k=l
i—Ic2
24. ^sed\2k k=l
25-26 Use the integral test to investigate the relationship between the value of p and the convergence of the series.
cos kn k=l
26. V
1
(b) k=\
7-8 Confirm that the integral test is applicable and use it to determine whether the series converges.
FOCUS ON CONCEPTS
27. Suppose that the series ~^uk converges and the series £] Dj: diverges. Show that the series ^(HI + vk) and ^(uk — vk) both diverge. [Hint: Assume that ^(Mfc + Vk) converges and use Theorem 9.4.3 to obtain a contradiction.]
630 Chapter 9 / Infinite Series 28. Find examples to show that if the series ]P u^ and J2 VK both diverge, then the series XX"* + vt) and £XMfr — Vk) may either converge or diverge.
;
29-30 Use the results of Exercises 27 and 28, if needed, to determine whether each series converges or diverges.
*:~1
n n+ 1
k=2
k=2
31-34 True-False Determine whether the statement is true or false. Explain your answer. 31. If X! uk converges to L, then ^(l/w^) converges to \/L. 32. If Y^ cuit diverges for some constant c, then J2 uk must diverge. 33. The integral test can be used to prove that a series diverges.
1
n n+ }
37. (a) It was stated in Exercise 35 that "
34. The series T^ —r is a o-series. I—I n*
_1__7T2
^ k2 ~ 6 k=\
35. Use a CAS to confirm that
A 1
-4 Figure Ex-36
Show that if sn is the «th partial sum of this series, then
n2
JT4
1
1
n +1 and then use these results in each part to find the sum of the series.
k4 36-40 Exercise 36 will show how a partial sum can be used to obtain upper and lower bounds on the sum of a series when the hypotheses of the integral test are satisfied. This result will be needed in Exercises 37^-0. 36. (a) Let ^°=1 w* be a convergent series with positive terms, and let / be a function that is decreasing and continuous on [n, +00) and such that HI = f(k) for k > n. Use an area argument and the accompanying figure to show that f(x)dx< Jn+\
L uk< I( k=n+l
/Jn
(b) Calculate ST, exactly, and then use the result in part (a) to show that 29 jr2 61_ < 18 T < 36 (c) Use a calculating utility to confirm that the inequalities in part (b) are correct. (d) Find upper and lower bounds on the error that results if the sum of the series is approximated by the 1 Oth partial sum. 38. In each part, find upper and lower bounds on the error that results if the sum of the series is approximated by the 1 Oth partial sum. '£
'-^H
39. It was stated in Exercise 35 that
— 90
f(x)dx
(b) Show that if S is the sum of the series Y^k=\ uk is the «th partial sum, then
an
d sn
(a) Let sn be the nth partial sum of the series above. Show that
" + 3(n + l)3
+00
/
<
J_
90 < S" + 3^
/•+«>
f ( x ) d x <S <sn+ +1
7T4
1 s
f(x)dx Jn
(b) We can use a partial sum of the series to approximate 7r4/90 to three decimal-place accuracy by capturing the
9.5 The Comparison, Ratio, and Root Tests sum of the series in an interval of length 0.001 (or less). Find the smallest value of n such that the interval containing jr4/90 in part (a) has a length of 0.001 or less. (c) Approximate ?r4/90 to three decimal places using the midpoint of an interval of width at most 0.001 that contains the sum of the series. Use a calculating utility to confirm that your answer is within 0.0005 of 7T4/90. 40. We showed in Section 9. 3 that the harmonic series XX=i l/^ diverges. Our objective in this problem is to demonstrate that although the partial sums of this series approach +00, they increase extremely slowly. (a) Use inequality (2) to show that for n > 2
(b) Use the inequalities in part (a) to find upper and lower bounds on the sum of the first million terms in the series.
631
(c) Show that the sum of the first billion terms in the series is less than 22. (d) Find a value of n so that the sum of the first n terms is greater than 100. 41. Use a graphing utility to confirm that the integral test applies to the series Y^=i k2e^k, and then determine whether the series converges. 42. (a) Show that the hypotheses of the integral test are satisfied by the series ELi l/(^ 3 + !)• (b) Use a CAS and the integral test to confirm that the series converges. (c) Construct a table of partial sums for n = 10, 20, 30, . . . , 100, showing at least six decimal places. (d) Based on your table, make a conjecture about the sum of the series to three decimal-place accuracy. (e) Use part (b) of Exercise 36 to check your conjecture.
UICK CHECK ANSWERS 9.4 1. lim Uk
2. —2; 7
3. integral; —=; diverges
fc-C+OO
fc
4. — ; p > l;0 < p < I kP
THE COMPARISON, RATIO, AND ROOT TESTS In this section we will develop some more basic convergence tests for series with nonnegative terms. Later, we will use some of these tests to study the convergence of Taylor series.
THE COMPARISON TEST We will begin with a test that is useful in its own right and is also the building block for other important convergence tests. The underlying idea of this test is to use the known convergence or divergence of a series to deduce the convergence or divergence of another series.
9.5.1 THEOREM (The Comparison Test) negative terms and suppose that It is not essential in Theorem 9.5.1 that the condition a^ < bk hold for all k, as stated; the conclusions of the theorem remain true if this condition is eventually true.
Let
and
be series with non-
\ (a) If the "bigger series" Tib^ converges, then the "smaller series" £<% also converges. (b) If the " smaller series" Saj; diverges, then the "bigger series" ££>£ also diverges.
We have left the proof of this theorem for the exercises; however, it is easy to visualize why the theorem is true by interpreting the terms in the series as areas of rectangles
632 Chapter 9 / Infinite Series (Figure 9.5.1). The comparison test states that if the total area ^ bk is finite, then the total area ^ ak must also be finite; and if the total area £] <% is infinite, then the total area ^ b^ must also be infinite. USING THE COMPARISON TEST
There are two steps required for using the comparison test to determine whether a series Y. uk with positive terms converges:
Step 1. Guess at whether the series ^ UK converges or diverges. For each rectangle, ak denotes the area of the blue portion and bk denotes the combined area of the white and blue portions. A Figure 9.5.1
Step 2. Find a series that proves the guess to be correct. That is, if we guess that ^ u% diverges, we must find a divergent series whose terms are "smaller" than the corresponding terms of J3 uk, and if we guess that ^ u^ converges, we must find a convergent series whose terms are "bigger" than the corresponding terms
In most cases, the series ^ u^ being considered will have its general term u^ expressed as a fraction. To help with the guessing process in the first step, we have formulated two principles that are based on the form of the denominator for M& . These principles sometimes suggest whether a series is likely to converge or diverge. We have called these "informal principles" because they are not intended as formal theorems. In fact, we will not guarantee that they always work. However, they work often enough to be useful.
9.5.2 INFORMAL PRINCIPLE Constant terms in the denominator of u c can usually be deleted without affecting the convergence or divergence of the series.
9.5.3 INFORMAL PRINCIPLE If a polynomial in k appears as a factor in the numerator or denominator of u^, all but the leading term in the polynomial can usually be discarded without affecting the convergence or divergence of the series.
Example 1 Use the comparison test to determine whether the following series converge or diverge.
£• k=\
i
1
2
2k + k
Solution (a). According to Principle 9.5.2, we should be able to drop the constant in the denominator without affecting the convergence or divergence. Thus, the given series is likely to behave like «, k=\
which is a divergent p-series (p — ^). Thus, we will guess that the given series diverges and try to prove this by finding a divergent series that is "smaller" than the given series. However, series (1) does the trick since >
=
for it = 1 , 2 , .
Thus, we have proved that the given series diverges.
9.5 The Comparison, Ratio, and Root Tests
633
Solution (b). According to Principle 9.5.3, we should be able to discard all but the leading term in the polynomial without affecting the convergence or divergence. Thus, the given series is likely to behave like x x
which converges since it is a constant times a convergent ^-series (p = 2). Thus, we will guess that the given series converges and try to prove this by finding a convergent series that is "bigger" than the given series. However, series (2) does the trick since
2k2 + k
W
for
*=1'2""
Thus, we have proved that the given series converges.
THE LIMIT COMPARISON TEST In the last example, Principles 9.5.2 and 9.5.3 provided the guess about convergence or divergence as well as the series needed to apply the comparison test. Unfortunately, it is not always so straightforward to find the series required for comparison, so we will now consider an alternative to the comparison test that is usually easier to apply. The proof is given in Appendix D.
9.5.4 THEOREM (The Limit Comparison Test) Let £} ak and J] bk be series with positive terms and suppose that a/c p = lim — k->+<»bk If p is finite and p > 0, then the series both converge or both diverge.
The cases where p = 0 or p = +00 are discussed in the exercises (Exercise 54). To use the limit comparison test we must again first guess at the convergence or divergence of J] ak and then find a series 53 bk that supports our guess. The following example illustrates this principle. > Example 2 Use the limit comparison test to determine whether the following series converge or diverge. (a)
E
-V+2
k=\
Solution (a). As in Example 1, Principle 9.5.2 suggests that the series is likely to behave like the divergent p-series (1). To prove that the given series diverges, we will apply the limit comparison test with
at = —=
and bk = —=
We obtain
ak
p = lim — = lim
=1
= lim i
Since p is finite and positive, it follows from Theorem 9.5.4 that the given series diverges.
634
Chapter 9 / Infinite Series
Solution (b). As in Example 1, Principle 9.5.3 suggests that the series is likely to behave like the convergent series (2). To prove that the given series converges, we will apply the limit comparison test with ak =
We obtain
1 2k + k
1
2
ak p — lim — = lim
2k2
— lim
=1 2+
k Since p is finite and positive, it follows from Theorem 9.5.4 that the given series converges, which agrees with the conclusion reached in Example 1 using the comparison test. Solution (c).
From Principle 9.5.3, the series is likely to behave like (3)
which converges since it is a constant times a convergent p-series. Thus, the given series is likely to converge. To prove this, we will apply the limit comparison test to series (3) and the given series. We obtain 3fc3 -2k2+ 4 =1 3 k* Since p is finite and nonzero, it follows from Theorem 9.5.4 that the given series converges, since (3) converges. -^
THE RATIO TEST
The comparison test and the limit comparison test hinge on first making a guess about convergence and then finding an appropriate series for comparison, both of which can be difficult tasks in cases where Principles 9.5.2 and 9.5.3 cannot be applied. In such cases the next test can often be used, since it works exclusively with the terms of the given series—it requires neither an initial guess about convergence nor the discovery of a series for comparison. Its proof is given in Appendix J.
9.5.5 that
THEOREM (The Ratio Test)
Let J] Uk be a series with positive terms and suppose ui+\ p = lim
(a) If p < 1, the series converges. (b) If p > 1 or p = +00, the series diverges. (c) If p — I , the series may converge or diverge, so that another test must be tried.
> Example 3 Each of the following series has positive terms, so the ratio test applies. In each part, use the ratio test to determine whether the following series converge or diverge. (2k)\ k=l
*=3
4k
(e)
1
9.5 The Comparison, Ratio, and Root Tests Solution (a).
The series converges, since
p = lim —i- = lim Solution (b).
= lim ——'-— = lim
= 0<1
The series converges, since ,. k+l 2k 1 k+ = hm —r-r • — = - lim —— k 2k^+™ k
p = hm Solution (c).
635
The series diverges, since
p = lim
= lim
= hm
• — = lim
.1 + i-
= e> 1
See Formula (4) of Section 6.1
Solution (d).
The series diverges, since
(2fe) = hm
Solution (e).
The ratio test is of no help since
,. p — hm
uk
1 = hm *->•+«>2(fc+l)-l
2k-I 2k-I = hm — =1 1
However, the integral test proves that the series diverges since b
2JC-1
dx
1 = lim - \n(2x - 1) I = +o>
b-M
Both the comparison test and the limit comparison test would also have worked here (verify).
THE ROOT TEST
In cases where it is difficult or inconvenient to find the limit required for the ratio test, the next test is sometimes useful. Since its proof is similar to the proof of the ratio test, we will omit it.
9.5.6 that
THEOREM (The Root Test)
Let ^u^bea series with positive terms and suppose .. .. , .i/if k/— p — hm yu k = hm (uk) ' k —> +00
k -> -fee
(a) I f p < \ , the series converges. (b) If p > 1 or p = +00, the series diverges. (c) Ifp = l, the series may converge or diverge, so that another test must be tried.
Example 4 diverge.
Use the root test to determine whether the following series converge or
k=2
4k-5 2/fc+l
(ln(Jt
636 Chapter 9 / Infinite Series Solution (a). The series diverges, since p = lim (uk)llk = lim Solution (b). The series converges, since p = lim (uk)l/k — lim
I/QUICK CHECK EXERCISES 9.5
= 0< 1
(See page 637 for answers.)
1 -4 Select between converges or diverges to fill the first blank.
3. Since lim
1. The series
A 2k2 + 1
= lim
the series JJ^=1 k\/3k. 4. Since
. by comparison with the ^-series Y^k=i
= +00
. by the.
. test.
,,, 1/A:
2. Since lim
(fc + i)W+' T-T-;
3
the series = lim
k
3
. by the
the series Y^k=i k /3 .
. by the.
. test.
3 test.
EXERCISE SET 9.5 1 -2 Make a guess about the convergence or divergence of the series, and confirm your guess using the comparison test.
, A
i
2. (a)
. A 3
(b)
11-16 Use the ratio test to determine whether the series converges. If the test is inconclusive, then say so. «
(b)
11. y^
12. f t
14.
15.
3. In each part, use the comparison test to show that the series converges.
(a) y\,...
10.
9. V
(b)y
4. In each part, use the comparison test to show that the series diverges.
;I
16. F-
1 7-20 Use the root test to determine whether the series converges. If the test is inconclusive, then say so. 88 17.
(a)f^
(b)V—^-r
5-10 Use the limit comparison test to determine whether the series converges. 5. V
8F + it - 8
6. V
8.
k=\
k=\
21 -24 True-False Determine whether the statement is true or false. Explain your answer.
9A: + 6
21. The limit comparison test decides convergence based on a limit of the quotient of consecutive terms in a series. 2)(Jfc + 5)
22. If linij.^ +aa(uk+i/M^) = 5, then JJ u^ diverges.
9.5 The Comparison, Ratio, and Root Tests
50 1 +
23. If limk^+ac(k Wfc) = 5, then ^Uk converges. 24. The root test decides convergence based on a limit of kth roots of terms in the sequence of partial sums for a series. 25-47 Use any method to determine whether the series converges.
'*-E^ / ' T,k
1
26.
25. > —
1-3-5-7
51. Show that In x < ~Jx if x > 0, and use this result to investigate the convergence of A In k ^ (a)E^ 0»I
52. (a) Make a conjecture about the convergence of the series X!r=i sinC77/^) by considering the local linear approximation of sin x at x = 0. (b) Try to confirm your conjecture using the limit comparison test. 53. (a) We will see later that the polynomial 1 — x2/2 is the "local quadratic" approximation for cos x at x = 0. Make a conjecture about the convergence of the series
30. T
29. 32-
(-\)k
L~l
k=l
36. > ~
37.
by considering this approximation. (b) Try to confirm your conjecture using the limit comparison test.
40. V
tan-'k
43. E 46.
44.
48. For what positive values of a does the series £^j°=1 (ak/ka) converge? 49-50 Find the general term of the series and use the ratio test to show that the series converges. 49 1 + 1 - 2 + 1 - 2 - 3 1 - 2 - 3 - 4
54. Let ]T ak and ^ bk be series with positive terms. Prove: (a) If lim^+oo (at/bit) = 0 and ^bk converges, then ^cik converges. (b) If \vct\k^+a> (ak/bk) = +oo and ^b^ diverges, then J] ak diverges. 55. Use Theorem 9.4.6 to prove the comparison test (Theorem 9.5.1). 56. Writing What does the ratio test tell you about the convergence of a geometric series? Discuss similarities between geometric series and series to which the ratio test applies. 57. Writing Given an infinite series, discuss a strategy for deciding what convergence test to use.
- TT3- n^ + 1-r^ + -"
I/QUICK CHECK ANSWERS 9.! 1. diverges; \/k213
1-3-5
- ^r + -5r- + - —- + •"
FOCUS ON CONCEPTS
k=\
35. V -
1-3
637
2. converges; ratio
3. diverges; ratio
4. converges; root
638 Chapter 9 / Infinite Series
ALTERNATING SERIES; ABSOLUTE AND CONDITIONAL CONVERGENCE Up to now we have focused exclusively on series with nonnegative terms. In this section we will discuss series that contain both positive and negative terms.
ALTERNATING SERIES
Series whose terms alternate between positive and negative, called alternating series, are of special importance. Some examples are
1 1 1 1
1
2+ 3~ 4+5
2
3
4
5
In general, an alternating series has one of the following two forms:
- a4
= -a\ + a2- 03+0.4-
(1)
(2)
k=l
where the a*'s are assumed to be positive in both cases. The following theorem is the key result on convergence of alternating series.
9.6.1 THEOREM (Alternating Series Test) An alternating series of either form (1) or form (2) converges if the following two conditions are satisfied: (a) (b)
a\ > a2 > 03 > • • • > Ok > • • • lim a/f = 0
PROOF We will consider only alternating series of form (1). The idea of the proof is to show that if conditions (a) and (b) hold, then the sequences of even-numbered and oddnumbered partial sums converge to a common limit S. It will then follow from Theorem 9.1.4 that the entire sequence of partial sums converges to S. Figure 9.6.1 shows how successive partial sums satisfying conditions (a) and (b) appear when plotted on a horizontal axis. The even-numbered partial sums A Figure 9.6.1
S2,S4, 56, 5g, . .., Sin, • • •
form an increasing sequence bounded above by a\, and the odd-numbered partial sums It is not essential for condition (a) in Theorem 9.6.1 to hold for all terms; an alternating series will converge if condition (b) is true and condition (a) holds eventually.
si, 5 3 ,s 5 ,... ,s2n-\, .. • form a decreasing sequence bounded below by 0. Thus, by Theorems 9.2.3 and 9.2.4, the even-numbered partial sums converge to some limit SE and the odd-numbered partial sums converge to some limit So• To complete the proof we must show that SE = So- But the
9.6 Alternating Series; Absolute and Conditional Convergence
639
(2n)-th term in the series is — «2«. so that s^n — £2/1-1 = — o-2n, which can be written as If an alternating series violates condition (b) of the alternating series test, then the series must diverge by the divergence test (Theorem 9.4.1).
However, 2n -> +00 and 2n — 1 —> +00 as n —*• +co, so that
S0 = n—lim 52n-i = n—lim (s2n + «2«) = SE + 0 — SE » -t-cc > +00 which completes the proof. •
Example 1 Use the alternating series test to show that the following series converge.
k=l
Solution (a). The series in part (a) of Example 1 is called the alternating harmonic series. Note that this series converges, whereas the harmonic series diverges.
The two conditions in the alternating series test are satisfied since ak = - >
k
Solution (b).
k+l
= ak+\
and
lim ak = lim - = 0
k^+™
k^+^k
The two conditions in the alternating series test are satisfied since k+4
ak
k=\
k2+ 4k
k(k+l) k +3
k2 + 4k <1 (k2 + 4k) + (k + 6)
so
ak > ak+] and
k+3 lim ak = lim - = lim
\
— 0
k APPROXIMATING SUMS OF ALTERNATING SERIES
The following theorem is concerned with the error that results when the sum of an alternating series is approximated by a partial sum.
9.6.2 THEOREM If an alternating series satisfies the hypotheses of the alternating series test, and ifS is the sum of the series, then: (a)
S lies between any two successive partial sums; that is, either sn _ »j _ snjr\
or
< S < sn
(3)
depending on which partial sum is larger, (b) If S is approximated by sn, then the absolute error \S — sn\ satisfies \S-sn\
(4) ofan+\.
640
Chapter 9 / Infinite Series PROOF We will prove the theorem for series of form (1). Referring to Figure 9.6.2 and keeping in mind our observation in the proof of Theorem 9.6. 1 that the odd-numbered partial sums form a decreasing sequence converging to S and the even-numbered partial sums form an increasing sequence converging to S, we see that successive partial sums oscillate from one side of S to the other in smaller and smaller steps with the odd-numbered partial sums being larger than S and the even-numbered partial sums being smaller than S. Thus, depending on whether n is even or odd, we have
or
< S <sn
A Figure 9.6.2
which proves (3). Moreover, in either case we have \S-Sn
(5)
< Sn+i -Sn\
But sn+i — sn= ±an+i (the sign depending on whether n is even or odd). Thus, it follows from (5) that \S — sn < an+\, which proves (4). Finally, since the odd-numbered partial sums are larger than S and the even-numbered partial sums are smaller than S, it follows that S — sn has the same sign as the coefficient of an+\ (verify). •
REMARK
> Example 2 series i s
1
0.8 In 2 0.6
M _•_ ± __•__•_
0.4 0.2
In words, inequality (4) states that for a series satisfying the hypotheses of the alternating series test, the magnitude of the error that results from approximating S by sn is at most that of the first term that is not included in the partial sum. Also, note that if a\ > a2 > • • • > ak > • • •, then inequality (4) can be strengthened to |5 - sn < a,!+1.
• » • • • « •15* * 2*0 *'
-0.2
Later in this chapter we will show that the sum of the alternating harmonic 1 1 1 , 1
2 This is illustrated in Figure 9.6.3.
3
k
4
(a) Accepting this to be so, find an upper bound on the magnitude of the error that results if In 2 is approximated by the sum of the first eight terms in the series. (b) Find a partial sum that approximates In 2 to one decimal-place accuracy (the nearest tenth).
-0.4 -0.6
Graph of the sequences of terms and nth partial sums for the alternating harmonic series A Figure 9.6.3
Solution (a).
It follows from the strengthened form of (4) that
= - < 0.12
(6)
As a check, let us compute sg exactly. We obtain
1
1
1
1
1
1
1
533
Thus, with the help of a calculator | l n 2 - j g | = In2-
533 840
• 0.059
This shows that the error is well under the estimate provided by upper bound (6). Solution (b). For one decimal-place accuracy, we must choose a value of n for which | In 2 - sn | < 0.05. However, it follows from the strengthened form of (4) that | I n 2 - sn\ < an+i so it suffices to choose n so that an+\ < 0.05.
9.6 Alternating Series; Absolute and Conditional Convergence
641
One way to find n is to use a calculating utility to obtain numerical values for a\, ai, « 3 , . . . until you encounter the first value that is less than or equal to 0.05. If you do this, you will find that it is 020 = 0.05; this tells us that partial sum s\^ will provide the desired accuracy. Another way to find n is to solve the inequality As Example 2 illustrates, the alternating harmonic series does not provide an efficient way to approximate In 2, since too many terms and hence too much computation is required to achieve reasonable accuracy. Later, we will develop better ways to approximate logarithms.
1 <0.05 n+1
algebraically. We can do this by taking reciprocals, reversing the sense of the inequality, and then simplifying to obtain n > 19. Thus, s19 will provide the required accuracy, which is consistent with the previous result. With the help of a calculating utility, the value of 519 is approximately s\g ^ 0.7 and the value of In 2 obtained directly is approximately In 2 & 0.69, which agrees with $19 when rounded to one decimal place. •<
ABSOLUTE CONVERGENCE
The series
^
1
2
2
2
^
I
i
3
2
^
^
4
5
2
26
2
i-...
does not fit in any of the categories studied so far—it has mixed signs but is not alternating. We will now develop some convergence tests that can be applied to such series.
9.6.3
DEFINITION A series Uk = MI + M 2 H ----- r- Uk H
k=\
is said to converge absolutely if the series of absolute values
k=\
converges and is said to diverge absolutely if the series of absolute values diverges.
>• Example 3
Determine whether the following series converge absolutely.
i Solution (a).
L J_ J_ _1
_!!_!!
The series of absolute values is the convergent geometric series I
—
l
_L
l
so the given series converges absolutely. Solution (b).
The series of absolute values is the divergent harmonic series
1 so the given series diverges absolutely.
1
1
1
642
Chapter 9 / Infinite Series
It is important to distinguish between the notions of convergence and absolute convergence. For example, the series in part (b) of Example 3 converges, since it is the alternating harmonic series, yet we demonstrated that it does not converge absolutely. However, the following theorem shows that if a series converges absolutely, then it converges.
9.6.4 Theorem 9.6.4 provides a way of inferring convergence of a series with positive and negative terms from a related series with nonnegative terms (the series of absolute values). This is important because most of the convergence tests that we have developed apply only to series with nonnegative terms.
THEOREM If the series
1 converges, then so does the series
\-uk-\ t=i
PROOF We will write the series J] uk as -\uk\\
(7)
We are assuming that ^ \uk\ converges, so thatifwecanshowthat^(wt + \uk\) converges, then it will follow from (7) and Theorem 9.4.3(a) that Y^ uk converges. However, the value of UK + \uk\ is either 0 or 21«,(. |, depending on the sign of uk. Thus, in all cases it is true that
1
0
- •
0.8 0.6 nU.4 A
0.2 0.2 0.4 0.6 -
But ^2\uk\ converges, since it is a constant times the convergent series ^(uk + \uk\) converges by the comparison test. •
• Kl . *i . 2
4
|; hence
n (
,
w
i -
£ _iQ
•
>
>• Example 4
{«*}
(a)
Show that the following series converge. 1
J_
_L
_L
J_
J_
2
2
2
2
2
2
2
3
4
5
•^ cosfe
6
it=l
1
Solution (a). Observe that this is not an alternating series because the signs alternate in pairs after the first term. Thus, we have no convergence test that can be applied directly. However, we showed in Example 3(a) that the series converges absolutely, so Theorem 9.6.4 implies that it converges (Figure 9.6.4a).
0.8 0.6 0.4
Kl
0.2 o
-0.2 -0.4
{«*}
-0.6
(b) Graphs of the sequences of j terms and nVn partial sums j for the series in Example 4
A Figure 9.6.4
Solution (b). With the help of a calculating utility, you will be able to verify that the signs of the terms in this series vary irregularly. Thus, we will test for absolute convergence. The series of absolute values is x •^—^ cos A:
However,
cos A: k2
1
F
9.6 Alternating Series; Absolute and Conditional Convergence
643
But J^ l/k2 is a convergent p-series (p — 2), so the series of absolute values converges by the comparison test. Thus, the given series converges absolutely and hence converges (Figure 9.6.4*). •*
CONDITIONAL CONVERGENCE
Although Theorem 9.6.4 is a useful tool for series that converge absolutely, it provides no information about the convergence or divergence of a series that diverges absolutely. For example, consider the two series -I
[
I
i /
i \&+l
I
(Q\
-1-H-J — - J - -
(9)
Both of these series diverge absolutely, since in each case the series of absolute values is the divergent harmonic series 1 1
1
However, series (8) converges, since it is the alternating harmonic series, and series (9) diverges, since it is a constant times the divergent harmonic series. As a matter of terminology, a series that converges but diverges absolutely is said to converge conditionally (or to be conditionally convergent). Thus, (8) is a conditionally convergent series. Example 5
In Example l(b) we used the alternating series test to show that the series v-», --t-i-i k + 3 k(k + 1) k=l
converges. Determine whether this series converges absolutely or converges conditionally. Solution. We test the series for absolute convergence by examining the series of absolute values: k+\ k + 3
E
k(k
k=l
Principle 9.5.3 suggests that the series of absolute values should behave like the divergent ^-series with p = 1. To prove that the series of absolute values diverges, we will apply the limit comparison test with ak =
We obtain
k +3
,. at ,. p = hm — = lim
and bk = k
— lim
k +3
1
Since p is finite and positive, it follows from the limit comparison test that the series of absolute values diverges. Thus, the original series converges and also diverges absolutely, and so converges conditionally. <
THE RATIO TEST FOR ABSOLUTE CONVERGENCE
Although one cannot generally infer convergence or divergence of a series from absolute divergence, the following variation of the ratio test provides a way of deducing divergence from absolute divergence in certain situations. We omit the proof.
644
Chapter 9 / Infinite Series
9.6.5 THEOREM (Ratio Test for Absolute Convergence) Let J] uk be a series with nonzero terms and suppose that \u i p = lim *+1 \uk\
(a)
If p < 1, then the series J] uk converges absolutely and therefore converges.
(b) If p > 1 or if p = +co, then the series ^ uk diverges. (c)
If p = 1 , no conclusion about convergence or absolute convergence can be drawn from this test.
> Example 6 converges.
Use the ratio test for absolute convergence to determine whether the series «, k (2k - 1)! '
k=\
k=\
Solution (a). Taking the absolute value of the general term uk we obtain (-1)*k\ Thus,
p = lim
\Uk+\
— lim —
M
d
2k+1 k\ — • —k = lim
2
k'-^'+oo (k + iy. 2
—0 < 1
which implies that the series converges absolutely and therefore converges. Solution (b). Taking the absolute value of the general term uk we obtain
(-1) Thus,
.(2k-iy.
(2k - 1)! 3*
\uk+i\ p
* =
lim
(2k -1)1
Uk
•
(2k— 1)!
=
lim (2k)(2k
which implies that the series diverges. -4
SUMMARY OF CONVERGENCE TESTS
We conclude this section with a summary of convergence tests that can be used for reference. The skill of selecting a good test is developed through lots of practice. In some instances a test may be inconclusive, so another test must be tried.
Summary of Convergence Tests NAME
Divergence Test (9.4.1)
Integral Test (9.4.4)
STATEMENT
COMMENTS
If lim «,. = 0, then t->+~ * may not converge.
If lim uk ^ 0, then YJ uk diverges.
Let y uk be a series with positive terms. If / is a function that is decreasing and continuous on an interval [a, +00) and such that uk = f(k) for all k > a, then
and k=\
both converge or both diverge. Let k=i ak and terms such that
Comparison Test (9.5.1)
k_f
r
f(x)dx
bk be series with nonnegative
u,.k mayJ or
This test only applies to series that have positive terms. Try this test when/(jc) is easy to integrate.
This test only applies to series with nonnegative terms.
flj < b\, a-i < b2, • • • , ak< bk, . . . If y bk converges, then y ak converges, and if y ak diverges, then y bk diverges. Let y ak and y bk be series with positive terms and let
at p= lim -^
Limit Comparison Test (9.5.4)
k—>+«> bk
Try this test as a last resort; other tests are often easier to apply.
This is easier to apply than the comparison test, but still requires some skill in choosing the series y bk for comparison.
If 0 < p < +00, then both series converge or both diverge. Let y Ratio Test (9.5.5)
uk be a series with positive terms and suppose that P=
fc^+oo
(a) Series converges if p < 1 . (b) Series diverges if p > 1 or p = +00. (c) The test is inconclusive if p = 1 .
Try this test when uk involves factorials or fcth powers.
Let Y^ uk be a series with positive terms and suppose that Root Test (9.5.6)
p = lim ^ k—>+oo
(a) The series converges if p < 1. (b) The series diverges if p > 1 or p = +00. (c) The test is inconclusive if p = 1.
Try this test when uk involves &th powers.
If ak > 0 for k = 1, 2, 3, . . . , then the series Alternating Series Test (9.6.1)
~ "3 + a4 ~ ' ' '
converge if the following conditions hold: (a) a\ > a-i > 03 > • • •
Let 2_]uk
Ratio Test for Absolute Convergence (9.6.5)
This test applies only to alternating series.
i series with nonzero terms and suppose that = lim K+i| \Uj,
(a) The series converges absolutely if p < 1. (b) The series diverges if p > 1 or p = +00. (c) The test is inconclusive if p = 1.
The series need not have positive terms and need not be alternating to use this test.
646
Chapter 9 / Infinite Series
I/QUICK CHECK EXERCISES 9.6
(See page 648 for answers.)
1. What characterizes an alternating series?
15
2. (a) The series
y\ ^ k2
converges by the alternating series test since. and (b) If
4. Given that
so =
and
lim then | S — sg \ <
k4/4k
= lim
is the series £2"=1(— l^VM* conditionally convergent, absolutely convergent, or divergent?
3. Classify each sequence as conditionally convergent, absolutely convergent, or divergent.
k=i CAS
EXERCISE SET 9.6
T -2 Show that the series converges by confirming that it satisfies the hypotheses of the alternating series test (Theorem 9.6.1).
3-6 Determine whether the alternating series converges; justify your answer. .
X "v .
, . - . k_i_i ^ "T" A
5.
=° \ ^
6.
t_i_i
16.
17.
. kn 21 v ssm
-E
t i l ^ "T" A
is. E sink
(-D*
i
>D t=3
7-T2 Use the ratio test for absolute convergence (Theorem 9.6.5) to determine whether the series converges or diverges. If the test is inconclusive, say so.
27.
29-32 True-False Determine whether the statement is true or false. Explain your answer. 29. An alternating series is one whose terms alternate between even and odd.
11.
30. If a series satisfies the hypothesis of the alternating series test, then the sequence of partial sums of the series oscillates between overestimates and underestimates for the sum of the series.
12.
13-28 Classify each series as absolutely convergent, conditionally convergent, or divergent.
13.
14.
15.
k=i
£2
31. If a series converges, then either it converges absolutely or it converges conditionally. 32. If XX"*:)2 converges, then 5Z ««: converges absolutely.
9.6 Alternating Series; Absolute and Conditional Convergence
33-36 Each series satisfies the hypotheses of the alternating series test. For the stated value of n, find an upper bound on the absolute error that results if the sum of the series is approximated by the nth partial sum. 00 / ii*+i °° i n*+i \~^
l—~*
V
'
~,
r-,
t-
\~^
^——/
k=\
k=\
V
*•)
k\
;„ = 5
*+'
35.
>/*
647
(c) According to the error bound in part (b) of Theorem 9.6.2, what value of n is required to ensure that -sn\ < 10~2? 48. Prove: If a series Y, o-k converges absolutely, then the series Y, a 2 converges. 49. (a) Find examples to show that if Y, ak converges, then Y^ a-l may diverge or converge. (b) Find examples to show that if Y^ a2 converges, then Y^, ak may diverge or converge.
36. 50. Let that
37-40 Each series satisfies the hypotheses of the alternating series test. Find a value of n for which the nth partial sum is ensured to approximate the sum of the series to the stated accuracy. -; |error| < 0.0001
37-
(-1)* -; |error| < 0.00001 k\
383940.
k=\
4k
-; two decimal places
(k+l)\n(k+l)
; one decimal place
41-42 Find an upper bound on the absolute error that results if SIQ is used to approximate the sum of the given geometric series. Compute S\Q rounded to four decimal places and compare this value with the exact sum of the series. 3 3 3 3 2 4 8 43-46 Each series satisfies the hypotheses of the alternating series test. Approximate the sum of the series to two decimal-place accuracy.
45. 46.
1 3 • 23 1
2! 4! 6! 1 4 • 24 1 1 55 + 4 • 5 75 + 4 • 7
be a series and define series
Pk = 0,
uk > 0 uk<0
£] 47. The purpose of this exercise is to show that the error bound in part (b) of Theorem 9.6.2 can be overly conservative in certain cases. (a) Use a CAS to confirm that TT 1 1 1 H-3 + 5 - 7 + -
(b) Use the CAS to show that \(n/4) - s25\ < 10~2.
0,
so
uk > 0 uk<0
(a) Show that ^ M* converges absolutely if and only if Y^ Pk and Y, Ik both converge. (b) Show that if one of ]T pk or J] qk converges and the other diverges, then Y, uk diverges. (c) Show that if Yl uk converges conditionally, then both k and Y, Ik diverge. 51. It can be proved that the terms of any conditionally convergent series can be rearranged to give either a divergent series or a conditionally convergent series whose sum is any given number S. For example, we stated in Example 2 that 1 1 1 1 1 Show that we can rearrange this series so that its sum is j In 2 by rewriting it as 1
il _ 1 _ 1
6
J.
s/ ' \5 ~ To ~ T~2
[Hint: Add the first two terms in each grouping. 52-54 Exercise 51 illustrates that one of the nuances of "conditional" convergence is that the sum of a series that converges conditionally depends on the order that the terms of the series are summed. Absolutely convergent series are more dependable, however. It can be proved that any series that is constructed from an absolutely convergent series by rearranging the terms will also be absolutely convergent and has the same sum as the original series. Use this fact together with parts (a) and (b) of Theorem 9.4.3 in these exercises. 52. It was stated in Exercise 35 of Section 9.4 that 7T2 _
FOCUS ON CONCEPTS
and
~6~~
Use this to show that jr2 _
+
1
1
1
¥ + y + ¥ +' 1 1
1
53. Use the series for ?r2/6 given in the preceding exercise to show that i r 1 jr™-2 42 12
648 Chapter 9 / Infinite Series 54. It was stated in Exercise 35 of Section 9.4 that 7t4 _
90 ~ Use this to show that
1
j_
Determine whether this series converges and use this series as an example in a discussion of the importance of hypotheses (a) and (b) of the alternating series test (Theorem 9.6.1). 56. Writing Discuss the ways that conditional convergence is "conditional." In particular, describe how one could rearrange the terms of a conditionally convergent series ]P u^ so that the resulting series diverges, either to +00 or to -co. [Hint: See Exercise 50.]
1
24 + 34 + 4 4 + '
— -•( — 1 1 96" *"& + 5* + 7** 55. Writing Consider the series 1 2 1 2 1 2 1
UICK CHECK ANSWERS 9.6 1. Terms alternate between positive and negative.
2 (a)1
-
^^5---F-(FTi)^
3. (a) conditionally convergent (b) divergent (c) absolutely convergent (d) conditionally convergent
4. absolutely convergent
MACLAURIN AND TAYLOR POLYNOMIALS In a local linear approximation the tangent line to the graph of a function is used to obtain a linear approximation of the function near the point oftangency. In this section we will consider how one might improve on the accuracy of local linear approximations by using higher-order polynomials as approximating functions. We will also investigate the error associated with such approximations.
LOCAL QUADRATIC APPROXIMATIONS
Recall from Formula (1) in Section 2.9 that the local linear approximation of a function /
at XQ is
f( ) ~ f(
(1)
)
In this formula, the approximating function f'(XQ)(x
Local linear approximation
A Figure 9.7.1
- XQ)
is a first-degree polynomial satisfying P(XQ) = f ( x o ) and P'(XQ) = f'(xo) (verify). Thus, the local linear approximation of / at XQ has the property that its value and the value of its first derivative match those of / at XQ. If the graph of a function / has a pronounced "bend" at XQ, then we can expect that the accuracy of the local linear approximation of / at XQ will decrease rapidly as we progress away from XQ (Figure 9.7.1). One way to deal with this problem is to approximate the function / at XQ by a polynomial p of degree 2 with the property that the value of p and the values of its first two derivatives match those of / at XQ. This ensures that the graphs of / and p not only have the same tangent line at XQ, but they also bend in the same direction at XQ (both concave up or concave down). As a result, we can expect that the graph of p will remain close to the graph of / over a larger interval around XQ than the graph of the local linear approximation. The polynomial p is called the local quadratic approximation off atx= XQ. To illustrate this idea, let us try to find a formula for the local quadratic approximation of a function / at x = 0. This approximation has the form - c2x2 where CQ, c\, and ci must be chosen so that the values of p(x) = CQ + CiX + C2X2
(2)
9.7 Maclaurin and Taylor Polynomials
649
and its first two derivatives match those of / at 0. Thus, we want
But the values of p(Q), p'(0), and p"(Q) are as follows: p(x) = c0 + c\x + c2x2
p(O) = GO
p'(x) =ci + 2c2x
p'(0) = ci
p"(jc) = 2c2
p"(0) = 2c2
Thus, it follows from (3) that co = /(O),
ci - /'(O),
c2 =
ffi-
and substituting these in (2) yields the following formula for the local quadratic approximation of / at x = 0:
f(x) « /(O) + /'(
(4)
> Example 1 Find the local linear and quadratic approximations of e* at x — 0, and graph ex and the two approximations together. Solution. If we let f ( x ) = ex, then /'(*) = /"(*) = ex; and hence
/(O) = /'(O) = /"(O) = e° = 1 y=\+x
Thus, from (4) the local quadratic approximation of e* at x = 0 is
and the local linear approximation (which is the linear part of the local quadratic approximation) is , ex « 1 + x
A Figure 9.7.2
The graphs of ex and the two approximations are shown in Figure 9.7.2. As expected, the local quadratic approximation is more accurate than the local linear approximation near x = 0. •« MACLAURIN POLYNOMIALS
It is natural to ask whether one can improve on the accuracy of a local quadratic approximation by using a polynomial of degree 3. Specifically, one might look for a polynomial of degree 3 with the property that its value and the values of its first three derivatives match
Colin Maclaurin (1698-1746) Scottish mathematician. Maclaurin's father, a minister, died when the boy was only six months old, and his mother when he was nine years old. He was then raised by an uncle who was also a minister. Maclaurin entered Glasgow University as a divinity student but switched to mathematics after one year. He received his Master's degree at age 17 and, in spite of his youth, began teaching at Marischal College in Aberdeen, Scotland. He met Isaac Newton during a visit to London in 1719 and from that time on became Newton's disciple. During that era, some of Newton's analytic methods were bitterly attacked by major
mathematicians and much of Maclaurin's important mathematical work resulted from his efforts to defend Newton's ideas geometrically. Maclaurin's work, A Treatise of Fluxions (1742), was the first systematic formulation of Newton's methods. The treatise was so carefully done that it was a standard of mathematical rigor in calculus until the work of Cauchy in 1821. Maclaurin was also an outstanding experimentalist; he devised numerous ingenious mechanical devices, made important astronomical observations, performed actuarial computations for insurance societies, and helped to improve maps of the islands around Scotland.
650
Chapter 9 / Infinite Series those of / at a point; and if this provides an improvement in accuracy, why not go on to polynomials of even higher degree? Thus, we are led to consider the following general problem.
9.7.1 PROBLEM Given a function / that can be differentiated n times at x = x0, find a polynomial p of degree n with the property that the value of p and the values of its first n derivatives match those of / at x0.
We will begin by solving this problem in the case where jtn = 0. Thus, we want a polynomial , , p(x) = c0 + cix + c2x2 + cix3 H ----- h cnx" (5) such that '(0) = //(0),
But
(6)
/"(O) = //'(O), . . . ,
p(x)
= C0 + C\X + C2X2 + C3JC3 H
p'(x)
— c{ + 2c2x + 3c3jc2 H
h CnXn
h ncnx"~l
p"(x) = 2c2 + 3 • 2c3x + - • • + « ( « - l)cnxnp'"(x) = 3 - 2c3 + • • • + n(n - l)(n - 2)cnx"p(n\x) = n(n - \)(n - 2) • • • (l)cn Thus, to satisfy (6) we must have
/(O) = p(0)
=c 0
/'(O) = p'(0) = c j /"(O) = p"(0) = 2c2 = 2\c2 /'"(O) = 77'"(0) = 3 • 2c3 = 3!c3 ; p<")(0) = n(n - l)(n - 2) • • • (l)c n = w!cn which yields the following values for the coefficients of p ( x ) : co = /(O),
ci = /'(O),
c2 =
/"(O) 2!
C3 =
/"'(O) 3!
n\
The polynomial that results by using these coefficients in (5) is called the nth Maclaurin polynomial for f.
Local linear approximations and local quadratic approximations at x = 0 of a function / are special cases of the MacLaurin polynomials for /. Verify that f(x) ^ pi (x) is the local linear approximation of / at x = 0, and f(x) <** P2(x) is the local quadratic approximation at x = 0.
9.7.2 DEFINITION If / can be differentiated n times at 0, then we define the nth Maclaurin polynomial for f to be
2!
3!
,7,
Note that the polynomial in (7) has the property that its value and the values of its first n derivatives match the values of / and its first n derivatives at x = 0.
9.7 Maclaurin and Taylor Polynomials *• Example 2 Solution.
651
Find the Maclaurin polynomials po, p\, p^, p^, and pn for ex .
Let f(x) = ex . Thus, f'(x) = f"(x) = f'"(x) = ••• = /<">(*) = ex
and /(O) = /'(O) = /"(O) = /'"(O) = • • • = / w (0) = e° = 1 Therefore,
po(x) = /(O) = 1
/"(O) 2!
/(O)
2!
/(O) *
Jt
- +-
2!
n\
JC
2!"
A Figure 9.7.3
n\
Figure 9.7.3 shows the graph of ex (in blue) and the graph of the first four Maclaurin polynomials. Note that the graphs of p\ (x), P2(x), and p$(x) are virtually indistinguishable from the graph of ex near x = 0, so these polynomials are good approximations of ex for x near 0. However, the farther x is from 0, the poorer these approximations become. This is typical of the Maclaurin polynomials for a function f(x); they provide good approximations of f(x) near 0, but the accuracy diminishes as x progresses away from 0. It is usually the case that the higher the degree of the polynomial, the larger the interval on which it provides a specified accuracy. Accuracy issues will be investigated later.
| Augustin Louis Cauchy (1789-1857) French mathematician. Cauchy's early education was acquired from his father, a barrister and master of the classics. Cauchy entered L'Ecole Polytechnique in 1805 to study engineer• ing, but because of poor health, was advised to concentrate on mathematics. His major mathematical work began in 1811 with a series of brilliant solutions to some difficult outstanding problems. In 1814 he wrote a treatise on integrals that was to become the basis for modern complex variable theory; in 1816 there followed a classic paper on wave propagation in liquids that won a prize from the French Academy; and in 1822 he wrote a paper that formed the basis of modern elasticity theory. Cauchy' s mathematical contributions for the next 35 years were brilliant and staggering in quantity, over 700 papers filling 26 modern volumes. Cauchy's work initiated the era of modern analysis. He brought to mathematics standards of precision and rigor undreamed of by Leibniz and Newton.
Cauchy's life was inextricably tied to the political upheavals of the time. A strong partisan of the Bourbons, he left his wife and children in 1830 to follow the Bourbon king Charles X into exile. For his loyalty he was made a baron by the ex-king. Cauchy eventually returned to France, but refused to accept a university position until the government waived its requirement that he take a loyalty oath. It is difficult to get a clear picture of the man. Devoutly Catholic, he sponsored charitable work for unwed mothers, criminals, and relief for Ireland. Yet other aspects of his life cast him in an unfavorable light. The Norwegian mathematician Abel described him as, "mad, infinitely Catholic, and bigoted." Some writers praise his teaching, yet others say he rambled incoherently and, according to a report of the day, he once devoted an entire lecture to extracting the square root of seventeen to ten decimal places by a method well known to his students. In any event, Cauchy is undeniably one of the greatest minds in the history of science.
652 Chapter 9 / Infinite Series *• Example 3
Find the nth Maclaurin polynomials for (a) sinjc
(b) cos x
Solution (a). In the Maclaurin polynomials for sin x, only the odd powers of x appear explicitly. To see this, let f ( x ) = sinjc; thus, f(x)
=sinjc
/(O) = 0
/'(*) = cosjt
/'(O) = 1
/"(*) = -sin;c
/"(O) = 0
f'"(x) = -cosx
/'"(0) = -1
(4)
Since f (x) = sinx = f ( x ) , the pattern 0, 1, 0, —1 will repeat as we evaluate successive derivatives at 0. Therefore, the successive Maclaurin polynomials for sin x are
PoW = 0 p2(x) = 0 + x + 0 X
~ ~3\ T.
P4(x)
X
=
3!
+0
JC3
" 3!
4
x5 ~~5\
3 P6(x)
=
V5
- — + 0 + — +0
x x x P7(*)=0 + jc + 0- — + 0 + — + 0 - — Because of the zero terms, each even-order Maclaurin polynomial [after po(x)] is the same as the preceding odd-order Maclaurin polynomial. That is,
A Figure 9.7.4
f2k+\
X1
X
= 0, 1 , 2 , . . . ) ~5\ 7!" The graphs of sinjc, p\(x), pi(x), p$(x), and pi(x) are shown in Figure 9.7.4. P2k+i(x) =
3!
Solution (b). In the Maclaurin polynomials for cos x, only the even powers of x appear explicitly; the computations are similar to those in part (a). The reader should be able to show that po(x) = pi(x) = i
x2 1- —
x2 x4 p4(x) = p$(x) = 1 - — + —
= Pi(x) =l-
x2
+
x4
~
x6
In general, the Maclaurin polynomials for cos x are given by 4
= 0,1,2,...) (2k)\ The graphs of cos x, p»(x), P2(x), P4(x), and P(,(X) are shown in Figure 9.7.5. (-!)*
A Figure 9.7.5
9.7 Maclaurin and Taylor Polynomials
653
TAYLOR POLYNOMIALS Up to now we have focused on approximating a function / in the vicinity of x = 0. Now we will consider the more general case of approximating / in the vicinity of an arbitrary domain value x0. The basic idea is the same as before; we want to find an nth-degree polynomial p with the property that its value and the values of its first n derivatives match those of / at XQ. However, rather than expressing p(x) in powers of x, it will simplify the computations if we express it in powers of x — x$\ that is,
- x0)2 H
p ( x ) = c0 + c\(x - x0)
cn(x - x0)n
(8)
We will leave it as an exercise for you to imitate the computations used in the case where x0 = 0 to show that = f(x0),
f'"(XQ)
c\ = f'(x0),
21
3!
n\
'
Substituting these values in (8) we obtain a polynomial called the nth Taylor polynomial about x — xofor f.
Local linear approximations and local quadratic approximations at x = XQ of a function / are special cases of the Taylor polynomials for /. Verify that f(x) =s p\ (x) is the local linear approximation of / at x = XQ, and f(x) « P2(x) is the local quadratic approximation at x — XQ.
9.7.3 DEFINITION If / can be differentiated n times at x0, then we define the nth \ Taylor polynomial for f about x = XQ to be
Pn(x) = f(Xo) + f\Xo)(x
3!
> Example 4 The Maclaurin polynomials are the special cases of the Taylor polynomials in which XQ = 0. Thus, theorems about Taylor polynomials also apply to Maclaurin polynomials.
- XQ) H
Solution.
(X -
X0)2
- x0y +
(X - Xoy
(9)
Find the first four Taylor polynomials for In x about x = 2.
Let f(x) — lux. Thus,
/"CO = - l / * 2
=ln2 /'(2) = 1/2 /"(2) = -1/4
/'"(AT) = 2/*3
/'"(2) = 1/4
/CO
=ln*
f ' ( x ) = l/x
Brook Taylor (1685-1731) English mathematician. Taylor was born of well-to-do parents. Musicians and artists were entertained frequently in the Taylor home, which undoubtedly had a lasting influence on him. In later years, Taylor published a definitive work on the mathematical theory of perspective and obtained major mathematical results about the vibrations of strings. There also exists an unpublished work, On Mustek, that was intended to be part of a joint paper with Isaac Newton. Taylor's life was scarred with unhappiness, illness, and tragedy. Because his first wife was not rich enough to suit his father, the two men argued bitterly and parted ways. Sub-
sequently, his wife died in childbirth. Then, after he remarried, his second wife also died in childbirth, though his daughter survived. Taylor's most productive period was from 1714 to 1719, during which time he wrote on a wide range of subjects—magnetism, capillary action, thermometers, perspective, and calculus. In his final years, Taylor devoted his writing efforts to religion and philosophy. According to Taylor, the results that bear his name were motivated by coffeehouse conversations about works of Newton on planetary motion and works of Halley ("Halley's comet") on roots of polynomials. Unfortunately, Taylor's writing style was so terse and hard to understand that he never received credit for many of his innovations.
654
Chapter 9 / Infinite Series
Substituting in (9) with XQ = 2 yields
>y
y = In x *
= /(2) + f'(2)(x - 2) = In 2 + \(x - 2)
-2) 2 = l n 2 + i ( x - 2 ) - l-(xP3(x)
f"(2) = f ( 2 ) + f'(2)(x - 2) + J-^(x - 2)2 + ^pC* - 2)3 = In2 + |(x - 2) - i(z - 2)2 + ±(x - 2)3
A Figure 9.7.6
The graph of In x (in blue) and its first four Taylor polynomials about x = 2 are shown in Figure 9.7.6. As expected, these polynomials produce their best approximations of In x near 2. -4 SIGMA NOTATION FOR TAYLOR AND MACLAURIN POLYNOMIALS
Frequently, we will want to express Formula (9) in sigma notation. To do this, we use the notation /®(^o) to denote the kth derivative of / at x = XQ, and we make the convention that /(0)(•*()) denotes /(XQ). This enables us to write ,
,,
- x0)k = f(x0) + f'(x0-)(x-
+ f"(xo) „. 21
(x - x0Y H
(10)
h
n In particular, we can write the nth Maclaurin polynomial for f ( x ) as kl
Example 5
t* = /(O) + f'(0)x + ^-~x2 H 2!
h
n
(H)
Find the «th Maclaurin polynomial for
1 1 -x and express it in sigma notation. Solution. Let f ( x ) — I/(I - x). The values of / and its first k derivatives at x = 0 are as follows: \ j w; J W 1 -x
/'(*) /"(*) TECHNOLOGY MASTERY Computer algebra systems have commands for generating Taylor polynomials of any specified degree. If you have a CAS, use it to find some of the Maclaurin and Taylor polynomials in Examples 3, 4, and 5.
fffff\ J (X)
f(4)(Vi
1
(I-*) 2
/'(O)
2 (1 - x)3 3 -2 (1-JC) 4
—4-3-2
= 1 = 1!
/"(O)
9
/"'(O)
ot
fCOrrvi
- 4!
91
9.7 Maclaurin and Taylor Polynomials 655 Thus, substituting /'^(O) = kl into Formula (11) yields the «th Maclaurin polynomial for !/(!-*): A-"
Pn (-»•) =
(n =0,1,2, ...)
k=0
*• Example 6 Find the nth Taylor polynomial for 1/jc about x = 1 and express it in sigma notation. Solution. Let f ( x ) — \/x. The computations are similar to those in Example 5. We leave it for you to show that /(I) = 1, 4
/'(I) = -1,
/< >(1)=4!, . . . ,
/
(t)
/"(I) = 2!,
/'"(I) = -3!,
(l) = (-!)**:!
(k)
Thus, substituting f ( l ) = (-!)*£! into Formula (10) with XQ = 1 yields the nth Taylor polynomial for 1 /x '. - 1)* = 1 - (x - 1) + (x - I) 2 - (x - I) 3 + • • • + (-!)"(* - 1)" £=0
THE fiTH REMAINDER It will be convenient to have a notation for the error in the approximation f ( x ) ~ pn(x). Accordingly, we will let Rn(x) denote the difference between f ( x ) and its nth Taylor polynomial; that is, fW
Rn(x) = f(x) - pn(x) = f(x) -
k\
(12)
This can also be written as
f(x) = pn(x) + Rn(x) =
k\
k=0
- x0)k + Rn(x)
(13)
The function Rn(x) is called the nth remainder for the Taylor series of /, and Formula (13) is called Taylor's formula with remainder. Finding a bound for Rn(x) gives an indication of the accuracy of the approximation pn(x) ^ f ( x ) . The following theorem, which is proved in Appendix D, provides such a bound.
9.7.4 THEOREM (The Remainder Estimation Theorem) If the function f can be differentiated n + 1 times on an interval containing the number XQ, and if M is an upper bound I for |/("+1)(*)l on the interval, that is, \f<-n+i)(x~)\ < M for all x in the interval, then The bound for \Rn(x)\ in (14) is called the Lagrange error bound.
M
n+l
1)!
for all x in the interval.
(14)
656 Chapter 9 / Infinite Series > Example 7 Use an nth Maclaurin polynomial for ex to approximate e to five decimalplace accuracy. Solution. We note first that the exponential function ex has derivatives of all orders for every real number x. From Example 2, the nth Maclaurin polynomial for ex is xjk
x" ~n\
X
E k\ =
2!~
k=0
from which we have e =e k=0
Thus, our problem is to determine how many terms to include in a Maclaurin polynomial for ex to achieve five decimal-place accuracy; that is, we want to choose n so that the absolute value of the nth remainder at x = 1 satisfies I/?„(!)I < 0.000005 To determine n we use the Remainder Estimation Theorem with f ( x ) = ex, x = I, XQ = 0, and the interval [0,1]. In this case it follows from (14) that
M
M (n+1)!
(15)
where M is an upper bound on the value of / ( " +1 ^(x) = ex for x in the interval [0, 1]. However, ex is an increasing function, so its maximum value on the interval [0, 1] occurs at x = 1; that is, ex < e on this interval. Thus, we can take M — e in (15) to obtain l#«(l)l < ~ (n + 1)!
(16)
Unfortunately, this inequality is not very useful because it involves e, which is the very quantity we are trying to approximate. However, if we accept that e < 3, then we can replace (16) with the following less precise, but more easily applied, inequality:
3 - (n + 1)! Thus, we can achieve five decimal-place accuracy by choosing n so that — < 0.000005 or (n + 1)! > 600,000 Since 9! = 362,880 and 10! = 3,628,800, the smallest value of n that meets this criterion is n =9. Thus, to five decimal-place accuracy 1
1
1
1
1
1
1
1
As a check, a calculator's 12-digit representation of e is e & 2.71828182846, which agrees with the preceding approximation when rounded to five decimal places. -^ *• Example 8 Use the Remainder Estimation Theorem to find an interval containing x = 0 throughout which f ( x ) = cos x can be approximated by p(x) = 1 — (x2/2!) to three decimal-place accuracy. Solution. We note first that f ( x ) = cos x has derivatives of all orders for every real number x, so the first hypothesis of the Remainder Estimation Theorem is satisfied over any interval that we choose. The given polynomial p(x) is both the second and the third
9.7 Maclaurin and Taylor Polynomials
657
Maclaurin polynomial for cos x; we will choose the degree n of the polynomial to be as large as possible, so we will take n — 3. Our problem is to determine an interval on which the absolute value of the third remainder at x satisfies \R3(x)\ < 0.0005 We will use the Remainder Estimation Theorem with f ( x ) = cosx, n — 3, and XQ = 0. It follows from (14) that
M ~ (3 + 1)!'
M\x\A 24
(17)
where Mis an upper bound for |/(4^(z) | = cos* . Since | cos x\ < 1 for every real number x, we can take M = 1 in (17) to obtain 0.0005
- 24 Thus we can achieve three decimal-place accuracy by choosing values of x for which
< 0.0005 or \x I < 0.3309 24 so the interval [—0.3309, 0.3309] is one option. We can check this answer by graphing |/(x) - p(x)\ over the interval [-0.3309, 0.3309] (Figure 9.7.7). •«
0.3309
-0.3309 A Figure 9.7.7
I/QUICK CH ECK EXERCISES 9.7
(See page 659 for answers.)
1. If / can be differentiated three times at 0, then the third Maclaurin polynomial for / is pi(x) = 2. The third Maclaurin polynomial for f ( x ) = e2* is
p3(x) =
x
+ +
(18)
2
X +
X3
3. If /(2) = 3, /'(2) = -4, and /"(2) = 10, then the second Taylor polynomial for / about x = 2 is pi(x) =
S. (a) If a function / has nth Taylor polynomial pn(x) about x = XQ, then the nth remainder Rn(x) is defined by Rn(x)= (b) Suppose that a function / can be differentiated five times on an interval containing XQ = 2 and that |/ (5) (jt)| < 20 for all x in the interval. Then the fourth remainder satisfies |/?4(;t)| < for all x in the interval.
4. The third Taylor polynomial for f ( x ) = x5 about x = — 1 is + (x + 1) P3(x) = +
EXERCISE SET 9.7
(x + I) 2 +
(x + I) 3
B Graphing utility
01-2 In each part, find the local quadratic approximation of / at x = XQ, and use that approximation to find the local linear approximation of / at XQ. Use a graphing utility to graph / and the two approximations on the same screen. 1. (a) /(jc) = e-*; x0 = 0 (b) f ( x ) = cosx; x0 = 0 2. (a) f ( x ) = sin*; x0 = n/2 (b) /(*) = ~Jx; x0 = 1 3. (a) Find the local quadratic approximation of */xat XQ = 1. (b) Use the result obtained in part (a) to approximate VlTl, and compare your approximation to that produced directly by your calculating utility. [Note: See Example 1 of Section 2.9.]
4. (a) Find the local quadratic approximation of cos* at (b) Use the result obtained in part (a) to approximate cos 2 °, and compare the approximation to that produced directly by your calculating utility. 5. Use an appropriate local quadratic approximation to approximate tan 61°, and compare the result to that produced directly by your calculating utility. 6. Use an appropriate local quadratic approximation to approximate V36.03, and compare the result to that produced directly by your calculating utility.
658
Chapter 9 / Infinite Series
7-16 Find the Maclaurin polynomials of orders n = 0,1,2,3, and 4, and then find the nth Maclaurin polynomials for the function in sigma notation. 7. e~x 8. eax 9. COS7DC 10. smnx
11. \n(l+x)
13. cosh x 16. xex
14. sinhjt;
M.*lt
l+x 15. x sin x
17-24 Find the Taylor polynomials of orders n = 0,1,2,3, and 4 about x = xo, and then find the nth Taylor polynomial for the function in sigma notation. 17. ex; x0 = 1 18. e~x; 19. 1; X
Xo
20.
= -1
21. sin TO:; XQ = -
22. cos*; XQ = —
23. Inx; x0 = 1
24. Inx; XQ = e
FOCUS ON CONCEPTS
37. Which of the functions graphed in the following figure is most likely to have p(x) = 1 — x + 2x2 as its secondorder Maclaurin polynomial? Explain your reasoning.
l
12
36. l/e; three decimal-place accuracy
\ III
IV
38. Suppose that the values of a function / and its first three derivatives at x = 1 are /(I) = 2,
25. (a) Find the third Maclaurin polynomial for
/'(I) = -3,
/"(1)=0,
/'"(I) = 6
Find as many Taylor polynomials for / as you can about
f ( x ) = \+2x-x2+x3 (b) Find the third Taylor polynomial about x = 1 for f ( x ) = l+2(x-l)-(x-
2
I) + (x - I)
3
26. (a) Find the nth Maclaurin polynomial for f ( x ) = c0 + C,A; + c2x2 -\
h cnxn
(b) Find the nth Taylor polynomial about x = I for
f(x) = c0 + a (x - 1) + c2(x -\}2 + --- + cn(x- 1)" 27-30 Find the first four distinct Taylor polynomials about x = XQ, and use a graphing utility to graph the given function and the Taylor polynomials on the same screen. 27. f ( x ) = e~2x; x0 = 0 29. f ( x ) = cosx; x0 = n
28. /(*) = sinx; x0 = n/2 30. ln(x + 1); x0 = 0
31-34 True-False Determine whether the statement is true or false. Explain your answer. 31. The equation of a tangent line to a differentiable function is a first-degree Taylor polynomial for that function. 32. The graph of a function / and the graph of its Maclaurin polynomial have a common ^-intercept. 33. If pf, (x) is the sixth-degree Taylor polynomial for a function / about x = x0, then p(*\x0) = 4!/<4> (*,,). 34. If p^(x) is the fourth-degree Maclaurin polynomial for ex, then 9 z
k -p4(2)|<-
35-36 Use the method of Example 7 to approximate the given expression to the specified accuracy. Check your answer to that produced directly by your calculating utility. > 35. */e; four decimal-place accuracy
39. Let p\ (x) and p2(x) be the local linear and local quadratic approximations of f ( x ) = esmx at x = 0. (a) Use a graphing utility to generate the graphs of f ( x ) , p \ ( x ) , and p2(x) on the same screen for -1 <x < 1. (b) Construct a table of values of f ( x ) , p\(x), and p2(x) for x = -1.00, -0.75, -0.50, -0.25, 0, 0.25, 0.50, 0.75, 1.00. Round the values to three decimal places. (c) Generate the graph of \f(x) — p \ ( x ) \ , and use the graph to determine an interval on which p i ( x ) approximates f ( x ) with an error of at most ±0.01. [Suggestion: Review the discussion relating to Figure 2.9.4.] (d) Generate the graph of |/(jc) — p2(x)\, and use the graph to determine an interval on which p2(x) approximates /(jc) with an error of at most ±0.01. 40. (a) The accompanying figure shows a sector of radius r and central angle 2a. Assuming that the angle a is small, use the local quadratic approximation of cos a at a = 0 to show that x ~ ra2/2. (b) Assuming that the Earth is a sphere of radius 4000 mi, use the result in part (a) to approximate the maximum amount by which a 100 mi arc along the equator will diverge from its chord.
< Figure Ex-40
9.8 Maclaurin and Taylor Series; Power Series 41. (a) Find an interval [0, b] over which e* can be approximated by 1 + x + (x2/2\) to three decimal-place accuracy throughout the interval. (b) Check your answer in part (a) by graphing
659
] 43-46 Use the Remainder Estimation Theorem to find an interval containing x = 0 over which f ( x ) can be approximated by p(x) to three decimal-place accuracy throughout the interval. Check your answer by graphing \f(x) — p(x)\ over the interval you obtained. x3 43. f ( x ) = smx; p(x) = x - —
over the interval you obtained.
X2
X4
44. f ( x ) = cosx; p(x) = 1 - — + — 42. Show that the nth Taylor polynomial for sinh x about x = In 4 is 16-(-I) 4 -(jE-ln4)*
45. f(x) =
E
1 l+x
:;
p(x) = 1 -x2+x4
46. /(*) = ln(l + x); p ( x ) = x-~ +
A=0
-
t/QUlCK CHECK ANSWERS 9.7
2.1; 2; 2; | 5. (a) f(x)-pn(x)
3. 3 -4(* - -2)2)2
4. -1; 5; -10; 10
(b) ||x-2| 5
MACLAURIN AND TAYLOR SERIES; POWER SERIES Recall from the last section that the nth Taylor polynomial pn (x) at x — x^for a function f was defined so its value and the values of its first n derivatives match those of f at XQ. This being the case, it is reasonable to expect that for values ofx near XQ the values of pn(x) will become better and better approximations of f ( x ) as n increases, and may possibly converge to f ( x ) as n-+ +co. We will explore this idea in this section.
MACLAURIN AND TAYLOR SERIES In Section 9.7 we defined the nth Maclaurin polynomial for a function / as
>"•
2!
and the nth Taylor polynomial for / about x = XQ as f(k) k=0
k\
= /(*)) + /'(*>)(*-• +
,
" 21
1
n\
It is not a big step to extend the notions of Maclaurin and Taylor polynomials to series by not stopping the summation index at n. Thus, we have the following definition.
660
Chapter 9 / Infinite Series
9.8.1
DEFINITION
If / has derivatives of all orders at XQ, then we call the series
k\
k=0
( 1- /
-I
+
(1)
the Taylor series for f about x = XQ. In the special case where XQ = 0, this series becomes
E
-I
f'(0)x +
XK +
+
k\
(2)
in which case we call it the Maclaurin series for f .
Note that the «th Maclaurin and Taylor polynomials are the «th partial sums for the corresponding Maclaurin and Taylor series. > Example 1
Find the Maclaurin series for (a) ex
(b) sinjc
(c) CGSJC
(d)
1 -x
Solution (a). In Example 2 of Section 9.7 we found that the nth Maclaurin polynomial for ex is « t „„ V2
Thus, the Maclaurin series for ex is 00
k
V^ k\ = i k=0
X
~k\
2!
Solution (b). In Example 3(a) of Section 9.7 we found that the Maclaurin polynomials for sin x are given by X
P2k+\(x) = P2/C+2W = X
---
X
1
3! 5! Thus, the Maclaurin series for sin x is QC
/
k=0
Y \
/
x3
2k+\
X,,
-----
7!
(-•
x5
-(-D*
. ... (2k + 1)!
(* = 0 , 1 , 2 , . . . )
x1
~
(2Jk+l)!
Solution (c). In Example 3(b) of Section 9.7 we found that the Maclaurin polynomials for cos x are given by P2k(x) = P2k+\(x) = 1 - — + — - — 2! 4! o!
(2*)!
(k = 0 , 1 , 2 , . . . )
Thus, the Maclaurin series for cos x is ; ivt _ _ J ___ I _____ I (2K)\ 2! 4! 6! X2
X4
X6
..2* l_
(2k)\
9.8 Maclaurin and Taylor Series; Power Series
Solution (rf). for
661
In Example 5 of Section 9.7 we found that the nth Maclaurin polynomial
= 0,1,2,...) Thus, the Maclaurin series for 1/(1 — x) is = 1 +x+x2-{ k=0
> Example 2
Find the Taylor series for l/jc about x — 1.
Solution. In Example 6 of Section 9.7 we found that the nth Taylor polynomial for 1 /x about x = 1 is
£(-!)*(* - If = 1 - (x - 1) + (x - I) 2 - (x - I) 3 + • • • + (-!)"(* - 1)" k=0
Thus, the Taylor series for 1/Jt about jt = 1 is GC
£(-!)*(* - 1)* = 1 - (* - 1) + (x - I) 2 - Oc - I) 3 + - • • + (-!)*(* - 1)* + • • • k=0
POWER SERIES IN x
Maclaurin and Taylor series differ from the series that we have considered in Sections 9.3 to 9.6 in that their terms are not merely constants, but instead involve a variable. These are examples of power series, which we now define. If CQ, c\, 02, • • • are constants and x is a variable, then a series of the form ckxk - c0 + c\x + c2x2
(3)
ckx
k=0
is called a power series in x. Some examples are
X
3
k=0
X2 k
(2fc)l
X2 =
" 2l
+
X4
X6
4[ ~ 6[
From Example 1, these are the Maclaurin series for the functions 1/(1 — x),e*, and cos x, respectively. Indeed, every Maclaurin series
E is a power series in x.
** = /(O) + /'(0)oc +
2!
k\
662
Chapter 9 / Infinite Series RADIUS AND INTERVAL OF CONVERGENCE
If a numerical value is substituted for A: in a power series Yl ctxk, then the resulting series of numbers may either converge or diverge. This leads to the problem of determining the set of z-values for which a given power series converges; this is called its convergence set. Observe that every power series in x converges at x — 0, since substituting this value in (3) produces the series CO + 0 + 0 + 0 + - - - + 0 + - - -
whose sum is CQ. In some cases x = 0 may be the only number in the convergence set; in other cases the convergence set is some finite or infinite interval containing x = 0. This is the content of the following theorem, whose proof will be omitted.
9.8.2
THEOREM For any power series in x, exactly one of the following is true:
(a)
The series converges only for x = 0.
(b)
The series converges absolutely (and hence converges) for all real values of x.
(c)
The series converges absolutely (and hence converges) for all x in some finite open interval (-R, R) and diverges ifx < — R or x > R. At either of the values x = R or x — — R, the series may converge absolutely, converge conditionally, or diverge, depending on the particular series.
This theorem states that the convergence set for a power series in x is always an interval centered at x — 0 (possibly just the value x = 0 itself or possibly infinite). For this reason, the convergence set of a power series in x is called the interval of convergence. In the case where the convergence set is the single value x = 0 we say that the series has radius of convergence 0, in the case where the convergence set is (—00, +00) we say that the series has radius of convergence +00, and in the case where the convergence set extends between —R and R we say that the series has radius of convergence R (Figure 9.8.1). Diverges
Diverges 0
!___ „
Converges
^
0 Diverges
> Figure 9.8.1
Diverges
Converges
0
-R
-
™
~ „-
R
FINDING THE INTERVAL OF CONVERGENCE
The usual procedure for finding the interval of convergence of a power series is to apply the ratio test for absolute convergence (Theorem 9.6.5). The following example illustrates how this works. >• Example 3 Find the interval of convergence and radius of convergence of the following power series. /
(t=0
i=0
*=0
1\fc
it
9.8 Maclaurin and Taylor Series; Power Series
663
Solution (a). Applying the ratio test for absolute convergence to the given series, we obtain p — lim = lim |jc| = \x\ = lim uk so the series converges absolutely if p = \x\ < 1 and diverges if p = \x\ > 1. The test is inconclusive if \x\ = 1 (i.e., if x = 1 or x — — 1), which means that we will have to investigate convergence at these values separately. At these values the series becomes
= 1 + 1 + 1 + 1 +. k=0
both of which diverge; thus, the interval of convergence for the given power series is (— 1 , 1 ), and the radius of convergence is R — 1 . Solution (b). Applying the ratio test for absolute convergence to the given series, we obtain fe! Uk+\ p = lim = lim — lim k^+xi k—>•+<* 1 uk Since p < 1 for all x, the series converges absolutely for all x. Thus, the interval of convergence is (-co, +00) and the radius of convergence is R = +00. Solution (c).
Ifx •£ 0, then the ratio test for absolute convergence yields
p = lim
Uk+]
uk
— lim
k\xk
= lim \(k+\)x\ =
Therefore, the series diverges for all nonzero values of x. Thus, the interval of convergence is the single value x = 0 and the radius of convergence is R — 0. Solution (d).
Since |(-l)fc| = |(-ir +1 1 = 1, we obtain p = lim fc^+oo
Uk+\
uk
— lim
3k(k+l)
1*1 3
The ratio test for absolute convergence implies that the series converges absolutely if | x \ < 3 and diverges if \x\ > 3. The ratio test fails to provide any information when |jc| =3, so the cases x = — 3 and x = 3 need separate analyses. Substituting x = — 3 in the given series yields « ,_itf,_^t i
= £•
which is the divergent harmonic series given series yields A (-1)*3*
E jfc+l
-. Substituting x = 3 in the
1 1 1 = l - r + T--r +
k=0
which is the conditionally convergent alternating harmonic series. Thus, the interval of convergence for the given series is (—3, 3] and the radius of convergence is R — 3. -^
664
Chapter 9 / Infinite Series POWER SERIES IN x - X0
If XQ is a constant, and if x is replaced by x — XQ in (3), then the resulting series has the form C2(X - X0)2 H ----- h Ck(x -
Ck(x - X0f — C0 k=0
This is called apower series in x — XQ. Some examples are |
(* - 1)| (x - I) 2
3)*
, . , ,. , = 1 - (* + 3) +
~ D* = l
|
(x - I) 3
|
Jt+1
k=0
E t=o
k\
2!
3!
+•
The first of these is a power series in x — 1 and the second is a power series in x + 3. Note that a power series in x is a power series in x — XQ in which XQ — 0. More generally, the Taylor series
k\ is a power series in x — XQ. The main result on convergence of a power series in x — XQ can be obtained by substituting x — XQ for x in Theorem 9.8.2. This leads to the following theorem.
9.8.3 THEOREM For a power series J^ck(x — x0)k, exactly one of the following statements is true: (a)
The series converges only for x = XQ.
(b)
The series converges absolutely (and hence converges) for all real values ofx.
(c) The series converges absolutely (and hence converges) for all x in some finite open interval (XQ — R,XQ + R) and diverges if x < XQ — R or x > XQ + R. At either of the values x = XQ — R or x = XQ + R, the series may converge absolutely, converge conditionally, or diverge, depending on the particular series.
It follows from this theorem that the set of values for which a power series in x — XQ converges is always an interval centered at x = XQ; we call this the interval of convergence (Figure 9.8.2). In part (a) of Theorem 9.8.3 the interval of convergence reduces to the single value x = XQ, in which case we say that the series has radius of convergence R = 0; in part
Diverges
Diverges Radius of convergence R = 0
Converges Radius of convergence R = +=
Diverges
K Figure 9.8.2
XO-R
Converges XQ
Diverges
XO + R
Radius of convergence R
9.8 Maclaurin and Taylor Series; Power Series
665
(b) the interval of convergence is infinite (the entire real line), in which case we say that the series has radius of convergence R = +00; and in part (c) the interval extends between XQ — R and XQ + R, in which case we say that the series has radius of convergence R.
Example 4 Find the interval of convergence and radius of convergence of the series
k=\ Solution. We apply the ratio test for absolute convergence.
p = lim
(x - 5)*+1 (k + I) 2
«t+l uk —
Mr
lim
SI 1 -
/k2
(* - 5)*
* VI = \x-5\
= |jc-5| lim
Thus, the series converges absolutely if |jc — 5| < 1, or -1 < x — 5 < I, or 4 < x < 6. The series diverges if x < 4 or x > 6. To determine the convergence behavior at the endpoints x = 4 and x — 6, we substitute these values in the given series. If x —6, the series becomes
i* _ A =i
~
'
+ i + i + i4
which is a convergent ^-series (p = 2). If x = 4, the series becomes 00
It will always be a waste of time to test for convergence at the endpoints of the interval of convergence using the ratio test, since p will always be i at those
since
(— l11*
\ "* v
>
^— '
2
1
1
^
^
1
__ _ i .__.______i__; __ 2
^ series converges absolutely, the interval of convergence for the given series is
[4,6]. The radius of convergence is R = 1 (Figure 9.8.3). •*
,.lim «t+i exists. Explain why this must be so.
Series diverges
Series converges absolutely
Series diverges
-5
> Figure 9.8.3
-R= 1-
-R= 1-
FUNCTIONS DEFINED BY POWER SERIES
If a function / is expressed as a power series on some interval, then we say that the power series represents f on that interval. For example, we saw in Example 4(a) of Section 9.3 that
if\x\ < 1, so this power series represents the function 1 /(I — x) on the interval — 1 < x < 1.
666 Chapter 9 / Infinite Series Sometimes new functions actually originate as power series, and the properties of the functions are developed by working with their power series representations. For example, the functions
(-1)** *2*
= y
j
22*(£!)2
22(1!)2
24(2!)2
26(3!)2
(4)
25(2!)(3!)
(5)
and X
2M
k=0
2 (kl)(k+l)l
2
which are called Bessel functions in honor of the German mathematician and astronomer Friedrich Wilhelm Bessel (1784-1846), arise naturally in the study of planetary motion and in various problems that involve heat flow. To find the domains of these functions, we must determine where their defining power series converge. For example, in the case of JQ(X) we have
p — lim
k—>+oo
— lim
Generated by Mathematica
= lim
A Figure 9.8.4
TECHNOLOGY MASTERY Many computer algebra systems have
= 0< 1
so the series converges for all x; that is, the domain of JQ(X) is (—00, +00). We leave it as an exercise (Exercise 59) to show that the power series for J\ (x) also converges for all x. Computer-generated graphs of JQ(X) and J\ (x) are shown in Figure 9.8.4.
the Bessel functions as part of their libraries. If you have a CAS with Bessel functions, use it to generate the graphs in Figure 9.8.4.
UICK CHECK EXERCISES 9.8
(See page 668 for answers.)
1. If / has derivatives of all orders at XQ, then the Taylor series for / about x = XQ is defined to be
the radius of convergence for the infinite series is
EtU (!/•%/*)(* -4)*
(b) When x = 3,
E2. Since
lim
2***
k=\ V^
= 2\x\
k=\ ^
Does this series converge or diverge? k k
the radius of convergence for the infinite series is __
(c) When* = 5,
3. Since
lim
t-V+oo
&xk)/k\
= lim
the interval of convergence for the series
3x
k+l
=0
is
Does this series converge or diverge? (d) The interval of convergence for the infinite series
4. (a) Since
lim
= lim
t-v+oo
= \x-4\
9.8 Maclaurin and Taylor Series; Power Series EXERCISE SET 9.8
B Graphing Utility
667
@ CAS
1-10 Use sigma notation to write the Maclaurin series for the function. 1. e'1
2. eax
5. l n ( l + A )
3. cos TTA
6. ,
1
4. sin nx
1. cosh A
1 +A
8. sinhA
9. A sin A
10. xex
/
*(*
11-18 Use sigma notation to write the Taylor series about A = AQ for the function. 12. = ln2 11. ex; A0 = 1 13. -; A0 = -l
14.
15. sin TTA ; AO = -
16. cos A; AQ = —
17. IHA; AO = 1
18. In*; XQ = e
9-\fc,,£
36.
35.
38.
A
19-22 Find the interval of convergence of the power series, and find a familiar function that is represented by the power series on that interval. 19. 1 — A + A (-l)V + • • 2
20. 1 + A
2
+A
2* 43.
45.
(A + 5)*
46- E
,3
4
21. 1 + (x - 2)
- 2Y +
+
+ • •• + (-!)*(*+ 3)* .... 23. Suppose that the function / is represented by the power
(a) Find the domain of /. (b) Find /(O) and /(I). 24. Suppose that the function / is represented by the power series A - 5 (x - 5)2 (x - 5)3 /(A) = 1 - —^ + 3 32 33 (a) Find the domain o f / . (b) Find/(3) and/(6). 25-28 True-False Determine whether the statement is true or false. Explain your answer. 25. If a power series in A converges conditionally at A = 3, then the series converges if x\ < 3 and diverges if \x > 3. 26. The ratio test is often useful to determine convergence at the endpoints of the interval of convergence of a power series. 27. The Maclaurin series for a polynomial function has radius of convergence +00.
-A xk
28. The series > — converges if \x\ < 1. k=0
29-48 Find the radius of convergence and the interval of convergence.
47.
- 3)k
~ I) 2*
49. Use the root test to find the interval of convergence of k
50. Find the domain of the function 1 3 5 2 1 m - Y - - -< *- >JC* £j (2k- 2)1 51. Show that the series
1 _ _ _l_ V
V2
V3
_ _l
is the Maclaurin series for the function cos~Jx, A> 0
{
cosh -J—x, A < 0 [Hint: Use the Maclaurin series for cos A and cosh A to obtain series for cos *Jx, where x > 0, and cosh ^/—A, where A <0.]
52. If a function / is represented by a power series on an interval, then the graphs of the partial sums can be used as approximations to the graph of /. (a) Use a graphing utility to generate the graph of 1/(1 — A) together with the graphs of the first four partial sums of its Maclaurin series over the interval (-1,1). (b) In general terms, where are the graphs of the partial sums the most accurate?
668 Chapter 9 / Infinite Series 53. Prove: (a) If / is an even function, then all odd powers of x in its Maclaurin series have coefficient 0. (b) If / is an odd function, then all even powers of x in its Maclaurin series have coefficient 0. 54. Suppose that the power series ]T Ck(x — xo)k has radius of convergence R and p is a nonzero constant. What can you say about the radius of convergence of the power series ^ pc^(x — *o)*? Explain your reasoning. [Hint: See Theorem 9.4.3.] 55. Suppose that the power series ^Ck(x — xo)k has a finite radius of convergence R, and the power series Yl dk(x — xo)k has a radius of convergence of +co. What can you say about the radius of convergence of XXct + dk)(x — xo)kl Explain your reasoning. 56. Suppose that the power series ^c^(x — xo)k has a finite radius of convergence R\ and the power series ]T) dk(x — xo)k has a finite radius of convergence R2. What can you say about the radius of convergence of XXC* + dk)(x — xo)kl Explain your reasoning. [Hint: The case RI = RI requires special attention.] 57. Show that if p is a positive integer, then the power series
has a radius of convergence of l/pp. 58. Show that if p and q are positive integers, then the power series
59. Show that the power series representation of the Bessel function Ji(x) converges for all x [Formula (5)]. 60. Approximate the values of the Bessel functions JQ(X) and J\ (x) at x = 1, each to four decimal-place accuracy. |"c"| 61. If the constant p in the general /^-series is replaced by a variable x for x > 1, then the resulting function is called the Riemann zeta function and is denoted by
(a) Let sn be the nth partial sum of the series for f (3.7). Find « such that sn approximates t; (3.7) to two decimalplace accuracy, and calculate sn using this value of n. [Hint: Use the right inequality in Exercise 36(b) of Section 9.4 with f ( x ) = I/A:3'7.] (b) Determine whether your CAS can evaluate the Riemann zeta function directly. If so, compare the value produced by the CAS to the value of sn obtained in part (a). 62. Prove: If lim^+00 \ck\llk = L, where L ^ 0, then 1/L is the radius of convergence of the power series 63. Prove: If the power series v^°°=o c~-kx•-* nas radius of convergence R, then the series ] =o Ckx2k has radius of convergence -J~R. 64. Prove: If the interval of convergence of the series Y^k=o ck(x ~ xo)k is (xo — R,XO + R], then the series converges conditionally at XQ + R. 65. Writing The sine function can be defined geometrically from the unit circle or analytically from its Maclaurin series. Discuss the advantages of each representation with regard to providing information about the sine function.
has a radius of convergence of +00.
I/QUICK CHECK ANSWERS 9.8 1.
k\
(jc-jto)*
2. -
3. (-00, +00)
4. (a) 1 (b) converges (c) diverges (d) [3,5)
CONVERGENCE OF TAYLOR SERIES In this section we will investigate when a Taylor series for a function converges to that function on some interval, and we will consider how Taylor series can be used to approximate values of trigonometric, exponential, and logarithmic functions.
THE CONVERGENCE PROBLEM FOR TAYLOR SERIES
Recall that the «th Taylor polynomial for a function / about x = XQ has the property that its value and the values of its first n derivatives match those of / at XQ. As n increases,
9.9 Convergence of Taylor Series
669
more and more derivatives match up, so it is reasonable to hope that for values of x near XQ the values of the Taylor polynomials might converge to the value of /(*); that is,
(1)
kl
fe=0
However, the wth Taylor polynomial for / is the nth partial sum of the Taylor series for /, so (1) is equivalent to stating that the Taylor series for / converges at x, and its sum is f ( x ) . Thus, we are led to consider the following problem.
Problem 9.9.1 is concerned not only with whether the Taylor series of a function / converges, but also whether it converges to the function / itself. Indeed, it is possible for a Taylor series of a function / to converge to values different from /(*) for certain values of x (Exercise 14).
9.9.1 PROBLEM Given a function / that has derivatives of all orders at x = XQ, determine whether there is an open interval containing XQ such that /(jc) is the sum of its Taylor series about x = XQ at each point in the interval; that is,
k=0
k\
, (x
(2)
for all values of x in the interval.
One way to show that (1) holds is to show that lim
n—> +GO
^/'
fft( x )\ -
=0
k=o
However, the difference appearing on the left side of this equation is the nth remainder for the Taylor series [Formula (12) of Section 9.7]. Thus, we have the following result.
9.9.2
THEOREM
The equality M k=0
holds at a point x if and only if lim Rn(x) — 0. 71^ +CO
ESTIMATING THE nTH REMAINDER
It is relatively rare that one can prove directly that Rn(x) —»•0 as n —>• +00. Usually, this is proved indirectly by finding appropriate bounds on |/? n (jc)| and applying the Squeezing Theorem for Sequences. The Remainder Estimation Theorem (Theorem 9.7.4) provides a useful bound for this purpose. Recall that this theorem asserts that if M is an upper bound for |/ ( " +l) (*)l on an interval containing XQ, then M \X (n + 1)!
-X0\
n+\
(3)
for all x in that interval. The following example illustrates how the Remainder Estimation Theorem is applied. Example 1
Show that the Maclaurin series for cos x converges to cos x for all x; that x4
x6
(-co < x < +co)
670
Chapter 9 / Infinite Series
Solution. From Theorem 9.9.2 we must show that Rn(x) -» 0 for all x as n -> +00. For this purpose let f ( x ) — cosx, so that for all x we have f(n+]\x) = icosjc
or / ( " +1) (jc) =
In all cases we have / (n+1) O)| < 1 , so we can apply (3) with M = lander, = 0 to conclude that „+] 0
The method of Example 1 can be easily modified to prove that the Taylor series for sin z and cos x about any point x = xo converge to sin x and cos x, respectively, for all x (Exercises 21 and 22). For reference, some of the most important Maclaurin series are listed in Table 9.9.1 at the end of this section.
1*1
(4)
However, it follows from Formula (5) of Section 9.2 with n + 1 in place of n and \x\ in place of x that in+l lim =0 (5)
(n + 1)!
Using this result and the Squeezing Theorem for Sequences (Theorem 9.1.5), it follows from (4) that |R n (x) \ ->• 0 and hence that Rn (x) -+ 0 as n -> +00 (Theorem 9.1.6). Since this is true for all x, we have proved that the Maclaurin series for COSJE converges to cosx for all x. This is illustrated in Figure 9.9.1, where we can see how successive partial sums approximate the cosine curve more and more closely. •*
> Figure 9.9.1
APPROXIMATING TRIGONOMETRIC FUNCTIONS
In general, to approximate the value of a function / at a point x using a Taylor series, there are two basic questions that must be answered: • About what point XQ should the Taylor series be expanded? • How many terms in the series should be used to achieve the desired accuracy? In response to the first question, XQ needs to be a point at which the derivatives of / can be evaluated easily, since these values are needed for the coefficients in the Taylor series. Furthermore, if the function / is being evaluated at x, then JCQ should be chosen as close as possible to x, since Taylor series tend to converge more rapidly near XQ. For example, to approximate sin 3° (— n/60 radians), it would be reasonable to take XQ = 0, since jr/60 is close to 0 and the derivatives of sin x are easy to evaluate at 0. On the other hand, to approximate sin 85° (— 177T/36 radians), it would be more natural to take jcn = n/2, since Yin / 36 is close to n/2 and the derivatives of sin* are easy to evaluate at n/2. In response to the second question posed above, the number of terms required to achieve a specific accuracy needs to be determined on a problem-by-problem basis. The next example gives two methods for doing this.
*• Example 2 Use the Maclaurin series for sin x to approximate sin 3° to five decimalplace accuracy.
9.9 Convergence of Taylor Series
671
Solution. In the Maclaurin series 2k+l
sinx —
(6)
(2k+l)\
the angle x is assumed to be in radians (because the differentiation formulas for the trigonometric functions were derived with this assumption). Since 3° = jz/60 radians, it follows from (6) that 1
(7T/60)3 (7T/60)5 (n/60) .re (?r\ sin 3° == sin — = — I — (7) 60 \60/ 3! 5! 7! We must now determine how many terms in the series are required to achieve five decimalplace accuracy. We will consider two possible approaches, one using the Remainder Estimation Theorem (Theorem 9.7.4) and the other using the fact that (7) satisfies the hypotheses of the alternating series test (Theorem 9.6.1).
Method 1. (The Remainder Estimation Theorem) Since we want to achieve five decimal-place accuracy, our goal is to choose n so that the absolute value of the nth remainder at x — n/60 does not exceed 0.000005 = 5 x 10~6; that is, — < 0.000005 (8) However, if we let f ( x ) = sinx, then /("+1 \x) is either isinjc or icosjc, and in either case \f(fl+^(x)\ < 1 for all x. Thus, it follows from the Remainder Estimation Theorem with M = 1, x0 = 0, and x = n/60 that (n/60) (n
B+l
Thus, we can satisfy (8) by choosing n so that (7T/60)"+1
(n + 1)!
< 0.000005
With the help of a calculating utility you can verify that the smallest value of n that meets this criterion is n — 3. Thus, to achieve five decimal-place accuracy we need only keep terms up to the third power in (7). This yields sin3 c (verify). As a check, a calculator gives sin3 when rounded to five decimal places.
3! c
• 0.05233595624, which agrees with (9)
Method 2. (The Alternating Series Test) We leave it for you to check that (7) satisfies the hypotheses of the alternating series test (Theorem 9.6.1). Let sn denote the sum of the terms in (7) up to and including the nth power of jr/60. Since the exponents in the series are odd integers, the integer n must be odd, and the exponent of the first term not included in the sum sn must be n + 2. Thus, it follows from part (b) of Theorem 9.6.2 that ,n /60)n+2 This means that for five decimal-place accuracy we must look for the first positive odd integer n such that I // / I H / I (jr < 0.000005 (n + 2)!
672
Chapter 9 / Infinite Series
With the help of a calculating utility you can verify that the smallest value of n that meets this criterion is n = 3. This agrees with the result obtained above using the Remainder Estimation Theorem and hence leads to approximation (9) as before. -4
ROUNDOFF AND TRUNCATION ERROR
There are two types of errors that occur when computing with series. The first, called truncation error, is the error that results when a series is approximated by a partial sum; and the second, called roundoff error, is the error that arises from approximations in numerical computations. For example, in our derivation of (9) we took n — 3 to keep the truncation error below 0.000005. However, to evaluate the partial sum we had to approximate TI, thereby introducing roundoff error. Had we not exercised some care in choosing this approximation, the roundoff error could easily have degraded the final result. Methods for estimating and controlling roundoff error are studied in a branch of mathematics called numerical analysis. However, as a rule of thumb, to achieve n decimal-place accuracy in a final result, all intermediate calculations must be accurate to at least n + 1 decimal places. Thus, in (9) at least six decimal-place accuracy in n is required to achieve the five decimal-place accuracy in the final numerical result. As a practical matter, a good working procedure is to perform all intermediate computations with the maximum number of digits that your calculating utility can handle and then round at the end. APPROXIMATING EXPONENTIAL FUNCTIONS
*•• Example 3
Show that the Maclaurin series for ex converges to ex for all x; that is, \k
k=0
ki ~
+x +
x2
+
x3
vk
X
+
~v. ^y. '
( — 00 < X <
Ii
+00)
Solution. Let f(x) = ex, so that
We want to show that Rn (x) -» 0 as n -» +00 for all x in the interval — oo < x < +00. However, it will be helpful here to consider the cases x < 0 and x > 0 separately. If x < 0, then we will take the interval in the Remainder Estimation Theorem (Theorem 9.7.4) to be [x, 0], and if x > 0, then we will take it to be [0, x ] . Since / (n+1) (jt) = e* is an increasing function, it follows that if c is in the interval [x, 0], then
and if c is in the interval [0, x], then
Thus, we can apply Theorem 9.7.4 with M = 1 in the case where x < 0 and with M = ex in the case where x > 0. This yields 0
ifx < 0
(n
0 < \Rn(x)\ < e*
\x\
n+]
(n + 1)!
if x > 0
Thus, in both cases it follows from (5) and the Squeezing Theorem for Sequences that I Rn (x) I -> 0 as n ->• +00, which in turn implies that Rn (x) ->• 0 as n -> +00. Since this is true for all x, we have proved that the Maclaurin series for ex converges to ex for all x. <
9.9 Convergence of Taylor Series
673
Since the Maclaurin series for ex converges to ex for all x, we can use partial sums of the Maclaurin series to approximate powers of e to arbitrary precision. Recall that in Example 7 of Section 9.7 we were able to use the Remainder Estimation Theorem to determine that evaluating the ninth Maclaurin polynomial for ex at x = 1 yields an approximation for e with five decimal-place accuracy:
APPROXIMATING LOGARITHMS
The Maclaurin series l n ( l + * ) = j e - y + y - - ^ + -"
In Example 2 of Section 9.6, we stated without proof that
«-,-l + l-l + l-... This result can be obtained by letting
X
X
X
x) = -x------
text discussion, this series converges
(-!<*
(ID
and on subtracting (11) from (10) we obtain
In ~] James Gregory ! (1638-1675) Scottish I mathematician and astronomer. Gregory, the son of a minister, was famous in his time as the inventor of the Gregorian reflecting telescope, so named in his honor. Although he is not generally ranked with the great mathematicians, much of his work relating to calculus was studied by Leibniz and Newton and undoubtedly influenced some of their discoveries. There is a manuscript, discovered posthumously, which shows that Gregory had anticipated Taylor series well before Taylor.
(10)
is the starting point for the approximation of natural logarithms. Unfortunately, the usefulness of this series is limited because of its slow convergence and the restriction— 1 < x < 1. However, if we replace x by — x in this series, we obtain
x = I in (10), but as indicated in the too slowly to be of practical use.
(~l<x
X l+x = 2 *+ — + — + — 7 l-xj V 3
(-1 < x < 1)
(12)
Series (12), first obtained by James Gregory in 1668, can be used to compute the natural logarithm of any positive number y by letting
y=
l+x 1 -X
or, equivalently, x =
y-i y'+i
(13)
and noting that —1 < x < 1. For example, to compute In 2 we let y = 2 in (13), which yields x = |. Substituting this value in (12) gives In2 =
G)3 , G)5 , G)7
(14)
In Exercise 19 we will ask you to show that five decimal-place accuracy can be achieved using the partial sum with terms up to and including the 13th power of |. Thus, to five decimal-place accuracy
G)1
(I)3
13
0.69315
(verify). As a check, a calculator gives In 2 ?& 0.69314718056, which agrees with the preceding approximation when rounded to five decimal places. APPROXIMATING n
In the next section we will show that *3 tan x = x h Letting x = I , we obtain 71
- - an -\
v •
—
(15)
674
Chapter 9 / Infinite Series
or
1 1 1 -- + --- + . This famous series, obtained by Leibniz in 1674, converges too slowly to be of computational value. A more practical procedure for approximating n uses the identity
(16)
-+tan-'-
which was derived in Exercise 87 of Section 6.7. By using this identity and series (15) to approximate tan"1 | and tan~' |, the value of n can be approximated efficiently to any degree of accuracy. BINOMIAL SERIES If m is a real number, then the Maclaurin series for (1 + x)m is called the binomial series; it is given by
m(m — 1) i m(m — \)(m — 2) , m(m — 1) • • • (m — k + 1) , 1 + mx + --- xz + --- x1" + ••• + --- x" + In the case where m is a nonnegative integer, the function f(x) — (1 + x)m is a polynomial f (m+1) (0) — f (m+2) (0) = f (m+3) (0) —
of degree m, so Let f(x) = (1 + x)m. Verify that /(0) = 1
/'(0)=m y»(0) =m(m _ i)
f '"(0) = m(m - l)(m - 2)
=0
and the binomial series reduces to the familiar binomial expansion m(m-l)
(1 + x) — 1 + mx -\
--
2!
m(m
2
x
- 2)
3
x +•
3!
• + x
which is valid for -co < x < +00. ^ can ^e Proved that if m is not a nonnegative integer, then the binomial series converges to (1 + x)m if x\ < 1. Thus, for such values of x (1 +x)n = l+mx
m(m — 1)
1
--L
mini — 1) • • • (m — k + 1) !:-'.-!:-!—L
k\
2!
or in sigma notation, (m — 1) • • • (m — k +
Example 4
Find binomial series for
Solution (a). Since the general term of the binomial series is complicated, you may find it helpful to write out some of the beginning terms of the series, as in Formula (17), to see developing patterns. Substituting m = — 2 in this formula yields
21 4
3!
-
2! 3! 4! 2 3 - 2x + 3x - 4x + 5x4
k=0
4!
9.9 Convergence of Taylor Series A V
Solution (b).
675
Substituting m = -\ in (17) yields
2
2! 1 3
I 2*
+
'
,
3, 3!
! -3'5
2
23 • 3!;
22^2!* '
Figure 9.9.2 shows the graphs of the functions in Example 4 compared to their thirddegree Maclaurin polynomials. SOME IMPORTANT MACLAURIN SERIES For reference, Table 9.9.1 lists the Maclaurin series for some of the most important functions, together with a specification of the intervals over which the Maclaurin series converge to those functions. Some of these results are derived in the exercises and others will be derived in the next section using some special techniques that we will develop.
Table 9.9.1 SOME IMPORTANT MACLAURIN SERIES INTERVAL OF CONVERGENCE
MACLAURIN SERIES
A Figure 9.9.2
' xk = 1 + x + x2 + x3 -
-\<x<\ -\<x<\
\+x2
—oo < X < +00
21 r2k-t
31
cos* =
x2 21
(2k)\
+
51
7!
x4 x6 4 T ~ 6!" + "
-oo < x < +00
—oo < x < +00
,n x3
tan"' x =
x5
x1
T + T ~T
t=o x3
sinh j: =
x5
x1
—oo < jf < +00
D!
cosh x = 2_] t=0
X2
(2t)!
X4 +
2l 4!"
X6 +
~6\+"
-oo < x < +00
m(m- 1) ••• (tn-k+l)
The behavior at the endpoints depends on m\ For m > 0 the series converges absolutely at both endpoints; for m < -1 the series diverges at both endpoints; and for -1 < m < 0 the series converges conditionally at x = 1 and diverges at x = -1.
676
Chapter 9 / Infinite Series
I/QUICK CHECK EXERCISES 9.9
L
(See page 677 for answers.) 4. If m is a real number but not a nonnegative integer, the binomial series °°
j
k=0
/ * k=\
k=o
converges to (1 + x)m if \x\ < for x in the interval.
3. k=\
EXERCISE SET 9.9
Graphing Utility
CAS
1. Use the Remainder Estimation Theorem and the method of Example 1 to prove that the Taylor series for sin x about x = ji/4 converges to sinx for all x. 2. Use the Remainder Estimation Theorem and the method of Example 3 to prove that the Taylor series for e* about x = 1 converges to ex for all x. 3-10 Approximate the specified function value as indicated and check your work by comparing your answer to the function value produced directly by your calculating utility. I 3. Approximate sin 4° to five decimal-place accuracy using both of the methods given in Example 2. 4. Approximate cos 3° to three decimal-place accuracy using both of the methods given in Example 2. 5. Approximate cos 0.1 to five decimal-place accuracy using the Maclaurin series for cos*. 6. Approximate tan~' 0.1 to three decimal-place accuracy using the Maclaurin series for tan^1 x. 7. Approximate sin 85 ° to four decimal-place accuracy using an appropriate Taylor series. 8. Approximate cos (—175°) to four decimal-place accuracy using a Taylor series. 9. Approximate sinh 0.5 to three decimal-place accuracy using the Maclaurin series for sinh x. 10. Approximate cosh 0.1 to three decimal-place accuracy using the Maclaurin series for cosh x. 11. (a) Use Formula (12) in the text to find a series that converges to In 1.25. (b) Approximate In 1.25 using the first two terms of the series. Round your answer to three decimal places, and compare the result to that produced directly by your calculating utility. 12. (a) Use Formula (12) to find a series that converges to In 3. (b) Approximate In 3 using the first two terms of the series. Round your answer to three decimal places, and compare the result to that produced directly by your calculating utility.
FOCUS ON CONCEPTS
13. (a) Use the Maclaurin series for tan ' x to approximate tan"1 jandtan" 1 | to three decimal-place accuracy. (b) Use the results in part (a) and Formula (16) to approximate jr. (c) Would you be willing to guarantee that your answer in part (b) is accurate to three decimal places? Explain your reasoning. (d) Compare your answer in part (b) to that produced by your calculating utility. 14. The purpose of this exercise is to show that the Taylor series of a function / may possibly converge to a value different from f ( x ) for certain values of x. Let /(*) =
0,
(a) Use the definition of a derivative to show that /'(0) = 0. (b) With some difficulty it can be shown that if n > 2 then / (n) (0) = 0. Accepting this fact, show that the Maclaurin series of / converges for all x, but converges to f ( x ) only at x = 0. 15. (a) Find an upper bound on the error that can result if cos x is approximated by 1 — (x2/2\) + (x4/4\) over the interval [-0.2, 0.2]. (b) Check your answer in part (a) by graphing cos*-
-^_+^
over the interval. 16. (a) Find an upper bound on the error that can result if ln(\ + x) is approximated by x over the interval [-0.01,0.01]. (b) Check your answer in part (a) by graphing +x)-x\
over the interval.
9.9 Convergence of Taylor Series 17. Use Formula (17) for the binomial series to obtain the Maclaurin series for
(a)
(e) Use the results in parts (c) and (d) to show that five decimal-place accuracy will be achieved if n satisfies
1
(c)
l+x 18. If m is any real number, and k is a nonnegative integer, then we define the binomial coefficient m k
by the formulas
n w
= 1 and
m(m — l)(m -- 2 ) . . . ( m - i f c + 1 ) ,m (k, for k > 1. Express Formula (17) in the text in terms of binomial coefficients. 19. In this exercise we will use the Remainder Estimation Theorem to determine the number of terms that are required in Formula (14) to approximate In 2 to five decimal-place accuracy. For this purpose let = In
1 -x (a) Show that
=ln(l
- x)
-t_o+*)"+'
(-1 <x
(-1)"
4 T
< 0.000005
n +1
and then show that the smallest value of n that satisfies this condition is n = 13. 20. UseFormula(12)andthemethodofExercise 19toapproximateln (|) to five decimal-place accuracy. Then check your work by comparing your answer to that produced directly by your calculating utility. 21. Prove: The Taylor series for cos x about any value x = XQ converges to cos x for all x. 22. Prove: The Taylor series for sin x about any value x = XQ converges to sin x for all x. 23. Research has shown that the proportion p of the population with IQs (intelligence quotients) between a and ft is approximately p=
+1
(i-*)" j
ft*
= /
_i rj= !0ay>
e " "i > ax
16V27T, 2nJa
Use the first three terms of an appropriate Maclaurin series to estimate the proportion of the population that has IQs between 100 and 110.
(b) Use the triangle inequality iequal [Theorem 0.1 A(d)] to show that l/ ( " + 1 ) «l<
677
l
[c] 24. (a) In 1706 the British astronomer and mathematician John Machin discovered the following formula for it/4, (c) Since we want to achieve five decimal-place accuracy, called Machin's formula: our goal is to choose n so that the absolute value of n , 1 1 the nth remainder at x = | does not exceed the value -=4tan-'--tan-' — 5 0.000005 = 0.5 x 10~ ; that is, \Rn (|)| < 0.000005. Use the Remainder Estimation Theorem to show that Use a CAS to approximate jr/4 using Machin's formula this condition will be satisfied if n is chosen so that to 25 decimal places. M / (b) In 1914 the brilliant Indian mathematician Srinivasa < 0.000005 Ramanujan (1887-1920) showed that (n + 1)! > 3, (n+[) where \ f ( x ) \ < M on the interval [0, |]. 1 -26,390*) (d) Use the result in part (b) to show that M can be taken 9801 as Use a CAS to compute the first four partial sums in M = n\ | 1 + B+l Ramanujan's formula. (I) (1 +*)"+'
(1-.
E
•QUICK CHECK ANSWERS 9.9 1. (-
(2k)\
22 X- k\
\k+] 3. (-1)*+' —;(-!,!] 4.
m(m — 1)
k\
678
Chapter 9 / Infinite Series
«l DIFFERENTIATING AND INTEGRATING POWER SERIES; MODELING WITH TAYLOR SERIES In this section we will discuss methods for finding power series for derivatives and integrals of functions, and we will discuss some practical methods for finding Taylor series that can be used in situations where it is difficult or impossible to find the series directly. DIFFERENTIATING POWER SERIES
We begin by considering the following problem. 9.10.1 PROBLEM Suppose that a function / is represented by a power series on an open interval. How can we use the power series to find the derivative of / on that interval? The solution to this problem can be motivated by considering the Maclaurin series for 3 sin*: xJf5 xX 1 x
sinx=x _ _ + _ _ _
( — 00 < X <
+00)
Of course, we already know that the derivative of sin x is cos x; however, we are concerned here with using the Maclaurin series to deduce this. The solution is easy—all we need to do is differentiate the Maclaurin series term by term and observe that the resulting series is the Maclaurin series for cos x:
x2 ~
~
+
x4
- —H
= cos x
Here is another example.
d x d -r[e ] =~
dx
x
— 2!
ax
x
x
— + 4!3! 7 X
The preceding computations suggest that if a function / is represented by a power series on an open interval, then a power series representation of /' on that interval can be obtained by differentiating the power series for / term by term. This is stated more precisely in the following theorem, which we give without proof. 9.10.2 THEOREM (Differentiation of Power Series) Suppose that a function f is represented by a power series in x — x$ that has a nonzero radius of convergence R; that is, °° f ( x ) = ^ ck(x - x0)k (x0 - R < x < x0 + R) k=o Then: (a)
The function f is differentiable on the interval (XQ — R, XQ + R).
(b)
If the power series representation for f is differentiated term by term, then the resulting series has radius of convergence R and converges to f on the interval (XQ - R, XQ + R); that is,
= f—' y^y-[<*(*-• dx k=0
-R <x
9.10 Differentiating and Integrating Power Series; Modeling with Taylor Series
679
This theorem has an important implication about the differentiability of functions that are represented by power series. According to the theorem, the power series for /' has the same radius of convergence as the power series for /, and this means that the theorem can be applied to /' as well as /. However, if we do this, then we conclude that /' is differentiable on the interval (x0 - R, x0 + R), and the power series for /" has the same radius of convergence as the power series for / and /'. We can now repeat this process ad infinitum, applying the theorem successively to /", /'",..., f(n\ . . . to conclude that / has derivatives of all orders on the interval (XQ — R, XQ + R). Thus, we have established the following result.
9.10.3 THEOREM If a function f can be represented by a power series in x — XQ with a nonzero radius of convergence R, then f has derivatives of all orders on the interval (XQ - R, XQ + R).
In short, it is only the most "well-behaved" functions that can be represented by power series; that is, if a function / does not possess derivatives of all orders on an interval (XQ — R,x0 + R), then it cannot be represented by a power series in x — x0 on that interval.
> Example 1 In Section 9.8, we showed that the Bessel function JQ(X), represented by the power series 00 / _ \ f c 2t
k=0
(1)
22k(k\)2
has radius of convergence +00 [see Formula (7) of that section and the related discussion]. Thus, JQ(X) has derivatives of all orders on the interval (—00, +00), and these can be obtained by differentiating the series term by term. For example, if we write (1) as
and differentiate term by term, we obtain See Exercise 45 for a relationship be-
tween JQ(X) and J\ (x).
J0(x) =
-A (-l)k(2k)x2k-} L—i
k=\
REMARK
22k(kl)2
i
(—1-i
- 1)!
The computations in this example use some techniques that are worth noting. First, when a power series is expressed in sigma notation, the formula for the general term of the series will often not be of a form that can be used for differentiating the constant term. Thus, if the series has a nonzero constant term, as here, it is usually a good idea to split it off from the summation before differentiating. Second, observe how we simplified the final formula by canceling the factor k from one of the factorials in the denominator. This is a standard simplification technique.
INTEGRATING POWER SERIES
Since the derivative of a function that is represented by a power series can be obtained by differentiating the series term by term, it should not be surprising that an antiderivative of a function represented by a power series can be obtained by integrating the series term by term. For example, we know that sin x is an antiderivative of cos x. Here is how this result
680 Chapter 9 / Infinite Series can be obtained by integrating the Maclaurin series for cos x term by term: X2
I
X
-
-y
X4
x3 3(2!)
X6
x5 5(4!)
x7 7(6!)
C = sinx
_
X
3! + 5!
+C
7!
The same idea applies to definite integrals. For example, by direct integration we have
/•' I Jo
dx
~ = tan"1 x \ = tan ' 1 - tan 0 = -- 0 =
1+* 2
J0
and we will show later in this section that 1
71
1
1
(2)
Thus,
A
dx
1 1 1 3+ 5 ~ 7
Jo i
Here is how this result can be obtained by integrating the Maclaurin series for 1/(1 + x2) term by term (see Table 9.9.1):
•]dx X3
X5
X1
1
1
1
The preceding computations are justified by the following theorem, which we give without proof.
Theorems 9.10.2 and 9.10.4 tell us how to use a power series representa-
9.10.4 THEOREM (Integration of Power Series) Suppose that a function f is represented by a power series in x — XQ that has a nonzero radius of convergence R; that is,
tion of a function / to produce power series representations of /'(*) and of convergence as /.
ries may not be the same because their convergence behavior may differ at the endpoints of the interval. cises 25 and 26.)
(See Exer-
(x0 - R < x < x0 + R)
/t=o
However, the
intervals of convergence for these se-
ck(x -
/(*) =
/ f(x) dx that have the same radius
(a) If the power series representation off is integrated term by term, then the resulting series has radius of convergence R and converges to an antiderivative for f ( x ) on the interval (XQ — R,XQ + R); that is, c I
- R < x < x0 + R)
f(x)dx =
J
(b) If a and ft are points in the interval (XQ — R, XQ + R), and if the power series representation of f is integrated term by term from a. to ft, then the resulting series converges absolutely on the interval (XQ — R, XQ + R) and rP
/ Ja
ck(x-xo?dx
9.10 Differentiating and Integrating Power Series; Modeling with Taylor Series
681
I POWER SERIES REPRESENTATIONS MUST BE TAYLOR SERIES
For many functions it is difficult or impossible to find the derivatives that are required to obtain a Taylor series. For example, to find the Maclaurin series for 1/(1 + x2) directly would require some tedious derivative computations (try it). A more practical approach is to substitute — x2 for x in the geometric series
1 — 1 + x + x2 + x3 + x4 + 1 -x to obtain
1)
1 = 1 - x2 + x4 - x6 + l+x2
However, there are two questions of concern with this procedure: • Where does the power series that we obtained for 1/(1 + x2) actually converge to How do we know that the power series we have obtained is actually the Maclaurin series for The first question is easy to resolve. Since the geometric series converges to 1/(1 — x) if \x\ < 1, the second series will converge to 1/(1 + x2) if |— x2\ < 1 or \x2\ < 1. However, this is true if and only if \x \ < 1 , so the power series we obtained for the function l / ( l + j c 2 ) converges to this function if —1 < x < 1. The second question is more difficult to answer and leads us to the following general problem.
9.10.5 PROBLEM Suppose that a function / is represented by a power series in x — XQ that has a nonzero radius of convergence. What relationship exists between the given power series and the Taylor series for / about x = XQ!
The answer is that they are the same; and here is the theorem that proves it.
Theorem 9.10.6 tells us that no matter how we arrive at a power series representation of a function /,
be it by
9.10.6 THEOREM If a function f is represented by a power series in x — XQ on some open interval containing XQ, then that power series is the Taylor series for f about x = x0.
substitution, by differentiation, by integration, or by some algebraic process, that series will be the Taylor series for / about x = xo, provided the series converges to / on some open interval con-
PROOF Suppose that
taining XQ.
f(x) = c0
H ----- h ck(x -
-\
for all x in some open interval containing XQ. To prove that this is the Taylor series for / about x = XQ, we must show that k\
for
k = 0,1,2,3,
However, the assumption that the series converges to f ( x ) on an open interval containing XQ ensures that it has a nonzero radius of convergence R; hence we can differentiate term
682
Chapter 9 / Infinite Series by term in accordance with Theorem 9.10.2. Thus,
f(x) f'(x)
= c0 + ci(x - *o) + c2(x - xo)2 + c3(x - x0)3
c4(x -
2
= ci + 2c2(x - XQ) + 3c3(x - xo} + 4c4(x -
/"(*) = 2!c2 + (3 • 2)c3(* - x0) + (4 - 3)c4(x - x0)2 f'"(x) = 3!c3 + (4 • 3 • 2)c4(* - *„) + • • •
On substituting x = XQ, all the powers of x — XQ drop out, leaving = 3!c3, . . .
from which we obtain f'"(x0) C3 = 2! 3! which shows that the coefficients CQ, c\, c2,03, ... are precisely the coefficients in the Taylor series about XQ for f(x). • CO =
c2 =
f(XQ),
f"(xo)
SOME PRACTICAL WAYS TO FIND TAYLOR SERIES
Example 2
Find Taylor series for the given functions about the given XQ
(a) e~*\
xQ = 0
x0 = 1
(b) hue,
x0 = I
(c) -, x
Solution (a). The simplest way to find the Maclaurin series for e for x in the Maclaurin series
x
is to substitute — x2
(3)
to obtain
r4
r6
r8
X
A
X
Since (3) converges for all values of x, so will the series fore Solution (b). Table 9.9.1:
K
.
We begin with the Maclaurin series for ln(l + x), which can be found in
=X
jc" xj JT ~ ~2~ " T ~ T
(-1 <x < 1)
Substituting x — 1 for x in this series gives ln(l + [x - 1]) = In* = (*-!)-
(x - I) 2 „ 2
+
(x - I) 3
(x - I) 4
*•
^ (4) 3 4 Since the original series converges when — 1 < x < 1, the interval of convergence for (4) will be — 1 < x — I < 1 or, equivalently, 0 < x < 2. Solution (c). Since \/x is the derivative of In x, we can differentiate the series for In x found in (b) to obtain
2(x- 1) 3(x - I) 2 4(x - I) 3 ~T : ~ 4~ = 1 - (x - 1) + (x - I) 2 - (x - I) 3 + • • •
1 _ x=
(5)
9.10 Differentiating and Integrating Power Series; Modeling with Taylor Series •
681
POWER SERIES REPRESENTATIONS MUST BE TAYLOR SERIES
For many functions it is difficult or impossible to find the derivatives that are required to obtain a Taylor series. For example, to find the Maclaurin series for 1/(1 + x2) directly would require some tedious derivative computations (try it). A more practical approach is to substitute -x2 for x in the geometric series
1 = 1 + x + x2 + x3 + x4 + 1 -x to obtain
(-!<*< 1)
1 _
JL
->v
|
^v
s\,
\
j\,
However, there are two questions of concern with this procedure: • Where does the power series that we obtained for 1/(1 + x2) actually converge to • How do we know that the power series we have obtained is actually the Maclaurin series for l / ( l + j c 2 ) ? The first question is easy to resolve. Since the geometric series converges tol/(l — x) if |jc| < 1, the second series will converge to 1/(1 + x2) if |— x2\ < 1 or \x2\ < 1. However, this is true if andonlyif \x < 1, so the power series we obtained for the function 1/(I +;c2) converges to this function if —1 < x < 1. The second question is more difficult to answer and leads us to the following general problem.
9.10.5 PROBLEM Suppose that a function / is represented by a power series in x — XQ that has a nonzero radius of convergence. What relationship exists between the given power series and the Taylor series for / about x = XQ!
The answer is that they are the same; and here is the theorem that proves it.
Theorem 9.10.6 tells us that no matter how we arrive at a power series representation of a function /,
be it by
9.10.6 THEOREM If a function f is represented by a power series in x — XQ on some open interval containing XQ, then that power series is the Taylor series for f about x = x0.
substitution, by differentiation, by integration, or by some algebraic process, that series will be the Taylor series for / about x — XQ, provided the series converges to / on some open interval con-
PROOF Suppose that
taining XQ.
/(*) = C0 + Ci (X - XQ) +C2(X - XQ)2 -\ ----- h Ck(x - X0f
-\ ----
for all x in some open interval containing XQ. To prove that this is the Taylor series for / about x = XQ, we must show that
k\
for fc = 0, 1,2, 3,
However, the assumption that the series converges to f ( x ) on an open interval containing XQ ensures that it has a nonzero radius of convergence R; hence we can differentiate term
682
Chapter 9 / Infinite Series
by term in accordance with Theorem 9.10.2. Thus, f(x)
= C0 + Ci(x - XQ) + C2(X - Xo)2 + C3(x - XG}3 + C4(x - X0)4
f(x)
= c, + 2c2(x - x0) + 3c3(x - xo)2 + 4c4(x - x0f + •••
f"(x) = 2!c2 + (3 • 2)c3(x - x0) + (4 • 3)c4(x - x0)2 + ••• f'"(x) = 3!c3 + (4 • 3 • 2)c4(x - x0) + • • •
On substituting x = XQ, all the powers of x — XQ drop out, leaving /(XQ) = CQ,
f'(x0) = C\,
f"(x0)—2\C2,
f'"(x0) = 3\C^,...
from which we obtain Co = f(Xo),
/"(*o)
Ci = f(X0),
C2 — —
,
f'"(x0)
C3 =
, . . .
which shows that the coefficients CQ,ci, C 2 , C 3 , . . . are precisely the coefficients in the Taylor series about x0 for f ( x ) . • SOME PRACTICAL WAYS TO FIND TAYLOR SERIES
Example 2
Find Taylor series for the given functions about the given XQ
(a) e~x\
x0=0
(b) hut, XQ = 1
(c) -,
x0 = I
X
Solution (a). The simplest way to find the Maclaurin series for e for x in the Maclaurin series x2 x3 2l+^]
to obtain
x4
+
x
is to substitute — x2
x4 4!"
x6
(3)
Xs
Since (3) converges for all values of x, so will the series f o r e
K
.
Solution (b). We begin with the Maclaurin series for ln(l + x), which can be found in Table 9.9.1: X2
X3
X4
2 3 Substituting x — 1 for x in this series gives
( - ! < * < 1)
4
, h
ln(l + [x - 1]) = Inx = (x - 1) -
, h ••
(4)
2 3 4 Since the original series converges when — 1 < x < 1, the interval of convergence for (4) will be —1 < x — 1 < 1 or, equivalently, 0 < x < 2.
Solution (c). Since 1/x is the derivative of In x, we can differentiate the series for In x found in (b) to obtain 1 _
— — 1 —
2(x - 1) z
H~
3(* - I) 2 ^
2
4(jc - I) 3 '.
3
H~ • * *
= 1 - (x - 1) + (x - I) - (x - I) + •
(5)
9.10 Differentiating and Integrating Power Series; Modeling with Taylor Series
683
By Theorem 9.10.2, we know that the radius of convergence for (5) is the same as that for (4), which is R = I . Thus the interval of convergence for (5) must be at least 0 < x < 2. Since the behaviors of (4) and (5) may differ at the endpoints x — 0 and x — 2, those must be checked separately. When x = 0, (5) becomes 1 - (-1) + (-1)2 - (-1)3 + • • • = 1 + 1 + 1 + 1 + • - •
which diverges by the divergence test. Similarly, when x =2, (5) becomes 1 - 1 + I7 - 1-
• • = 1-1 + 1-1 +
which also diverges by the divergence test. Thus the interval of convergence for (5) is 0 < x < 2. +
> Examples Find the Maclaurin series for tan lx. Solution, It would be tedious to find the Maclaurin series directly. A better approach is r j to start with the formula I
J
l+
ClX
=
X ~~r C-
and integrate the Maclaurin series 1
(-1 <x < 1)
term by term. This yields
tan"1 x + C =
] dx
or
X3
X5
X1
X9
c
The constant of integration can be evaluated by substituting x = 0 and using the condition tan"1 0 = 0. This gives C = 0, so that x5
REMARK
x1
x9
( - ! < * < 1)
(6)
Observe that neither Theorem 9.10.2 nor Theorem 9.10.3 addresses what happens at the endpoints of the interval of convergence. However, it can be proved that if the Taylor series for / about x = x0 converges to f(x) for all x in the interval (x0 - R, x0 + R), and if the Taylor series converges at the right endpoint x0 + R, then the value that it converges to at that point is the limit of f(x) as x -> xa + R from the left; and if the Taylor series converges at the left endpoint x0 — R, then the value that it converges to at that point is the limit of f(x) as x —>• x0 — R from the right. For example, the Maclaurin series for tan"1 x given in (6) converges at both x = — 1 and x = \, since the hypotheses of the alternating series test (Theorem 9.6.1) are satisfied at those points. Thus, the continuity of tan"' x on the interval [—1,1] implies that at x = 1 the Maclaurin series converges to hm tan—i x = tan—i 1 = —
and at x = -1 it converges to lim tan"1 x = tan~'(—1) = :->-!+
4
1
This shows that the Maclaurin series for tan" x actually converges to tan"1 x on the closed interval — 1 < x < 1. Moreover, the convergence at x = 1 establishes Formula (2).
684 Chapter 9 / Infinite Series •
APPROXIMATING DEFINITE INTEGRALS USING TAYLOR SERIES
Taylor series provide an alternative to Simpson's rule and other numerical methods for approximating definite integrals.
Example 4 Approximate the integral /"'
/
2
e
~*~
dx
Jo
to three decimal-place accuracy by expanding the integrand in a Maclaurin series and integrating term by term. Solution. We found in Example 2(a) that the Maclaurin series for e~* is _X2 _
6
_
^ _ ^
2
~ ~X
+
*?_
2[ " 3i + 4[
Therefore, /•'
2
/•' f
/ e~x dx = I 1 -. Jo Jo I. _ 3
3
+
5(2!) ~ 7(3!)
5-2!
+
9(4!)
1 1 + 7-3! 9-4!
(-1)" Since this series clearly satisfies the hypotheses of the alternating series test (Theorem 9.6.1), it follows from Theorem 9.6.2 that if we approximate the integral by 5,, (the nth partial sum of the series), then 1 [2(n + l) + !](» + !)!
e * dx — sn
1 (2n + 3)(n + 1)1
Thus, for three decimal-place accuracy we must choose n such that 1
(2n + 3)(n + 1)!
< 0.0005 = 5 x 10-4
With the help of a calculating utility you can show that the smallest value of n that satisfies this condition is n = 5. Thus, the value of the integral to three decimal-place accuracy is 1 What advantages does the method of Example 4 have over Simpson's rule? What are its disadvantages?
/'-'' Jo
1
1
1
1
9-4!
11-5!
: 0.747
As a check, a calculator with a built-in numerical integration capability produced the approximation 0.746824, which agrees with our result when rounded to three decimal places. <
FINDING TAYLOR SERIES BY MULTIPLICATION AND DIVISION
The following examples illustrate some algebraic techniques that are sometimes useful for finding Taylor series.
9.10 Differentiating and Integrating Power Series; Modeling with Taylor Series
685
> Example 5 Find the first three nonzero terms in the Maclaurin series for the function
Solution. Using the series for e * and tan ' x obtained in Examples 2 and 3 gives tan
/, 22 + — ---- U - — + — x=ll-x
1
Multiplying, as shown in the margin, we obtain e
l-^-+i— 2 24
. 6
120
*-*!+*!-. 2 24
tan
4 3 31 5 x = x - -AT + — x — • • •
More terms in the series can be obtained by including more terms in the factors. Moreover, one can prove that a series obtained by this method converges at each point in the intersection of the intervals of convergence of the factors (and possibly on a larger interval). Thus, we can be certain that the series we have obtained converges for all x in the interval — 1 < x < 1 (why?). -4
,
T
30
15
TECHNOLOGY MASTERY If you have a CAS. use its capability for ... , . , .. ... . .- . multiplying and dividing polynomials to perform the computations in Examples 5 and 6.
Example 6
Find the first three nonzero terms in the Maclaurin series for tanx.
Using the first three terms in the Maclaurin series for sin x and cos x, we can express t a n A: a s 3 5 x— sin x 3! 5! tanjc = cos* Dividing, we obtain 6 as shown in the margin, 6 ' x3 2x5 tan x = x + —- + —— +
MODELING PHYSICAL LAWS WITH TAYLOR SERIES
Taylor series provide an important way of modeling physical laws. To illustrate the idea we will consider the problem of modeling the period of a simple pendulum (Figure 9.10.1). As explained in Chapter 7 Making Connections Exercise 5, the period T of such a pendulum is given by
( A Figure 9.10.1
\
T =4.1- [
l . :dd) k2sin2(/)
(7)
where L — length of the supporting rod g = acceleration due to gravity k = sin($o/2), where #o is the initial angle of displacement from the vertical The integral, which is called a complete elliptic integral of the first kind, cannot be expressed in terms of elementary functions and is often approximated by numerical methods. Unfortunately, numerical values are so specific that they often give little insight into general physical principles. However, if we expand the integrand of (7) in a series and integrate term by term, then we can generate an infinite series that can be used to construct various mathematical models for the period T that give a deeper understanding of the behavior of the pendulum.
686
Chapter 9 / Infinite Series
To obtain a series for the integrand, we will substitute — k2 sin2 for x in the binomial series for 1/Vl + x that we derived in Example 4(b) of Section 9.9. If we do this, then we can rewrite (7) as
1-3-5
T = 4J -
2 3 3!
sin6
\d
(8)
If we integrate term by term, then we can produce a series that converges to the period T. However, one of the most important cases of pendulum motion occurs when the initial displacement is small, in which case all subsequent displacements are small, and we can assume that k — sin(0o/2) ^ 0. In this case we expect the convergence of the series for T to be rapid, and we can approximate the sum of the series by dropping all but the constant term in (8). This yields /—
«) ACE STOCK LIMITED/Alamy Understanding the motion of a pendulum played a critical role in the advance of accurate timekeeping with the development of the pendulum clock in the 17th century.
T = 2nl-
(9)
v#
which is called the first-order model of T or the model for small vibrations. This model can be improved on by using more terms in the series. For example, if we use the first two terms in the series, we obtain the second-order model
(10) (verify). I/QUICK CHECK EXERCISES 9.10
(See page 689 for answers.)
1. The Maclaurin series for e * obtained by substituting —x2 for x in the series
ls e
y2
3.
V
E.x2 +
.x+.
= Eit=o r~>CC
4. Suppose that /(I) = 4 and f ' ( x ) = 2. A
. (x - I) 3 + . k=0
EXERCISE SET 9.10
CAS
1. In each part, obtain the Maclaurin series for the function by making an appropriate substitution in the Maclaurin series for 1/(1 — x). Include the general term in your answer, and state the radius of convergence of the series. (a) ~ (b)
1-2* l+x 2-x 2. In each part, obtain the Maclaurin series for the function by making an appropriate substitution in the Maclaurin series for ln(l + x). Include the general term in your answer, and
state the radius of convergence of the series, (a) In(l-jt) (b) ln(l + x2) (c) ln(l +2x) (d) ln(2 + ;t) 3. In each part, obtain the first four nonzero terms of the Maclaurin series for the function by making an appropriate substitution in one of the binomial series obtained in Example 4 of Section 9.9. 2 2 -1/2 (a) (2 + jc)
(b) (1 - * )-
9.10 Differentiating and Integrating Power Series; Modeling with Taylor Series 4. (a) Use the Maclaurin series for 1/(1 — x) to find the Maclaurin series for I/(a — x), where a ^ 0, and state the radius of convergence of the series, (b) Use the binomial series for 1/(1 + x)2 obtained in Example 4 of Section 9.9 to find the first four nonzero terms in the Maclaurin series for l/(a + x)2, where a j^ 0, and state the radius of convergence of the series. 5-8 Find the first four nonzero terms of the Maclaurin series for the function by making an appropriate substitution in a known Maclaurin series and performing any algebraic operations that are required. State the radius of convergence of the series. 5. (a) sin 2* 6. (a) cos 2*
7. (a) 8. (a)
l+3x
x x- 1
(b) e~2x (b) jtV
(c) e*1
(d) jc2cos:o: x
(d) sin (A: ) 2 3/2
(b) * sinh 2*
(c) *(1 - * )
(b) 3cosh(* 2 )
(c)
(b)
2
(b) In
10. (a) cos *
(1 + 2xy
1
(b) Find the interval of convergence of the Taylor series. 12. Use the method of Exercise 11 to find the Taylor series of l/x about * = XQ, and state the interval of convergence of the Taylor series. 13-14 Find the first four nonzero terms of the Maclaurin series for the function by multiplying the Maclaurin series of the factors.
13. (a) ex sin * 14. (a) e~" cos* 15-16 Find the first four nonzero terms of the Maclaurin series for the function by dividing appropriate Maclaurin series. 1 sin* 15. (a) sec x \ = (b) COS*
16. (a)
l+x
(b)
4x-2
x3 + x2 + 2x - 2 x2-l
20.
21-22 Confirm the derivative formula by differentiating the appropriate Maclaurin series term by term. 21. (a) —[cos*] = — sin*
(b) —
22. (a) — [sinh *] = cosh *
(b) — |
dx
dx
dx
l+x I
dx
23-24 Confirm the integration formula by integrating the appropriate Maclaurin series term by term.
23. (a)
11. (a) Use a known Maclaurin series to find the Taylor series of l/x about * = 1 by expressing this function as
!_ _
19.
ex dx = ex + C
*
9-10 Find the first four nonzero terms of the Maclaurin series for the function by using an appropriate trigonometric identity or property of logarithms and then substituting in a known Maclaurin series. 9. (a) sin *
19-20 Find the first five nonzero terms of the Maclaurin series for the function by using partial fractions and a known Maclaurin series.
2
(c) xe~
2
687
l-x
17. Use the Maclaurin series for e* and e~x to derive the Maclaurin series for sinh* and cosh*. Include the general terms in your answers and state the radius of convergence of each series. 18. Use the Maclaurin series for sinh x and cosh x to obtain the first four nonzero terms in the Maclaurin series for tanh x.
(b) / sinh x dx = cosh x + C
24. 4. (a) / si sin * dx = — cos * + C
I
dx = l n ( l +x) 1 +* 25. Consider the series
(b)
E fc=0
Determine the intervals of convergence for this series and for the series obtained by differentiating this series term by term. 26. Consider the series
E
(-3)*
Determine the intervals of convergence for this series and for the series obtained by integrating this series term by term. 27. (a) Use the Maclaurin series for \/(\—x) to find the Maclaurin series for
(b) Use the Maclaurin series obtained in part (a) to find /< 5 >(0)and/< 6 >(0). (c) What can you say about the value of / (n) (0)? 28. Let /(*) = x2 cos 2*. Use the method of Exercise 27 to find/(99)(0). 29-30 The limit of an indeterminate form as x —> XQ can sometimes be found by expanding the functions involved in Taylor series about x = XQ and taking the limit of the series term by term. Use this method to find the limits in these exercises. sinjc 29. (a) lim (b) lim x
30. (a) lim <• o
I — COS X
sin x
(b) lim
688 Chapter 9 / Infinite Series 31 -34 Use Maclaurin series to approximate the integral to three decimal-place accuracy. -1
.
/-1/2
sin(.x >c2)2)dx
32. / Jo
r°
tan'l(2x2)dx
1/2
33. /
dx.
JO
FOCUS ON CONCEPTS
35. (a) Find the Maclaurin series for e* . What is the radius of convergence? (b) Explain two different ways to use the Maclaurin series for ex to find a series for x3e* . Confirm that both methods produce the same series. 36. (a) Differentiate the Maclaurin series for 1/(1 — x), and use the result to show that GO
> kx =
°° xk ]T — = -\n(l-x) for - ! < * < ! t=i * (c) Use the result in part (b) to show that — = ln(l + x) k
for - ! < * < !
(d) Show that the series in part (c) converges if x = 1. (e) Use the remark following Example 3 to show that "
xk
7 (—1)*+1—=ln(l+jr) k tt
for— l<x
37. Use the results in Exercise 36 to find the sum of the series. ,A k 1 2 3 4 *=1 CO
V
1 1 2 4 ' 2(4 ) ' 3(43)
1 4(44)
38. Use the results in Exercise 36 to find the sum of each series.
(e-Y)k _ e - \
(e - I) 2
(e - I)3
39. (a) Use the relationship
1
: dx = sin x + C
to find the first four nonzero terms in the Maclaurin series for sin"1 x. (b) Express the series in sigma notation. (c) What is the radius of convergence? 41. We showed by Formula (19) of Section 8.2 that if there are >>o units of radioactive carbon- 1 4 present at time r = 0, then the number of units present f years later is y(t) = yoe-0'000™'
(a) Express y(t) as a Maclaurin series. (b) Use the first two terms in the series to show that the number of units present after 1 year is approximately (0.999879);yo. (c) Compare this to the value produced by the formula for
for — 1 < x < 1
(b) Integrate the Maclaurin series for 1/(1 — x), and use the result to show that
l
40. (a) Use the relationship 1
: dx = sinh ' x + C
to find the first four nonzero terms in the Maclaurin series for sinh"1 x. (b) Express the series in sigma notation. (c) What is the radius of convergence?
42. Suppose that a simple pendulum with a length of L = 1 meter is given an initial displacement of OQ = 5° from the vertical. (a) Approximate the period T of the pendulum using Formula (9) for the first-order model of T. [Note: Take g = 9.8 m/s2.] (b) Approximate the period of the pendulum using Formula (10) for the second-order model. (c) Use the numerical integration capability of a CAS to approximate the period of the pendulum from Formula (7), and compare it to the values obtained in parts (a) and (b). 43. Use the first three nonzero terms in Formula (8) and the Wallis sine formula in the Endpaper Integral Table (Formula 122) to obtain a model for the period of a simple pendulum. 44. Recall that the gravitational force exerted by the Earth on an object is called the object's weight (or more precisely, its Earth weight). If an object of mass m is on the surface of the Earth (mean sea level), then the magnitude of its weight is mg, where g is the acceleration due to gravity at the Earth's surface. A more general formula for the magnitude of the gravitational force that the Earth exerts on an object of mass mis ^ ^ mgR2
~ (R + h)2 where R is the radius of the Earth and h is the height of the object above the Earth's surface. (a) Use the binomial series for 1/(1 + x)2 obtained in Example 4 of Section 9.9 to express F as a Maclaurin series in powers of hi R. (b) Show that if h = 0, then F — mg. (c) Show that if h/R « 0, then F « mg - (2mgh/R). [Note: The quantity 2mgh/R can be thought of as a "correction term" for the weight that takes the object's height above the Earth's surface into account.] (d) If we assume that the Earth is a sphere of radius R = 4000 mi at mean sea level, by approximately what
Chapter 9 Review Exercises percentage does a person's weight change in going from mean sea level to the top of Mt. Everest (29,028 ft)? 45. (a) Show that the Bessel function JQ(X) given by Formula (4) of Section 9.8 satisfies the differential equation 1 Xy" -|- y -)_ Xy = Q. (This is called the Bessel equation of order zero.) (b) Show that the Bessel function J\(x) given by Formula (5) of Section 9.8 satisfies the differential equation x2y" + xy' + (x2 - l)y = 0. (This is called theBessel equation of order one.) (c) Show that J^(x) = -J\(x).
689
46. Prove: If the power series ]TJ=0 akxk and £^_0 bkxk have the same sum on an interval (—r, r), then ak = bk for all values of k. 47. Writing Evaluate the limit x — sin x lim in two ways: using L'Hopital's rule and by replacing sin x by its Maclaurin series. Discuss how the use of a series can give qualitative information about how the value of an indeterminate limit is approached.
UICK CHECK ANSWERS 9.10
U^
(x -
4. (a) -i (b)4;l;-~;-L;4;(-l)*- J
CHAPTER 9 REVIEW EXERCISES
1. What is the difference between an infinite sequence and an infinite series? 2. What is meant by the sum of an infinite series? 3. (a) What is a geometric series? Give some examples of convergent and divergent geometric series, (b) What is a p-series? Give some examples of convergent and divergent p-series. 4. State conditions under which an alternating series is guaranteed to converge. 5. (a) What does it mean to say that an infinite series converges absolutely? (b) What relationship exists between convergence and absolute convergence of an infinite series? 6. State the Remainder Estimation Theorem, and describe some of its uses. 7. If a power series in x — XQ has radius of convergence R, what can you say about the set of *-values at which the series converges? 8. (a) Write down the formula for the Maclaurin series for / in sigma notation. (b) Write down the formula for the Taylor series for / about x = XQ in sigma notation. 9. Are the following statements true or false? If true, state a theorem to justify your conclusion; if false, then give a counterexample. (a) If J2 uk converges, then u^ —> 0 as k —>• +°o. (b) If HI —>• 0 as k -» +00, then J] uk converges. (c) If /(«) = an for n = 1, 2, 3 , . . . , and if an-*L as n -» +00, then f(x) —>• L as x -»• +00. (d) If f(n) = an for n = 1, 2, 3 , . . . , and if /(*) -» L as x ->• +00, then an -» L as n —>• +00. (e) If 0 < an < 1, then {«„} converges. (f) If 0 < Hi < 1, then J^Mjt converges.
(g) If £ uk and £ vk converge, then J^(uk + vk) diverges. (h) If Y^ Uk and J2 vk diverge, then J2(uk — vk) converges. (i) If 0 < u/f < vk and £ vk converges, then J^ uk converges. ( j) If 0 < uk < vt and ^ uk diverges, then J] vk diverges. (k) If an infinite series converges, then it converges absolutely. (1) If an infinite series diverges absolutely, then it diverges. 10. State whether each of the following is true or false. Justify your answers. (a) The function f(x) = z ll/3 has a Maclaurin series. (b) 1 + 1-1 + 1-1 + 1-1 + .. . = 1 (c) 1 + 1-1 + 1-1 + 1-1 + ... = ! 11. Find the general term of the sequence, starting with n = 1, determine whether the sequence converges, and if so find its limit. 3 4 5 2 2 2 2 2 2 -l ' 3 -2 '4 -32"" I 2 3 4 (b) 3 ' 5 ' 7'~9"" 12. Suppose that the sequence {ak } is defined recursively by a0 = c,
= V«/fc
Assuming that the sequence converges, find its limit if (a)c=i (b)c=|. 13. Show that the sequence is eventually strictly monotone.
14. (a) Give an example of a bounded sequence that diverges, (b) Give an example of a monotonic sequence that diverges. 15-20 Use any method to determine whether the series converge, a
690
Chapter 9 / Infinite Series
15. (a) YJ-I
29. (a) Find the first five Maclaurin polynomials of the function p(x) = 1 -Ix + 5x2 + 4x3. (b) Make a general statement about the Maclaurin polynomials of a polynomial of degree n. 30. Show that the approximation
(b;
k=l CO
16. (a) y(-i)*_:—
(b:
17. (a)
(b)
18. (a)
(b) C S(1/A:)
°
3! 5! is accurate to four decimal places if 0 < x < n/4. 31. Use a Maclaurin series and properties of alternating series to show that | ln(l + x) - x\ < x2/2 if 0 < x < 1. 32. Use Maclaurin series to approximate the integral
"' 1 -cos* •dx
I— 1/2
/Jo 21. Find a formula for the exact error that results when the sum of the geometric series J^=0(l/5)k is approximated by the sum of the first 100 terms in the series. " 1 22. Suppose that YJ Uk = 2 . Find (b) lim uk
(a) MIOO
(c)
e2
23. In each part, determine whether the series converges; if so, find its sum. E jk ~ 1
to three decimal-place accuracy. 33. In parts (a)-(d), find the sum of the series by associating it with some Maclaurin series. 4 8 16
+ 1) k=l
E[
(d) 1
e4
e6
+
2! 3! 34. In each part, write out the first four terms of the series, and then find the radius of convergence.
: + !)-!
k=l
24. It can be proved that lim \/nl = +00 and
n -> +co
k?\
lim -
n -> +os
In each part, use these limits and the root test to determine whether the series converges. (a) YJ 2 ^
(b)
E^T k=0
£=0
25. Let a, b, and p be positive constants. For which values of °° 1 p does the series y converge? ^ (a + bk)P 26. Find the interval of convergence of
l ' 3 - 5 - - - ( 2 / t - 1)
35. Use an appropriate Taylor series for %/x to approximate 4/28 to three decimal-place accuracy, and check your answer by comparing it to that produced directly by your calculating utility. 36. Differentiate the Maclaurin series for xex and use the result to show that T^k+l
2^ k=0
k\
37. Use the supplied Maclaurin series for sin x and cos x to find the first four nonzero terms of the Maclaurin series for the given functions. Y2k+l
sin x =
k=0
27. (a) Show that kk >k\.
Y2k CO
k
(b) Use the comparison test to show that Y^ ]c converges. k=l
(c) Use the root test to show that the series converges. 28. Does the series 1 — f + f — f + f + • • • converge? Justify your answer.
(a) sin x cos x
(b) | sin 2x
Chapter 9 Making Connections
691
CHAPTER 9 MAKING CONNECTIONS 1. As shown in the accompanying figure, suppose that lines L\ and LI form an angle 9, 0 < 6 < ?r/2, at their point of intersection P. A point PO is chosen that is on LI and a units from P. Starting from P0 a zigzag path is constructed by successively going back and forth between L\ and L 2 along a perpendicular from one line to the other. Find the following sums in terms of 9 and a.
(a) Give upper and lower bounds on the distance between the bug and point A when it finally stops. [Hint: As stated in Example 2 of Section 9.6, assume that the sum of the alternating harmonic series is In 2.] (b) Give upper and lower bounds on the total distance that the bug has traveled when it finally stops. [Hint: Use inequality (2) of Section 9.4.]
(a) P0P} + PtP2 + P2P?, + ••• (b) P0Pl
180cm
(c) P1P2
PO
A Figure Ex-4
P
2 P4 P,
•« Figure Ex-1
5. In Section 5.6 we defined the kinetic energy A" of a particle with mass in and velocity v to be K = ^mv2 [see Formula (7) of that section]. In this formula the mass m is assumed to be constant, and K is called the Newtonian kinetic energy. However, in Albert Einstein's relativity theory the mass m increases with the velocity and the kinetic energy K is given by the formula
2. (a) Find A and B such that 6*
K = m0c2
2kB
4
(b) Use the result in part (a) to find a closed form for the nth partial sum of the series — 2*)
and then find the sum of the series. Source: This exercise is adapted from a problem that appeared in the Forty-Fifth Annual William Lowell Putnam Competition.
3. Show that the alternating p-series
in which mo is the mass of the particle when its velocity is zero, and c is the speed of light. This is called the relativistic kinetic energy. Use an appropriate binomial series to show that if the velocity is small compared to the speed of light (i.e., v/c « 0), then the Newtonian and relativistic kinetic energies are in close agreement. 6. In Section 8.4 we studied the motion of a falling object that has mass m and is retarded by air resistance. We showed that if the initial velocity is i>o and the drag force FK is proportional to the velocity, that is, FK = —cv, then the velocity of the object at time t is
mg converges absolutely if p > 1, converges conditionally if 0 < p < 1, and diverges if p < 0. 4. As illustrated in the accompanying figure, a bug, starting at point A on a 180 cm wire, walks the length of the wire, stops and walks in the opposite direction for half the length of the wire, stops again and walks in the opposite direction for onethird the length of the wire, stops again and walks in the opposite direction for one-fourth the length of the wire, and so forth until it stops for the 1000th time.
where g is the acceleration due to gravity [see Formula (16) of Section 8.4]. (a) Use a Maclaurin series to show that if ct/m ~ 0, then the velocity can be approximated as
v(t) ~ DO - f — +g}t (b) Improve on the approximation in part (a).
EXPANDING THE CALCULUS HORIZON To learn how ecologists use mathematical models based on the process of iteration to study the growth and decline of animal populations, see the module entitled Iteration and Dynamical Systems at: www.wiley.com/college/anton
*»»*•••
PARAMETRIC AND POLAR CURVES; CONIC SECTIONS
Dwight R. Kutm
Mathematical curves, such as the
In this chapter we will study alternative ways of expressing curves in the plane. We will begin
spirals in the center of a sunflower,
by studying parametric curves: curves described in terms of component functions. This study
can be described conveniently using
will include methods for finding tangent lines to parametric curves. We will then introduce
ideas developed in this chapter.
polar coordinate systems and discuss methods for finding tangent lines to polar curves, arc length of polar curves, and areas enclosed by polar curves. Our attention will then turn to a review of the basic properties of conic sections: parabolas, ellipses, and hyperbolas. Finally, we will consider conic sections in the context of polar coordinates and discuss some applications in astronomy.
PARAMETRIC EQUATIONS; TANGENT LINES AND ARC LENGTH FOR PARAMETRIC CURVES Graphs of functions must pass the vertical line test, a limitation that excludes curves with self-intersections or even such basic curves as circles. In this section we will study an alternative method for describing curves algebraically that is not subject to the severe restriction of the vertical line test. We will then derive formulas required to find slopes, tangent lines, and arc lengths of these parametric curves. We will conclude with an investigation of a classic parametric curve known as the cycloid.
PARAMETRIC EQUATIONS
Suppose that a particle moves along a curve C in the zy-plane in such a way that its x- and y-coordinates, as functions of time, are
x = f(t),
y = g(t)
We call these the parametric equations of motion for the particle and refer to C as the trajectory of the particle or the graph of the equations (Figure 10.1.1). The variable t is called the parameter for the equations.
A moving particle with trajectory C
> Example 1 Sketch the trajectory over the time interval 0 < t < 10 of the particle whose parametric equations of motion are x = t — 3 sin ?,
692
y = 4 — 3 cos t
(1)
Chapter 9 Making Connections
691
CHAPTER 9 MAKING CONNECTIONS 1. As shown in the accompanying figure, suppose that lines L\ and L2 form an angle 9, 0 < 9 < jr/2, at their point of intersection P. A point P0 is chosen that is on LI and a units from P. Starting from PQ a zigzag path is constructed by successively going back and forth between L\ and LI along a perpendicular from one line to the other. Find the following sums in terms of 9 and a. (a) (b) (c)
P
0
P
2
P
4 P6
< Figure Ex-1
2. (a) Find A and B such that
(a) Give upper and lower bounds on the distance between the bug and point A when it finally stops. [Hint: As stated in Example 2 of Section 9.6, assume that the sum of the alternating harmonic series is In 2.] (b) Give upper and lower bounds on the total distance that the bug has traveled when it finally stops. [Hint: Use inequality (2) of Section 9.4.] 180cm A Figure Ex-4
5. In Section 5.6 we defined the kinetic energy K of a particle with mass m and velocity v to be K = ^mv2 [see Formula (7) of that section]. In this formula the mass m is assumed to be constant, and K is called the Newtonian kinetic energy. However, in Albert Einstein's relativity theory the mass m increases with the velocity and the kinetic energy K is given by the formula
6k
2kA 2kB k k —2) 3* — 2 (b) Use the result in part (a) to find a closed form for the nth partial sum of the series
f-^ (3*+1 -2*+')(3*-2*) and then find the sum of the series. Source: This exercise is adapted from a problem that appeared in the Forty-Fifth Annual William Lowell Putnam Competition.
3. Show that the alternating p-series J_ 2/>
J_ 3/>
_L 4/>
*+i_L kP
converges absolutely if p > 1, converges conditionally if 0 < p < 1, and diverges if p < 0. 4. As illustrated in the accompanying figure, a bug, starting at point A on a 180 cm wire, walks the length of the wire, stops and walks in the opposite direction for half the length of the wire, stops again and walks in the opposite direction for onethird the length of the wire, stops again and walks in the opposite direction for one-fourth the length of the wire, and so forth until it stops for the 1000th time.
K =
1
-1
in which mo is the mass of the particle when its velocity is zero, and c is the speed of light. This is called the relativistic kinetic energy. Use an appropriate binomial series to show that if the velocity is small compared to the speed of light (i.e., v/c ~ 0), then the Newtonian and relativistic kinetic energies are in close agreement. 6. In Section 8.4 we studied the motion of a falling object that has mass m and is retarded by air resistance. We showed that if the initial velocity is VQ and the drag force FR is proportional to the velocity, that is, FR = —cv, then the velocity of the object at time t is
v(t) = where g is the acceleration due to gravity [see Formula (16) of Section 8.4]. (a) Use a Maclaurin series to show that if ct/m % 0, then the velocity can be approximated as v(t) w DO - (— + V m (b) Improve on the approximation in part (a).
EXPANDING THE CALCULUS HORIZON To learn how ecologists use mathematical models based on the process of iteration to study the growth and decline of animal populations, see the module entitled Iteration and Dynamical Systems at:
www.wiley.com/college/anton
PARAMETRIC AND POLAR CURVES; CONIC SECTIONS Dwight R. Kuhn
Mathematical curves, such as the spirals in the center of a sunflower, can be described conveniently using ideas developed in this chapter.
In this chapter we will study alternative ways of expressing curves in the plane. We will begin by studying parametric curves: curves described in terms of component functions. This study will include methods for finding tangent lines to parametric curves. We will then introduce polar coordinate systems and discuss methods for finding tangent lines to polar curves, arc length of polar curves, and areas enclosed by polar curves. Our attention will then turn to a review of the basic properties of conic sections: parabolas, ellipses, and hyperbolas. Finally, we will consider conic sections in the context of polar coordinates and discuss some applications in astronomy.
PARAMETRIC EQUATIONS; TANGENT LINES AND ARC LENGTH FOR PARAMETRIC CURVES Graphs of functions must pass the vertical line test, a limitation that excludes curves with self-intersections or even such basic curves as circles. In this section we will study an alternative method for describing curves algebraically that is not subject to the severe restriction of the vertical line test. We will then derive formulas required to find slopes, tangent lines, and arc lengths of these parametric curves. We will conclude with an investigation of a classic parametric curve known as the cycloid.
PARAMETRIC EQUATIONS
Suppose that a particle moves along a curve C in the ;ry-plane in such a way that its x- and ^-coordinates, as functions of time, are
* = /(*), y = g(t) (x,y)
A moving particle with trajectory C \ A Figure 10.1.1
692
We call these the parametric equations of motion for the particle and refer to C as the trajectory of the particle or the graph of the equations (Figure 10.1.1). The variable t is called the parameter for the equations. Example 1 Sketch the trajectory over the time interval 0 < t < 10 of the particle whose parametric equations of motion are
x = t — 3 sin f,
y = 4 — 3 cos t
(1)
10.1 Parametric Equations; Tangent Lines and Arc Length for Parametric Curves
693
Solution. One way to sketch the trajectory is to choose a representative succession of times, plot the (x, y) coordinates of points on the trajectory at those times, and connect the points with a smooth curve. The trajectory in Figure 10.1.2 was obtained in this way from the data in Table 10.1.1 in which the approximate coordinates of the particle are given at time increments of 1 unit. Observe that there is no r-axis in the picture; the values of t appear only as labels on the plotted points, and even these are usually omitted unless it is important to emphasize the locations of the particle at specific times. -4 Table 10.1.1
t
X
y
0
0.0 -1.5 -0.7 2.6 6.3 7.9 6.8 5.0 5.0 7.8 11.6
1.0 2.4 5.2 7.0 6.0 3.1 1.1 1.7 4.4 6.7 6.5
1
2 3 4 5 6 7 8 9 10
TECHNOLOGY MASTERY Read the documentation for your graphing utility to learn how to graph parametric equations, and then generate the trajectory in Example 1. Explore the behavior of the particle beyond time t = 10.
2
4
6
8
10
12
A Figure 10.1.2
Although parametric equations commonly arise in problems of motion with time as the parameter, they arise in other contexts as well. Thus, unless the problem dictates that the parameter t in the equations
x = f(t),
y = g(t)
represents time, it should be viewed simply as an independent variable that varies over some interval of real numbers. (In fact, there is no need to use the letter t for the parameter; any letter not reserved for another purpose can be used.) If no restrictions on the parameter are stated explicitly or implied by the equations, then it is understood that it varies from -co to +00. To indicate that a parameter t is restricted to an interval [a, b], we will write
x = f(t),
> Example 2
(a
Find the graph of the parametric equations
x — cos t, Solution.
y = g(t)
y = sin t
(0 < t < 2jr)
(2)
One way to find the graph is to eliminate the parameter t by noting that x1 + y2 = sin21 + cos21 = 1
Thus, the graph is contained in the unit circle x2 + y2 = 1. Geometrically, the parameter t can be interpreted as the angle swept out by the radial line from the origin to the point (x, y) = (cosf, sinr) on the unit circle (Figure 10.1.3). As t increases from 0 to 2jr, the point traces the circle counterclockwise, starting at (1,0) when t — 0 and completing one full revolution when t — 2n. One can obtain different portions of the circle by varying the interval over which the parameter varies. For example,
x = cos t, A Figure 10.1.3
y = sin t
represents just the upper semicircle in Figure 10.1.
(0 < t < if)
(3)
694
Chapter 10 / Parametric and Polar Curves; Conic Sections ORIENTATION
The direction in which the graph of a pair of parametric equations is traced as the parameter increases is called the direction of increasing parameter or sometimes the orientation imposed on the curve by the equations. Thus, we make a distinction between a curve, which is a set of points, and a parametric curve, which is a curve with an orientation imposed on it by a set of parametric equations. For example, we saw in Example 2 that the circle represented parametrically by (2) is traced counterclockwise as t increases and hence has counterclockwise orientation. As shown in Figures 10.1.2 and 10.1.3, the orientation of a parametric curve can be indicated by arrowheads. To obtain parametric equations for the unit circle with clockwise orientation, we can replace t by —t in (2) and use the identities cos(—t) = cos t and sin(—t) — — sin t. This x = cos t,
x = cos(-/), y = sin(-f) (0 < t < 271)
y — — sin t
(0 < t < 2n)
Here, the circle is traced clockwise by a point that starts at (1, 0) when t = 0 and completes one full revolution when t — 2n (Figure 10.1.4).
A Figure 10.1.4
TECHNOLOGY MASTERY
When parametric equations are graphed using a calculator, the orientation can often be determined by watching the direction in which the graph is traced on the screen. However, many computers graph so fast that it is often hard to discern the orientation. See if you can use your graphing utility to confirm that (3) has a counterclockwise orientation.
> Example 3
Graph the parametric curve
x = 2t - 3,
y = 6t -1
by eliminating the parameter, and indicate the orientation on the graph. Solution. To eliminate the parameter we will solve the first equation for t as a function of x, and then substitute this expression for t into the second equation:
f = (i)(* + 3) y = 6 (§)(* + 3 ) - 7
y = 3x + 2 Thus, the graph is a line of slope 3 and y-intercept 2. To find the orientation we must look to the original equations; the direction of increasing t can be deduced by observing that x increases as / increases or by observing that y increases as t increases. Either piece of information tells us that the line is traced left to right as shown in Figure 10.1.5. -^
: = 2t-3, y= i A Figure 10.1.5 REMARK
(1,1)
Not all parametric equations produce curves with definite orientations; if the equations are badly behaved, then the point tracing the curve may leap around sporadically or move back and forth, failing to determine a definite direction. For example, if x = sinf,
y = sin2 t
then the point (x, y) moves along the parabola y = x2. However, the value of x varies periodically between -] and 1, so the point (x, y) moves periodically back and forth along the parabola between the points (—1,1) and (1, 1) (as shown in Figure 10.1.6). Later in the text we will discuss restrictions that eliminate such erratic behavior, but for now we will just avoid such complications. A Figure 10.1.6
EXPRESSING ORDINARY FUNCTIONS PARAMETRICALLY
An equation y — f ( x ) can be expressed in parametric form by introducing the parameter t = x; this yields the parametric equations
y = f(t)
10.1 Parametric Equations; Tangent Lines and Arc Length for Parametric Curves
695
For example, the portion of the curve y = cos x over the interval [—2n, 2n] can be expressed parametrically as x — t, y — cos t (—7,n < t < In) (Figure 10.1.7).
>• Figure 10.1.7
If a function / is one-to-one, then it has an inverse function / ' . In this case the equation y = /"' (jc) is equivalent to x = f ( y ) . We can express the graph of /"' in parametric form by introducing the parameter y = t ; this yields the parametric equations
x = f(t),
y =t
For example, Figure 10.1.8 shows the graph of /(*) = x5 + x + 1 and its inverse. The graph of / can be represented parametrically as y = f 5 +t + 1
x = t,
and the graph of f~l can be represented parametrically as
x = t5 + t + l,
A Figure 10.1.8
y =t
TANGENT LINES TO PARAMETRIC CURVES
We will be concerned with curves that are given by parametric equations
x = f(t),
y = g(t)
in which f ( t ) and g(t) have continuous first derivatives with respect to t. It can be proved that if dx/dt ^ 0, then y is a differentiable function of jc, in which case the chain rule implies that _ dx dx/dt This formula makes it possible to find dy/dx directly from the parametric equations without eliminating the parameter. *• Example 4
Find the slope of the tangent line to the unit circle x = cos t,
y = sin t
(0 < t < In}
at the point where t = jt/6 (Figure 10.1.9). Solution. From (4), the slope at a general point on the circle is dy dx
dy/dt dx/dt
cos t = - cot t — sin t
Thus, the slope at t — n/6 is A Figure 10.1.9
dy_ dx
= -cot- = -\/3 <* 6
(5)
696
Chapter 10 / Parametric and Polar Curves; Conic Sections
Note that Formula (5) makes sense geometrically because the radius from the origin to the point P(cost, sint) has slope m = tan t. Thus the tangent line at P, being perpendicular to the radius, has slope
1
1 = - cot t tanf
It follows from Formula (4) that the tangent line to a parametric curve will be horizontal at those points where dy/dt = 0 and dx/dt ^ 0, since dy/dx = 0 at such points. Two different situations occur when dx/dt = 0. At points where dx/dt = 0 and dy/dt ^ 0, the right side of (4) has a nonzero numerator and a zero denominator; we will agree that the curve has infinite slope and a vertical tangent line at such points. At points where dx/dt and dy/dt are both zero, the right side of (4) becomes an indeterminate form; we call such points singular points. No general statement can be made about the behavior of parametric curves at singular points; they must be analyzed case by case.
(Figure 10.1.10).
>• Example 5 In a disastrous first flight, an experimental paper airplane follows the trajectory of the particle in Example 1: x = t — 3 sin ?,
P(cos t, sin t)
y — 4 — 3 cos t
(t >
but crashes into a wall at time t = 10 (Figure 10.1.11). (a) At what times was the airplane flying horizontally? (b) At what times was it flying vertically? Solution (a). The airplane was flying horizontally at those times when dy/dt — 0 and dx/dt ^ 0. From the given trajectory we have
1
Radius OP has slope m = tant.
A Figure 10.1.10
dy
dx — = 3 sin t and — = 1 — 3 cos t (6) dt dt Setting dy/dt = 0 yields the equation 3 sin t = 0, or, more simply, sin t = 0. This equation has four solutions in the time interval 0 < t < 10:
X=-Q,. X — -s., X — In., \=^T*. Since dx/dt = 1 — 3 cos t ^ 0 for these values of t (verify), the airplane was flying horizontally at times f=0,
t=jr^3.U,
t
6.28,
and t = 3n ^ 9.42
which is consistent with Figure 10.1.11. Stamsfovas Kairys/iStockphoto The complicated motion of a paper airplane is best described mathematicallyusing parametric equations.
Solution (b). The airplane was flying vertically at those times when dx/dt = 0 and dy/dt ^ 0. Setting dx/dt = 0 in (6) yields the equation 1 — 3 cos t = 0 or cos t = |
This equation has three solutions in the time interval 0 < t < 10 (Figure 10.1.12): t—cos~1^,
t = 2n—cos~1^,
t = 2n + co&~l~
y = cos t
10.1 Parametric Equations; Tangent Lines and Arc Length for Parametric Curves
697
Since dy/dt = 3 sin t is not zero at these points (why?), it follows that the airplane was flying vertically at times
t — cos
1.23,
f « 2^-1.23 « 5.05,
t ^ 2n+ 1.23 « 7.51
which again is consistent with Figure 10.1.11. •< Example 6
The curve represented by the parametric equations 1
(-co < t < +00)
is called a semicubical parabola. The parameter t can be eliminated by cubing x and squaring y, from which it follows that y2 — x3. The graph of this equation, shown in Figure 10.1.13, consists of two branches: an upper branch obtained by graphing y = x3/2 and a lower branch obtained by graphing y = — x3/2. The two branches meet at the origin, which corresponds to t = 0 in the parametric equations. This is a singular point because the derivatives dx/dt = 2t and dy/dt = 3t2 are both zero there. ^
> Example 7 Without eliminating the parameter, find dy/dx and d2y/dx2 at (1, 1) and (1, —1) on the semicubical parabola given by the parametric equations in Example 6. A Figure 10.1.13
Solution. From (4) we have
WARNING
dy dy/dt 3 dx dx/dt It = 2* and from (4) applied to y' = dy/dx we have
Although it is true that
dy dy/dt ~dic ~ dx/dt you cannot conclude that d2y/dx2 is the quotient of d2y/dt2 and d2x/dt2. To illustrate that this conclusion is erroneous, show that for the parametric curve in Example 7,
d2y/dt2 d2x/dt2
(7)
d2y = _ dy' " ~ ~dx
dy'/dt 3 3/2 (8) dx/dt It 47 Since the point (1, 1) on the curve corresponds to t = 1 in the parametric equations, it follows
from (7) and (8) that
dy_ dx
= - and
3
dx2
4
Similarly, the point (1, — 1) corresponds to t = — 1 in the parametric equations, so applying (7) and (8) again yields
dy = -- and dx t~-\
d2y dx2
3 "4
Note that the values we obtained for the first and second derivatives are consistent with the graph in Figure 10.1.13, since at (1,1) on the upper branch the tangent line has positive slope and the curve is concave up, and at (1, —1) on the lower branch the tangent line has negative slope and the curve is concave down. Finally, observe that we were able to apply Formulas (7) and (8) for both t = 1 and t = — 1, even though the points (1,1) and (1, —1) lie on different branches. In contrast, had we chosen to perform the same computations by eliminating the parameter, we would have had to obtain separate derivative formulas for y = jc3/2 and y = -x3/2. -4
ARC LENGTH OF PARAMETRIC CURVES
The following result provides a formula for finding the arc length of a curve from parametric equations for the curve. Its derivation is similar to that of Formula (3) in Section 5.4 and will be omitted.
698
Chapter 10 / Parametric and Polar Curves; Conic Sections
Formulas (4) and (5) in Section 5.4 can be viewed as special cases of (9). For example, Formula (4) in Section 5.4 can be obtained from (9) by writing y = f(x) parametrically as
x = t, y = f(t)
10.1.1
ARC LENGTH FORMULA FOR PARAMETRIC CURVES
If HO Segment of the
curve represented by the parametric equations
x = x(t),
y = y(t)
(a < t < b)
is traced more than once as t increases from a to b, and ifdx/dt and dy/dt are continuous functions for a < t < b, then the arc length L of the curve is given by
and Formula (5) in Section 5.4 can be obtained by writing x = g(y) parametrically as
x = g(f),
(9)
y=t
Example 8 equations
Use (9) to find the circumference of a circle of radius a from the parametric x=acost,
y = asint
(0
Solution. 2n
// j \2
= Jo/ \l\-f\ -r \\dtj + (\dt JT
o
-|27r
a dt = at \ — 2na -4 Jo
THE CYCLOID (THE APPLE OF DISCORD)
The results of this section can be used to investigate a curve known as a cycloid. This curve, which is one of the most significant in the history of mathematics, can be generated by a point on a circle that rolls along a straight line (Figure 10.1.14). This curve has a fascinating history, which we will discuss shortly; but first we will show how to obtain parametric equations for it. For this purpose, let us assume that the circle has radius a and rolls along the positive jc-axis of a rectangular coordinate system. Let P(x, y) be the point on the circle that traces the cycloid, and assume that P is initially at the origin. We will take as our parameter the angle d that is swept out by the radial line to P as the circle rolls (Figure 10.1.14). It is standard here to regard 9 as positive, even though it is generated by a clockwise rotation. The motion of P is a combination of the movement of the circle's center parallel to the A;-axis and the rotation of P about the center. As the radial line sweeps out an angle 9, the point P traverses an arc of length a6, and the circle moves a distance aO along the jc-axis. Thus, as suggested by Figure 10.1.15, the center moves to the point (ad, a), and the coordinates of P are x = a9 — a sin 9,
}> = a — a cos 9
(10)
These are the equations of the cycloid in terms of the parameter 9. One of the reasons the cycloid is important in the history of mathematics is that the study of its properties helped to spur the development of early versions of differentiation and integration. Work on the cycloid was carried out by some of the most famous names in seventeenth century mathematics, including Johann and Jakob Bernoulli, Descartes, L'Hopital, Newton, and Leibniz. The curve was named the "cycloid" by the Italian mathematician and astronomer, Galileo, who spent over 40 years investigating its properties. An early problem of interest was that of constructing tangent lines to the cycloid. This problem was first solved by Descartes, and then by Fermat, whom Descartes had challenged with the question. A modern solution to this problem follows directly from the parametric equations (10) and Formula (4). For example, using Formula (4), it is straightforward to show that the *-intercepts of the cycloid are cusps and that there is a horizontal tangent line to the cycloid halfway between adjacent ^-intercepts (Exercise 60).
10.T Parametric Equations; Tangent Lines and Arc Length for Parametric Curves
A Figure 10.1.14
A Figure 10.1.16
699
A Figure 10.1.15
Another early problem was determining the arc length of an arch of the cycloid. This was solved in 1658 by the famous British architect and mathematician, Sir Christopher Wren. He showed that the arc length of one arch of the cycloid is exactly eight times the radius of the generating circle. [For a solution to this problem using Formula (9), see Exercise 71.] The cycloid is also important historically because it provides the solution to two famous mathematical problems—the brachistochrone problem (from Greek words meaning "shortest time") and the tautochrone problem (from Greek words meaning "equal time"). The brachistochrone problem is to determine the shape of a wire along which a bead might slide from a point P to another point Q, not directly below, in the shortest time. The tautochrone problem is to find the shape of a wire from P to Q such that two beads started at any points on the wire between P and Q reach Q in the same amount of time. The solution to both problems turns out to be an inverted cycloid (Figure 10.1.16). In June of 1696, Johann Bernoulli posed the brachistochrone problem in the form of a challenge to other mathematicians. At first, one might conjecture that the wire should form a straight line, since that shape results in the shortest distance from P to Q. However, the inverted cycloid allows the bead to fall more rapidly at first, building up sufficient speed to reach Q in the shortest time, even though it travels a longer distance. The problem was solved by Newton, Leibniz, and L'Hopital, as well as by Johann Bernoulli and his older brother Jakob; it was formulated and solved incorrectly years earlier by Galileo, who thought the answer was a circular arc. In fact, Johann was so impressed with his brother Jakob's solution that he claimed it to be his own. (This was just one of many disputes about the cycloid that eventually led to the curve being known as the "apple of discord.") One solution of the brachistochrone problem leads to the differential equation y = 2a
(U)
where a is a positive constant. We leave it as an exercise (Exercise 72) to show that the cycloid provides a solution to this differential equation.
Newton's solution of the brachistochrone problem in his own handwriting
700
Chapter TO / Parametric and Polar Curves; Conic Sections
I/QUICK CHECK EXERCISES 10.1
(See page 705 for answers.)
1. Find parametric equations for a circle of radius 2, centered at (3, 5). 2. The graph of the curve described by the parametric equations x = 4t — 1, y = 3t + 2 is a straight line with slope and y-intercept 3. Suppose that a parametric curve C is given by the equations x = f ( t ) , y = g(t) for 0 < t < 1. Find parametric equations for C that reverse the direction the curve is traced as the parameter increases from 0 to 1. EXERCISE SET 10.1
S Graphing Utility
~*~f~.
x = f(t),
y = g(r)
we can use the formula dy/dx = 5. Let L be the length of the curve
x = In/,
y = sin/
(1 < t < TT)
An integral expression for L is
@ CAS
1. (a) By eliminating the parameter, sketch the trajectory over the time interval 0 < t < 5 of the particle whose parametric equations of motion are
x = t -1,
4. To find dy/dx directly from the parametric equations
y = t +1
I Johann (left) and Jakob (right) Bernoulli Members of an amazing Swiss family that included several genii!; •; erations of outstanding mathematicians and scientists. ..-*? Nikolaus Bernoulli (1623-1708), a druggist, fled from Antwerp to escape religious persecution and ultimately Is settled in Basel, Switzerland. There he had three sons, Jakob I (also called Jacques or James), Nikolaus, and Johann I (also called Jean or John). The Roman numerals are used to distinguish family members with identical names (see the family tree below). Following Newton and Leibniz, the Bernoulli brothers, Jakob I and Johann I, are considered by some to be the two most important founders of calculus. Jakob I was self-taught in mathematics. His father wanted him to study for the ministry, but he turned to mathematics and in 1686 became a professor at the University of Basel. When he started working in mathematics, he knew nothing of Newton's and Leibniz' work. He eventually became familiar with Newton's results, but because so little of Leibniz' work was published, Jakob duplicated many of Leibniz' results. Jakob's younger brother Johann I was urged to enter into business by his father. Instead, he turned to medicine and studied mathematics under the guidance of his older brother. He eventually became a mathematics professor at Groningen in Holland, and then, when Jakob died in 1705, Johann succeeded him as mathematics professor at Basel. Throughout their lives, Jakob 1 and Johann I had a mutual passion for criticizing each other's work, which frequently erupted into ugly confrontations. Leibniz tried to mediate the disputes, but Jakob, who resented Leibniz' superior intellect, accused him of siding with Johann, and thus Leibniz became entangled in the arguments. The brothers often worked on common problems that they posed as challenges to one another. Johann, interested in gaining fame, often used unscrupulous means to make himself appear the originator of his brother's results; Jakob occasionally retaliated. Thus, it is often difficult to determine who deserves credit for many results. However, both men made major contributions
(b) Indicate the direction of motion on your sketch. (c) Make a table of x- and y-coordinates of the particle at times/ = 0, 1 , 2 , 3 , 4 , 5 . (d) Mark the position of the particle on the curve at the times in part (c), and label those positions with the values o f t . to the development of calculus. In addition to his work on calculus, Jakob helped establish fundamental principles in probability, including the Law of Large Numbers, which is a cornerstone of modern probability theory. Among the other members of the Bernoulli family, Daniel, son of Johann I, is the most famous. He was a professor of mathematics at St. Petersburg Academy in Russia and subsequently a professor of anatomy and then physics at Basel. He did work in calculus and probability, but is best known for his work in physics. A basic law of fluid flow, called Bernoulli's principle, is named in his honor. He won the annual prize of the French Academy 10 times for work on vibrating strings, tides of the sea, and kinetic theory of gases. Johann II succeeded his father as professor of mathematics at Basel. His research was on the theory of heat and sound. Nikolaus I was a mathematician and law scholar who worked on probability and series. On the recommendation of Leibniz, he was appointed professor of mathematics at Padua and then went to Basel as a professor of logic and then law. Nikolaus II was professor of jurisprudence in Switzerland and then professor of mathematics at St. Petersburg Academy. Johann III was a professor of mathematics and astronomy in Berlin and Jakob II succeeded his uncle Daniel as professor of mathematics at St. Petersburg Academy in Russia. Truly an incredible family! Nikolaus Bernou; li (1623- -1708) 1 Jakob I (1654-1705) (Jacques, Jamesi
Nike laus
Niko aus I (1687--1759)
~1 Johann 1 (1667-1748) (Jean, John) 1 1 1 Nikolaus II Daniel (1695-1726) (1700-1782)
Johann II (1710-1790) 1
Johann III Jakob II (1744-1807) (1759-1789)
10.1 Parametric Equations; Tangent Lines and Arc Length for Parametric Curves 2. (a) By eliminating the parameter, sketch the trajectory over the time interval 0 < ? < 1 of the particle whose parametric equations of motion are x = cos(nt),
y = sin(jr?)
(b) Indicate the direction of motion on your sketch. (c) Make a table of x- and ^-coordinates of the particle at times? =0,0.25,0.5,0.75, 1. (d) Mark the position of the particle on the curve at the times in part (c), and label those positions with the values of?. 3-12 Sketch the curve by eliminating the parameter, and indicate the direction of increasing ?.
701
(b) Assuming that the plane flies in a room in which the floor is at y = 0, explain why the plane will not crash into the floor. [For simplicity, ignore the physical size of the plane by treating it as a particle.] (c) How high must the ceiling be to ensure that the plane does not touch or crash into it? 21-22 Graph the equation using a graphing utility.
21. (a) x (b) x 22. (a) x (b) x
= y2 + 2y + 1 = sin y, -2n
3. x = 3? - 4, y = 6? + 2 FOCUS ON CONCEPTS
4. x = t - 3, y = 3? - 7 (0 < ? < 3) 5. x = 2 cos ?, y = 5 sin ? (0 < ? < 2n)
6. x = V/, y = 2? + 4 7. x = 3 + 2 cos ?, y = 2 + 4 sin ? (0 < ? < 2jt)
23. In each part, match the parametric equation with one of the curves labeled (I)-(VI), and explain your reasoning. (a) x = V?, y = sin 3? (b) x = 2 cos ?, y = 3 sin ? (c) x = t cos ?, y = ? sin ?
8. x = sec ?, y = tan ? (n < ? < 3jr/2) 9. * = cos2?, y = sin?
(d) x =
(-jr/2 < t < n/2)
1+?3 2?2 iTT^ (f) x = 4 cos ?, y = sin 2?
2
10. x = 4? + 3, y = 16? - 9 11. x = 2 sin2 t, y = 3 cos2 ? 12. x = sec2 ?, y = tan2 ?
3?2
3?
(0 < ? < 7T/2)
(0 < ? < jr/2)
l+t 3'
I 13-18 Find parametric equations for the curve, and check your work by generating the curve with a graphing utility. 13. A circle of radius 5, centered at the origin, oriented clockwise. 14. The portion of the circle x2 + y2 = 1 that lies in the third quadrant, oriented counterclockwise. 15. A vertical line intersecting the jc-axis at x = 2, oriented upward. 16. The ellipse x2/4 + y2/9 = 1, oriented counterclockwise. 17. The portion of the parabola x = y2 joining (1, —1) and (1, 1), oriented down to up. 18. The circle of radius 4, centered at (1, —3), oriented counterclockwise. 19. (a) Use a graphing utility to generate the trajectory of a particle whose equations of motion over the time interval 0 < ? < 5 are 3
X = 6t-^t ,
V
24. (a) Identify the orientation of the curves in Exercise 23. (b) Explain why the parametric curve x = t2,
y = t4
(-!
does not have a definite orientation.
2
y = \ + ±t
(b) Make a table of x- and y -coordinates of the particle at times? = 0, 1,2, 3,4, 5. (c) At what times is the particle on the y-axis? (d) During what time interval is y < 5? (e) At what time does the jc-coordinate of the particle reach a maximum? 20. (a) Use a graphing utility to generate the trajectory of a paper airplane whose equations of motion for ? > 0 are x = ?-2sin?,
IV A Figure Ex-23
y = 3-2cos?
25. (a) Suppose that the line segment from the point P(x0, y0) to Q(x\ , y } ) is represented parametrically by x =XQ + (XI — x0)t,
y = yo + (y\ - yo)t
(0 < ? < 1)
and that R(x, y) is the point on the line segment corresponding to a specified value of? (see the accompanying figure on the next page). Show that / = r/q, where r is the distance from P to R and q is the distance from P tO Q-
ton'.)
702 Chapter 10 / Parametric and Polar Curves; Conic Sections (b) What value of t produces the midpoint between points P and Ql (c) What value of t produces the point that is three-fourths of the way from P to Ql
35. For the parametric curve x = x(t),y = 3?4 — 2f 3 , the derivative of y with respect to x is computed by dy_ _ 12f3 - 6t2 dx x'(t) 36. The curve represented by the parametric equations x = t3, y = t + t6 (—00 < t < +00) is concave down for t < 0.
•4 Figure Ex-25
26. Find parametric equations for the line segment joining P(2, —1) and 2(3, 1), and use the result in Exercise 25 to find (a) the midpoint between P and Q (b) the point that is one-fourth of the way from P to 2 (c) the point that is three-fourths of the way from P to Q. 27. (a) Show that the line segment joining the points (XQ, yo) and (x\, y\) can be represented parametrically as t-t0 X =X0 t\ -to (to
t-t0 (b) Which way is the line segment oriented? (c) Find parametric equations for the line segment traced from (3, —1) to (1,4) as t varies from 1 to 2, and check your result with a graphing utility. 28. (a) By eliminating the parameter, show that if a and c are not both zero, then the graph of the parametric equations x=at + b, y = ct+d (to < t < t\) is a line segment. (b) Sketch the parametric curve x = 2t-\, y = t + l (1
37. Parametric curves can be defined piecewise by using different formulas for different values of the parameter. Sketch the curve that is represented piecewise by the parametric equations (x = 2t, y = 4r2 (0 < t < ~) \x = 2-2t, y = 2t Q
•^ Figure Ex-38
0 39. (a) Find parametric equations for the ellipse that is centered at the origin and has intercepts (4, 0), (-4, 0), (0, 3), and (0, -3). (b) Find parametric equations for the ellipse that results by translating the ellipse in part (a) so that its center is at (-1,2). (c) Confirm your results in parts (a) and (b) using a graphing utility. 0 40. We will show later in the text that if a projectile is fired from ground level with an initial speed of DO meters per second at an angle a with the horizontal, and if air resistance is neglected, then its position after t seconds, relative to the coordinate system in the accompanying figure on the next page is x = (DO cos a)?, y = (Dosina)r — ~gt2 where g ~ 9.8 m/s 2 . (a) By eliminating the parameter, show that the trajectory lies on the graph of a quadratic polynomial. (b) Use a graphing utility to sketch the trajectory if a = 30° and DO = 1000 m/s. (c) Using the trajectory in part (b), how high does the shell rise? (mm.)
10.1 Parametric Equations; Tangent Lines and Arc Length for Parametric Curves (d) Using the trajectory in part (b), how far does the shell travel horizontally?
703
53-54 Find all values of t at which the parametric curve has (a) a horizontal tangent line and (b) a vertical tangent line. -S 53. x = 2 sin t, y = 4 cos t (0 < t < 2n) 54. x = 2t3 - I5t2 + 24t + 7, y = t2 + t + 1 55. In the mid- 1850s the French physicist Jules Antoine Lissajous (1822-1880) became interested in parametric equations of the form
•< Figure Ex-40
FOCUS ON CONCEPTS
41. (a) Find the slope of the tangent line to the parametric curve x = t/2, y = r 2 + l a t f = -l and at t = 1 without eliminating the parameter, (b) Check your answers in part (a) by eliminating the parameter and differentiating an appropriate function of x. 42. (a) Find the slope of the tangent line to the parametric curve x = 3 cos t, y = 4 sin t at t = jt/4 and at t = In/4 without eliminating the parameter, (b) Check your answers in part (a) by eliminating the parameter and differentiating an appropriate function of x. 43. For the parametric curve in Exercise 41, make a conjecture about the sign of d2y/dx2 at t = — 1 and at t = 1, and confirm your conjecture without eliminating the parameter.
x = sin a?,
y = sinbt
in the course of studying vibrations that combine two perpendicular sinusoidal motions. I f a / b is a rational number, then the combined effect of the oscillations is a periodic motion along a path called a Lissajous curve. (a) Use a graphing utility to generate the complete graph of the Lissajous curves corresponding to a = 1, b — 2; a = 2, b = 3; a = 3, b = 4; and a = 4, b = 5. (b) The Lissajous curve
x = sin t,
y = sin It
(0 < t < 2n)
crosses itself at the origin (see Figure Ex-55). Find equations for the two tangent lines at the origin. 56. The prolate cycloid = 2 — jicost,
= 2t — 7tsmt
crosses itself at a point on the x-axis (see the accompanying figure). Find equations for the two tangent lines at that point.
44. For the parametric curve in Exercise 42, make a conjecture about the sign of d2y/dx2 at t = n/4 and at t = ln/4, and confirm your conjecture without eliminating the parameter. 45-50 Find dy/dx and d2y/dx2 at the given point without eliminating the parameter.
45. x = v/F, y = It + 4; t = 1 46. x = \t2 + \, y = ±t3-t; t = 2 47. x = sect, y = tan?; t = n/3
A Figure Ex-55
A Figure Ex-56
57. Show that the curve x = t2, y = t3 — 4t intersects itself at the point (4, 0), and find equations for the two tangent lines to the curve at the point of intersection. 58. Show that the curve with parametric equations
48. x = sinh;, y = cosh/; t = 0 49. x = & + co&9, y = 1 + s i n # ; 9 = n/6 50. x = cos0, y = 3 sin<£;
x = e',
y = e~'
at t = 1 without eliminating the parameter, (b) Find the equation of the tangent line in part (a) by eliminating the parameter. 52. (a) Find the equation of the tangent line to the curve
x = It + 4,
y = 8r2 - 2t + 4
at t = 1 without eliminating the parameter, (b) Find the equation of the tangent line in part (a) by eliminating the parameter.
intersects itself at the point (3 , 1 ) , and find equations for the two tangent lines to the curve at the point of intersection. 59. (a) Use a graphing utility to generate the graph of the parametric curve
x = cos t,
y = sin t
(0 < t < 2n)
and make a conjecture about the values of t at which singular points occur. (b) Confirm your conjecture in part (a) by calculating appropriate derivatives. 60. Verify that the cycloid described by Formula (10) has cusps at its x -intercepts and horizontal tangent lines at midpoints between adjacent x-intercepts (see Figure 10.1.14).
10.1 Parametric Equations; Tangent Lines and Arc Length for Parametric Curves (d) Using the trajectory in part (b), how far does the shell travel horizontally?
703
53-54 Find all values of t at which the parametric curve has (a) a horizontal tangent line and (b) a vertical tangent line, li 53. x = 2 sin t, y = 4 cos t (0 < t < 2n) 54. x = 2t3 - I5t2 + 24? + 7, y = t2 + t+\ 55. In the mid-1850s the French physicist Jules Antoine Lissajous (1822-1880) became interested in parametric equations of the form
•4 Figure Ex-40
FOCUS ON CONCEPTS
41. (a) Find the slope of the tangent line to the parametric curve x = t/2, y = t2 + 1 at t = -1 and at t = 1 without eliminating the parameter, (b) Check your answers in part (a) by eliminating the parameter and differentiating an appropriate function of x. 42. (a) Find the slope of the tangent line to the parametric curve x = 3 cos;, y = 4 sin tatt = n/4 and at t = 7;r/4 without eliminating the parameter, (b) Check your answers in part (a) by eliminating the parameter and differentiating an appropriate function of x. 43. For the parametric curve in Exercise 41, make a conjecture about the sign of d2y/dx2 at t = — 1 and at t = 1, and confirm your conjecture without eliminating the parameter.
x = sinar,
in the course of studying vibrations that combine two perpendicular sinusoidal motions. If alb is a rational number, then the combined effect of the oscillations is a periodic motion along a path called a Lissajous curve. (a) Use a graphing utility to generate the complete graph of the Lissajous curves corresponding to a = l,b = 2; a = 2, b = 3; a = 3, b = 4; and a = 4, b = 5. (b) The Lissajous curve
x = sin t,
y = sin 2t
45. x = ~Jt, y = 2t + 4; t = 1 46. x = \t2 + \, y = \t3-t; t = 2 47. x = sect, y = tanf; t = n/3
56. The prolate cycloid x = 2 — n cos t,
y = 2t — n sin t
49. x =9 + cos6», y = 1 + sin#; 9 = n/6 50. x = co&
y = e~'
at t = 1 without eliminating the parameter, (b) Find the equation of the tangent line in part (a) by eliminating the parameter. 52. (a) Find the equation of the tangent line to the curve
x = It + 4,
(—n
crosses itself at a point on the ;t-axis (see the accompanying figure). Find equations for the two tangent lines at that point.
y
A Figure Ex-55
A Figure Ex-56
57. Show that the curve x = t2, y = t3 — 4t intersects itself at the point (4, 0), and find equations for the two tangent lines to the curve at the point of intersection. 58. Show that the curve with parametric equations
48. x = sinh t, y = cosh t; t = 0
x — e*,
(0 < t < In)
crosses itself at the origin (see Figure Ex-55). Find equations for the two tangent lines at the origin.
44. For the parametric curve in Exercise 42, make a conjecture about the sign of d2y/dx2 at t = jt/4 and at t = ln/4, and confirm your conjecture without eliminating the parameter. 45-50 Find dy/dx and d2y/dx2 at the given point without eliminating the parameter. S
y = smbt
y = 8r2 - 2t + 4
at t = 1 without eliminating the parameter, (b) Find the equation of the tangent line in part (a) by eliminating the parameter.
jc = r 2 - 3r + 5,
y = t3+t2- 10?+ 9
intersects itself at the point (3, 1), and find equations for the two tangent lines to the curve at the point of intersection. 59. (a) Use a graphing utility to generate the graph of the parametric curve x = cos 3 1,
y = sin 3 1
(0 < t < 2n)
and make a conjecture about the values of t at which singular points occur. (b) Confirm your conjecture in part (a) by calculating appropriate derivatives. 60. Verify that the cycloid described by Formula (10) has cusps at its x -intercepts and horizontal tangent lines at midpoints between adjacent ;t-intercepts (see Figure 10.1.14).
704
Chapter 10 / Parametric and Polar Curves; Conic Sections
61. (a) What is the slope of the tangent line at time t to the trajectory of the paper airplane in Example 5? (b) What was the airplane's approximate angle of inclination when it crashed into the wall? 62. Suppose that a bee follows the trajectory x=t-2cost,
y = 2-2sint
(0 < t < 10)
(a) At what times was the bee flying horizontally? (b) At what times was the bee flying vertically? 63. Consider the family of curves described by the parametric equations x = a cost + h,
y = bsint + k
(0 < t < lit)
where a ^ 0 and b ^ 0. Describe the curves in this family if (a) h and k are fixed but a and b can vary (b) a and b are fixed but h and k can vary (c) a = 1 and b = 1, but h and k vary so that h = k + 1. 0 64. (a) Use a graphing utility to study how the curves in the family x = 2acos?t,
y = 2acostsint
(—2n
change as a varies from 0 to 5. (b) Confirm your conclusion algebraically. (c) Write a brief paragraph that describes your findings.
FOCUS ON CONCEPTS
73. The amusement park rides illustrated in the accompanying figure consist of two connected rotating arms of length 1—an inner arm that rotates counterclockwise at 1 radian per second and an outer arm that can be programmed to rotate either clockwise at 2 radians per second (the Scrambler ride) or counterclockwise at 2 radians per second (the Calypso ride). The center of the rider cage is at the end of the outer arm. (a) Show that in the Scrambler ride the center of the cage has parametric equations x = cos t + cos 2t,
y = sin t — sin 2t
(b) Find parametric equations for the center of the cage in the Calypso ride, and use a graphing utility to confirm that the center traces the curve shown in the accompanying figure. (c) Do you think that a rider travels the same distance in one revolution of the Scrambler ride as in one revolution of the Calypso ride? Justify your conclusion.
65-70 Find the exact arc length of the curve over the stated interval.
65. x = t2, y = |?3
(0 < t < 1)
66. x = 41 - 2, y = 2t3/4 (1 < t < 16) 67. x = cos 3t , y = sin 3t (0 < t < ii)
68. 69. 70. [c] 71.
x = sin t + cos t , y = sin t — cos t (0 < t < JT) x = e2'(sint + cosf), y = e2l(sint -cost)(-l < t < 1) * = 2sin- 1 r, y = \n(l-t2) (0 < t < \) (a) Use Formula (9) to show that the length L of one arch of a cycloid is given by /•27T /•
-cos6)d9
L =a / Jo
(b) Use a CAS to show that L is eight times the radius of the wheel that generates the cycloid (see the accompanying figure).
A Figure Ex-73
74. (a) If a thread is unwound from a fixed circle while being held taut (i.e., tangent to the circle), then the end of the thread traces a curve called an involute of a circle. Show that if the circle is centered at the origin, has radius a, and the end of the thread is initially at the point (a, 0), then the involute can be expressed parametrically as x = a(cosO + 9sir\9),
In
< Figure Ex-71
72. Use the parametric equations in Formula (10) to verify that the cycloid provides one solution to the differential equation
where a is a positive constant.
y = a(sin9 — 9cos9)
where 9 is the angle shown in part (a) of the accompanying figure on the next page. (b) Assuming that the dog in part (b) of the accompanying figure on the next page unwinds its leash while keeping it taut, for what values of 9 in the interval 0 < 9 < 2n will the dog be walking North? South? East? West? (c) Use a graphing utility to generate the curve traced by the dog, and show that it is consistent with your answer in part (b).
10.2 Polar Coordinates
705
75. Find the area of the surface generated by revolving x = t2, y = 3? (0 < t < 2) about the x-axis. 76. Find the area of the surface generated by revolving the curve x = e' cos t, y = e' sin t (0 < t < 7T/2) about the *-axis. 77. Find the area of the surface generated by revolving the curve x = cos2 t,y = sin21 (0 < t < n/2) about the y-axis. 78. Find the area of the surface generated by revolving x = 6t, y = 4t2 (0 < t < 1) about the y-axis. 79. By revolving the semicircle (a)
x = rcost,
A Figure Ex-74
75-80 If f ' ( t ) and g'(f) are continuous functions, and if no segment of the curve x = f ( t ) , y = g(t) (a
T)* dt
o
and the area of the surface generated by revolving the curve about the y-axis is
'=/W(£) ya v \ ^ /
+ - -
y = rsint
(0 < t < TI)
about the x -axis, show that the surface area of a sphere of radius r is 47zr2. 80. The equations x = a(j) — a sin 0,
y = a — a cos
(0 < 0 < 2n)
represent one arch of a cycloid. Show that the surface area generated by revolving this curve about the x-axis is given by 5 = 64TO2/3. 81. Writing Consult appropriate reference works and write an essay on American mathematician Nathaniel Bowditch (1773-1838) and his investigation of Bowditch curves (better known as Lissajous curves; see Exercise 55). 82. Writing What are some of the advantages of expressing a curve parametrically rather than in the form y = f ( x ) l
[The derivations are similar to those used to obtain Formulas (4) and (5) in Section 5.5.] Use the formulas above in these exercises.
I/QUICK CHECK ANSWERS 10.1 (Q
2. f ; 2.75 3. x = /(I - t), y = g(l - t) 4.
+ cos 2 1 dt
POLAR COORDINATES Up to now we have specified the location of a point in the plane by means of coordinates relative to two perpendicular coordinate axes. However, sometimes a moving point has a special affinity for some fixed point, such as a planet moving in an orbit under the central attraction of the Sun. In such cases, the path of the panicle is best described by its angular direction and its distance from the fixed point. In this section we will discuss a new kind of coordinate system that is based on this idea. POLAR COORDINATE SYSTEMS
Apolar coordinate system in a plane consists of a fixed point O, called Ihspole (or origin), and a ray emanating from the pole, called the polar axis. In such a coordinate system
706
Chapter 10 / Parametric and Polar Curves; Conic Sections P(r, &)
Pole A Figure 10.2.1
Polar axis
we can associate with each point P in the plane a pair of polar coordinates (r, 9), where r is the distance from P to the pole and 9 is an angle from the polar axis to the ray OP (Figure 10.2.1). The number r is called the radial coordinate of P and the number 9 the angular coordinate (orpolar angle) of P. In Figure 10.2.2, the points (6, n/4), (5, 2n/3), (3, 571/4), and (4, lln/6) are plotted in polar coordinate systems. If P is the pole, then r — 0, but there is no clearly defined polar angle. We will agree that an arbitrary angle can be used in this case; that is, (0,9) are polar coordinates of the pole for all choices of 9. . (5, 27t/3)
(6, 71/4)
A Figure 10.2.2
The polar coordinates of a point are not unique. For example, the polar coordinates (l,7jr/4),
(I,-Ti/4),
and (1, 15;r/4)
all represent the same point (Figure 10.2.3). 157T/4
77T/4
(1,15*74)
> Figure 10.2.3
In general, if a point P has polar coordinates (r, &), then
(r, 0 + 2nn)
5TC/4 Polar axis
(-3,7i/4) A Figure 10.2.4
and (r, 6 - Imt)
are also polar coordinates of P for any nonnegative integer n. Thus, every point has infinitely many pairs of polar coordinates. As defined above, the radial coordinate r of a point P is nonnegative, since it represents the distance from P to the pole. However, it will be convenient to allow for negative values of r as well. To motivate an appropriate definition, consider the point P with polar coordinates (3, 5n/4). As shown in Figure 10.2.4, we can reach this point by rotating the polar axis through an angle of 5n/4 and then moving 3 units from the pole along the terminal side of the angle, or we can reach the point P by rotating the polar axis through an angle of 7T/4 and then moving 3 units from the pole along the extension of the terminal side. This suggests that the point (3, 5n/4) might also be denoted by (—3, jr/4), with the minus sign serving to indicate that the point is on the extension of the angle's terminal side rather than on the terminal side itself. In general, the terminal side of the angle 9 + JT is the extension of the terminal side of 9, so we define negative radial coordinates by agreeing that
(-r, 9)
and
(r, 9 + n)
are polar coordinates of the same point. RELATIONSHIP BETWEEN POLAR AND RECTANGULAR COORDINATES
Frequently, it will be useful to superimpose a rectangular jcv-coordinate system on top of a polar coordinate system, making the positive *-axis coincide with the polar axis. If this is done, then every point P will have both rectangular coordinates ( x , y ) and polar coordinates
10.2 Polar Coordinates
707
(r, 6). As suggested by Figure 10.2.5, these coordinates are related by the equations x = rcos9,
y = rsin9
(1)
These equations are well suited for finding x and y when r and 9 are known. However, to find r and 9 when x and y are known, it is preferable to use the identities sin2 6 + cos2 9 = 1 and tan 0 — sin $/ cos 6 to rewrite (1) as 1
y
x = r cos 0
tan 9 = -
(2)
A Figure 10.2.5
> Example 1 Find the rectangular coordinates of the point P whose polar coordinates are (r, 9) = (6, 2jr/3) (Figure 10.2.6). Solution.
A Figure 10.2.6
Substituting the polar coordinates r = 6 and 9 = 2n/3 in (1) yields
Thus, the rectangular coordinates of P are (x, y) = (—3, 3\/3). 7T/2
> Example 2 Find polar coordinates of the point P whose rectangular coordinates are (-2, -273) (Figure 10.2.7). -»— 0
Solution. We will find the polar coordinates (r, 9) of P that satisfy the conditions r > 0 and 0 < 9 < 2n. From the first equation in (2), r2 = x2 + y2 = (-2)2 + (-273 ) 2 = 4 + 12 = 16
so r = 4. From the second equation in (2), />(-2,-2V3) '
-273 = 73 —2
y
A Figure 10.2.7
From this and the fact that (—2, — 2V3) lies in the third quadrant, it follows that the angle satisfying the requirement 0 < 9 < In is 9 = 4n/3. Thus, (r, 9) = (4, 4n/3) are polar coordinates of P. All other polar coordinates of P are expressible in the form [4, — + Inn)
or (-4, - + 2njr\
where n is an integer. GRAPHS IN POLAR COORDINATES
We will now consider the problem of graphing equations in r and 9, where 9 is assumed to be measured in radians. Some examples of such equations are r=l,
0=n/4,
r = 9,
r = sin#,
r = cos26>
In a rectangular coordinate system the graph of an equation in x and y consists of all points whose coordinates (x, y) satisfy the equation. However, in a polar coordinate system, points have infinitely many different pairs of polar coordinates, so that a given point may have some polar coordinates that satisfy an equation and others that do not. Given an equation
708
Chapter 10 / Parametric and Polar Curves; Conic Sections
in r and 8, we define its graph in polar coordinates to consist of all points with at least one pair of coordinates (r, 9) that satisfy the equation. > Example 3
Sketch the graphs of (a) r = 1
in polar coordinates. Solution (a). For all values of 9, the point ( 1 , 6 ) is 1 unit away from the pole. Since 9 is arbitrary, the graph is the circle of radius 1 centered at the pole (Figure 10.2.8a). Solution (b). For all values of r, the point (r, n/4) lies on a line that makes an angle of ji/4 with the polar axis (Figure 10.2.8&). Positive values of r correspond to points on the line in the first quadrant and negative values of r to points on the line in the third quadrant. Thus, in absence of any restriction on r, the graph is the entire line. Observe, however, that had we imposed the restriction r > 0, the graph would have been just the ray in the first quadrant. A
> Figure 10.2.8
Equations r = f ( 9 ) that express r as a function of 9 are especially important. One way to graph such an equation is to choose some typical values of 9, calculate the corresponding values of r, and then plot the resulting pairs (r, 9) in a polar coordinate system. The next two examples illustrate this process. Example 4
A Figure 10.2.9
Graph the spiral r — 6 (0 < 0). Compare your graph to that in Figure 10.2.9.
Sketch the graph of r
> 0) in polar coordinates by plotting points.
Solution. Observe that as 9 increases, so does r; thus, the graph is a curve that spirals out from the pole as 9 increases. A reasonably accurate sketch of the spiral can be obtained by plotting the points that correspond to values of 9 that are integer multiples of n/2, keeping in mind that the value of r is always equal to the value of 9 (Figure 10.2.9). -^ > Example 5 points.
Sketch the graph of the equation r = sin 9 in polar coordinates by plotting
Solution. Table 10.2.1 shows the coordinates of points on the graph at increments of 71/6.
These points are plotted in Figure 10.2.10. Note, however, that there are 13 points listed in the table but only 6 distinct plotted points. This is because the pairs from 9 = n on yield
10.2
Polar Coordinates
709
duplicates of the preceding points. For example, (—1/2, ln/6) and (1/2, n/6) represent the same point. -^
Table 10.2.1
e
n 6
JT 3
I
Vs"
V5"
i
2
2
2
(~"f)
(l'f)
(RADIANS)
0
r = sin d
o
2
<,.«
(0,0)
f|'|)
(2 ' 3 )
JT 2
('' f )
2^ 3
5s 6
1
7w "6~
o
-1 1 2
47C
17C
5)r
3
2
3
UJT 6
V3 2
2
V3~ 2
6/
/ V3~ 4;r\ ("T' T)
Cr1i' 32M) /r ^2
1
2n
Q
57rl
y)i
(-!' T)
(0, 2s)
Observe that the points in Figure 10.2.10 appear to lie on a circle. We can confirm that this is so by expressing the polar equation r = sin 6 in terms of x and y. To do this, we multiply the equation through by r to obtain
r 2 = r sin 9 which now allows us to apply Formulas (1) and (2) to rewrite the equation as
x2 + y2 = y Rewriting this equation as x2 + y2 — y = 0 and then completing the square yields A Figure 10.2.10
r = sinfl
which is a circle of radius ^ centered at the point (0, {) in the xy-plane. It is often useful to view the equation r — f(9) as an equation in rectangular coordinates (rather than polar coordinates) and graphed in a rectangular 0r-coordinate system. For example, Figure 10.2.11 shows the graph of r = sin 6* displayed using rectangular 9rcoordinates. This graph can actually help to visualize how the polar graph in Figure 10.2. 10 is generated: • At 9 — 0 we have r = 0, which corresponds to the pole (0, 0) on the polar graph. • As 9 varies from 0 to n/2, the value of r increases from 0 to 1, so the point (r, 9) moves along the circle from the pole to the high point at (1 , n/2). • As 9 varies from jt/2 to n, the value of r decreases from 1 back to 0, so the point (r, 9) moves along the circle from the high point back to the pole. • As 9 varies from n to 3jr/2, the values of r are negative, varying from 0 to — 1 . Thus, the point (r, 9) moves along the circle from the pole to the high point at (1, n/2), which is the same as the point (— 1, 3n/2). This duplicates the motion that occurred for 0 < 9 < n/2. •
As 9 varies from 3n/2 to 2n, the value of r varies from — 1 to 0. Thus, the point (r, 9) moves along the circle from the high point back to the pole, duplicating the motion that occurred for n/2 <9
> Examples
Sketch the graph of r — cos 20 in polar coordinates.
Solution. Instead of plotting points, we will use the graph of r = cos 26 in rectangular coordinates (Figure 10.2.12) to visualize how the polar graph of this equation is generated. The analysis and the resulting polar graph are shown in Figure 10.2.13. This curve is called a four-petal rose. ^
710
Chapter 10 / Parametric and Polar Curves; Conic Sections
r varies from 1 to 0 as e varies from 0 to Ti/4.
r varies from 0 to -1 as 6 varies from
r varies from 0 to 1 as G varies from 3rc/4 to n.
71/4 tO n!2.
r varies from ; i r varies from
1 to 0 as 0
0 to -1 as 0
varies from j ] varies from : n to 571/4. 5n/4 to 37T/2. i
r varies from ; -1 to 0 as 0 j varies from ; 37T/2 to 7/T/4. j
1
r varies from '• 0 to 1 as varies from
A Figure 10.2.13
SYMMETRY TESTS
Observe that the polar graph of r — cos 28 in Figure 10.2.13 is symmetric about the x-axis and the y-axis. This symmetry could have been predicted from the following theorem, which is suggested by Figure 10.2.14 (we omit the proof).
10.2.1
THEOREM (Symmetry Tests)
(a) A curve in polar coordinates is symmetric about the x-axis if replacing 6 by —6 in its equation produces an equivalent equation (Figure 10.2.14a). The converse of each part of Theorem 10.2.1 is false. See Exercise 79.
(b) A curve in polar coordinates is symmetric about the y-axis if replacing 9 by n — 9 in its equation produces an equivalent equation (Figure 10.2.14b). (c) A curve in polar coordinates is symmetric about the origin if replacing 6 by 0 + 7i, or replacing r by ~r in its equation produces an equivalent equation (Figure 10.2.14c).
nil
nil
nil
0,0)
(r,B)
' (r,
(r, -e
(-r, 0) (a)
> Figure 10.2.14
(b)
(c)
> Example 7 Use Theorem 10.2.1 to confirm that the graph of r = cos 29 in Figure 10.2.13 is symmetric about the x-axis and y-axis. Solution, A graph that is symmetric about both the z-axis and the y-axis is also symmetric about the origin. Use Theorem 10.2.1(c) to verify that the curve in Example 7 is symmetric about the origin.
To test for symmetry about the x-axis, we replace 9 by -0. This yields r = cos(-20) = cos 29
Thus, replacing 9 by —9 does not alter the equation. To test for symmetry about the y-axis, we replace 9 by n — 9. This yields r - cos2(;r -9) = cos(2n - 29) = cos(-2#) = cos 29 Thus, replacing 9 by n — 9 does not alter the equation. •«!
10.2 Polar Coordinates
711
*• Example 8 Sketch the graph of r — a(l — cos 9) in polar coordinates, assuming a to be a positive constant. Solution. Observe first that replacing 9 by -9 does not alter the equation, so we know in advance that the graph is symmetric about the polar axis. Thus, if we graph the upper half of the curve, then we can obtain the lower half by reflection about the polar axis. As in our previous examples, we will first graph the equation in rectangular 8 r -coordinates. This graph, which is shown in Figure 10.2.15a, can be obtained by rewriting the given equation as r = a - a cos 9, from which we see that the graph in rectangular 9r -coordinates can be obtained by first reflecting the graph of r — a cos 9 about the jc-axis to obtain the graph of r = —a cos 9, and then translating that graph up a units to obtain the graph of r — a — a cos 9. Now we can see the following: • As 0 varies from 0 to n/3, r increases from 0 to a/2. • As 9 varies from n/3 to n/2, r increases from a/2 to a. • As 9 varies from n/2 to 2n/3, r increases from a to 3a/2. • As 9 varies from 2n/3 to n, r increases from 3a/2 to 2a. This produces the polar curve shown in Figure 10.2.15b. The rest of the curve can be obtained by continuing the preceding analysis from n to 2n or, as noted above, by reflecting the portion already graphed about the x-axis (Figure 10.2.15c). This heart-shaped curve is called a cardioid (from the Greek word kardia meaning "heart" ). -4
la 3a 2 a a 2
j
>"
A
2
I
71 n2n 3 2 3
\. \
I
__™™™™__\™__________™.
71
V, ?
(la, n)
In
(a)
(c)
A Figure 10.2.15
*• Example 9
Sketch the graph of r2 = 4 cos 29 in polar coordinates.
Solution. This equation does not express r as a function of 9, since solving for r in terms of 9 yields two functions: r = 2vcos 29 2
and
r — -2Vcos20
Thus, to graph the equation r = 4 cos 29 we will have to graph the two functions separately and then combine those graphs. We will start with the graph of r = 2\/cos29. Observe first that this equation is not changed if we replace 9 by —9 or if we replace 9 by n — 9. Thus, the graph is symmetric about the ;t-axis and the y-axis. This means that the entire graph can be obtained by graphing the portion in the first quadrant, reflecting that portion about the y-axis to obtain the portion in the second quadrant, and then reflecting those two portions about the Jt-axis to obtain the portions in the third and fourth quadrants.
712
Chapter 10 / Parametric and Polar Curves; Conic Sections
To begin the analysis, we will graph the equation r = 2\/cos 20 in rectangular Orcoordinates (see Figure 10.2.16a). Note that there are gaps in that graph over the intervals jr/4 < 6 < 3n/4 and 5n/4 < 9 < In I'4 because cos 29 is negative for those values of 6. From this graph we can see the following: • As 9 varies from 0 to n/4, r decreases from 2 to 0. • As 0 varies from n/4 to jr/2, no points are generated on the polar graph. This produces the portion of the graph shown in Figure 10.2.l6b. As noted above, we can complete the graph by a reflection about the y-axis followed by a reflection about the x-axis (Figure 10.2.16c). The resulting propeller-shaped graph is called a lemniscate (from the Greek word lemniscos for a looped ribbon resembling the number 8). We leave it for you to verify that the equation r = 2T/cos29 has the same graph as r = —2Vcos26>, but traced in a diagonally opposite manner. Thus, the graph of the equation r2 = 4 cos 29 consists of two identical superimposed lemniscates. •<
nil
4
4
4
4
(a)
(b)
A Figure 10.2.16
FAMILIES OF LINES AND RAYS THROUGH THE POLE
If #o is a fixed angle, then for all values of r the point (r, OQ) lies on the line that makes an angle of 9 = 9o with the polar axis; and, conversely, every point on this line has a pair of polar coordinates of the form (r, 00)- Thus, the equation 9 = 90 represents the line that passes through the pole and makes an angle of #o with the polar axis (Figure 10.2.17
We will consider three families of circles in which a is assumed to be a positive constant: r = 2a cos 6
(*) A Figure 10.2.17
r = 2a sin 9
(3-5)
The equation r = a represents a circle of radius a centered at the pole (Figure 10.2.18a). Thus, as a varies, this equation produces a family of circles centered at the pole. For families (4) and (5), recall from plane geometry that a triangle that is inscribed in a circle with a diameter of the circle for a side must be a right triangle. Thus, as indicated in Figures 10.2.18& and 10.2.18c, the equation r = 2a cos 9 represents a circle of radius a, centered on the jc-axis and tangent to the y-axis at the origin; similarly, the equation r — 2a sin 9 represents a circle of radius a, centered on the y-axis and tangent to the ;c-axis at the origin. Thus, as a varies, Equations (4) and (5) produce the families illustrated in Figures 10.2.18^ and 10.2.18e.
10.2 Polar Coordinates
713
nil
7T/2
(a)
(e)
A Figure 10.2.18 •
FAMILIES OF ROSE CURVES
In polar coordinates, equations of the form Observe that replacing 9 by —9 does not change the equation r = la cos 0
r = a sin n6
(6-7)
r = a cos nO
and that replacing 9 by JT — 9 does not change the equation r = la sin 9. This explains why the circles in Figure 10.2.18d are symmetric about the x-axis and those in Figure 10.2.18e are symmetric about the j>-axis.
in which a > 0 and n is a positive integer represent families of flower-shaped curves called roses (Figure 10.2.19). The rose consists of n equally spaced petals of radius a if n is odd and 2n equally spaced petals of radius a if n is even. It can be shown that a rose with an even number of petals is traced out exactly once as 9 varies over the interval 0 < 9 < 2n and a rose with an odd number of petals is traced out exactly once as 9 varies over the interval 0 < 9 < n (Exercise 78). A four-petal rose of radius 1 was graphed in Example 6.
What do the graphs of one-petal roses look like? ROSE CURVES
n =2
r = a sin nd
n =6
0
o
r = a cos n9
M A Figure 10.2.19
FAMILIES OF CARDIOIDS AND LIMACONS
Equations with any of the four forms
r = a ± b sin 9
r = a±b cos 9
(8-9)
in which a > 0 and b > 0 represent polar curves called limagons (from the Latin word Umax for a snail-like creature that is commonly called a "slug"). There are four possible shapes for a limagon that are determined by the ratio a/b (Figure 10.2.20). If a — b (the case a/b = 1), then the lima§on is called a cardioid because of its heart-shaped appearance, as noted in Example 8.
714
Chapter 10 / Parametric and Polar Curves; Conic Sections
alb< 1
alb= 1
1 < alb < 2
alb > 2
Dimpled limagon
Convex limagon
• Figure 10.2.20
K Example 10 Figure 10.2.21 shows the family of limagons r = a +cos0 with the constant a varying from 0.25 to 2.50 in steps of 0.25. In keeping with Figure 10.2.20, the limagons evolve from the loop type to the convex type. As a increases from the starting value of 0.25, the loops get smaller and smaller until the cardioid is reached at a = 1. As a increases further, the lima§ons evolve through the dimpled type into the convex type. -4
a = 0.25
a = 0.5
a = 0.75
a=l
a = 1.25
a = 1.5
a = 1.75
a=2
a = 2.25
a = 2.5
A Figure 10.2.21
FAMILIES OF SPIRALS
A spiral is a curve that coils around a central point. Spirals generally have "left-hand" and "right-hand" versions that coil in opposite directions, depending on the restrictions on the polar angle and the signs of constants that appear in their equations. Some of the more common types of spirals are shown in Figure 10.2.22fornonnegative values of 6, a, andb. nil
Lituus spiral A Figure 10.2.22
SPIRALS IN NATURE
Spirals of many kinds occur in nature. For example, the shell of the chambered nautilus (below) forms a logarithmic spiral, and a coiled sailor's rope forms an Archimedean spiral. Spirals also occur in flowers, the tusks of certain animals, and in the shapes of galaxies.
10.2 Polar Coordinates
715
Thomas Taylor/Photo Researchers
Rex Ziak/Stone/Getty Images
Courtesy NASA & The Hubble Heritage Team
The shell of the chambered nautilus reveals a logarithmic spiral. The animal lives in the outermost chamber.
A sailor's coiled rope forms an Archimedean spiral,
A spiral galaxy.
GENERATING POLAR CURVES WITH GRAPHING UTILITIES
For polar curves that are too complicated for hand computation, graphing utilities can be used. Although many graphing utilities are capable of graphing polar curves directly, some are not. However, if a graphing utility is capable of graphing parametric equations, then it can be used to graph a polar curve r = f ( 6 ) by converting this equation to parametric form. This can be done by substituting /(0) for r in (1). This yields jr; = /(0)cos0,
y = /(0)sin0
(10)
which is a pair of parametric equations for the polar curve in terms of the parameter 9.
> Example 11
Express the polar equation
se
r = 2 + cos — parametrically, and generate the polar graph from the parametric equations using a graphing utility. Solution. Substituting the given expression for r in x = r cos 6 and y = r sin 6 yields the parametric equations 50' 501 y — 2 + cos — sin 6
501 2 + cos — cos 0,
TECHNOLOGY MASTERY Use a graphing utility to duplicate the curve in Figure 10.2.23. If your graphing utility requires that t be used as the parameter, then you will have to replace 0 by ( in (10) to generate the graph.
T
Next, we need to find an interval over which to vary 0 to produce the entire graph. To find such an interval, we will look for the smallest number of complete revolutions that must occur until the value of r begins to repeat. Algebraically, this amounts to finding the smallest positive integer n such that 2 +cos
5(0 + 2«7T)
= 2 + cos
or
50
r
\
50
cos — + 5nn\ = cos —
50
716
Chapter 10 / Parametric and Polar Curves; Conic Sections
For this equality to hold, the quantity 5nn must be an even multiple of n; the smallest n for which this occurs is n = 2. Thus, the entire graph will be traced in two revolutions, which means it can be generated from the parametric equations 5(91 f 5(9"1 56» I cos9, y = \ 2 + ccos o s — | sin# (0 < 0 < 4n) sm 9 I 2 +cos— 2 J This yields the graph in Figure 10.2.23. -4
A Figure 10.2.23
I/QUICK CHECK EXERCISES 10.2
(See page 719 for answers.)
1. (a) Rectangular coordinates of a point (x, y) may be recovered from its polar coordinates (r, 9) by means of the equations x = and y = (b) Polar coordinates (r, 0) may be recovered from rectangular coordinates (x, y) by means of the equations r2 = and tan 9 = 2. Find the rectangular coordinates of the points whose polar coordinates are given. (a) (4.7T/3) (b) (2,-7T/6) (c) (6, -2;r/3) EXERCISE SET 10.2
(d) (4,5jr/4)
3. In each part, find polar coordinates satisfying the stated conditions for the point whose rectangular coordinates are (a) r > 0 and 0 < 6 < 2rc (b) r < 0 and 0 < 9 < 2n 4. In each part, state the name that describes the polar curve most precisely: a rose, a line, a circle, a Iima9on, a cardioid, a spiral, a lemniscate, or none of these. (a) r = I - d (b) r = 1 + 2 sin 0 (c) r = sin 20 (d) r = cos2 9 (e) r = csc# (f) r = 2 + 2cos6> (g) r = -2sin<9
Graphing Utility
1 -1 Plot the points in polar coordinates. 1. (a) (3.7T/4) (b) (5,27r/3) (d) (4,7^/6) (e) (-6,-jr) 2. (a) (2, -jr/3) (b) (3/2, -7;r/4) (d) (-5,-7T/6) (e) (2.47T/3)
(c) (f) (c) (f)
(l,jr/2) (-l,9;r/4) (-3, 3;r/2) (0, TT)
3-4 Find the rectangular coordinates of the points whose polar coordinates are given. 3. (a) (6,71/6) (c) (-6,-5;r/6) (b) (7.27T/3) (d) (0, -TT) (e) (7, 177T/6) (f) (-5,0) 4. (a) (-2,7r/4) (b) (6, -jr/4) (c) (4,9jr/4) (e) (-4, -37T/2) (f) (0, 3ir) (d) (3,0) 5. In each part, a point is given in rectangular coordinates. Find two pairs of polar coordinates for the point, one pair satisfying r > 0 and 0 < 9 < 2n, and the second pair satisfying r > 0 and -2n < 9 < 0. (a) (-5,0) (b) (2V3, -2) (c) (0, -2) (d) (-8,-8) (e) (-3,373) (f) (1,1) 6. In each part, find polar coordinates satisfying the stated conditions for the point whose rectangular coordinates are (-V3, I).
(a) r > 0 (b) r < 0 (c) r > 0 (d) r < 0
and 0 < 9 < 2n and 0 < 0 < 2n and -2;r < 9 < 0 and -n < 9 < n
7-8 Use a calculating utility, where needed, to approximate the polar coordinates of the points whose rectangular coordinates are given. 7. (a) (3,4) (b) (6,-8) (c) (-1, tan"1 1) 8. (a) (-3,4) (b) (-3, 1.7) (c) (2, siiT'i) 9-10 Identify the curve by transforming the given polar equation to rectangular coordinates. 9. (a) r = 2 (b) r sin 0 = 4 f, (c) r = 3 cos e (d) r = 3 cos 9 + 2 sin 9 10. (a) r = 5 sec 6» (b) r = 2 sin 9 (c) r = 4 cos 0 + 4 sin 9 (d) r = sec Q tan 9 11-12 Express the given equations in polar coordinates. 11. (a) x = 3 (b) x2 + y2 = 7 2 2 (c) x + y + 6y = 0 (d) 9xy = 4
10.2 Polar Coordinates
12. (a) y = -3 (c) x2 + y2 + 4x = 0
(b) x2 + y1 = 5 (d) X2(x2 + y2)
27. r = 3(1+ sinfl)
28. r = 5-5sin0
29. r = 4 - 4 c o s 0
30. r = 1 + 2 sin 0
31. r = -1 -cos0
32. r = 4 + 3 cos 0
FOCUS ON CONCEPTS
33. r = 3 — sin 0
34. r = 3 + 4cos0
13-16 A graph is given in a rectangular 0r-coordinate system. Sketch the corresponding graph in polar coordinates.
35. r - 5 = 3 sin 9
36. r = 5 -2cos0
37. r = -3 - 4 sin 9
38. r2 = cos 29
13.
39. r 2 = 16 sin 29
40. r = 40
41. r = 40
42. r = 40
3 -/•*-
(0 < 0)
717
(0 > 0)
43. r = -2 cos 20
44. r = 3 sin 20
45. r = 9 sin 40
46. r = 2 cos 30
47-50 True-False Determine whether the statement is true or false. Explain your answer. 47. The polar coordinate pairs (—1, :r/3) and (1, —2jr/3) describe the same point.
2
4
48. If the graph of r = /(0) drawn in rectangular 0rcoordinates is symmetric about the r-axis, then the graph of r = f ( 9 ) drawn in polar coordinates is symmetric about the jc-axis.
6
17-20 Find an equation for the given polar graph. [Note: Numeric labels on these graphs represent distances to the origin.]
49. The portion of the polar graph of r = sin 20 for values of 0 between jr/2 and JT is contained in the second quadrant.
17. (a)
50. The graph of a dimpled limac,on passes through the polar origin. | 51-55 Determine a shortest parameter interval on which a complete graph of the polar equation can be generated, and then use a graphing utility to generate the polar graph. Circle
Circle
18. (a)
Cardioid
(b)
(c)
CN
^ (\
19(
Limacon
"
Three-petal rose
Four-petal rose 20. (a)
Limagon
x—4-~%- (b)
Lemniscate
„
(c)
Five-petal rose
r\
r\
51. r = cos — 2
52. r = sin -
0 53. r = 1 - 2 sin 4 0 55. r = cos -
0 54. r = 0.5 + cos -
56. The accompanying figure shows the graph of the "butterfly curve" a r = e c o s e -2cos40 + sin 3 4 Determine a shortest parameter interval on which the complete butterfly can be generated, and then check your answer using a graphing utility.
4
Circle
21-46 Sketch the curve in polar coordinates.
21.0 = -
22. 0 = ~
23. r = 3
24. r = 4cos0
25. r = 6sin0
26. r - 2 = 2cos0
•< Figure Ex-56
718
Chapter 10 / Parametric and Polar Curves; Conic Sections
57. The accompanying figure shows the Archimedean spiral r = 9/2 produced with a graphing calculator. (a) What interval of values for 9 do you think was used to generate the graph? (b) Duplicate the graph with your own graphing utility.
61-62 A polar graph of r = f ( 9 ) is given over the stated interval. Sketch the graph of (a) r = f(-6) (b) r = (d) r = -f(0).
(c) r = f(e +
61. 0 < 9 < 7T/2
62. 71/2 < 0 < 7T
7C/2
(1, 37T/4)
[-9, 9] x [-6, 6] *Scl = l.yScI = 1
•4 Figure Ex-57
58. Find equations for the two families of circles in the accompanying figure.
A Figure Ex-61
A Figure Ex-62
63-64 Use the polar graph from the indicated exercise to sketch the graph of
(a) r = f(9) + 1 63. Exercise 61
A Figure Ex-58
59. (a) Show that if a varies, then the polar equation r = a sec 9
(-?r/2 < 9 < jt/2)
describes a family of lines perpendicular to the polar axis, (b) Show that if b varies, then the polar equation r = bcsc9
(0 < 9 < Tt)
describes a family of lines parallel to the polar axis. FOCUS ON CONCEPTS
60. The accompanying figure shows graphs of the Archimedean spiral r = 6 and the parabolic spiral r = *J(j. Which is which? Explain your reasoning.
(b) r = 2f(9) - 1. 64. Exercise 62
65. Show that if the polar graph of r = f ( 9 ) is rotated counterclockwise around the origin through an angle a, then r = f(9 — a) is an equation for the rotated curve. [Hint: If (/o, 60) is any point on the original graph, then (ro, $o + a ) is a point on the rotated graph.] 66. Use the result in Exercise 65 to find an equation for the lemniscate that results when the lemniscate in Example 9 is rotated counterclockwise through an angle of n/2. 67. Use the result in Exercise 65 to find an equation for the cardioid r = 1 + cos 9 after it has been rotated through the given angle, and check your answer with a graphing utility. (a) n-
(b) |
(c) ji
(d) y
68. (a) Show that if A and B are not both zero, then the graph of the polar equation
r = A sin 0 + B cos 9 is a circle. Find its radius. (b) Derive Formulas (4) and (5) from the formula given in part (a). 69. Find the highest point on the cardioid r = 1 + cos 9. 70. Find the leftmost point on the upper half of the cardioid r = 1 +cos6>. 71. Show that in a polar coordinate system the distance d between the points (r\, 9\) and fa, $2) is - 2r,r 2 cos(<9] - 02) 72-74 Use the formula obtained in Exercise 71 to find the distance between the two points indicated in polar coordinates.
A Figure Ex-60
72. (3,7T/6) and (2,7T/3)
10.3 Tangent Lines, Arc Length, and Area for Polar Curves
73. Successive tips of the four-petal rose r = cos 29. Check your answer using geometry. 74. Successive tips of the three-petal rose r = sin 39. Check your answer using trigonometry. 75. In the late seventeenth century the Italian astronomer Giovanni Domenico Cassini (1625-1712) introduced the family of curves (x1 + y2 + a2)2 -b4 - 4a2x2 = 0
(a > 0, b > 0)
in his studies of the relative motions of the Earth and the Sun. These curves, which are called Cassini ovals, have one of the three basic shapes shown in the accompanying figure. (a) Show that if a = b, then the polar equation of the Cassini oval is r2 = 2a2 cos 29, which is a lemniscate. (b) Use the formula in Exercise 71 to show that the lemniscate in part (a) is the curve traced by a point that moves in such a way that the product of its distances from the polar points (a, 0) and (a, n) is a2.
719
76. Show that the hyperbolic spiral r = 1 /9 (9 > 0) has a horizontal asymptote at y = 1 by showing that y —>• 1 and x —>• +co as 9 —>• 0 + . Confirm this result by generating the spiral with a graphing utility. 77. Show that the spiral r = I/O2 does not have any horizontal asymptotes. 78. Prove that a rose with an even number of petals is traced out exactly once as 9 varies over the interval 0 < 9 < 2n and a rose with an odd number of petals is traced out exactly once as 9 varies over the interval 0 < 9 < n. 79. (a) Use a graphing utility to confirm that the graph of r = 2 - sin (61/2) (0 < 9 < 4n) is symmetric about the x -axis. (b) Show that replacing 9 by —9 in the polar equation r = 2 — sin($/2) does not produce an equivalent equation. Why does this not contradict the symmetry demonstrated in part (a)? 80. Writing Use a graphing utility to investigate how the family of polar curves r = 1 + a cos n6 is affected by changing the values of a and n, where a is a positive real number and n is a positive integer. Write a brief paragraph to explain your conclusions. 81. Writing Why do you think the adjective "polar" was chosen in the name "polar coordinates"?
a
A Figure Ex-75
76-77 Vertical and horizontal asymptotes of polar curves can sometimes be detected by investigating the behavior of x = r cos 9 and y = r sin 9 as 9 varies. This idea is used in these exercises.
I/QUICK CHECK ANSWERS 10.2 1. (a) rcos0; rsin6» (b) x2 + y2; y/x 2. (a) (2,273) (b) (V3,-1) (c) (-3, -3
TANGENT LINES, ARC LENGTH, AND AREA FOR POLAR CURVES In this section we will derive the formulas required to find slopes, tangent lines, and arc lengths of polar curves. We will then show how to find areas of regions that are bounded by polar curves.
TANGENT LINES TO POLAR CURVES
Our first objective in this section is to find a method for obtaining slopes of tangent lines to polar curves of the form r = f ( 9 ) in which r is a differentiable function of 6. We showed in the last section that a curve of this form can be expressed parametrically in terms of the parameter 9 by substituting f ( 0 ) for r in the equations x — r cos 9 and y = r sin 9. This ylddS
x = f ( 9 ) cos 9,
y = f ( 9 ) sin 9
720
Chapter 10 / Parametric and Polar Curves; Conic Sections
from which we obtain — = -/(0) sin 6 + f ' ( 6 ) cos 6 = -r sin 6 + — cos 9 d6 dO
(1) — do Thus, if dx/dO and dy/d9 are continuous and if dx/d0 ^ 0, then y is a differentiable function of x, and Formula (4) in Section 10.1 with 6 in place of t yields
- = f ( 0 ) cos 9 + f ' ( 9 ) sin^ =
dy
dylde
dx
dx/dO
.dr_ r cos 9+sine ~d9 . Q , adr —r sin 9 + cos 0 — d9
(2)
> Example 1 Find the slope of the tangent line to the circle r = 4 cos 0 at the point where 0 = n/4. Solution. From (2) with r = 4 cos 0, so that dr/dO = —4 sin 0, we obtain
Tangent
dy _ 4cos 2 0-4sin 2 0 _ cos20 - sin29 dx —8 sin 0 cos 0 2 sin 0 cos 0 Using the double-angle formulas for sine and cosine, dy cos 20 — = =-cot20 dx sm 20 Thus, at the point where 0 = n/4 the slope of the tangent line is
m—
A Figure 10.3.1
dy dx
= -cot^=0
which implies that the circle has a horizontal tangent line at the point where 9 = 7T/4 (Figure 10.3.1). + > Example 2 Find the points on the cardioid r — 1 — cos 0 at which there is a horizontal tangent line, a vertical tangent line, or a singular point. Solution. A horizontal tangent line will occur where dy/dO = 0 and dx/d9 ^ 0, a vertical tangent line where dy/d& ^ 0 and dx/d9 = 0, and a singular point where dy/d9 = 0 and dxIdO = 0. We could find these derivatives from the formulas in (1). However, an alternative approach is to go back to basic principles and express the cardioid parametrically by substituting r — 1 — cos 0 in the conversion formulas x = r cos 0 and y = r sin 0. This yie
S
x = (1 -cos0)cos0,
y = (l -cos0)sin0
(0 < 0 < 2n)
Differentiating these equations with respect to 0 and then simplifying yields (verify) — = sin 0(2 cos 0-1), dO
— = (l-cos0)(l+2cos0) dO
Thus,dx/d9 — Oif sin0 = Oorcos0 = \, anddy/d9 = Oif cos0 = 1 orcos0 = — i. We leave it for you to solve these equations and show that the solutions of dx/dO = 0 on the interval 0 < 0 < 2n are re 5n -, d9
10.3 Tangent Lines, Arc Length, and Area for Polar Curves
721
and the solutions of dy/dO = 0 on the interval 0 < 9 < 2n are 2n 3
de
4* 3
Thus, horizontal tangent lines occur at 6 = 2n/3 and 0 = 4n/3; vertical tangent lines occur at 0 = jt/3, n, and 5n/3; and singular points occur at 9 — 0 and 0 — 2jt (Figure 10.3.2). Note, however, that r = 0 at both singular points, so there is really only one singular point on the cardioid—the pole. -^
TANGENT LINES TO POLAR CURVES AT THE ORIGIN Formula (2) reveals some useful information about the behavior of a polar curve r = f ( 9 ) that passes through the origin. If we assume that r = 0 and dr/d6 ^ 0 when 9 — OQ, then it follows from Formula (2) that the slope of the tangent line to the curve at 9 = OQ is
A Figure 10.3.2
dy_ dx
0 + sin 6»o 0 + cos 6>o
d0__ dr
cos
= tan 90
(Figure 10.3.3). However, tan 0$ is also the slope of the line 9 = OQ, so we can conclude that this line is tangent to the curve at the origin. Thus, we have established the following result.
71/2
10.3.1 THEOREM If the polar curve r — f ( 9 ) passes through the origin at 0 — 6>o and ifdr/d9 ^ 0 at 9 = 0Q, then the line 9 = 0Q is tangent to the curve at the origin. Slope = tan 00
This theorem tells us that equations of the tangent lines at the origin to the curve r = f ( 9 ) can be obtained by solving the equation /(<9) = 0. It is important to keep in mind, however, that r = f ( 9 ) may be zero for more than one value of 9, so there may be more than one tangent line at the origin. This is illustrated in the next example.
A Figure 10.3.3
6 = 27T/3
nil
8 = Ti/3
> Example 3 The three-petal rose r = sin 39 in Figure 10.3.4 has three tangent lines at the origin, which can be found by solving the equation sin 39 = 0
It was shown in Exercise 78 of Section 10.2 that the complete rose is traced once as 9 varies over the interval 0 < 9 < TC, so we need only look for solutions in this interval. We leave it for you to confirm that these solutions are = 0, 0--.
and
9=
2;r
Since dr/d9 — 3 cos 39 ^ 0 for these values of 9, these three lines are tangent to the rose at the origin, which is consistent with the figure. < A Figure 10.3.4
ARC LENGTH OF A POLAR CURVE A formula for the arc length of a polar curve r — f ( 9 ) can be derived by expressing the curve in parametric form and applying Formula (9) of Section 10.1 for the arc length of a parametric curve. We leave it as an exercise to show the following.
722
Chapter 10 / Parametric and Polar Curves; Conic Sections
10.3.2 ARC LENGTH FORMULA FOR POLAR CURVES If no segment of the polar curve r = f ( 6 ) is traced more than once as 9 increases from a to ft, and if dr/d6 is continuous for a < 0 < ft, then the arc length L from 6 = a to 8 = ft is
-jf
ff>
Ja
Ja
nil
> Example 4 and 6 = it.
/ dr\
= I
p + ( — ) dO V
(3)
\UU /
Find the arc length of the spiral r = ee in Figure 10.3.5 between
=0
Solution. n
(n, e ) A Figure 10.3.5 n
= I Jo > Example 5
- 1) * 31.3
Find the total arc length of the cardioid r — 1 + cos 9.
Solution. The cardioid is traced out once as 9 varies from 9 = 0 to 9 = 2n. Thus, P
/
f2
/dr\
Jr2+( — \ d9= I V \dv J Jo
+cos6dd Id e ntity( 45) A H BR off Appendix
=2 /*27T
r- 1 + cos 0
=2/ Jo
\cos\e\d6
Since cos ^9 changes sign at n, we must split the last integral into the sum of two integrals: the integral from 0 to n plus the integral from it to 2jt. However, the integral from it to lit is equal to the integral from 0 to it, since the cardioid is symmetric about the polar axis (Figure 10.3.6). Thus, L=2 I Jo
A Figure 10.3.6
cos \9 d9 = 4 I cos ± 9 d6 = 8 sin ±0 | = 8 Jo o
AREA IN POLAR COORDINATES
We begin our investigation of area in polar coordinates with a simple case. r=f(6)
10.3.3 AREA PROBLEM IN POLAR COORDINATES Suppose that a and ft are angles that satisfy the condition a < p < a + 2n j and suppose that f ( 9 ) is continuous and nonnegative for a < 9 < ft. Find the area of the region R enclosed by the polar curve r — f ( 9 ) and the rays 9 = a and 9 = ft (Figure 10.3.7).
A Figure 10.3.7
10.3 Tangent Lines, Arc Length, and Area for Polar Curves
723
In rectangular coordinates we obtained areas under curves by dividing the region into an increasing number of vertical strips, approximating the strips by rectangles, and taking a limit. In polar coordinates rectangles are clumsy to work with, and it is better to partition the region into wedges by using rays
A0, A Figure 10.3.8
such that
a < 9\ < 6*2 < • • • < dn-\ < ft
(Figure 10.3.8). As shown in that figure, the rays divide the region R into n wedges with areas A\, A2,..., An and central angles A0], A02, - . . , A0n. The area of the entire region can be written as n A = Al+A2 + - - - + An = J^Ak (4) k=\ /•=/«?)
If A0* is small, then we can approximate the area Ak of the A:th wedge by the area of a sector with central angle A0* and radius /(#£)> where 0 = 0£ is any ray that lies in the A:th wedge (Figure 10.3.9). Thus, from (4) and Formula (5) of Appendix B for the area of a sector, we obtain „ n A = £ Ak « £ £[/(0t*)]2 A0t (5)
A Figure 10.3.9
k=\
k=l
If we now increase n in such a way that max A#t —>• 0, then the sectors will become better and better approximations of the wedges and it is reasonable to expect that (5) will approach the exact value of the area A (Figure 10.3.10); that is,
A —
A Figure 10.3.10
A
lim
max A$i -
Ja
Note that the discussion above can easily be adapted to the case where /(#) is nonpositive for a < 9 < ft. We summarize this result below.
10.3.4 AREA IN POLAR COORDINATES
If a and ft are angles that satisfy the condition
a < ft < a + In and if /(#) is continuous and either nonnegative or nonpositive for a < 0 < ft, then the area A of the region R enclosed by the polar curve r = f ( 9 ) (a < 9 < ft) and the lines 9 — a and 9 = 6 is rP =
I Ja
rfi £[/(0)] 2 <»= / Ja
2
5r
dO
(6)
The hardest part of applying (6) is determining the limits of integration. This can be done as follows: Area in Polar Coordinates: Limits of Integration Step 1. Sketch the region R whose area is to be determined. Step 2. Draw an arbitrary "radial line" from the pole to the boundary curve r — /(#). Step 3. Ask, "Over what interval of values must 0 vary in order for the radial line to sweep out the region /??" Step 4. Your answer in Step 3 will determine the lower and upper limits of integration.
724
Chapter 10 / Parametric and Polar Curves; Conic Sections
> Example 6 r = 1 — cos#. r= 1 - cos 6
Find the area of the region in the first quadrant that is within the cardioid
Solution. The region and a typical radial line are shown in Figure 10.3.11. Forthe radial line to sweep out the region, 9 must vary from 0 to nil. Thus, from (6) with a = 0 and ft = n/2, we obtain fTtll j
A=
Jo
i
i-TC/2
-r2 d8 = 2 2 J0
i
fir/2
(1 -cos6)2 d9 = - / 2 JQ
(1 - 2cos6 + cos2 9) dO
With the help of the identity cos2 0 = |(1 + cos 20), this can be rewritten as
A Figure 10.3.11
2 -1JT/2 1 \ 1 [3 1 Y' 3 2cos6» + -cos29 \dQ = - \ -9 -2sin9 + -sin29 = -jr-l < 2 / 2 2 4 Jo
3
The shaded region is swept out by the radial line as 0 varies from 0 to nil.
2 > Example 7
Find the entire area within the cardioid of Example 6.
Solution. For the radial line to sweep out the entire cardioid, 9 must vary from 0 to 2n. Thus, from (6) with a = 0 and ft = 2n,
A= I Jo
-r2dO = ~l 2 2 JQ
(l-cos0fd9
If we proceed as in Example 6, this reduces to 3n 2
A= -
Alternative Solution. Since the cardioid is symmetric about the jc-axis, we can calculate the portion of the area above the Jt-axis and double the result. In the portion of the cardioid above the jc-axis, 9 ranges from 0 to n, so that
A = 2 I \r2d9= I ( l Jo 2 JQ
=T^
USING SYMMETRY
Although Formula (6) is applicable if r = f ( 9 ) is negative, area computations can sometimes be simplified by using symmetry to restrict the limits of integration to intervals where r > 0. This is illustrated in the next example. > Example 8
Find the area of the region enclosed by the rose curve r = cos 29.
Solution. Referring to Figure 10.2.13 and using symmetry, the area in the first quadrant that is swept out for 0 < 9 < n/4 is one-eighth of the total area inside the rose. Thus, from Formula (6) r-JT/4 i /•7T/4 1
rn/4 /"3T/4
-r2dO =4 I 2 J0
= 8/ Jo
n/4 i
/
cos220d9 /-ir/4
-(l+cos46»)J6> = 2 / 2 JQ /4 r =-sin49 \
i
~
o
( +cos49)dO
10.3 Tangent Lines, Arc Length, and Area for Polar Curves
725
Sometimes the most natural way to satisfy the restriction a < /6 < a + 2n required by Formula (6) is to use a negative value for a. For example, suppose that we are interested in finding the area of the shaded region in Figure 10.3.12a. The first step would be to determine the intersections of the cardioid r = 4 + 4 cos 9 and the circle r = 6, since this information is needed for the limits of integration. To find the points of intersection, we can equate the two expressions for r. This yields 4 + 4 cos 0—6
or
cos 6 = -
which is satisfied by the positive angles 5?r T However, there is a problem here because the radial lines to the circle and cardioid do not sweep through the shaded region shown in Figure 10.3.12b as 9 varies over the interval 7T/3 < 9 < 5:r/3. There are two ways to circumvent this problem—one is to take advantage of the symmetry by integrating over the interval 0 < 0 < n/3 and doubling the result, and the second is to use a negative lower limit of integration and integrate over the interval —7T/3 < 0 < jr/3 (Figure 10.3.12c). The two methods are illustrated in the next example. = - and
nil
nil
nil
7C/2
(a)
A Figure 10.3.12
*• Example 9 Find the area of the region that is inside of the cardioid r = 4 + 4 cos 6 and outside of the circle r — 6. Solution Using a Negative Angle. The area of the region can be obtained by subtracting the areas in Figures 10.3. YLd and 10.3.12e: A=
— (f>Ydd 2
2
I-,13 2
'*irea 'ns't'e cardioid minus area inside circle.
n/3
•£
/32
1
-36]d6
/
(16cos6» + 8cos 2 6»- 10) dO •JT/3
= [16 sin 9 + (49 + 2 sin 29) - 10 <9]*^/3 = 18^3 - 4n Solution Using Symmetry. Using symmetry, we can calculate the area above the polar axis and double it. This yields (verify) fTT/3
A =2 I Jo
-[(4 + 4cos<9) 2 -36]d<9 = 2(9^/3-2:r) = 18^3-4;r 2
which agrees with the preceding result. -4
10.3 Tangent Lines, Arc Length, and Area for Polar Curves
725
Sometimes the most natural way to satisfy the restriction a < ft < a + In required by Formula (6) is to use a negative value for a. For example, suppose that we are interested in finding the area of the shaded region in Figure 10.3.12a. The first step would be to determine the intersections of the cardioid r = 4 + 4 cos 9 and the circle r = 6, since this information is needed for the limits of integration. To find the points of intersection, we can equate the two expressions for r. This yields 4 + 4 cos ( 9 = 6 or
cos6> = -
2
which is satisfied by the positive angles — —
57T
and
T
However, there is a problem here because the radial lines to the circle and cardioid do not sweep through the shaded region shown in Figure W.3.l2b as 9 varies over the interval 7i/3 <9< 5n/3. There are two ways to circumvent this problem—one is to take advantage of the symmetry by integrating over the interval 0 < 6 < n/3 and doubling the result, and the second is to use a negative lower limit of integration and integrate over the interval —n/3 < 9 < n/3 (Figure 10.3.12c). The two methods are illustrated in the next example. nil
(a)
(c)
(d)
(e)
A Figure 10.3.12
> Example 9 Find the area of the region that is inside of the cardioid r = 4 + 4 cos 0 and outside of the circle r — 6. Solution Using a Negative Angle. The area of the region can be obtained by subtracting the areas in Figures 10.3.12d and 10.3.12e:
r i2 = 7-W3
Area inside cardioid minus area inside circle. 1
JT/3 1
/
-.732 2
/•w/3 LV.
f 4cos0r-36]d0 = J-W3 J-jt/3
= [16 sin 0 + (40 + 2 sin 20) - 10 6»]^3/3 = 18V3 - 4n Solution Using Symmetry. Using symmetry, we can calculate the area above the polar axis and double it. This yields (verify) /"fir/3 1
A =2 I Jo
-[(4 + 4cos6») 2 -36]Jc9 = 2(9^3 - 2:r) = 18\/3-47T 2
which agrees with the preceding result. •*
726
Chapter 10 / Parametric and Polar Curves; Conic Sections INTERSECTIONS OF POLAR GRAPHS
In the last example we found the intersections of the cardioid and circle by equating their expressions for r and solving for 9. However, because a point can be represented in different ways in polar coordinates, this procedure will not always produce all of the intersections. For example, the cardioids
r = 1 +cos 8
r = 1 — cos 9
and r = 1 + cos 9
(7)
intersect at three points: the pole, the point (1, ;r/2), and the point (1, 3:r/2) (Figure 10.3.13). Equating the right-hand sides of the equations in (7) yields 1 - cos 9 — 1 + cos 8 or cos 9 = 0, so n 0 = -+kjr, k = 0, ±1,±2, ...
A Figure 10.3.13
Substituting any of these values in (7) yields r = 1, so that we have found only two distinct points of intersection, (1,7T/2) and (1, 3jt/2); the pole has been missed. This problem occurs because the two cardioids pass through the pole at different values of 9—the cardioid r — 1 — cos9 passes through the pole at 0 = 0, and the cardioid r = 1 + cos9 passes through the pole at 9 — n. The situation with the cardioids is analogous to two satellites circling the Earth in intersecting orbits (Figure 10.3.14). The satellites will not collide unless they reach the same point at the same time. In general, when looking for intersections of polar curves, it is a good idea to graph the curves to determine how many intersections there should be.
The orbits intersect, but the satellites do not collide.
A Figure 10.3.14
•QUICK CHECK EXERCISES 10.3
(See page 729 for answers.)
1. (a) To obtain dy/dx directly from the polar equation r = f ( 9 ) , we can use the formula dy dx
dy/d9 dx/dO
(b) Use the formula in part (a) to find dy/dx directly from the polar equation r = esc 6. 2. (a) What conditions on f(0o) and /'(#o) guarantee that the line 9 = OQ is tangent to the polar curve r = f ( 9 ) at the origin?
EXERCISE SET 10.3
B Graphing Utility
(b) What are the values of OQ in [0, 2jt] at which the lines 0 = QQ are tangent at the origin to the four-petal rose r = cos 26>? 3. (a) To find the arc length L of the polar curve r = f ( 9 ) (a < 9 < ft), we can use the formula L = (b) The polar curve r = sec 9 (0 < 6 < jr/4) has arc length L= 4. The area of the region enclosed by a nonnegati ve polar curve r = f(0) (a < 9 < ft) and the lines 9 = a and 9 = ft is given by the definite integral 5. Find the area of the circle r = a by integration.
@ CAS
1 -6 Find the slope of the tangent line to the polar curve for the given value of 9. 1. r = 2 sin 9; 9 = n/6 3. r = \/9; 9 = 2
5. r = sin30; 0 = n/4
2. r = 1 + co&9; 9 = n/2 4. r = a sec 20; 9 = x/6 6. r = 4 - 3 s i n 0 ; 9 = n
7-8 Calculate the slopes of the tangent lines indicated in the accompanying figures. 7. r = 2 + 2 sin 9
8. r = 1 - 2 sin 0
A Figure Ex-7
A Figure Ex-8
9-10 Find polar coordinates of all points at which the polar curve has a horizontal or a vertical tangent line. 9. r = a(l + cos 9) 10. r = a sin 9
10.3 Tangent Lines, Arc Length, and Area for Polar Curves 11-12 Use a graphing utility to make a conjecture about the number of points on the polar curve at which there is a horizontal tangent line, and confirm your conjecture by finding appropriate derivatives. 11. r = sin 9 cos2 9
12. r = 1 - 2 sin 6
13-18 Sketch the polar curve and find polar equations of the tangent lines to the curve at the pole. 13. r = 2cos30
14. r = 4 s i n 6 »
15. r = 4Vcos20
16. r = sin26>
17. r = 1 - 2cos6>
18. r = 29
27. In each part, find the area of the circle by integration. (a) r = 2a sin 9 (b) r = la cos 9 28. (a) Show that r = 1 sin 9 + 2 cos 9 is a circle. (b) Find the area of the circle using a geometric formula and then by integration. 29-34 Find the area of the region described. 29. The region that is enclosed by the cardioid r = 1 + 2 sin &. 30. The region in the first quadrant within the cardioid r = 1 +cos0.
19-22 Use Formula (3) to calculate the arc length of the polar curve.
31. The region enclosed by the rose r = 4 cos 39.
19. The entire circle r = a
33. The region enclosed by the inner loop of the limac,on r — 1 + 2 cos 9. [Hint: r < 0 over the interval of integration.]
20. The entire circle r = 2a cos 0 21. The entire cardioid r = a(l — cos 9) 22. r = e30 from 9 = 0 to 9 = 2 23. (a) Show that the arc length of one petal of the rose r = cos n9 is given by
727
32. The region enclosed by the rose r = 2 sin 29.
34. The region swept out by a radial line from the pole to the curve r = 2/9 as 9 varies over the interval 1 < 9 < 3. 35-38 Find the area of the shaded region.
/* TCI (2n
'•I
(b) Use the numerical integration capability of a calculating utility to approximate the arc length of one petal of the four-petal rose r = cos 29. (c) Use the numerical integration capability of a calculating utility to approximate the arc length of one petal of the n-petal rose r = cosn9 for n = 2, 3, 4, . . . , 20; then make a conjecture about the limit of these arc lengths as n —> +<». 24. (a) Sketch the spiral r = e~0/g (0 < 9 < +00). (b) Find an improper integral for the total arc length of the spiral. (c) Show that the integral converges and find the total arc length of the spiral.
39-46 Find the area of the region described.
25. Write down, but do not evaluate, an integral for the area of each shaded region.
39. The region inside the circle r = 3 sin 9 and outside the cardioid r = 1 + sin6>. 40. The region outside the cardioid r = 2 — 2 cos 9 and inside the circle r = 4. 41. The region inside the cardioid r = 2 + 2 cos 9 and outside the circle r = 3.
r= 1 — cos 9
r = 2 cos 6
• = sin 20
42. The region that is common to the circles r = 2 cos 6 and r = 2 sinfl. 43. The region between the loops of the limac,on r = i + cos 9 . 44. The region inside the cardioid r = 2 + 2 cos 9 and to the right of the line r cos 9 = \ . 45. The region inside the circle r = 2 and to the right of the line
r = 1 - sin 0
r = cos 26
26. Find the area of the shaded region in Exercise 25(d).
46. The region inside the rose r = 2a cos 29 and outside the circle r = a-\/2.
728
Chapter 10 / Parametric and Polar Curves; Conic Sections
47-50 True-False Determine whether the statement is true or false. Explain your answer. 47. The *-axis is tangent to the polar curve r = cos(0/2) at 48. The arc length of the polar curve r = -JO for 0 < 9 < nil is given by fit/2
L
-fJo
49. The area of a sector with central angle 9 taken from a circle of radius r is 9r2.
•4 Figure Ex-54
[c] 55. (a) Show that the Folium of Descartes x3 — 3xy + y3 = 0 can be expressed in polar coordinates as 3 sin 9 cos 0
50. The expression
computes the area enclosed by the inner loop of the Iima9on r =1-
FOCUS ON CONCEPTS
51. (a) Find the error: The area that is inside the lemniscate r2 = a2 cos 29 is
A=
t= 5s
/
1 2J
Jo l
= \a
_
I2* sin 29
/
Jo
=0
Jo (b) Find the correct area. (c) Find the area inside the lemniscate r2 = 4 cos 20 and outside the circle r = \/2.
cos3 9 + sin3 0 (b) Use a CAS to show that the area inside of the loop is | (Figure 2.13d). 56. (a) What is the area that is enclosed by one petal of the rose r = a cos nO if n is an even integer? (b) What is the area that is enclosed by one petal of the rose r = a cos n9 if n is an odd integer? (c) Use a CAS to show that the total area enclosed by the rose r = a cos nO is jta2/2 if the number of petals is even. [Hint: See Exercise 78 of Section 10.2.] (d) Use a CAS to show that the total area enclosed by the rose r = a cos nO is na2/4 if the number of petals is odd. 57. One of the most famous problems in Greek antiquity was "squaring the circle," that is, using a straightedge and compass to construct a square whose area is equal to that of a given circle. It was proved in the nineteenth century that no such construction is possible. However, show that the shaded areas in the accompanying figure are equal, thereby "squaring the crescent."
52. Find the area inside the curve r2 = sin 20. 53. A radial line is drawn from the origin to the spiral r = aO (a > 0 and 9 > 0). Find the area swept out during the second revolution of the radial line that was not swept out during the first revolution. 54. As illustrated in the accompanying figure, suppose that a rod with one end fixed at the pole of a polar coordinate system rotates counterclockwise at the constant rate of 1 rad/s. At time / = 0 a bug on the rod is 10 mm from the pole and is moving outward along the rod at the constant speed of 2 mm/s. (a) Find an equation of the form r = f ( 0 ) for the path of motion of the bug, assuming that 0 = 0 when f = 0. (b) Find the distance the bug travels along the path in part (a) during the first 5 s. Round your answer to the nearest tenth of a millimeter.
•^ Figure Ex-57
E3 58. Use a graphing utility to generate the polar graph of the equation r = cos 30 + 2, and find the area that it encloses. E3 59. Use a graphing utility to generate the graph of the bifolium r = 2 cos 9 sin2 6, and find the area of the upper loop. 60. Use Formula (9) of Section 10.1 to derive the arc length formula for polar curves, Formula (3). 61. As illustrated in the accompanying figure, let P(r, 0) be a point on the polar curve r = f(0), let \jr be the smallest counterclockwise angle from the extended radius OP to the
10.3 Tangent Lines, Arc Length, and Area for Polar Curves
tangent line at P, and let be the angle of inclination of the tangent line. Derive the formula r tan \lr = —;— dr/de by substituting tan 4> for dy/dx in Formula (2) and applying the trigonometric identity tan 0 — tan 9 tan(4> -9) = 1 + tan
729
surface of revolution that is generated by revolving a parametric curve about the x-axis or y-axis. Use those formulas to derive the following formulas for the areas of the surfaces of revolution that are generated by revolving the portion of the polar curve r = f ( 9 ) from 9 = a to 9 = ft about the polar axis and about the line 9 = Tt/2: A
S=
N
— } d8 d9 .
r
S=
Ja
fdrY 2m cos 9 r2 + [— d9 V
About e = 0
About 0 = jr/
\«0/
(b) State conditions under which these formulas hold.
•^ Figure Ex-61
62-63 Use the formula for -ty obtained in Exercise 61. 8 62. (a) Use the trigonometric identity 9 l-cosfl tan - = —: to show that if (r, 9) is a point on the cardioid r = 1 - cos 6 (0 < 9 < 2jr) then ir =9/2. (b) Sketch the cardioid and show the angle ^ at the points where the cardioid crosses the y-axis. (c) Find the angle fy at the points where the cardioid crosses the y-axis. 63. Show that for a logarithmic spiral r = aebe, the angle from the radial line to the tangent line is constant along the spiral (see the accompanying figure). [Note: For this reason, logarithmic spirals are sometimes called equiangular spirals.]
•^ Figure Ex-63
64. (a) In the discussion associated with Exercises 75-80 of Section 10.1, formulas were given for the area of the
•QUICK CHECK ANSWERS
65-68 Sketch the surface, and use the formulas in Exercise 64 to find the surface area. 65. The surface generated by revolving the circle r = cos 9 about the line 9 = n/2. 66. The surface generated by revolving the spiral r = ee (0 < B < a/2) about the line 9 = n/2. 67. The "apple" generated by revolving the upper half of the cardioid r = 1 — cos 9 (0 < 9 < n) about the polar axis. 68. The sphere of radius a generated by revolving the semicircle r = a in the upper half -plane about the polar axis. 69. Writing
(a) Show that if 0 < 6\ <92
A=i (b) Use the formula obtained in part (a) to describe an approach to answer Area Problem 10.3.3 that uses an approximation of the region R by triangles instead of circular wedges. Reconcile your approach with Formula (6). 70. Writing In order to find the area of a region bounded by two polar curves it is often necessary to determine their points of intersection. Give an example to illustrate that the points of intersection of curves r = f ( 9 ) and r = g(9) may not coincide with solutions to f ( 9 ) = g(9). Discuss some strategies for determining intersection points of polar curves and provide examples to illustrate your strategies.
10.3
dr r cos # + sin $ — j ^ g 7 1. (a) -SSU (b) •+ = 0 2. (a) f(90) = 0, f'(90) / 0 (b) 90 = -, —, — , — -r sin 9 + cos 9 —d9
/dr\2
rP
'.(a) / Jr2+ — ) Ja
V
\**"/
(b)
f2" . / \a2 dB = na Jo
730
Chapter 10 / Parametric and Polar Curves; Conic Sections
CONIC SECTIONS In this section we will discuss some of the basic geometric properties of parabolas, ellipses, and hyperbolas. These curves play an important role in calculus and also arise naturally in a broad range of applications in such fields as planetary motion, design of telescopes and antennas, geodetic positioning, and medicine, to name a few.
CONIC SECTIONS
Circles, ellipses, parabolas, and hyperbolas are called conic sections or conies because they can be obtained as intersections of a plane with a double-napped circular cone (Figure 10.4.1). If the plane passes through the vertex of the double-napped cone, then the intersection is a point, a pair of intersecting lines, or a single line. These are called degenerate conic sections.
i
,v Circle
Ellipse
', Parabola i
Some students may already be familiar with the material in this section, in which case it can be treated as a review. Instructors who want to spend some additional time on precalculus review may want to allocate more than one lecture on this material.
10.4 Conic Sections
731
DEFINITIONS OF THE CONIC SECTIONS
Although we could derive properties of parabolas, ellipses, and hyperbolas by defining them as intersections with a double-napped cone, it will be better suited to calculus if we begin with equivalent definitions that are based on their geometric properties.
10.4.1 DEFINITION A parabola is the set of all points in the plane that are equidistant from a fixed line and a fixed point not on the line.
All points on the parabola are equidistant from the focus and directrix.
The line is called the directrix of the parabola, and the point is called the focus (Figure 10.4.2). A parabola is symmetric about the line that passes through the focus at right angles to the directrix. This line, called the axis or the axis of symmetry of the parabola, intersects the parabola at a point called the vertex.
10.4.2 DEFINITION An ellipse is the set of all points in the plane, the sum of whose distances from two fixed points is a given positive constant that is greater than the distance between the fixed points. Directrix A Figure 10.4.2
The two fixed points are called the foci (plural of "focus") of the ellipse, and the midpoint of the line segment joining the foci is called the center (Figure 10.4.3a). To help visualize Definition 10.4.2, imagine that two ends of a string are tacked to the foci and a pencil traces a curve as it is held tight against the string (Figure 10.4.3&). The resulting curve will be an ellipse since the sum of the distances to the foci is a constant, namely, the total length of the string. Note that if the foci coincide, the ellipse reduces to a circle. For ellipses other than circles, the line segment through the foci and across the ellipse is called the major axis (Figure 10.4.3c), and the line segment across the ellipse, through the center, and perpendicular to the major axis is called the minor axis. The endpoints of the major axis are called vertices. The sum of the distances to the foci is constant.
Vertex
Figure 10.4.3
10.4.3 DEFINITION A hyperbola is the set of all points in the plane, the difference of whose distances from two fixed distinct points is a given positive constant that is less than the distance between the fixed points.
The two fixed points are called the foci of the hyperbola, and the term "difference" that is used in the definition is understood to mean the distance to the farther focus minus the distance to the closer focus. As a result, the points on the hyperbola form two branches, each
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Chapter 10 / Parametric and Polar Curves; Conic Sections
"wrapping around" the closer focus (Figure 10.4.4a). The midpoint of the line segment joining the foci is called the center of the hyperbola, the line through the foci is called the focal axis, and the line through the center that is perpendicular to the focal axis is called the conjugate axis. The hyperbola intersects the focal axis at two points called the vertices. Associated with every hyperbola is a pair of lines, called the asymptotes of the hyperbola. These lines intersect at the center of the hyperbola and have the property that as a point P moves along the hyperbola away from the center, the vertical distance between P and one of the asymptotes approaches zero (Figure 10.4.4&).
Conjugate
The distance from the • farther focus minus the j distance to the closer focus is constant.
z. / Focal ' ax j s ,i / \ Focus Vertex
Vertex
These distances approach zero as the point moves away f rom t ne center.
These distances approach *• zero as the point moves away from the center.
(a) A Figure 10.4.4
EQUATIONS OF PARABOLAS IN STANDARD POSITION
It is traditional in the study of parabolas to denote the distance between the focus and the vertex by p. The vertex is equidistant from the focus and the directrix, so the distance between the vertex and the directrix is also p; consequently, the distance between the focus and the directrix is 2p (Figure 10.4.5). As illustrated in that figure, the parabola passes through two of the corners of a box that extends from the vertex to the focus along the axis of symmetry and extends 2p units above and 2p units below the axis of symmetry. The equation of a parabola is simplest if the vertex is the origin and the axis of symmetry is along the x-axis or j-axis. The four possible such orientations are shown in Figure 10.4.6. These are called the standard positions of a parabola, and the resulting equations are called the standard equations of a parabola.
Directrix A Figure 10.4.5
PARABOLAS IN STANDARD POSITION
= -p
A Figure 10.4.6
10.4 Conic Sections
733
To illustrate how the equations in Figure 10.4.6 are obtained, we will derive the equation for the parabola with focus (/?, 0) and directrix x — —p. Let P(x, y) be any point on the parabola. Since P is equidistant from the focus and directrix, the distances PF and PD in Figure 10.4.7 are equal; that is,
I
rr = rU
D(-p,y)*
(1)
where D(—p, y) is the foot of the perpendicular from P to the directrix. From the distance formula, the distances PF and PD are
PF = V(* - p)2 + y2
and PD =
(2)
p)2
(3)
Substituting in (1) and squaring yields
(x - p)2 and after simplifying
y12 = 4pX
(4)
The derivations of the other equations in Figure 10.4.6 are similar. A TECHNIQUE FOR SKETCHING PARABOLAS
Parabolas can be sketched from their standard equations using four basic steps:
Sketching a Parabola from Its Standard Equation Step 1. Determine whether the axis of symmetry is along the x-axis or the y-axis. Referring to Figure 10.4.6, the axis of symmetry is along the jc-axis if the equation has a y2-term, and it is along the y-axis if it has an x2-term. Step 2. Determine which way the parabola opens. If the axis of symmetry is along the x-axis, then the parabola opens to the right if the coefficient of x is positive, and it opens to the left if the coefficient is negative. If the axis of symmetry is along the y-axis, then the parabola opens up if the coefficient of y is positive, and it opens down if the coefficient is negative. 2P
X A Figure 10.4.8
Step 3. Determine the value of p and draw a box extending p units from the origin along the axis of symmetry in the direction in which the parabola opens and extending 2p units on each side of the axis of symmetry. Step 4. Using the box as a guide, sketch the parabola so that its vertex is at the origin and it passes through the corners of the box (Figure 10.4.8).
*• Example 1
Sketch the graphs of the parabolas
(a) x2 = 12y
(b) y 2 + 8x = 0
and show the focus and directrix of each.
A Figure 10.4.9
Solution (a). This equation involves x2, so the axis of symmetry is along the y-axis, and the coefficient of y is positive, so the parabola opens upward. From the coefficient of y, we obtain 4p = 12 or p = 3. Drawing a box extending p = 3 units up from the origin and 2p = 6 units to the left and 2p — 6 units to the right of the y-axis, then using corners of the box as a guide, yields the graph in Figure 10.4.9. The focus is p = 3 units from the vertex along the axis of symmetry in the direction in which the parabola opens, so its coordinates are (0, 3). The directrix is perpendicular to the axis of symmetry at a distance of p = 3 units from the vertex on the opposite side from the focus, so its equation is y = —3.
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Chapter TO / Parametric and Polar Curves; Conic Sections
Solution (ft).
We first rewrite the equation in the standard form y2 = -8x
This equation involves y2, so the axis of symmetry is along the x-axis, and the coefficient of x is negative, so the parabola opens to the left. From the coefficient of x we obtain 4p = 8, so p — 2. Drawing a box extending p — 2 units left from the origin and 2p = 4 units above and 2p = 4 units below the ;t-axis, then using corners of the box as a guide, yields the graph in Figure 10.4.10. •« > Example 2 Find an equation of the parabola that is symmetric about the ;y-axis, has its vertex at the origin, and passes through the point (5,2). Solution, Since the parabola is symmetric about the y-axis and has its vertex at the origin, the equation is of the form x2 = 4py
A Figure 10.4.10
or x2 = —4py
where the sign depends on whether the parabola opens up or down. But the parabola must open up since it passes through the point (5,2), which lies in the first quadrant. Thus, the equation is of the form x2 = 4py (5) Since the parabola passes through (5, 2), wemusthaveS 2 = 4p • 2or4p = ~. Therefore, (5) becomes
EQUATIONS OF ELLIPSES IN STANDARD POSITION
\
A Figure 10.4.11
It is traditional in the study of ellipses to denote the length of the major axis by 2a, the length of the minor axis by 2b, and the distance between the foci by 2c (Figure 10.4.11). The number a is called the semimajor axis and the number b the semiminor axis (standard but odd terminology, since a and b are numbers, not geometric axes). There is a basic relationship between the numbers a, b, and c that can be obtained by examining the sum of the distances to the foci from a point P at the end of the major axis and from a point Q at the end of the minor axis (Figure 10.4.12). From Definition 10.4.2, these sums must be equal, so we obtain c2 = (a-c)
from which it follows that
a=
(6)
or, equivalently, (7) A Figure 10.4.12
A Figure 10.4.13
From (6), the distance from a focus to an end of the minor axis is a (Figure 10.4.13), which implies that for all points on the ellipse the sum of the distances to the foci is 2a. It also follows from (6) that a > b with the equality holding only when c = 0. Geometrically, this means that the major axis of an ellipse is at least as large as the minor axis and that the two axes have equal length only when the foci coincide, in which case the ellipse is a circle. The equation of an ellipse is simplest if the center of the ellipse is at the origin and the foci are on the x-axis or y-axis. The two possible such orientations are shown in Figure 10.4.14.
T 0.4 Conic Sections
735
These are called the standard positions of an ellipse, and the resulting equations are called the standard equations of an ellipse.
ELLIPSES IN STANDARD POSITION
(0,C)
> Figure 10.4.14
To illustrate how the equations in Figure 10.4.14 are obtained, we will derive the equation for the ellipse with foci on the ;t-axis. Let P(x, y ) be any point on that ellipse. Since the sum of the distances from P to the foci is 2a, it follows (Figure 10.4.15) that PF' +PF=2a so VGc + c)2 + y2 + v/Ot - c)2 + y2 = 2a
Transposing the second radical to the right side of the equation and squaring yields A Figure 10.4.15
(x + c)2 + y2= 4a2 -
- c)2 + y2 + (x - c)2 + y2
and, on simplifying, - c)2 + y2 = a -- x a
(8)
Squaring again and simplifying yields v-2
aL9
a9L — C9L
which, by virtue of (6), can be written as 2
2
X
y 2-+ -
a
b*
(9)
Conversely, it can be shown that any point whose coordinates satisfy (9) has la as the sum of its distances from the foci, so that such a point is on the ellipse.
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Chapter 10 / Parametric and Polar Curves; Conic Sections A TECHNIQUE FOR SKETCHING ELLIPSES
Ellipses can be sketched from their standard equations using three basic steps:
Sketching an Ellipse from Its Standard Equation
ty
Step 1. Determine whether the major axis is on the x-axis or the y-axis. This can be ascertained from the sizes of the denominators in the equation. Referring to Figure 10.4.14, and keeping in mind that a2 > b2 (since a > b), the major axis is along the x-axis if x2 has the larger denominator, and it is along the y-axis if y2 has the larger denominator. If the denominators are equal, the ellipse is a circle. Step 2. Determine the values of a and b and draw a box extending a units on each side of the center along the major axis and b units on each side of the center along the minor axis.
Rough sketch
A Figure 10.4.16
Step 3. Using the box as a guide, sketch the ellipse so that its center is at the origin and it touches the sides of the box where the sides intersect the coordinate axes (Figure 10.4.16).
Example 3
Sketch the graphs of the ellipses 7T 9
7^ 16
showing the foci of each. Solution (a). Since y2 has the larger denominator, the major axis is along the y-axis. Moreover, since a2 > b2, we must have a2 = 16 and b2 = 9, so a = 4 and b = 3
Drawing a box extending 4 units on each side of the origin along the y-axis and 3 units on each side of the origin along the x-axis as a guide yields the graph in Figure 10.4.17. The foci lie c units on each side of the center along the major axis, where c is given by (7). From the values of a2 and b2 above, we obtain A Figure 10.4.17
c = Ja2 - b2 = V16-9 = V? RS 2.6
Thus, the coordinates of the foci are (0, V"7) and (0, — \/7), since they lie on the y-axis. Solution (b).
We first rewrite the equation in the standard form
Since x2 has the larger denominator, the major axis lies along the x-axis, and we have a2 = 4 and b2 — 2. Drawing a box extending a = 1 units on each side of the origin along the x-axis and extending b = \/2 « 1 .4 units on each side of the origin along the y-axis as a guide yields the graph in Figure 10.4.18. From (7), we obtain _ c = Ja2-b2 = -fl^ 1.4 A Figure 10.4.18
Thus, the coordinates of the foci are (\/2, 0) and (— \/2, 0), since they lie on the x-axis. •<
10.4 Conic Sections
*• Example 4 points (0, ±4). Solution.
737
Find an equation for the ellipse with foci (0, ±2) and major axis with end-
From Figure 10.4. 14, the equation has the form x2 y2 __ L L- — 1
Va2 ~ and from the given information, a = 4 and c = 2. It follows from (6) that
b2 =a2 -c2 = 16-4= 12 so the equation of the ellipse is
2
2
EQUATIONS OF HYPERBOLAS IN STANDARD POSITION
It is traditional in the study of hyperbolas to denote the distance between the vertices by 2a, the distance between the foci by 2c (Figure 10.4.19), and to define the quantity b as
b=
(10)
This relationship, which can also be expressed as + b2
c= A Figure 10.4.19
is pictured geometrically in Figure 10.4.20. As illustrated in that figure, and as we will show later in this section, the asymptotes pass through the corners of a box extending b units on each side of the center along the conjugate axis and a units on each side of the center along the focal axis. The number a is called the semifocal axis of the hyperbola and the number b the semiconjugate axis. (As with the semimajor and semiminor axes of an ellipse, these are numbers, not geometric axes.) If V is one vertex of a hyperbola, then, as illustrated in Figure 10.4.21, the distance from V to the farther focus minus the distance from V to the closer focus is [(c -
A Figure 10.4.20
2a] -(c-d) =
Thus, for all points on a hyperbola, the distance to the farther focus minus the distance to the closer focus is 2a. The equation of a hyperbola has an especially convenient form if the center of the hyperbola is at the origin and the foci are on the Jt-axis or 3>-axis. The two possible such orientations are shown in Figure 10.4.22. These are called the standard positions of a hyperbola, and the resulting equations are called the standard equations of a hyperbola. The derivations of these equations are similar to those already given for parabolas and ellipses, so we will leave them as exercises. However, to illustrate how the equations of the asymptotes are derived, we will derive those equations for the hyperbola b2
We can rewrite this equation as A Figure 10.4.21
(11)
b2 which is equivalent to the pair of equations and
y =
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Chapter 10 / Parametric and Polar Curves; Conic Sections
HYPERBOLAS IN STANDARD POSITION A?
*• Figure 10.4.22
Thus, in the first quadrant, the vertical distance between the line y — (b/a)x and the hyperbola can be written as
a
a
(Figure 10.4.23). But this distance tends to zero as x —> +00 since i
( -x a
\
2
a
j
\/* — a ) = lim - f x — \A:2 — a2 j 2
(x - Jx2 - a2 ) (j* = lim — *^+» a
x + V x2 - a2 ab
= lim
=0
A Figure 10.4.23
The analysis in the remaining quadrants is similar.
A QUICK WAY TO FIND ASYMPTOTES
There is a trick that can be used to avoid memorizing the equations of the asymptotes of a hyperbola. They can be obtained, when needed, by replacing 1 by 0 on the right side of the hyperbola equation, and then solving for y in terms of x . For example, for the hyperbola
a2
_
b2
we would write x2 y2 2 b2 22 — x2 - — 2 = 0 or / = — aba1
which are the equations for the asymptotes.
b
or y = ±-x a
10.4 Conic Sections
739
A TECHNIQUE FOR SKETCHING HYPERBOLAS
Hyperbolas can be sketched from their standard equations using four basic steps:
Sketching a Hyperbola front Its Standard Equation Step 1. Determine whether the focal axis is on the x-axis or the y-axis. This can be ascertained from the location of the minus sign in the equation. Referring to Figure 10.4.22, the focal axis is along the jc-axis when the minus sign precedes the y2-term, and it is along the y-axis when the minus sign precedes the z2-term. Step 2. Determine the values of a and b and draw a box extending a units on either side of the center along the focal axis and b units on either side of the center along the conjugate axis. (The squares of a and b can be read directly from the equation.) Step 3. Draw the asymptotes along the diagonals of the box. Step 4. Using the box and the asymptotes as a guide, sketch the graph of the hyperbola (Figure 10.4.24).
Example 5
Sketch the graphs of the hyperbolas (a) j - - = 1 (b) y2 - x2 = 1
showing their vertices, foci, and asymptotes. Solution (a). The minus sign precedes the y2-term, so the focal axis is along the .x-axis. From the denominators in the equation we obtain a2 = 4
and b2 = 9
Since a and b are positive, we must have a = 2 and b = 3. Recalling that the vertices lie a units on each side of the center on the focal axis, it follows that their coordinates in this case are (2, 0) and (—2, 0). Drawing a box extending a = 2 units along the x-axis on each side of the origin and b — 3 units on each side of the origin along the y-axis, then drawing the asymptotes along the diagonals of the box as a guide, yields the graph in Figure 10.4.25. To obtain equations for the asymptotes, we replace 1 by 0 in the given equation; this yields 2 2 o A Figure 10.4.25
=0
or
y = ±-jc
4 9 2 The foci lie c units on each side of the center along the focal axis, where c is given by (11). From the values of a1 and b2 above we obtain
c = Va + b2 = V4 + 9 = V13 % 3.6 Since the foci lie on the x-axis in this case, their coordinates are (\/13, 0) and (—\/T3, 0). Solution (b). The minus sign precedes the *2-term, so the focal axis is along the y-axis. From the denominators in the equation we obtain a2 — 1 and b1 — 1, from which it follows that a — 1 and b = 1
A Figure 10.4.26
Thus, the vertices are at (0, -1) and (0, 1). Drawing a box extending a — 1 unit on either side of the origin along the y-axis and b = 1 unit on either side of the origin along the je-axis, then drawing the asymptotes, yields the graph in Figure 10.4.26. Since the box is actually
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Chapter 10 / Parametric and Polar Curves; Conic Sections
—^^^^^^^—^^^^^^^— A hyperbola in which a = b, as in part (b) of Example 5, is called an equilateral hyperbola. Such hyperbolas always have perpendicular asymptotes.
a square, the asymptotes are perpendicular and have equations y = ±x. This can also be by replacing 1 by 0 in the given equation, which yields y 2 — x2 = 0 or y = ±x. Also,
seen
c = \ a2 + b2 = v l + 1 = v 2 ,— .— so the foci, which lie on the y-axis, are (0, — V2) and (0, V2). •* Example 6
Find the equation of the hyperbola with vertices (0, ±8) and asymptotes
Solution. Since the vertices are on the y-axis, the equation of the hyperbola has the form (y 2 /a 2 ) - (x2/b2) = 1 and the asymptotes are
From the locations of the vertices we have a — 8, so the given equations of the asymptotes yield a 8 4 y = ±-x = ±-x — ±-x b b 3 from which it follows that b = 6. Thus, the hyperbola has the equation
z
64
36
=1 ,
TRANSLATED COMICS
Equations of conies that are translated from their standard positions can be obtained by replacing x by x — h and y by y — k in their standard equations. For a parabola, this translates the vertex from the origin to the point (h,k); and for ellipses and hyperbolas, this translates the center from the origin to the point (h, k). Parabolas with vertex (h, k) and axis parallel to x-axis (y — k)2 =
4p(x — h)
[Opens right]
2
(y — k) — —4p(x — h)
[Opens left]
(12) (13)
Parabolas with vertex (h, k) and axis parallel to y-axis (X — h)2 =
4p(y — k)
2
(X — h) = —4p(y — K)
[Opens up] [Opens down]
(14) (15)
Ellipse with center (h, k) and major axis parallel to x-axis (x-h)2 a 5—
H
(y-fc) 2 z ^— =! \-b
(16)
Ellipse with center (h, k) and major axis parallel to y-axis (x - h)2 b2
(y - k)2 a2
=i
(17)
Hyperbola with center (h, k) and focal axis parallel to x-axis (x - h)2 a2
(y - k)2 b2
(18)
Hyperbola with center (h, k) and focal axis parallel to y-axis (y - k)2 a2
(x - h)2 b2
(19)
10.4 Conic Sections
t* Example 7 at (4, 2).
741
Find an equation for the parabola that has its vertex at (1, 2) and its focus
Solution* Since the focus and vertex are on a horizontal line, and since the focus is to the right of the vertex, the parabola opens to the right and its equation has the form (y - k)2 = 4p(x - h)
Since the vertex and focus are 3 units apart, we have p = 3, and since the vertex is at ( h , k ) — (1,2), we obtain (y - 2)2 = 12(:t - 1)
Sometimes the equations of translated conies occur in expanded form, in which case we are faced with the problem of identifying the graph of a quadratic equation in x and y:
Ax2 + Cy2 + Dx + Ey + F = 0
(20)
The basic procedure for determining the nature of such a graph is to complete the squares of the quadratic terms and then try to match up the resulting equation with one of the forms of a translated conic. > Example 8
Describe the graph of the equation
y2 - BJC - 6y - 23 = 0 Solution. The equation involves quadratic terms in y but none in x, so we first take all of the y -terms to one side: , y2 - 6y = &x + 23 (-4,3)-
Next, we complete the square on the y-terms by adding 9 to both sides:
(y - 3)2 = 8* + 32 Finally, we factor out the coefficient of the jc-term to obtain Directrix
A Figure 10.4.27
This equation is of form (12) with h = —4, k = 3, and p = 2, so the graph is a parabola with vertex (—4, 3) opening to the right. Since p = 2, the focus is 2 units to the right of the vertex, which places it at the point (—2,3); and the directrix is 2 units to the left of the vertex, which means that its equation is x = —6. The parabola is shown in Figure 10.4.27. •<
*• Example 9
Describe the graph of the equation 16jc2 + 9y2 - 64x - 54y + 1 = 0
Solution. This equation involves quadratic terms in both x and y, so we will group the x-terms and the y-terms on one side and put the constant on the other: (16.x2 - 64jc) + (9/ - 54;y) = -1 Next, factor out the coefficients of x2 and y2 and complete the squares: I6(x2 - 4x + 4) + 9(y2 - 6y + 9) = -1 + 64 + 81
16(x - 2)2 + 9(y - 3)2 = 144
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Chapter 10 / Parametric and Polar Curves; Conic Sections
Finally, divide through by 144 to introduce a 1 on the right side: (*-2) 2 (y-3) 2 9 16 This is an equation of form (17), with h = 2, k = 3, a2 = 16, and b2 = 9. Thus, the graph of the equation is an ellipse with center (2, 3) and major axis parallel to the y-axis. Since a = 4, the major axis extends 4 units above and 4 units below the center, so its endpoints are (2, 7) and (2, —1) (Figure 10.4.28). Since b = 3, the minor axis extends 3 units to the left and 3 units to the right of the center, so its endpoints are (—1, 3) and (5, 3). Since
(-1, 3)
(2,-D
(2,3-V?)
c = Va2 - b2 = Vl6 - 9 = V7 the foci lie \/7 units above and below the center, placing them at the points (2, 3 + \/7) and (2, 3-77). +
Example 10 Describe the graph of the equation 2 X
- y2 - 4X + 8y - 21 = 0
Solution. This equation involves quadratic terms in both x and y, so we will group the x -terms and the y-terms on one side and put the constant on the other:
(x2 - 4x) - (y2 - 8y) = 21 We leave it for you to verify by completing the squares that this equation can be written as
(x - 2)2 (y - 4)2 =1 (21) 9 9 This is an equation of form (18) with h = 2, k = 4, a2 = 9, and b2 = 9. Thus, the equation represents a hyperbola with center (2, 4) and focal axis parallel to the jc-axis. Since a = 3, the vertices are located 3 units to the left and 3 units to the right of the center, or at the points (-1, 4) and (5, 4).From(ll),c = Va2 + b2 = V9 + 9 = 3^2, so the foci are located 3^2 units to the left and right of the center, or at the points (2 - 3^2, 4) and (2 + 3\/2, 4). The equations of the asymptotes may be found using the trick of replacing 1 by 0 in (21) to obtain (^ _ 2)2 (y — 4)2 9 9 This can be written as y — 4 = ±(x — 2), which yields the asymptotes y = x + 2 and y = — x + 6 -2l =0 A Figure 10.4.29
With the aid of a box extending a — 3 units left and right of the center and b — 3 units above and below the center, we obtain the sketch in Figure 10.4.29. «i
REFLECTION PROPERTIES OF THE CONIC SECTIONS
Parabolas, ellipses, and hyperbolas have certain reflection properties that make them extremely valuable in various applications. In the exercises we will ask you to prove the following results. 10.4.4 THEOREM (Reflection Property of Parabolas) The tangent line at a point P on na parabola makes equal angles with the line through P parallel to the axis of symmetry.. and the line through P and the focus (Figure 10.4.30a).
10.4 Conic Sections
743
10.4.5 THEOREM (Reflection Property of Ellipses) A line tangent to an ellipse at a point P makes equal angles with the lines joining P to the foci (Figure 10.4.30b).
10.4.6 THEOREM (Reflection Property of Hyperbolas) A line tangent to a hyperbola at a point P makes equal angles with the lines joining P to the foci (Figure 10.4.30c).
• Tangent line at P
(a)
Figure 10.4.30
John MeadyScience Photo Library/Photo Researchers Incoming signals are reflected by the parabolic antenna to the receiver at the focus.
Tangent line at P
(b)
(c)
APPLICATIONS OF THE CONIC SECTIONS Fermat's principle in optics implies that light reflects off of a surface at an angle equal to its angle of incidence. (See Exercise 62 in Section 3.5.) In particular, if a reflecting surface is generated by revolving a parabola about its axis of symmetry, it follows from Theorem 10.4.4 that all light rays entering parallel to the axis will be reflected to the focus (Figure 10.4.3 la); conversely, if a light source is located at the focus, then the reflected rays will all be parallel to the axis (Figure 10.4.31ft). This principle is used in certain telescopes to reflect the approximately parallel rays of light from the stars and planets off of a parabolic mirror to an eyepiece at the focus; and the parabolic reflectors in flashlights and automobile headlights utilize this principle to form a parallel beam of light rays from a bulb placed at the focus. The same optical principles apply to radar signals and sound waves, which explains the parabolic shape of many antennas.
Figure 10.4.31
(a)
Visitors to various rooms in the United States Capitol Building and in St. Paul's Cathedral in London are often astonished by the "whispering gallery" effect in which two people at opposite ends of the room can hear one another's whispers very clearly. Such rooms have ceilings with elliptical cross sections and common foci. Thus, when the two people stand at the foci, their whispers are reflected directly to one another off of the elliptical ceiling. Hyperbolic navigation systems, which were developed in World War II as navigational aids to ships, are based on the definition of a hyperbola. With these systems the ship receives
744
Chapter 10 / Parametric and Polar Curves; Conic Sections
synchronized radio signals from two widely spaced transmitters with known positions. The ship's electronic receiver measures the difference in reception times between the signals and then uses that difference to compute the difference 2a between its distances from the two transmitters. This information places the ship somewhere on the hyperbola whose foci are at the transmitters and whose points have 2a as the difference in their distances from the foci. By repeating the process with a second set of transmitters, the position of the ship can be approximated as the intersection of two hyperbolas (Figure 10.4.32). (The modern global positioning system (GPS) is based on the same principle.)
A Figure 10.4.32
UICK CHECK EXERCISES 10.4
(See page 748 for answers.)
1. Identify the conic. (a) The set of points in the plane, the sum of whose distances to two fixed points is a positive constant greater than the distance between the fixed points is (b) The set of points in the plane, the difference of whose distances to two fixed points is a positive constant less than the distance between the fixed points is (c) The set of points in the plane that are equidistant from a fixed line and a fixed point not on the line is 2. (a) The equation of the parabola with focus (p, 0) and directrix x = —p is (b) The equation of the parabola with focus (0, p) and directrix y = — p is 3. (a) Suppose that an ellipse has semimajor axis a and semiminor axis b. Then for all points on the ellipse, the sum of the distances to the foci is equal to (b) The two standard equations of an ellipse with semimajor axis a and semiminor axis b are and
EXERCISE SET 10.4
(c) Suppose that an ellipse has semimajor axis a, semiminor axis b, and foci (±c, 0). Then c may be obtained from a and b by the equation c = 4. (a) Suppose that a hyperbola has semifocal axis a and semiconjugate axis b. Then for all points on the hyperbola, the difference of the distance to the farther focus minus the distance to the closer focus is equal to (b) The two standard equations of a hyperbola with semifocal axis a and semiconjugate axis b are and (c) Suppose that a hyperbola in standard position has semifocal axis a, semiconjugate axis b, and foci (±c, 0). Then c may be obtained from a and b by the equation c= The equations of the asymptotes of this hyperbola are y = ±
Graphing Utility
FOCUS ON CONCEPTS
1. In parts (a)-(f), find the equation of the conic.
(C)
-3
-3-2-1 0
1
2
3
4
1 2 3
-3 -2 -1
1
2 3
10.4 Conic Sections
745
18. Vertex (5,—3); axis parallel to the y-axis; passes through (9,5).
-3-2-1
0
-3-2-1 0
1 2 3
1 2 3
2. (a) Find the focus and directrix for each parabola in Exercise 1. (b) Find the foci of the ellipses in Exercise 1. (c) Find the foci and the equations of the asymptotes of the hyperbolas in Exercise 1. 3-6 Sketch the parabola, and label the focus, vertex, and directrix. 8 3. (a) y2 = 4x (b) x2 = -8y 4. (a) / = -10jc (b) x2=4y 2 5. (a) (y - I) = -12(x + 4) (b) (x -l)2 = 2(y~ \] 6. (a) y2 - 6y - 2x + 1 = 0
(b) y = 4x2 + &x + 5
7-10 Sketch the ellipse, and label the foci, vertices, and ends of the minor axis. x2 v2 (b) 9x2 + y2 = 9 X
-
(b) 4x2 ± y2 = 36
9. (a) U + 3) 2 +4(y-5) 2 = 16 (b) ijc2 + i(y + 2 ) 2 - l = 0 10. (a) 9x2 + 4y2 - 18* + 24y + 9 = 0 (b) 5x2 + 9y2 + 2(k - 54y = -56
11-14 Sketch the hyperbola, and label the vertices, foci, and asymptotes. x y (b) 9y2 ~x2 = 36 12- (a)
-- =
(b) 16*2 - 25y2 = 400
13. (a) (b) 16(* + I)2 - 8(y - 3)2 = 16 14. (a) x2 - 4y2 + 2x + 8y - 7 = 0 (b) I6x2 - y2 - 32;c - 6y = 57 15-18 Find an equation for the parabola that satisfies the given conditions, a 15. (a) Vertex (0, 0); focus (3,0). (b) Vertex (0, 0); directrix y = ~. 16. (a) Focus (6,0); directrix x = —6. (b) Focus (1, 1); directrix y = —2. 17. Axis y = 0; passes through (3, 2) and (2, -\/2).
19-22 Find an equation for the ellipse that satisfies the given conditions. 19. (a) Ends of major axis (±3, 0); ends of minor axis (0, ±2). (b) Length of minor axis 8; foci (0, ±3). 20. (a) Foci (±1,0); b = VI (b) c = 2\/3; a = 4; center at the origin; foci on a coordinate axis (two answers). 21. (a) Ends of major axis (0, ±6); passes through (—3, 2). (b) Foci (—1, 1) and (—1, 3); minor axis of length 4. 22. (a) Center at (0,0); major and minor axes along the coordinate axes; passes through (3, 2) and (1,6). (b) Foci (2,1) and (2, —3); major axis of length 6. 23-26 Find an equation for a hyperbola that satisfies the given conditions. [Note: In some cases there may be more than one hyperbola.] 23. (a) Vertices (±2,0); foci (±3,0). (b) Vertices (0, ±2); asymptotes y = ±|*. 24. (a) (b) 25. (a) (b) 26. (a) (b)
Asymptotes y = ±\x; b = 4. Foci (0, ±5); asymptotes y = ±2x. Asymptotes y = ±|x; c = 5. Foci (±3, 0); asymptotes y = ±2x. Vertices (0, 6) and (6, 6); foci 10 units apart, Asymptotes y = x — 2 and y = — x + 4; passes through the origin.
27-30 True-False Determine whether the statement is true or false. Explain your answer. 27. A hyperbola is the set of all points in the plane that are equidistant from a fixed line and a fixed point not on the line. 28. If an ellipse is not a circle, then the foci of an ellipse lie on the major axis of the ellipse. 29. If a parabola has equation y2 = 4px, where p is a positive constant, then the perpendicular distance from the parabola's focus to its directrix is p. 30. The hyperbola (y2/a2) — x2 = 1 has asymptotes the lines y = ±x/a. 31. (a) As illustrated in the accompanying figure, a parabolic arch spans a road 40 ft wide. How high is the arch if a center section of the road 20 ft wide has a minimum clearance of 12 ft? (b) How high would the center be if the arch were the upper half of an ellipse?
-20ft—H -40ft
•* Figure Ex-31
746
Chapter 10 / Parametric and Polar Curves; Conic Sections
32. (a) Find an equation for the parabolic arch with base b and height h, shown in the accompanying figure, (b) Find the area under the arch.
•^ Figure Ex-32
39. Find an equation of the ellipse traced by a point that moves so that the sum of its distances to (4, 1) and (4, 5) is 12. 40. Find the equation of the hyperbola traced by a point that moves so that the difference between its distances to (0, 0) and (1,1) is 1. 41. Show that an ellipse with semimajor axis a and semiminor axis b has area A = nab. FOCUS ON CONCEPTS
33. Show that the vertex is the closest point on a parabola to the focus. [Suggestion: Introduce a convenient coordinate system and use Definition 10.4.1.] 34. As illustrated in the accompanying figure, suppose that a comet moves in a parabolic orbit with the Sun at its focus and that the line from the Sun to the comet makes an angle of 60° with the axis of the parabola when the comet is 40 million miles from the center of the Sun. Use the result in Exercise 33 to determine how close the comet will come to the center of the Sun. 35. For the parabolic reflector in the accompanying figure, how far from the vertex should the light source be placed to produce a beam of parallel rays?
42. Show that if a plane is not parallel to the axis of a right circular cylinder, then the intersection of the plane and cylinder is an ellipse (possibly a circle). [Hint: Let 9 be the angle shown in the accompanying figure, introduce coordinate axes as shown, and express x' and y' in terms of x and y.]
•4 Figure Ex-42
A Figure Ex-34
A Figure Ex-35
36. (a) Show that the right and left branches of the hyperbola 2
b2 can be represented parametrically as
x— a cosh t, x = —acosht,
y = bsinht y = bsinht
(— oo < t < +°o) (—00 < t < +00)
(b) Use a graphing utility to generate both branches of the hyperbola x2 — y2 = 1 on the same screen. 37. (a) Show that the right and left branches of the hyperbola
43. As illustrated in the accompanying figure, a carpenter needs to cut an elliptical hole in a sloped roof through which a circular vent pipe of diameter D is to be inserted vertically. The carpenter wants to draw the outline of the hole on the roof using a pencil, two tacks, and a piece of string (as in Figure 10.4.3ft). The center point of the ellipse is known, and common sense suggests that its major axis must be perpendicular to the drip line of the roof. The carpenter needs to determine the length L of the string and the distance T between a tack and the center point. The architect's plans show that the pitch of the roof is p (pitch = rise over run; see the accompanying figure). Find T and L in terms of D and p. Source.' This exercise is based on an article by William H. Enos, which appeared in the Mathematics Teacher, Feb. 1991, p. 148.
can be represented parametrically as x= a sect, y = btant x = -a sec t, y = b tan t
(—n/2
(b) Use a graphing utility to generate both branches of the hyperbola x2 — y2 = 1 on the same screen. 38. Find an equation of the parabola traced by a point that moves so that its distance from (2, 4) is the same as its distance to the x -axis.
< Figure Ex-43
44. As illustrated in the accompanying figure on the next page, suppose that two observers are stationed at the points FI(C, 0) and F2(—c, 0) in an xy-coordinate system. Suppose also that the sound of an explosion in the xy-plane is heard by the F\ observer / seconds before it
10.4 Conic Sections 747 is heard by the F2 observer. Assuming that the speed of sound is a constant v, show that the explosion occurred somewhere on the hyperbola x2
y2
2 2
2
c - (v2t2/4) ~
v t /4
F2(-c, 0)
FI(C, 0)
•4 Figure Ex-44
45. As illustrated in the accompanying figure, suppose that two transmitting stations are positioned 100 km apart at points Fi(50, 0) and F2(—50, 0) on a straight shoreline in an jry-coordinate system. Suppose also that a ship is traveling parallel to the shoreline but 200 km at sea. Find the coordinates of the ship if the stations transmit a pulse simultaneously, but the pulse from station F\ is received by the ship 100 microseconds sooner than the pulse from station F2. [Hint: Use the formula obtained in Exercise 44, assuming that the pulses travel at the speed of light (299,792,458 m/s).]
F2(-50, 0)
at the point (XQ, yo) has the equation XXQ +,yyo_, ~# ~& ~ 50. Prove: The line tangent to the hyperbola b2 at the point (XQ, yo) has the equation XXQ yy0 _ a2 " b2 51. Use the results in Exercises 49 and 50 to show that if an ellipse and a hyperbola have the same foci, then at each point of intersection their tangent lines are perpendicular.
52. Consider the second-degree equation Ax2 + Cy2 + Dx + Ey + F = 0 where A and C are not both 0. Show by completing the square: (a) If AC > 0, then the equation represents an ellipse, a circle, a point, or has no graph. (b) If AC < 0, then the equation represents a hyperbola or a pair of intersecting lines. (c) If AC = 0, then the equation represents a parabola, a pair of parallel lines, or has no graph. 53. In each part, use the result in Exercise 52 to make a statement about the graph of the equation, and then check your conclusion by completing the square and identifying the graph. (a) x2 -5y2-2x-\0y-9 = 0 (b) x2 - 3y2 - 6y - 3 = 0 (c) 4x2 + 8y2 + 16* + 16y + 20 = 0 (d) 3x2 + y2 + l2x + 2y + l3 = 0 (e) x2 + 8x + 2y + 14 = 0 (f ) 5x2 + 4Qx + 2y + 94 = 0 54. Derive the equation x2 = 4py in Figure 10.4.6.
•* Figure Ex-45
46. A nuclear cooling tower is to have a height of h feet and the shape of the solid that is generated by revolving the region R enclosed by the right branch of the hyperbola 1521jc2 - 225y2 = 342,225 and the lines x = 0, y = —h/2, and y = h/2 about the y-axis. (a) Find the volume of the tower. (b) Find the lateral surface area of the tower.
47. Let R be the region that is above the x-axis and closed between the curve b2x2 — a2y2 = a2b2 and the line X = Vfl 2 + b2.
(a) Sketch the solid generated by revolving R about the *-axis, and find its volume. (b) Sketch the solid generated by revolving R about the y-axis, and find its volume. 48. Prove: The line tangent to the parabola x2 = 4py at the point (x0, yo) is x0x = 2p(y + y 0 ). 49. Prove: The line tangent to the ellipse x2
y2
55. Derive the equation (x2/b2) + (y2/a2) = 1 given in Figure 10.4.14. 56. Derive the equation (x2/a2) — (y2/b2) = 1 given in Figure 10.4.22. 57. Prove Theorem 10.4.4. [Hint: Choose coordinate axes so that the parabola has the equation x2 = 4py. Show that the tangent line at P(XQ, yo) intersects the y-axis at <2(0, — yo) and that the triangle whose three vertices are at P, Q, and the focus is isosceles.] 58. Given two intersecting lines, let L2 be the line with the larger angle of inclination fa, and let L \ be the line with the smaller angle of inclination 4>\ . We define the angle 0 between L\ and LI by 9 = fa — i- (See the accompanying figure on the next page.) (a) Prove: If LI and L2 are not perpendicular, then m2-ml where L\ and L2 have slopes m\ and m2. (b) Prove Theorem 10.4. 5. [Hint: Introduce coordinates so that the equation (x2/a2) + (y2/b2) = 1 describes the ellipse, and use part (a).] (mm.)
748 Chapter 10 / Parametric and Polar Curves; Conic Sections (c) Prove Theorem 10.4.6. [Hint: Introduce coordinates so that the equation (x2/a2) — (y2/b2) = 1 describes the hyperbola, and use part (a).]
59. Writing Suppose that you want to draw an ellipse that has given values for the lengths of the major and minor axes by using the method shown in Figure 10.4.36. Assuming that the axes are drawn, explain how a compass can be used to locate the positions for the tacks. 60. Writing List the forms for standard equations of parabolas, ellipses, and hyperbolas, and write a summary of techniques for sketching conic sections from their standard equations.
•^ Figure Ex-58
I/QUICK CHECK ANSWERS 10.4 1. (a) an ellipse (b) a hyperbola (c) a parabola Y
2
V
2
2 X
? V
2. (a) y2 = 4px (b) x2 = 4py
3. (a) 2a (b)z - + t? = 1; — + i- = 1 (c) Va2 - b2 a o/ oz a/
X
2
V
2
2 V
9 JC
4. (a) 2a (b) — - yz— = 11; i- _._ = 1 (c) az oz a b
ROTATION OF AXES; SECOND-DEGREE EQUATIONS In the preceding section -we obtained equations of conic sections with axes parallel to the coordinate axes. In this section we will study the equations of conies that are "tilted" relative to the coordinate axes. This will lead us to investigate rotations of coordinate axes.
QUADRATIC EQUATIONS IN x AND y We saw in Examples 8 to 10 of the preceding section that equations of the form
Ax2 + Cy2 + Dx + Ey + F = 0
(1)
can represent conic sections. Equation (1) is a special case of the more general equation
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
(2)
which, if A, B, and C are not all zero, is called a quadratic equation in x and y. It is usually the case that the graph of any second-degree equation is a conic section. If B — 0, then (2) reduces to (1) and the conic section has its axis or axes parallel to the coordinate axes. However, if B ^ 0, then (2) contains a cross-product term Bxy, and the graph of the conic section represented by the equation has its axis or axes "tilted" relative to the coordinate axes. As an illustration, consider the ellipse with foci F\(l, 2) and FI(—\, —2) and such that the sum of the distances from each point P(x, y) on the ellipse to the foci is 6 units. Expressing this condition as an equation, we obtain (Figure 10.5.1)
P(x,y)
V(jc - I) 2 + (y - 2)2 + J(x + I) 2 + (y + 2)2 = 6
Squaring both sides, then isolating the remaining radical, then squaring again ultimately -3 A Figure 10.5.1
as the equation of the ellipse. This is of form (2) with A — 8, B — —4, C = 5, D — 0, E = 0, and F = -36.
10.5 Rotation of Axes; Second-Degree Equations 749 ROTATION OF AXES
To study conies that are tilted relative to the coordinate axes it is frequently helpful to rotate the coordinate axes, so that the rotated coordinate axes are parallel to the axes of the conic. Before we can discuss the details, we need to develop some ideas about rotation of coordinate axes. In Figure 10.5.2a the axes of an .ry-coordinate system have been rotated about the origin through an angle 9 to produce a new jc'y'-coordinate system. As shown in the figure, each point P in the plane has coordinates (x', y') as well as coordinates (x, y). To see how the two are related, let r be the distance from the common origin to the point P, and let a be the angle shown in Figure lQ.5.2b. It follows that x — r cos(0 + a),
y = r sin($ + a)
(3)
and
(4)
Using familiar trigonometric identities, the relationships in (3) can be written as
x = r cos 9 cos a — r sin 9 sin a y = r sin 9 cos a + r cos 9 sin a and on substituting (4) in these equations we obtain the following relationships called the rotation equations: x = x' cos 9 — y' sin 9 y = x' sin 9 + y' cos 9
*• Figure 10.5.2
(b)
(a)
> Example 1 Suppose that the axes of an ry-coordinate system are rotated through an angle of 9 — 45 ° to obtain an jc'y'-coordinate system. Find the equation of the curve x2 - xy + y2 - 6 - 0
in ^'^'-coordinates. Solution. Substituting sin# = sin45° = l/\/2 and cos9 =cos45° = l/\/2 in (5) yields the rotation equations x'
x = —p V2
y'
-p V2
x'
y'
and y = — + — 72 V2
Substituting these into the given equation yields x' y' \2 ( *' y'\( x' y'\ ( x' y' \2 — - ^ = ) -I —--*-)[ — + -?-) + [—+ ^-\ -6 = 0 72 V2/ VV2 >/2/ VV2 x/27 \V2 V2.J
750
Chapter TO / Parametric and Polar Curves; Conic Sections
or x'2 - 2x'y' + y'2 - x'2 + y'2 + x'2 + 2x'y' + y'2
or x'2
y'2
12
4
=1
which is the equation of an ellipse (Figure 10.5.3). <
A Figure 10.5.3
If the rotation equations (5) are solved for x' and y' in terms of x and y, one obtains (Exercise 16): x' = xcos6 + y sind (6) y' = —x sin 6 + y cos 9
*• Example 2 Find the new coordinates of the point (2, 4) if the coordinate axes are rotated through an angle of 9 = 30°. Solution. Using the rotation equations in (6) with x = 2, y — 4, cos 9 = cos 30° = \/3/2, and sin$ — sin 30° — 1/2, we obtain x' = 2(73/2) + 4(1/2) = 73 + 2 y' = -2(1/2) + 4(73/2) = -1 + 2^3 Thus, the new coordinates are (V3 + 2, — 1 + 2-\/3). -^
ELIMINATING THE CROSS-PRODUCT TERM In Example 1 we were able to identify the curve x2 — x y + y2 — 6 — 0 as an ellipse because the rotation of axes eliminated the ry-term, thereby reducing the equation to a familiar form. This occurred because the new x'y'-ax&s were aligned with the axes of the ellipse. The following theorem tells how to determine an appropriate rotation of axes to eliminate the cross-product term of a second-degree equation in x and y.
10.5.1 THEOREM
If the equation
Ax2 + Bxy + Cy2 + Dx + Ey + F — 0
It is always possible to satisfy (8) with an angle 9 in the interval
0 < 9 < ji/2
(7)
is such that B ^ 0, and if an x'y'-coordinate system is obtained by rotating the xy-ax.es ' through an angle 6 satisfying A-C cot 29 = " 7 ^ (8) B , in x' y' -coordinates, Equation (7 ) will have the form A' x'2 + C'y'2 + D'x' + E'y' + F' = 0
We will always choose 6 in this way.
PROOF
Substituting (5) into (7) and simplifying yields A'x'2 + B'x'y' + C'y'2 + D'x' + E'y' + F' = 0
10.5 Rotation of Axes; Second-Degree Equations where
751
A' = A cos2 9 + B cos 9 sin 9 + C sin2 9 B' = B(cos2 9 ~ sin2 0) + 2(C - A) sin 6 cos 9 C" = A sin2 9 - B sin 6 cos 9 + C cos2 9
(9)
D' = D cos 9 + E sin 9
E' - -Dsm9 + Ecos9 F' = F
(Verify.) To complete the proof we must show that B' — 0 if cot 29 =
or, equivalently,
A-C B
A-C B
cos 29
sin 29
(10)
However, by using the trigonometric double-angle formulas, we can rewrite B' in the form 5' = ficos26>-(A-C)sin26>
Thus, B' = 0 if 9 satisfies (10). • *• Example 3
Identify and sketch the curve xy = 1.
Solution. As a first step, we will rotate the coordinate axes to eliminate the cross-product term. Comparing the given equation to (7), we have A = 0,
5 = 1,
C=0
Thus, the desired angle of rotation must satisfy cot2# =
A-C
=
B
0-0 1
=0
This condition can be met by taking 29 = n/2 or 9 = jr/4 = 45 °. Making the substitutions =cos45° - l/v^andsinfl - sin45° = 1/^/2 in (5) yields x' x = —= 72
y' -= 72
and
x' y' y = —= -\—— 72 72
Substituting these in the equation xy — 1 yields ,,'2
A Figure 10.5.4
which is the equation in the *'/ -coordinate system of an equilateral hyperbola with vertices at (\/2, 0) and (— \/2, 0) in that coordinate system (Figure 10.5.4). < In problems where it is inconvenient to solve cot 29 =
A-C B
for 9, the values of sin 9 and cos 0 needed for the rotation equations can be obtained by first calculating cos 29 and then computing sin 9 and cos 9 from the identities /1 - cos 29 sin 9 = , /
and
cos $ =
11 + cos 29
752
Chapter 10 / Parametric and Polar Curves; Conic Sections
Example 4
Identify and sketch the curve 153;t2 - I92xy + 91y2 - 30x - 40y - 200 = 0
Solution. We have A = 153, B =. -192, and C = 97, so 56 ~£~ -~~I92 ~~~'
A
cot 20 =
A Figure 10.5.5
24 Since 0 is to be chosen in the range 0 < 9 < n/2, this relationship is represented by the triangle in Figure 10.5.5. From that triangle we obtain cos 20 = — ^, which implies that / I + cos 29 co& 9 = ,/
sin 9 =
1 - cos 26» 2
V
2
5
Substituting these values in (5) yields the rotation equations
x = \x'- f /
and y = f jc' + |/
and substituting these in turn in the given equation yields
-f (3xr - 4y') - f (4xr + 3y') -200 = 0
which simplifies to A Figure 10.5.6
^+ ^^ _50x, _ ^ = Q
or
x'2 + 9y'2 - 2x' - 8 = 0
Completing the square yields /2
1
+7=1 There is a method for deducing the
kind of curve represented by a seconddegree equation directly from the equation itself without rotating coordinate axes. For a discussion of this topic, see the section on the discriminant that appears in Web Appendix K.
which is the equation in the jc'j'-coordinate system of an ellipse with center (1, 0) in that coordinate system and semiaxes a = 3 and b = 1 (Figure 10.5.6). •*
UICK CHECK EXERCISES 10.5
(See page 754 for answers.)
1. Suppose that an ry-coordinate system is rotated 9 radians to produce a new x'y'-coordinate system. (a) x and y may be obtained from x', y', and 9 using the rotation equations x = and y = (b) x' and y' may be obtained from x, y, and 6 using the equations x' = and y' = 2. If the equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 is such that B ^ 0, then the jry-term in this equation can be
eliminated by a rotation of axes through an angle 6 satisfying cot 29 = __ 3. In each part, determine a rotation angle 9 that will eliminate the xy-term. (a) 2x2 + xy + 2y2 + x - y = 0 (b) x2 + 2V3xy + 3y2-2x + y = (c) 3x2 4. Express 2x2 + xy + 2y2 = 1 in the ^'/-coordinate system obtained by rotating the xy-coordinate system through the angle 9 = jr/4.
10.5 Rotation of Axes; Second-Degree Equations
753
EXERCISE SET 10.5
1. Let an ^'/-coordinate system be obtained by rotating an .ry-coordinate system through an angle of 9 = 60°. (a) Find the ^'^'-coordinates of the point whose xy-coordinates are (—2, 6). (b) Find an equation of the curve V3xy + y2 = 6 in ^'/-coordinates. (c) Sketch the curve in part (b), showing both jry-axes and x'y'-axes. 2. Let an .^'/-coordinate system be obtained by rotating an xy-coordinate system through an angle of 9 = 30°. (a) Find the ^'/-coordinates of the point whose xy-coordinates are (1, —\/3). (b) Find an equation of the curve 2x2 + 2\^3xy = 3 in ^'/-coordinates. (c) Sketch the curve in part (b), showing both xy-axes and x'/-axes. 3-12 Rotate the coordinate axes to remove the xy-term. Then identify the type of conic and sketch its graph. S 2
3. xy = -9
2
4. x - xy + y - 2 = 0
5. x2 + 4xy-2y2-6 = 0 6. 31x2 + lOVsYy + 2ly2 -144 = 0 2
2
7. x + 2^3xy + 3y + 2^3x -2y = 0 8. 34x2 - 24xy + 4ly2 -25 = 0 9. 9x2 - 24xy + 16y2 - 80x - 60y + 100 = 0 10. 5x2 - 6xy + 5y2 - &j2x + &^2y = 8
18. Let an x'/-coordinate system be obtained by rotating an xy-coordinate system through an angle 9. Explain how to find the xy-equation of a line whose ^'/-equation is known.
19-22 Show that the graph of the given equation is a parabola. Find its vertex, focus, and directrix, a 19. x2 + 2xy + y2 + 4^/2x - 4^2y = 0 20. x2 - 2V3xy + 3y2 - 8^3x - 8y = 0 21. 9x2 - 24xy + 16y2 - 80x - 60y + 100 = 0 22. x2 + 2^3xy + 3y2 + I6^3x -I6y-96 = 0 23-26 Show that the graph of the given equation is an ellipse. Find its foci, vertices, and the ends of its minor axis. 23. 288;t2 - 168xy + 337y2 - 3600 = 0 24. 25x2 - I4xy + 25y2 - 288 = 0 25. 31.x2 + lOV3xy + 21y2 - 32* + 32^3y - 80 = 0 26. 43x2 - l4-V3xy + 51 y2 - 36^3x - 36y - 540 = 0 27-30 Show that the graph of the given equation is a hyperbola. Find its foci, vertices, and asymptotes. 27. x2 - \0-V3xy + Uy2 + 64 = 0 28. 17x2 - 3\2xy + 108y2 - 900 = 0 29. 32y2 - 52xy - lx2 + 12^5x - l44^/5y + 900 = 0
11. 52x2 - 72xy + 73y2 + 40x + 30y-15 = 0
30. 2^2y2 + 5^2xy + 2^2x2 + l&x + 18y + 36^2 = 0
12. 6x2 + 24xy - y2 - Ux + 26y + 11 = 0
31. Show that the graph of the equation
13. Let an x'/-coordinate system be obtained by rotating anxycoordinate system through an angle of 45 °. Use (6) to find an equation of the curve 3x'2 + y'2 = 6 in ry-coordinates. 14. Let an x'/-coordinate system be obtained by rotating an xy-coordinate system through an angle of 30°. Use (5) to find an equation in .^/-coordinates of the curve y = x2.
is a portion of a parabola. [Hint: First rationalize the equation and then perform a rotation of axes.] FOCUS ON CONCEPTS
FOCUS ON CONCEPTS
32. Derive the expression for B' in (9).
15. Let an ^'/-coordinate system be obtained by rotating an xy-coordinate system through an angled. Prove: For every value of 9, the equation x2 + y2 = r2 becomes the equation x'2 + y'2 = r2. Give a geometric explanation.
33. Use (9) to prove that B2 - 4AC = B'2 - 4A'C' for all values of 9. 34. Use (9) to prove that A + C = A' + C' for all values of 9.
16. Derive (6) by solving the rotation equations in (5) for x' and y' in terms of x and y.
35. Prove: If A = C in (7), then the cross-product term can be eliminated by rotating through 45 °.
17. Let an ^'/-coordinate system be obtained by rotating an xy-coordinate system through an angle 9. Explain how to find the ry-coordinates of a point whose x'y'coordinates are known.
36. Prove: If B ^ 0, then the graph of x2 + Bxy + F = 0 is a hyperbola if F / 0 and two intersecting lines if F = 0.
754
Chapter TO / Parametric and Polar Curves; Conic Sections
•QUICK CHECK ANSWERS 10.5 1. (a) x'cosO — y ' s i n O ; x' sinO + y'cosO (b) xcos9 + y sin (3; — x sin 9 + y cos 9 2
2
4. 5x' + 3/ = 2
2.
B
3. (a) — (b) — (c) — 4 3 6
CONIC SECTIONS IN POLAR COORDINATES It will be shown later in the text that if an object moves in a gravitational field that is directed toward a fixed point (such as the center of the Sun), then the path of that object must be a conic section with the fixed point at a focus. For example, planets in our solar system move along elliptical paths with the Sun at a focus, and the comets move along parabolic, elliptical, or hyperbolic paths with the Sun at a focus, depending on the conditions under which they were born. For applications of this type it is usually desirable to express the equations of the conic sections in polar coordinates with the pole at a focus. In this section we will show how to do this. THE FOCUS-DIRECTRIX CHARACTERIZATION OF CONICS To obtain polar equations for the conic sections we will need the following theorem.
It is an unfortunate historical accident that the letter e is used for the base of the natural logarithm as well as for the eccentricity of conic sections. However, as a practical matter the appropriate in-
10.6.1 THEOREM (Focus-Directrix Property of Conies) Suppose that a point P moves in the plane determined by a fixed point (called the focus) and a fixed line (called the directrix), where the focus does not lie on the directrix. If the point moves in such a way that its distance to the focus divided by its distance to the directrix is some constant e (called the eccentricity), then the curve traced by the point is a conic section. Moreover, the conic is
terpretation will usually be clear from the context in which the letter is used.
(a) a parabola if e = 1
(b) an ellipse ifO < e < 1
(c) a hyperbola ife > 1.
We will not give a formal proof of this theorem; rather, we will use the specific cases in Figure 10.6.1 to illustrate the basic ideas. For the parabola, we will take the directrix to be x = —p, as usual; and for the ellipse and the hyperbola we will take the directrix to be x = a2/c. We want to show in all three cases that if P is a point on the graph, F is the focus, and D is the directrix, then the ratio PF/PD is some constant e, where e — 1 for the parabola, 0 < e < 1 for the ellipse, and e > 1 for the hyperbola. We will give the arguments for the parabola and ellipse and leave the argument for the hyperbola as an exercise.
10.6 Conic Sections in Polar Coordinates 755 For the parabola, the distance PF to the focus is equal to the distance PD to the directrix, so that PF/PD — 1, which is what we wanted to show. For the ellipse, we rewrite Equation (8) of Section 10.4 as
c Ia j(x - c}2 + y2 =a- -x = x a a\ c But the expression on the left side is the distance PF, and the expression in the parentheses on the right side is the distance PD, so we have shown that PF = -PD a
Thus, PF/PD is constant, and the eccentricity is (1)
If we rule out the degenerate case where a — 0 or c = 0, then it follows from Formula (7) of Section 10.4 that 0 < c < a , s o O < < ? < 1, which is what we wanted to show. We will leave it as an exercise to show that the eccentricity of the hyperbola in Figure 10.6.1 is also given by Formula (1), but in this case it follows from Formula (11) of Section 10.4 that c > a, so e > 1. ECCENTRICITY OF AN ELLIPSE AS A MEASURE OF FLATNESS
The eccentricity of an ellipse can be viewed as a measure of its flatness—as e approaches 0 the ellipses become more and more circular, and as e approaches 1 they become more and more flat (Figure 10.6.2). Table 10.6.1 shows the orbital eccentricities of various celestial objects. Note that most of the planets actually have fairly circular orbits. Table 10.6.1 CELESTIAL BODY
= 0.9
Ellipses with a common focus ; and equal semimajor axes
A Figure 10.6.2
Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto Halley's comet
ECCENTRICITY
0.206 0.007 0.017 0.093 0.048 0.056 0.046 0.010 0.249 0.970
POLAR EQUATIONS OF CONICS
Directrix
A Figure 10.6.3
Our next objective is to derive polar equations for the conic sections from their focusdirectrix characterizations. We will assume that the focus is at the pole and the directrix is either parallel or perpendicular to the polar axis. If the directrix is parallel to the polar axis, then it can be above or below the pole; and if the directrix is perpendicular to the polar axis, then it can be to the left or right of the pole. Thus, there are four cases to consider. We will derive the formulas for the case in which the directrix is perpendicular to the polar axis and to the right of the pole. As illustrated in Figure 10.6.3, let us assume that the directrix is perpendicular to the polar axis and d units to the right of the pole, where the constant d is known. If P is a point
756
Chapter 10 / Parametric and Polar Curves; Conic Sections
on the conic and if the eccentricity of the conic is e, then it follows from Theorem 10.6.1 that PF/PD = e or, equivalently, that PF = ePD
(2)
However, it is evident from Figure 10.6.3 that PF = r and PD = d — r cos 9. Thus, (2) can be written as r = e(d — r cos9) which can be solved for r and expressed as ed 1 + e cos 6
(verify). Observe that this single polar equation can represent a parabola, an ellipse, or a hyperbola, depending on the value of e. In contrast, the rectangular equations for these conies all have different forms. The derivations in the other three cases are similar.
10.6.2 THEOREM If a conic section with eccentric ity e is positioned in a polar coordinate system so that its focus is at the pole and the c orresponding directrix is d units from the pole and is either paral lei or perpendicular to the polar axis, then the equation of the conic has one offourpos sible forms, depending on its orientation: Directrix
Directrix
Focus \ /
/ Focus
1
ed 1 + ecosO Directrix ri ght of pole
ed 1 - e cos 9 Directrix left of pole
v°<
Directrix ^«- Foe us\
\
\
US/ ^^
Directrix
ed
ed
1 + e sin 0 Directrix above pole
(3-4)
1 — e sin 9
(5-6)
Directrix below pole
SKETCHING CONICS IN POLAR COORDINATES
Precise graphs of conic sections in polar coordinates can be generated with graphing utilities. However, it is often useful to be able to make quick sketches of these graphs that show their orientations and give some sense of their dimensions. The orientation of a conic relative to the polar axis can be deduced by matching its equation with one of the four forms in Theorem 10.6.2. The key dimensions of a parabola are determined by the constant p (Figure 10.4.5) and those of ellipses and hyperbolas by the constants a, b, and c (Figures 10.4.11 and 10.4.20). Thus, we need to show how these constants can be obtained from the polar equations.
10.6 Conic Sections in Polar Coordinates Directrix
Tr/2
Example 1 Sketch the graph of r =
I-cos9
757
in polar coordinates.
Solution. The equation is an exact match to (4) with d = 2 and e = 1. Thus, the graph is a parabola with the focus at the pole and the directrix 2 units to the left of the pole. This tells us that the parabola opens to the right along the polar axis and p = 1. Thus, the parabola looks roughly like that sketched in Figure 10.6.4. < All of the important geometric information about an ellipse can be obtained from the values o f a , b , and c in Figure 10.6.5. One way to find these values from the polar equation of an ellipse is based on finding the distances from the focus to the vertices. As shown in the figure, let ro be the distance from the focus to the closest vertex and r\ the distance to the farthest vertex. Thus,
A Figure 10.6.4
r0 = a — c
and
r\ = a
(7)
from which it follows that
(8-9) Moreover, it also follows from (7) that
A Figure 10.6.5
r0r, =a2-c2=b2 In words, Formula (8) states that a is the arithmetic average (also called the arithmetic mean) of ro and r\, and Formula (10) states that b is the geometric mean of ro and r\.
Thus,
(10)
b—
Example 2
Find the constants a, b, and c for the ellipse r —
2 + cos 9
Directrix
Solution. This equation does not match any of the forms in Theorem 10.6.2 because they all require a constant term of 1 in the denominator. However, we can put the equation into one of these forms by dividing the numerator and denominator by 2 to obtain
1 + \ cos 9 A Figure 10.6.6
This is an exact match to (3) with d — 6 and e — \, so the graph is an ellipse with the directrix 6 units to the right of the pole. The distance ro from the focus to the closest vertex can be obtained by setting 9 = 0 in this equation, and the distance r\ to the farthest vertex can be obtained by setting 9 = n. This yields 1 + icosO
\
1
= cos n
i
Thus, from Formulas (8), (10), and (9), respectively, we obtain
a - \(r\ + r0) = 4,
b — ^7^ — 2-$,
c — \(r\ - r0) — 2
The ellipse looks roughly like that sketched in Figure 10.6.6. -4
A Figure 10.6.7
All of the important information about a hyperbola can be obtained from the values of a, b, and c in Figure 10.6.7. As with the ellipse, one way to find these values from the polar equation of a hyperbola is based on finding the distances from the focus to the vertices. As
758
Chapter TO / Parametric and Polar Curves; Conic Sections
shown in the figure, let ro be the distance from the focus to the closest vertex and r\ the distance to the farthest vertex. Thus, — c —a
r\ = c + a
and
(11)
from which it follows that (12-13)
c= Un +r 0 )
))
Moreover, it also follows from (11) that In words, Formula (13) states that c is the arithmetic mean of rg and n, and Formula (14) states that b \s the geometric mean of rg and r\.
r(>r\ = c2 — a 2 = bj 2
from which it follows that
> Example 3
(14)
Sketch the graph of r =
2
:— in polar coordinates. 1 + 2 sin 9
Solution. This equation is an exact match to (5) with d — 1 and e — 2. Thus, the graph is a hyperbola with its directrix 1 unit above the pole. However, it is not so straightforward to compute the values of ro and r\, since hyperbolas in polar coordinates are generated in a strange way as 9 varies from 0 to 2n. This can be seen from Figure 10.6.8a, which is the graph of the given equation in rectangular Or -coordinates. It follows from this graph that the corresponding polar graph is generated in pieces (see Figure 10.6.8&): • As 9 varies over the interval 0 < 9 < In/6, the value of r is positive and varies from 2 down to 2/3 and then to +00, which generates part of the lower branch. • As 9 varies over the interval In/6 < 9 < 37T/2, the value of r is negative and varies from —oo to —2, which generates the right part of the upper branch. • As 9 varies over the interval 3jr/2 < 9 < 1 ljr/6, the value of r is negative and varies from —2 to —oo, which generates the left part of the upper branch. • As 9 varies over the interval llTT/6 < 9 < 2n, the value of r is positive and varies from +oc to 2, which fills in the missing piece of the lower right branch.
7k 3* 1 to
2 "T
-3
1 + 2 sin 0
Rough sketch
(a)
(b)
A Figure 10.6.8
To obtain a rough sketch of a hyperbola, it is generally sufficient to locate the center, the asymptotes, and the points where e = 0,8 = nil, 6 — n, and 6 = 3jr/2.
It is now clear that we can obtain A-Q by setting 6 = n/2 and r\ by setting 9 = 3n/2. Keeping in mind that ro and r\ are positive, this yields l+2sin(7r/2)
2 3'
2
+2sin(37T/2)
=2
10.6 Conic Sections in Polar Coordinates
759
Thus, from Formulas (12), (14), and (13), respectively, we obtain
a = ~(r\ -r0) = -,
b=
1
c=
4
_(ri+ro) = -
Thus, the hyperbola looks roughly like that sketched in Figure 10.6.8c. •<
APPLICATIONS IN ASTRONOMY In 1609 Johannes Kepler published a book known as Astronomia Nova (or sometimes Commentaries on the Motions of Mars) in which he succeeded in distilling thousands of years of observational astronomy into three beautiful laws of planetary motion (Figure 10.6.9).
10.6.3
Equal areas are swept out in equal times, and the square of the period T is proportional to a 3
Apogee
A Figure 10.6.10
—• Perigee
KEPLER'S LAWS
•
First law (Law of Orbits). Each planet moves in an elliptical orbit with the Sun at a focus.
•
Second law (Law of Areas). The radial line from the center of the Sun to the center of a planet sweeps out equal areas in equal times.
•
Third law (Law of Periods). The square of a planet's period (the time it takes the planet to complete one orbit about the Sun) is proportional to the cube of the semimajor axis of its orbit.
Kepler's laws, although stated for planetary motion around the Sun, apply to all orbiting celestial bodies that are subjected to a single central gravitational force—artificial satellites subjected only to the central force of Earth's gravity and moons subjected only to the central gravitational force of a planet, for example. Later in the text we will derive Kepler's laws from basic principles, but for now we will show how they can be used in basic astronomical computations. In an elliptical orbit, the closest point to the focus is called the perigee and the farthest point the apogee (Figure 10.6.10). The distances from the focus to the perigee and apogee
Johannes Kepler (1571-1630) German astronomer and physicist. Kepler, whose work provided our contemporary view of planetary motion, led a fascinating but ill-starred life. His alcoholic father made him work in a family-owned tavern as a child, later withdrawing him from elementary school and hiring him out as a field laborer, where the boy contracted smallpox, permanently crippling his hands and impairing his eyesight. In later years, Kepler's first wife and several children died, his mother was accused of witchcraft, and being a Protestant he was often subjected to persecution by Catholic authorities. He was often impoverished, eking out a living as an astrologer and prognosticator. Looking back on his unhappy childhood, Kepler described his father as "criminally inclined" and "quarrelsome" and his mother as "garrulous" and "bad-tempered." However, it was his mother who left an indelible mark on the sixyear-old Kepler by showing him the comet of 1577; and in later life he personally prepared her defense against the witchcraft charges. Kepler became acquainted with the work of Copernicus as a student at the University of Tubingen, where he received his master's de-
gree in 1591. He continued on as a theological student, but at the urging of the university officials he abandoned his clerical studies and accepted a position as a mathematician and teacher in Graz, Austria. However, he was expelled from the city when it came under Catholic control, and in 1600 he finally moved on to Prague, where he became an assistant at the observatory of the famous Danish astronomer Tycho Brahe. Brahe was a brilliant and meticulous astronomical observer who amassed the most accurate astronomical data known at that time; and when Brahe died in 1601 Kepler inherited the treasure-trove of data. After eight years of intense labor, Kepler deciphered the underlying principles buried in the data and in 1609 published his monumental work, Astronomia Nova, in which he stated his first two laws of planetary motion. Commenting on his discovery of elliptical orbits, Kepler wrote, "I was almost driven to madness in considering and calculating this matter. I could not find out why the planet would rather go on an elliptical orbit (rather than a circle). Oh ridiculous me!" It ultimately remained for Isaac Newton to discover the laws of gravitation that explained the reason for elliptical orbits.
760
Chapter 10 / Parametric and Polar Curves; Conic Sections
are called the perigee distance and apogee distance, respectively. For orbits around the Sun, it is more common to use the terms perihelion and aphelion, rather than perigee and apogee, and to measure time in Earth years and distances in astronomical units (AU), where 1 AU is the semimajor axis a of the Earth's orbit (approximately 150 x 106 km or 92.9 x 106 mi). With this choice of units, the constant of proportionality in Kepler's third law is 1, since a = 1 AU produces a period of T — 1 Earth year. In this case Kepler's third law can be expressed as T = a372 (15)
Directrix
Shapes of elliptical orbits are often specified by giving the eccentricity e and the semimajor axis a, so it is useful to express the polar equations of an ellipse in terms of these constants. Figure 10.6.11, which can be obtained from the ellipse in Figure 10.6.1 and the relationship c = ea, implies that the distance d between the focus and the directrix is d =
a e
c=
a e
ea —
06)
from which it follows that ed = a(l — e 2 ). Thus, depending on the orientation of the ellipse, the formulas in Theorem 10.6.2 can be expressed in terms of a and e as r^
A Figure 10.6.11
a(l-e2) l i e cos 9
+: Directrix right of pole —: Directrix left of pole
a(\-e2) ~ 1 ± e sin 6
r=
(17-18)
+: Directrix above pole —: Directrix below pole
Moreover, it is evident from Figure 10.6.11 that the distances from the focus to the closest and farthest vertices can be expressed in terms of a and e as = a — ea = a(l — e) and r\ = a + ea = a(\ + e)
Halley's comet
A Figure 10.6.12
(19-20)
> Example 4 Halley's comet (last seen in 1986) has an eccentricity of 0.97 and a semimajor axis of a — 18.1 AU. (a) Find the equation of its orbit in the polar coordinate system shown in Figure 10.6.12. (b) Find the period of its orbit. (c) Find its perihelion and aphelion distances. Solution (a).
From (17), the polar equation of the orbit has the form
a(l-e2) 1 + e cos 9 1 2 But a(\ - e ) = 18.1[1 - (0.97) ] ~ 1.07. Thus, the equation of the orbit is 1.07 r= 1 +0.97 cos 9 Science Photo Library/Photo Researchers Halley's comet photographed April 21, 1910 in Peru.
Solution (b).
From (15), with a = 18.1, the period of the orbit is T = (18.1)3/2?»77years
Solution (c). Since the perihelion and aphelion distances are the distances to the closest and farthest vertices, respectively, it follows from (19) and (20) that r0=a-ea=a(l-e) = 18.1(1 - 0.97) % 0.543 AU n =a + ea = a(l+e) = 18.1(1 + 0.97) % 35.7 AU
10.6 Conic Sections in Polar Coordinates
761
or since 1 AU & 150 x 106 km, the perihelion and aphelion distances in kilometers are r0 = 18.1(1 - 0.97)(150 x 106) % 81,500,000 km n = 18.1(1 + 0.97)(150 x 106) « 5,350,000,000 km + Minimum distance
*• Example 5 An Apollo lunar lander orbits the Moon in an elliptic orbit with eccentricity e = 0.12 and semimajor axis a = 2015 km. Assuming the Moon to be a sphere of radius 1740 km, find the minimum and maximum heights of the lander above the lunar surface (Figure 10.6.13). Solution. If we let rn and rl denote the minimum and maximum distances from the center of the Moon, then the minimum and maximum distances from the surface of the Moon will be
4nin = r0 - 1740 dmax = ri - 1740
or from Formulas (19) and (20) Maximum distance A Figure 10.6.13
dmin = r0 - 1740 = a(l - e) - 1740 = 2015(0.88) - 1740 = 33.2 km rfmax = n - 1740 = a(l + e) - 1740 = 2015(1.12) - 1740 = 516.8 km +
QUICK CHECK EXERCISES 10.6
(See page 763 for answers.)
1. In each part, name the conic section described. (a) The set of points whose distance to the point (2, 3) is half the distance to the line x + y = 1 is (b) The set of points whose distance to the point (2, 3) is equal to the distance to the line x + y = I is (c) The set of points whose distance to the point (2, 3) is twice the distance to the line x + y = 1 is 2. In each part: (i) Identify the polar graph as a parabola, an ellipse, or a hyperbola; (ii) state whether the directrix is above, below, to the left, or to the right of the pole; and (iii) find the distance from the pole to the directrix. (a) r = —J—4 + cos 9
EXERCISE SET 10.6
(b) r =
-4cos6>
B Graphing Utility
1 -2 Find the eccentricity and the distance from the pole to the directrix, and sketch the graph in polar coordinates. 1. (a) r =
2-2cos6> 4 2. (a) r = 2 + 3 cos 9
(b) r =
2 + sin 9 5 (b) r = 3+ 3 sin0
| 3-4 Use Formulas (3)-(6) to identify the type of conic and its orientation. Check your answer by generating the graph with a graphing utility. • 3. (a) r =
1 - sin 9
1 (d) r = 4 + 4 sin 9 1 - sin 9 3. If the distance from a vertex of an ellipse to the nearest focus is r0, and if the distance from that vertex to the farthest focus is n, then the semimajor axis is a = and the semiminor axis is b = 4. If the distance from a vertex of a hyperbola to the nearest focus is rn, and if the distance from that vertex to the farthest focus is r\, then the semifocal axis is a = and the semiconjugate axis is b = (c) r =
(b) r =
4 + 3 sin 9
4. (a) r =
2-3sin0
(b) r =
12 4 + cos 9
5-6 Find a polar equation for the conic that has its focus at the pole and satisfies the stated conditions. Points are in polar coordinates and directrices in rectangular coordinates for simplicity. (In some cases there may be more than one conic that satisfies the conditions.) 5. (a) Ellipse; e = |; directrix x = 2. (b) Parabola; directrix x = 1. (c) Hyperbola; e = |; directrix y = 3.
762
Chapter 10 / Parametric and Polar Curves; Conic Sections
6. (a) Ellipse; ends of major axis (2, jr/2) and (6, 3;r/2). (b) Parabola; vertex (2, JT). (c) Hyperbola; e = A/2; vertex (2, 0). 7-8 Find the distances from the pole to the vertices, and then apply Formulas (8)-(10) to find the equation of the ellipse in rectangular coordinates. 6 1 7. (a) r = ——(b) r = 2 + sin 9 2-cos9 6 8. (a) r = (b) r = 5 + 2 cos 9 4 - 3 sin 6 9-10 Find the distances from the pole to the vertices, and then apply Formulas (12)-(14) to find the equation of the hyperbola in rectangular coordinates.
9. (a) r = , , ^ . „ 1 + 2 sin 9 4 10. (a) r = -2sm8
(b) r =
2-3cos6» 15 (b) r = 2 + 8 cos 0
11-12 Find a polar equation for the ellipse that has its focus at the pole and satisfies the stated conditions. • 11. (a) Directrix to the right of the pole; a = 8; e=\. (b) Directrix below the pole; a = 4; e = |. 12. (a) Directrix to the left of the pole; ft = 4; e = f . (b) Directrix above the pole; c = 5; e = |. 13. Find the polar equation of an equilateral hyperbola with a focus at the pole and vertex (5, 0). FOCUS ON CONCEPTS
14. Prove that a hyperbola is an equilateral hyperbola if and only if e = \/2. 15. (a) Show that the coordinates of the point P on the hyperbola in Figure 10.6.1 satisfy the equation
J(x - c)2 + y2 = -x -a a (b) Use the result obtained in part (a) to show that PF/PD = da. 16. (a) Show that the eccentricity of an ellipse can be expressed in terms of r$ and r\ as r\ -ro r\ +r0
(b) Show that
l-e 17. (a) Show that the eccentricity of a hyperbola can be expressed in terms of ro and r\ as
r\ +r0 r\ (b) Show that
e+1 e- 1
18. (a) Sketch the curves
1 +cos6»
and
r =
1
1 - cos 0
(b) Find polar coordinates of the intersections of the curves in part (a). (c) Show that the curves are orthogonal, that is, their tangent lines are perpendicular at the points of intersection. 19-22 True-False Determine whether the statement is true or false. Explain your answer. S; 19. If an ellipse is not a circle, then the eccentricity of the ellipse is less than one. 20. A parabola has eccentricity greater than one. 21. If one ellipse has foci that are farther apart than those of a second ellipse, then the eccentricity of the first is greater than that of the second. 22. If d is a positive constant, then the conic section with polar equation ^
1 + cos 0 is a parabola. 23-28 Use the following values, where needed: radius of the Earth = 4000 mi = 6440 km 1 year (Earth year) = 365 days (Earth days) 1 AU = 92.9 x 106 mi = 150 x 106 km 23. The dwarf planet Pluto has eccentricity e = 0.249 and semimajor axis a = 39.5 AU. (a) Find the period T in years. (b) Find the perihelion and aphelion distances. (c) Choose a polar coordinate system with the center of the Sun at the pole, and find a polar equation of Pluto's orbit in that coordinate system. (d) Make a sketch of the orbit with reasonably accurate proportions. 24. (a) Let a be the semimajor axis of a planet's orbit around the Sun, and let T be its period. Show that if T is measured in days and a is measured in kilometers, then T = (365 x I(r 9 )(a/150) 3/2 . (b) Use the result in part (a) to find the period of the planet Mercury in days, given that its semimajor axis is a = 57.95 x 106 km. (c) Choose a polar coordinate system with the Sun at the pole, and find an equation for the orbit of Mercury in that coordinate system given that the eccentricity of the orbit is e = 0.206. (d) Use a graphing utility to generate the orbit of Mercury from the equation obtained in part (c). 25. The Hale-Bopp comet, discovered independently on July 23, 1995 by Alan Hale and Thomas Bopp, has an orbital eccentricity of e = 0.9951 and a period of 2380 years. (a) Find its semimajor axis in astronomical units (AU). (b) Find its perihelion and aphelion distances. (««<.)
Chapter 10 Review Exercises (c) Choose a polar coordinate system with the center of the Sun at the pole, and find an equation for the Hale-Bopp orbit in that coordinate system. (d) Make a sketch of the Hale-Bopp orbit with reasonably accurate proportions.
763
On November 6, 1997 the spacecraft Galileo was placed in a Jovian orbit to study the moon Europa. The orbit had eccentricity 0.814580 and semimajor axis 3,514,918.9 km. Find Galileo's minimum and maximum heights above the molecular hydrogen layer (see the accompanying figure).
26. Mars has a perihelion distance of 204,520,000 km and an aphelion distance of 246,280,000 km. (a) Use these data to calculate the eccentricity, and compare your answer to the value given in Table 10.6.1. (b) Find the period of Mars. (c) Choose a polar coordinate system with the center of the Sun at the pole, and find an equation for the orbit of Mars in that coordinate system. (d) Use a graphing utility to generate the orbit of Mars from the equation obtained in part (c). 27. Vanguard 1 was launched in March 1958 into an orbit around the Earth with eccentricity e = 0.21 and semimajor axis 8864.5 km. Find the minimum and maximum heights of Vanguard 1 above the surface of the Earth. 28. The planet Jupiter is believed to have a rocky core of radius 10,000 km surrounded by two layers of hydrogen— a 40,000 km thick layer of compressed metallic-like hydrogen and a 20,000 km thick layer of ordinary molecular hydrogen. The visible features, such as the Great Red Spot, are at the outer surface of the molecular hydrogen layer.
Not to scale A Figure Ex-28
29. Writing Discuss how a hyperbola's eccentricity e affects the shape of the hyperbola. How is the shape affected as e approaches 1 ? As e approaches +00? Draw some pictures to illustrate your conclusions. 30. Writing Discuss the relationship between the eccentricity e of an ellipse and the distance z between the directrix and center of the ellipse. For example, if the foci remain fixed, what happens to z as e approaches 0?
•QIUICK CHECK ANSWERS 10.6 1. (a) an ellipse (b) a parabola (c) a hyperbola 2. (a) (i) ellipse (ii) to the right of the pole (iii) distance = 1 (b) (i) hyperbola (ii) to the left of the pole (iii) distance = j (c) (i) parabola (ii) above the pole (iii) distance = i (d) (i) parabola (ii) below the pole (iii) distance = 4 3. \(r\ + ro); Von 4. \(r\ —TO); «JrQr\
CHAPTER 10 REVIEW EXERCISES
Graphing Utility
0 1. Find parametric equations for the portion of the circle x2 + y2 = 2 that lies outside the first quadrant, oriented clockwise. Check your work by generating the curve with a graphing utility. 02. (a) Suppose that the equations * = f ( t ) , y = g(t) describe a curve C as t increases from 0 to 1. Find parametric equations that describe the same curve C but traced in the opposite direction as t increases from 0 to 1. (b) Check your work using the parametric graphing feature of a graphing utility by generating the line segment between (1,2) and (4, 0) in both possible directions as t increases from 0 to 1. 3. (a) Find the slope of the tangent line to the parametric curve x = t2 + 1, y = t/2 at t = -1 and t = 1 without eliminating the parameter. (b) Check your answers in part (a) by eliminating the parameter and differentiating a function of x.
4. Find dy/dx and d2y/dx2 at t = 2 for the parametric curve x
= \t\y = \t\
5. Find all values of t at which a tangent line to the parametric curve x = 2 cos t, y = 4 sin t is (a) horizontal (b) vertical. 6. Find the exact arc length of the curve
= l- 5t\ y = 4r5 -
(0 < t < 1)
7. In each part, find the rectangular coordinates of the point whose polar coordinates are given. (a) (-8,jr/4) (b) (7, -7T/4) (c) (8, 9ir/4) (d) (5,0) (e) (-2,-3;r/2) (f) (Q,n) 8. Express the point whose ^-coordinates are (— 1, 1) in polar coordinates with (a) r > 0, 0 < e < 2n (b) r < 0, 0 < 0 < 2n (c) r > 0, -TT < 6 < 7t (d) r < 0, -n < 9 < n.
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Chapter 10 / Parametric and Polar Curves; Conic Sections
9. In each part, use a calculating utility to approximate the polar coordinates of the point whose rectangular coordinates are given. (a) (4,3) (b) (2,-5) (c) (l.tair 1 !)
(d) Find a polar equation r = f(9) for the conchoid in part (a), and then find polar equations for the tangent lines to the conchoid at the pole.
10. In each part, state the name that describes the polar curve most precisely: a rose, a line, a circle, a lima§on, a cardioid, a spiral, a lemniscate, or none of these, (a) r = 3cos0 (b) r = cos36>
(d) r = 3 — cos 6 cos 9 (e) r = 1 - 3 cos 9 (f) r2 = 3cos9 2 (g) r = (3cos6>) (h) r = 1+39 11. In each part, identify the curve by converting the polar equation to rectangular coordinates. Assume that a > 0.
^ Figure Ex-21
(c) r =
(a) r = a sec2 -
(c) r = 4 esc
j
(b) r2 cos 29= a1
23. Find the area of the region that is enclosed by the cardioid r = 2 + 2cos9.
(d) r = 4 cos 6 + 8 sin 9
24. Find the area of the region in the first quadrant within the cardioid r = 1 + sin 9.
12. In each part, express the given equation in polar coordinates. (a) x = 1 (b) x2 + y2 = 9 2 2 (c) x + y - 6y = 0 (d) 4xy = 9 13-17 Sketch the curve in polar coordinates. *
13. 0 =
6 15. r = 3(1 - sin 9)
22. (a) Find the arc length of the polar curve r = 1/9 for 7T/4 < 9 < 7i/2. (b) What can you say about the arc length of the portion of the curve that lies inside the circle r = 1 ?
14. r = 6 cos 9
25. Find the area of the region that is common to the circles r = 1, r = 2 cos 9, and r = 2sin0. 26. Find the area of the region that is inside the cardioid r = a(\ + sin 9) and outside the circle r = a sin 9. 27-30 Sketch the parabola, and label the focus, vertex, and directrix, H
16. r2 = sin 29
17. r = 3 - cos 9 18. (a) Show that the maximum value of the y-coordinate of points on the curve r = 1 /\/0 for 9 in the interval (0, ji] occurs when tan 9 = 29. (b) Use a calculating utility to solve the equation in part (a) to at least four decimal-place accuracy. (c) Use the result of part (b) to approximate the maximum value of y for 0 < 9 < it. 19. (a) Find the minimum and maximum x -coordinates of points on the cardioid r = 1 — cos 9. (b) Find the minimum and maximum ^-coordinates of points on the cardioid in part (a).
27. y2 = 6x
28. x2 = -9y
29. (y + I) 2 = -l(x - 4)
30. (x - ±) 2 = 2(y - 1)
31-34 Sketch the ellipse, and label the foci, the vertices, and the ends of the minor axis. • x2 v2 31. — + — = 1 32. 4x2 + 9y2 = 36
33. 9(x - I) 2 + 16(y - 3)2 = 144 34. 3(x + 2)2 + 4(y + I) 2 = 12 35-37 Sketch the hyperbola, and label the vertices, foci, and asymptotes. B v-2
20. Determine the slope of the tangent line to the polar curve r = I + sin 9 at 9 = n/4.
(Q
is called a conchoid ofNicomedes (see the accompanying figure for the case 0 < a < b). (a) Describe how the conchoid
x = cot t + 4 cos t,
16
,,2
4
— i
36. 9y 2 - 4x2 = 36
37.
21. A parametric curve of the form x = acotf +bcost,
— —
y = 1 + 4 sin t
is generated as t varies over the interval 0 < t < lit. (b) Find the horizontal asymptote of the conchoid given in part (a). (c) For what values of / does the conchoid in part (a) have a horizontal tangent line? A vertical tangent line?
38. In each part, sketch the graph of the conic section with reasonably accurate proportions. (a) x2 - 4x + 8y + 36 = 0 (b) 3x2 + 4y2 - 30x - 8y + 67 = 0 (c) 4x2 - 5y2 - 8x - 30y - 21 = 0 39-41 Find an equation for the conic described.
•
39. A parabola with vertex (0, 0) and focus (0, —4). 40. An ellipse with the ends of the major axis (0, ±V5) and the ends of the minor axis (±1,0). 41. A hyperbola with vertices (0, ±3) and asymptotes y = ±x.
Chapter 10 Review Exercises 42. It can be shown in the accompanying figure that hanging cables form parabolic arcs rather than catenaries if they are subjected to uniformly distributed downward forces along their length. For example, if the weight of the roadway in a suspension bridge is assumed to be uniformly distributed along the supporting cables, then the cables can be modeled by parabolas. (a) Assuming a parabolic model, find an equation for the cable in the accompanying figure, taking the y-axis to be vertical and the origin at the low point of the cable. (b) Find the length of the cable between the supports.
46. 7;t2 + 2Vlxy + 5y2 - 4 = 0 47. 4V5.*:2 + 475*3; + ^5y2 + 5x - \0y = 0 48. Rotate the coordinate axes to show that the graph of \7x2 - 3l2xy + 108.V2 + 1080* - 1440^+4500 = 0 is a hyperbola. Then find its vertices, foci, and asymptotes. 49. In each part: (i) Identify the polar graph as a parabola, an ellipse, or a hyperbola; (ii) state whether the directrix is above, below, to the left, or to the right of the pole; and (iii) find the distance from the pole to the directrix.
(a) r = (c) r =
3 + cos 6
1 3(1+sin 61)
(b) r = (d) r =
1
- 3 cos 6» 3 -sin6>
50-51 Find an equation in ^-coordinates for the conic section that satisfies the given conditions. B
- 4200 ft A Figure Ex-42
43. It will be shown later in this text that if a projectile is launched with speed VQ at an angle a with the horizontal and at a height yo above ground level, then the resulting trajectory relative to the coordinate system in the accompanying figure will have parametric equations
x = (DO cos a)?,
765
(VQ sin a)? — ^
where g is the acceleration due to gravity. (a) Show that the trajectory is a parabola. (b) Find the coordinates of the vertex.
50. (a) Ellipse with eccentricity e = | and ends of the minor axis at the points (0, ±3). (b) Parabola with vertex at the origin, focus on the y-axis, and directrix passing through the point (7,4). (c) Hyperbola that has the same foci as the ellipse 3*2 + 16>>2 = 48 and asymptotes y = ±2x/3. 51. (a) Ellipse with center (—3, 2), vertex (2, 2), and eccentricity e = |. (b) Parabola with focus (—2, —2) and vertex (—2, 0). (c) Hyperbola with vertex (—1,7) and asymptotes y - 5 = ±8(* + 1). 52. Use the parametric equations x = a cost, y = bsint to show that the circumference C of an ellipse with semimajor axis a and eccentricity e is fir/2 V 1 — e2s'm2 udu C = 4a I
•« Figure Ex-43
44. Mickey Mantle is recognized as baseball's unofficial king of long home runs. OnApril 17, 1953 Mantle blasted a pitch by Chuck Stobbs of the hapless Washington Senators out of Griffith Stadium, just clearing the 50 ft wall at the 391 ft marker in left center. Assuming that the ball left the bat at a height of 3 ft above the ground and at an angle of 45 °, use the parametric equations in Exercise 43 with g = 32 ft/s 2 to find (a) the speed of the ball as it left the bat (b) the maximum height of the ball (c) the distance along the ground from home plate to where the ball struck the ground. 45-47 Rotate the coordinate axes to remove the xy-term, and then name the conic. •
45. x2 + y2 - 3xy - 3 = 0
Jo 53. Use Simpson's rule or the numerical integration capability of a graphing utility to approximate the circumference of the ellipse 4;c2 + 9y2 = 36 from the integral obtained in Exercise 52. 54. (a) Calculate the eccentricity of the Earth's orbit, given that the ratio of the distance between the center of the Earth and the center of the Sun at perihelion to the distance between the centers at aphelion is |y. (b) Find the distance between the center of the Earth and the center of the Sun at perihelion, given that the average value of the perihelion and aphelion distances between the centers is 93 million miles. (c) Use the result in Exercise 52 and Simpson's rule or the numerical integration capability of a graphing utility to approximate the distance that the Earth travels in 1 year (one revolution around the Sun).
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Chapter 10 / Parametric and Polar Curves; Conic Sections
CHAPTER 10 MAKING CONNECTIONS
0 CAS
[c] 1. Recall from Section 6.6 that the Fresnel sine and cosine functions are defined as
S(x) =
f sin• ('I -
Jo
\ 2 /
dt
and C(x) =
cos cosI|— -^- I| rff V 2/
/./o
triangle and the other end to a fixed point F. A pencil holds the string taut against the base of the triangle as the edge opposite Q slides along a horizontal line L below F. Show that the pencil traces an arc of a parabola with focus F and directrix L.
The following parametric curve, which is used to study amplitudes of light waves in optics, is called a clothoid or Cornu spiral in honor of the French scientist Marie Alfred Cornu (1841-1902):
/"
x = C(t) = I cos Jo
(nu2\
du
\ 2 )
y = S(t) =
( — 00 < t <
+00)
•4 Figure Ex-3
du
(a) Use a CAS to graph the Cornu spiral. (b) Describe the behavior of the spiral as t -> +co and as t-+ —oo. (c) Find the arc length of the spiral for — 1 < t < 1. 2. (a) The accompanying figure shows an ellipse with semimajor axis a and semiminor axis b. Express the coordinates of the points P, Q, and R in terms of t. (b) How does the geometric interpretation of the parameter t differ between a circle x = a cost,
y = a smf
x=acost,
y=
4. The accompanying figure shows a method for constructing a hyperbola. A corner of a ruler is pinned to a fixed point FI and the ruler is free to rotate about that point. A piece of string whose length is less than that of the ruler is tacked to a point F2 and to the free corner Q of the ruler on the same edge as FI . A pencil holds the string taut against the top edge of the ruler as the ruler rotates about the point FI . Show that the pencil traces an arc of a hyperbola with foci FI and ¥^.
and an ellipse
•4 Figure Ex-4
5. Consider an ellipse E with semimajor axis a and semiminor axis b, and set c = ^/a2 — b2. (a) Show that the ellipsoid that results when E is revolved about its major axis has volume V = \jtab1 and surface S = 2jtab I - + - sin"1 a c a •4 Figure Ex-2
3. The accompanying figure shows Kepler's method for constructing a parabola. A piece of string the length of the left edge of the drafting triangle is tacked to the vertex Q of the
(b) Show that the ellipsoid that results when E is revolved about its minor axis has volume V = ^7ta2b and surface area / a b a+c S = Inab - + - In \b c b
EXPANDING THE CALCULUS HORIZON To learn how polar coordinates and conic sections can be used to analyze the possibility of a collision between a comet and Earth, see the module entitled Comet Collision at: www.wiley.com/college/anton
GRAPHING FUNCTIONS USING CALCULATORS AND COMPUTER ALGEBRA SYSTEMS GRAPHING CALCULATORS AND COMPUTER ALGEBRA SYSTEMS
The development of new technology has significantly changed how and where mathematicians, engineers, and scientists perform their work, as well as their approach to problem solving. Among the most significant of these developments are programs called Computer Algebra Systems (abbreviated CAS), the most common being Mathematica and Maple* Computer algebra systems not only have graphing capabilities, but, as their name suggests, they can perform many of the symbolic computations that occur in algebra, calculus, and branches of higher mathematics. For example, it is a trivial task for a CAS to perform the factorization 23x5 + I41x4 - I39x3 - 3464x2 - 2112* + 23040 = (x
5)(x - 3)2(x + 8)3
or the exact numerical computation 63456 V 3 177295
43907 22854377 J
2251912457164208291259320230122866923 382895955819369204449565945369203764688375
Technology has also made it possible to generate graphs of equations and functions in seconds that in the past might have taken hours. Figure A. 1 shows the graphs of the function f ( x ) = x4 — x3 — 2x2 produced with various graphing utilities; the first two were generated with the CAS programs, Mathematica and Maple, and the third with a graphing calculator. Graphing calculators produce coarser graphs than most computer programs but have the advantage of being compact and portable.
Generated by Mathematica
Generated by Maple
Generated by a graphing calculator
A Figure A.I Mathematica is a product of Wolfram Research, Inc.; Maple is a product of Waterloo Maple Software, Inc.
Al
A2
Appendix A: Graphing Functions Using Calculators and Computer Algebra Systems
(b,d)
(a,d)
[c,d]
(b,c)
(a,c)
The window [a, b] x [c, d}
A Figure A.2
VIEWING WINDOWS Graphing utilities can only show a portion of the xy-plane in the viewing screen, so the first step in graphing an equation is to determine which rectangular portion of the ry-plane you want to display. This region is called the viewing window (or viewing rectangle). For example, in Figure A.I the viewing window extends over the interval [—3, 3] in the redirection and over the interval [—4,4] in the y-direction, so we denote the viewing window by [—3, 3] x [—4,4] (read "[—3, 3] by [—4, 4]"). In general, if the viewing window is [a, b] x [c, d], then the window extends between x = a and x = b in the ;t-direction and between y = c and y — d in the y-direction. We will call [a, b] the x-interval for the window and [c, d] the y-interval for the window (Figure A.2). Different graphing utilities designate viewing windows in different ways. For example, the first two graphs in Figure A. 1 were produced by the commands Plot[x"4 - x~3 -2*x~2, {x, -3, (Mathematica)
TECHNOLOGY MASTERY Use your graphing utility to generate the graph of the function 4
f(x) = x
- x^ - 2x2
in the window [—3, 3] x [—4,4].
-2 -4L
- 3 - 2 - 1 0 1 2 : Generated by Mathematica
plot(x~4 - x~3
3}, PlotRange->{-4, 4}]
-2*x~2, x = - 3 . . 3 , y = - 4 . . 4 ) ;
(Maple) and the last graph was produced on a graphing calculator by pressing the GRAPH button after setting the values for the variables that determine the x -interval and y-interval to be xMax = 3,
yMin = —4,
yMax = 4
B TICK MARKS AND GRID LINES To help locate points in a viewing window, graphing utilities provide methods for drawing tick marks (also called scale marks). With computer programs such as Mathematica and Maple, there are specific commands for designating the spacing between tick marks, but if the user does not specify the spacing, then the programs make certain default choices. For example, in the first two parts of Figure A. 1 , the tick marks shown were the default choices. On some graphing calculators the spacing between tick marks is determined by two scale variables (also called scale factors), which we will denote by xScl
and
yScl
(The notation varies among calculators.) These variables specify the spacing between the tick marks in the x- and y-directions, respectively. For example, in the third part of Figure A. 1 the window and tick marks were designated by the settings
Generated by a graphing calculator A Figure A.3
[-5, 5] x [-5, 5] *Scl = 0.5, yScl = 10
A Figure A.4
= —3
jcMax = 3
yMin = -4
yMax = 4
jcScl = 1
yScl = 1
Most graphing utilities allow for variations in the design and positioning of tick marks. For example, Figure A.3 shows two variations of the graphs in Figure A.I; the first was generated on a computer using an option for placing the ticks and numbers on the edges of a box, and the second was generated on a graphing calculator using an option for drawing grid lines to simulate graph paper.
> Example 1 Figure A.4 shows the window [—5, 5] x [—5, 5] with the tick marks spaced 0.5 unit apart in the x-direction and 10 units apart in the y-direction. No tick marks are actually visible in the y-direction because the tick mark at the origin is covered by the jc-axis, and all other tick marks in that direction fall outside of the viewing window.
Appendix A: Graphing Functions Using Calculators and Computer Algebra Systems
A3
*• Example 2 Figure A.5 shows the window [-10, 10] x [-10, 10] with the tick marks spaced 0.1 unit apart in the x- and y-directions. In this case the tick marks are so close together that they create thick lines on the coordinate axes. When this occurs you will usually want to increase the scale factors to reduce the number of tick marks to make them legible. -4
TECHNOLOGY MASTERY
Graphing calculators provide a way of clearing all settings and returning them to default values. For example, on one calculator the default window is [-10, 10] x [—10, 10] and the default scale factors arexScI = landyScI = 1. Read your documentation to determine the default values for your calculator and how to restore the default settings. If you are using a CAS, read your documentation to determine the commands for specifying the spacing between tick marks.
CHOOSING A VIEWING WINDOW [-10, 10] x [-10, 10] xSc\ = 0.1,yScl = 0.1
A Figure A.5
When the graph of a function extends indefinitely in some direction, no single viewing window can show it all. In such cases the choice of the viewing window can affect one's perception of how the graph looks. For example, Figure A.6 shows a computer-generated graph of y — 9 — x2, and Figure A.7 shows four views of this graph generated on a calculator. • In part (a) the graph falls completely outside of the window, so the window is blank (except for the ticks and axes). • In part (b) the graph is broken into two pieces because it passes in and out of the window. • In part (c) the graph appears to be a straight line because we have zoomed in on a very small segment of the curve. • In part (d) we have a more revealing picture of the graph shape because the window encompasses the high point on the graph and the intersections with the x-axis.
A Figure A.6
[-2, 2] x [-2, 2] *Scl = \,ySc\ = 1
[-4,4] x [-2,5] *Scl = l.yScI = 1
(a)
(*)
[2.5, 3.5] x [-1, 1] *Scl =0.1,yScl= 1
[-4, 4] x [-3, 10] *Scl = l , y S c l = 1
(c) A Figure A.7 Four views of y = 9 — x2
A4
Appendix A: Graphing Functions Using Calculators and Computer Algebra Systems
The following example illustrates how the domain and range of a function can be used to find a good viewing window when the graph of the function does not extend indefinitely in both the x- and y-directions. > Example 3 Use the domain and range of the function f(x) = V l 2 — 3x2 to determine a viewing window that contains the entire graph. Solution. The natural domain of / is [—2, 2] and the range is [0, +/12] (verify), so the entire graph will be contained in the viewing window [—2, 2] x [0, V^2]- For clarity, it is desirable to use a slightly larger window to avoid having the graph too close to the edges of the screen. For example, taking the viewing window to be [—3, 3] x [—1,4] yields the graph in Figure A.8. -4 [-3, 3] x [-1,4] *Scl = \,ySc\ = 1 A Figure A.8
Sometimes it will be impossible to find a single window that shows all important features of a graph, in which case you will need to decide what is most important for the problem at hand and choose the window appropriately. > Example 4 Graph the equation y = x3 - ]2x2 + 18 in the following windows and discuss the advantages and disadvantages of each window. (a) [-10, 10] x [-10, 10] with *Scl = 1, yScl = 1 (b) [-20, 20] x [-20, 20] with *Scl = 1, yScl = 1 (c) [-20, 20] x [-300, 20] with jcScl = 1, yScl = 20 (d) [-5, 15] x [-300, 20] with zScl = 1, yScl = 20 (e) [1,2] x [-1, 1] with jtScl = 0.1, yScl = 0.1 Solution (a). The window in Figure A.9a has chopped off the portion of the graph that intersects the y-axis, and it shows only two of three possible real roots for the given cubic polynomial. To remedy these problems we need to widen the window in both the x- and y-directions. Solution (b). The window in Figure A.9b shows the intersection of the graph with the y-axis and the three real roots, but it has chopped off the portion of the graph between the two positive roots. Moreover, the ticks in the y-direction are nearly illegible because they are so close together. We need to extend the window in the negative y-direction and increase yScl. We do not know how far to extend the window, so some experimentation will be required to obtain what we want. Solution (c). The window in Figure A.9c shows all of the main features of the graph. However, we have some wasted space in the x-direction. We can improve the picture by shortening the window in the jc-direction appropriately. Solution (d). The window in Figure A.9d shows all of the main features of the graph without a lot of wasted space. However, the window does not provide a clear view of the roots. To get a closer view of the roots we must forget about showing all of the main features of the graph and choose windows that zoom in on the roots themselves. Solution (e). The window in Figure A.9e displays very little of the graph, but it clearly shows that the root in the interval [1, 2] is approximately 1.3. •<
Appendix A: Graphing Functions Using Calculators and Computer Algebra Systems
AS
[-10, 10] x [-10, 10] jtScI = \,ySc\ = 1
[-20, 20] x [-20, 20] jtScI = \,ySc\ = 1
[-20, 20] x [-300, 20] *Scl = l,ySc\ = 20
(a)
(ft)
(C)
[-5, 15] x [-300, 20] xScI = l.ySd = 20
l = 0.1,jScl =0.1
(d)
Figure A.9
(e)
Sometimes you will want to determine the viewing window by choosing the ^-interval and allowing the graphing utility to determine a y-interval that encompasses the maximum and minimum values of the function over the ^-interval. Most graphing utilities provide some method for doing this, so read your documentation to determine how to use this feature. Allowing the graphing utility to determine the ^-interval of the window takes some of the guesswork out of problems like that in part (b) of the preceding example.
TECHNOLOGY MASTERY
• ZOOMING
The process of enlarging or reducing the size of a viewing window is called zooming. If you reduce the size of the window, you see less of the graph as a whole, but more detail of the part shown; this is called zooming in. In contrast, if you enlarge the size of the window, you see more of the graph as a whole, but less detail of the part shown; this is called zooming out. Most graphing calculators provide menu items for zooming in or zooming out by fixed factors. For example, on one calculator the amount of enlargement or reduction is controlled by setting values for two zoom factors, denoted by xFact and yFact. If jcFact — 10 and
yFact = 5
then each time a zoom command is executed the viewing window is enlarged or reduced by a factor of 10 in the jc-direction and a factor of 5 in the y-direction. With computer programs such as Mathematica and Maple, zooming is controlled by adjusting the ;e-interval and y-interval directly; however, there are ways to automate this by programming.
[-5, 5] x [-1000, 1000] *Scl = l,}-Scl = 500
(a)
•
COMPRESSION
Enlarging the viewing window for a graph has the geometric effect of compressing the graph, since more of the graph is packed into the calculator screen. If the compression is sufficiently great, then some of the detail in the graph may be lost. Thus, the choice of the viewing window frequently depends on whether you want to see more of the graph or more of the detail. Figure A. 10 shows two views of the equation [-5, 5]x[-10, 10] *Scl = l.vScI = 1 A Figure A.10
y = x5(x-2) In part (a) of the figure the y-interval is very large, resulting in a vertical compression that obscures the detail in the vicinity of the jc-axis. In part (b) the y-interval is smaller, and
A6
Appendix A: Graphing Functions Using Calculators and Computer Algebra Systems
consequently we see more of the detail in the vicinity of the jc-axis but less of the graph in the ^-direction. > Example 5 The function f ( x ) = x + 0.01 sin(507T.r) is the sum of f\ (x) — x, whose graph is the line y = x, and /aOc) — 0.01 sin(507r;c), whose graph is a sinusoidal curve with amplitude 0.01 and period 2jr/50jt — 0.04. This suggests that the graph of f ( x ) will follow the general path of the line y — x but will have bumps resulting from the contributions of the sinusoidal oscillations, as shown in part (c) of Figure A. 11. Generate the four graphs shown in Figure A. 11 and explain why the oscillations are visible only in part (c). Solution. To generate the four graphs, you first need to put your utility in radian mode.* Because the windows in successive parts of the example are decreasing in size by a factor of 10, calculator users can generate successive graphs by using the zoom feature with the zoom factors set to 10 in both the x- and v-directions. (a) In Figure A. 1 la the graph appears to be a straight line because the vertical compression has hidden the small sinusoidal oscillations (their amplitude is only 0.01). (b) In Figure A.life small bumps begin to appear on the line because there is less vertical compression. (c) In Figure A.lie the oscillations have become clear because the vertical scale is more in keeping with the amplitude of the oscillations. (d) In Figure A.I la" the graph appears to be a straight line because we have zoomed in on such a small portion of the curve. -^
[-10, 10] x [-10, 10] xSc\ = \,ySc\ = 1
(a)
, l ] x [-1,1] l = 0.1,;yScl = 0.1 (b)
[-0.1, 0.1] x [-0.1, 0.1] ;tScl = 0.01,yScl =0.01
[-0.01, 0.01] x [-0.01, 0.01] jcScI = 0.001,yScl = 0.001
(c)
(d)
A Figure A.ll ASPECT RATIO DISTORTION
Figure A. 12a shows a circle of radius 5 and two perpendicular lines graphed in the window [-10, 10] x [-10, 10] with *Scl = 1 and yScl = 1. However, the circle is distorted and the lines do not appear perpendicular because the calculator has not used the same length for 1 unit on the x-axis and 1 unit on the y-axis. (Compare the spacing between the ticks on the axes.) This is called aspect ratio distortion. Many calculators provide a menu item for automatically correcting the distortion by adjusting the viewing window appropriately. For example, one calculator makes this correction to the viewing window [-10, 10] x [-10, 10] by changing it to [-16.9970674487,16.9970674487] x [-10, 10] (Figure A.\2b). With computer programs such as Mathematica and Maple, aspect ratio distortion is controlled with adjustments to the physical dimensions of the viewing window on the computer screen, rather than altering the x- and y-intervals of the viewing window. In this text we follow the convention that angles are measured in radians unless degree measure is specified.
Appendix A: Graphing Functions Using Calculators and Computer Algebra Systems
[-10, 10] x [-10, 10] >• Figure A. 12
A7
ArScI = l,ySc\ = 1
[-16.9970674487, 16.9970674487] x [-10, 10] xSc\ = l,ySc\ = 1
(a)
(*)
SAMPLING ERROR
63 Pixels
127 Pixels A viewing window with 63 rows of 127 pixels A Figure A.13
TECHNOLOGY MASTERY If you are using a graphing calculator, read the documentation to determine its resolution.
The viewing window of a graphing utility is composed of a rectangular grid of small rectangular blocks called pixels. For black-and-white displays each pixel has two states— an activated (or dark) state and a deactivated (or light state). A graph is formed by activating appropriate pixels to produce the curve shape. In one popular calculator the grid of pixels consists of 63 rows of 127 pixels each (Figure A. 13), in which case we say that the screen has a resolution of 127 x 63 (pixels in each row x number of rows). A typical resolution for a computer screen is 1280 x 1024. The greater the resolution, the smoother the graphs tend to appear on the screen. The procedure that a graphing utility uses to generate a graph is similar to plotting points by hand: When an equation is entered and a window is chosen, the utility selects the xcoordinates of certain pixels (the choice of which depends on the window being used) and computes the corresponding ^-coordinates. It then activates the pixels whose coordinates most closely match those of the calculated points and uses a built-in algorithm to activate additional intermediate pixels to create the curve shape. This process is not perfect, and it is possible that a particular window will produce a false impression about the graph shape because important characteristics of the graph occur between the computed points. This is called sampling error. For example, Figure A.14 shows the graph of y = cos(lO^jc) produced by a popular calculator in four different windows. (Your calculator may produce different results.) The graph in part (a) has the correct shape, but the other three do not because of sampling error: • In part (b) the plotted pixels happened to fall at the peaks of the cosine curve, giving the false impression that the graph is a horizontal line. • In part (c) the plotted pixels fell at successively higher points along the graph. • In part (d) the plotted points fell in some regular pattern that created yet another misleading impression of the graph shape.
[-1,1] x [-1,1] *Scl=0.5,;yScl = 0
(a)
[-12.6, 12.6] x[-l, 1] ;tScl = l.yScI =0.5
[-12.5, 12.6] x[-l, 1] *Scl = l,>Scl = 0.5
[-6, 6] x [-1,1] *Scl = l.yScI = 0.5
(b)
(c)
(d)
A Figure A.14 REMARK | For trigonometric graphs with rapid oscillations, Figure A.14 suggests that restricting the ^-interval to a few periods is likely to produce a more accurate representation about the graph shape.
A8
Appendix A: Graphing Functions Using Calculators and Computer Algebra Systems FALSE GAPS
Sometimes graphs that are continuous appear to have gaps when they are generated on a calculator. These false gaps typically occur where the graph rises so rapidly that vertical space is opened up between successive pixels. >• Example 6 Figure A. 15 shows the graph of the semicircle y = \/9 — x1 in two viewing windows. Although this semicircle has x -intercepts at the points x = ±3, part (a) of the figure shows false gaps at those points because there are no pixels with jc-coordinates ±3 in the window selected. In part (b) no gaps occur because there are pixels with x -coordinates x = ±3 in the window being used. •*
[-5, 5] x [-5, 5] ;tScl = 1,3-Scl = 1
[-6.3, 6.3] x [-5, 5] A:Scl = !,>>Scl = 1
(a)
(b)
> Figure A.15
[-10,10] x [-10,10] *Scl = l.yScI = 1
FALSE LINE SEGMENTS
y = I/O:- 1) with false line segments
In addition to creating false gaps in continuous graphs, calculators can err in the opposite direction by placing/a/se line segments in the gaps of discontinuous curves. *• Example 7 Figure A.16a shows the graph of y = 1 / ( x — 1) in the default window on a calculator. Although the graph appears to contain vertical line segments near x = 1, they should not be there. There is actually a gap in the curve at x — 1, since a division by zero occurs at that point (Figure A.I6b). •«
ERRORS OF OMISSION
| Actual curve shape of y = \l(x-
1)
A Figure A.16
Most graphing utilities use logarithms to evaluate functions with fractional exponents such as f ( x ) — x2/3 — \fx^. However, because logarithms are only defined for positive numbers, many (but not all) graphing utilities will omit portions of the graphs of functions with fractional exponents. For example, one calculator graphs y — x2^ as in Figure A.17a, whereas the actual graph is as in Figure A.I Ib. (For a way to circumvent this problem, see the discussion preceding Exercise 23.)
TECHNOLOGY MASTERY Determine whether your graphing utility produces the graph of the equation y = x1^ for both positive and negative values of x.
[-4, 4] x [-1, 4]
1
-4 -3 -2 -I
xSd = l , y S c l = 1 > Figure A.17
(a)
(b)
2
3
4
Appendix A: Graphing Functions Using Calculators and Computer Algebra Systems
[-10, 10] x [-10, 10] ;tScl = l,yScl = 1
A7
[-16.9970674487, 16.9970674487] x [-10, 10] *Scl = l,ySc\ = 1
> Figure A.12
(*)
SAMPLING ERROR
The viewing window of a graphing utility is composed of a rectangular grid of small rectangular blocks called pixels. For black-and-white displays each pixel has two states— an activated (or dark) state and a deactivated (or light state). A graph is formed by activating appropriate pixels to produce the curve shape. In one popular calculator the grid of pixels consists of 63 rows of 127 pixels each (Figure A. 13), in which case we say that the screen has a resolution of 127 x 63 (pixels in each row x number of rows). Atypical resolution for a computer screen is 1280 x 1024. The greater the resolution, the smoother the graphs tend to appear on the screen. The procedure that a graphing utility uses to generate a graph is similar to plotting points by hand: When an equation is entered and a window is chosen, the utility selects the xcoordinates of certain pixels (the choice of which depends on the window being used) and computes the corresponding y-coordinates. It then activates the pixels whose coordinates most closely match those of the calculated points and uses a built-in algorithm to activate additional intermediate pixels to create the curve shape. This process is not perfect, and it is possible that a particular window will produce a false impression about the graph shape because important characteristics of the graph occur between the computed points. This is called sampling error. For example, Figure A. 14 shows the graph of y — cos(107r;c) produced by a popular calculator in four different windows. (Your calculator may produce different results.) The graph in part (a) has the correct shape, but the other three do not because of sampling error:
63 Pixels
-127 PixelsA viewing window with 63 rows of 127 pixels A Figure A.13
TECHNOLOGY MASTERY If you are using a graphing calculator, read the documentation to determine its resolution.
• In part (b) the plotted pixels happened to fall at the peaks of the cosine curve, giving the false impression that the graph is a horizontal line. • In part (c) the plotted pixels fell at successively higher points along the graph. • In part (d) the plotted points fell in some regular pattern that created yet another misleading impression of the graph shape. '.
i A
JL
1
<•
\
-:\
'•'.
\\
i";
ill
1 II li ii ) fil ,! : 1 1
i !
i
i iJ
i • !'
: ::
j
! - i
1 1
•!•! III
M i '•
i '. \ l { :. : _ _ • : : :
1 1 1
: ' ! !
MM
!
1
i '•
1 Hi
1 : 111
! 1 ! := = i i i l l ' II 1 J 1 ! ;; 1 i: •' ! !.• II !! I-' :l ! ::
".
'•'•
\
'•
.
r
i • i_
•:
.•
i | j
::
il fi r:
II - 1 1 ii i i ii • i i •- i .!
j i
i
i : • •
! i. i
! :! ,t ||
!i !:. ij
I.
'••'
[-12.6, 12.6] X [-1, 1] jcScI = l,yScl = 0.5
[-12.5, 12.6] x [-1, 1] xScI = l.yScI = 0.5
[-6,6] x [-1,1]
*Scl = 0.5, yScI = 0.5
(a)
00
(C)
(d)
= l,.yScl = 0.5
A Figure A.14 REMARK I For trigonometric graphs with rapid oscillations, Figure A.14 suggests that restricting the ^-interval to I a few periods is likely to produce a more accurate representation about the graph shape.
10.2 Polar Coordinates
705
75. Find the area of the surface generated by revolving x = t2, y = 3? (0 < t < 2) about the x-axis. 76. Find the area of the surface generated by revolving the curve x = e' cos t, y = e' sin t (0 < t < 7T/2) about the *-axis. 77. Find the area of the surface generated by revolving the curve x = cos2 t,y = sin21 (0 < t < n/2) about the y-axis. 78. Find the area of the surface generated by revolving x = 6t, y = 4t2 (0 < t < 1) about the y-axis. 79. By revolving the semicircle (a)
x = rcost,
A Figure Ex-74
75-80 If f ' ( t ) and g'(f) are continuous functions, and if no segment of the curve x = f ( t ) , y = g(t) (a
T)* dt
o
and the area of the surface generated by revolving the curve about the y-axis is
'=/W(£) ya v \ ^ /
+ - -
y = rsint
(0 < t < TI)
about the x -axis, show that the surface area of a sphere of radius r is 47zr2. 80. The equations x = a(j) — a sin 0,
y = a — a cos
(0 < 0 < 2n)
represent one arch of a cycloid. Show that the surface area generated by revolving this curve about the x-axis is given by 5 = 64TO2/3. 81. Writing Consult appropriate reference works and write an essay on American mathematician Nathaniel Bowditch (1773-1838) and his investigation of Bowditch curves (better known as Lissajous curves; see Exercise 55). 82. Writing What are some of the advantages of expressing a curve parametrically rather than in the form y = f ( x ) l
[The derivations are similar to those used to obtain Formulas (4) and (5) in Section 5.5.] Use the formulas above in these exercises.
I/QUICK CHECK ANSWERS 10.1 (Q
2. f ; 2.75 3. x = /(I - t), y = g(l - t) 4.
+ cos 2 1 dt
POLAR COORDINATES Up to now we have specified the location of a point in the plane by means of coordinates relative to two perpendicular coordinate axes. However, sometimes a moving point has a special affinity for some fixed point, such as a planet moving in an orbit under the central attraction of the Sun. In such cases, the path of the panicle is best described by its angular direction and its distance from the fixed point. In this section we will discuss a new kind of coordinate system that is based on this idea. POLAR COORDINATE SYSTEMS
Apolar coordinate system in a plane consists of a fixed point O, called Ihspole (or origin), and a ray emanating from the pole, called the polar axis. In such a coordinate system
706
Chapter 10 / Parametric and Polar Curves; Conic Sections P(r, &)
Pole A Figure 10.2.1
Polar axis
we can associate with each point P in the plane a pair of polar coordinates (r, 9), where r is the distance from P to the pole and 9 is an angle from the polar axis to the ray OP (Figure 10.2.1). The number r is called the radial coordinate of P and the number 9 the angular coordinate (orpolar angle) of P. In Figure 10.2.2, the points (6, n/4), (5, 2n/3), (3, 571/4), and (4, lln/6) are plotted in polar coordinate systems. If P is the pole, then r — 0, but there is no clearly defined polar angle. We will agree that an arbitrary angle can be used in this case; that is, (0,9) are polar coordinates of the pole for all choices of 9. . (5, 27t/3)
(6, 71/4)
A Figure 10.2.2
The polar coordinates of a point are not unique. For example, the polar coordinates (l,7jr/4),
(I,-Ti/4),
and (1, 15;r/4)
all represent the same point (Figure 10.2.3). 157T/4
77T/4
(1,15*74)
> Figure 10.2.3
In general, if a point P has polar coordinates (r, &), then
(r, 0 + 2nn)
5TC/4 Polar axis
(-3,7i/4) A Figure 10.2.4
and (r, 6 - Imt)
are also polar coordinates of P for any nonnegative integer n. Thus, every point has infinitely many pairs of polar coordinates. As defined above, the radial coordinate r of a point P is nonnegative, since it represents the distance from P to the pole. However, it will be convenient to allow for negative values of r as well. To motivate an appropriate definition, consider the point P with polar coordinates (3, 5n/4). As shown in Figure 10.2.4, we can reach this point by rotating the polar axis through an angle of 5n/4 and then moving 3 units from the pole along the terminal side of the angle, or we can reach the point P by rotating the polar axis through an angle of 7T/4 and then moving 3 units from the pole along the extension of the terminal side. This suggests that the point (3, 5n/4) might also be denoted by (—3, jr/4), with the minus sign serving to indicate that the point is on the extension of the angle's terminal side rather than on the terminal side itself. In general, the terminal side of the angle 9 + JT is the extension of the terminal side of 9, so we define negative radial coordinates by agreeing that
(-r, 9)
and
(r, 9 + n)
are polar coordinates of the same point. RELATIONSHIP BETWEEN POLAR AND RECTANGULAR COORDINATES
Frequently, it will be useful to superimpose a rectangular jcv-coordinate system on top of a polar coordinate system, making the positive *-axis coincide with the polar axis. If this is done, then every point P will have both rectangular coordinates ( x , y ) and polar coordinates
10.2 Polar Coordinates
707
(r, 6). As suggested by Figure 10.2.5, these coordinates are related by the equations x = rcos9,
y = rsin9
(1)
These equations are well suited for finding x and y when r and 9 are known. However, to find r and 9 when x and y are known, it is preferable to use the identities sin2 6 + cos2 9 = 1 and tan 0 — sin $/ cos 6 to rewrite (1) as 1
y
x = r cos 0
tan 9 = -
(2)
A Figure 10.2.5
> Example 1 Find the rectangular coordinates of the point P whose polar coordinates are (r, 9) = (6, 2jr/3) (Figure 10.2.6). Solution.
A Figure 10.2.6
Substituting the polar coordinates r = 6 and 9 = 2n/3 in (1) yields
Thus, the rectangular coordinates of P are (x, y) = (—3, 3\/3). 7T/2
> Example 2 Find polar coordinates of the point P whose rectangular coordinates are (-2, -273) (Figure 10.2.7). -»— 0
Solution. We will find the polar coordinates (r, 9) of P that satisfy the conditions r > 0 and 0 < 9 < 2n. From the first equation in (2), r2 = x2 + y2 = (-2)2 + (-273 ) 2 = 4 + 12 = 16
so r = 4. From the second equation in (2), />(-2,-2V3) '
-273 = 73 —2
y
A Figure 10.2.7
From this and the fact that (—2, — 2V3) lies in the third quadrant, it follows that the angle satisfying the requirement 0 < 9 < In is 9 = 4n/3. Thus, (r, 9) = (4, 4n/3) are polar coordinates of P. All other polar coordinates of P are expressible in the form [4, — + Inn)
or (-4, - + 2njr\
where n is an integer. GRAPHS IN POLAR COORDINATES
We will now consider the problem of graphing equations in r and 9, where 9 is assumed to be measured in radians. Some examples of such equations are r=l,
0=n/4,
r = 9,
r = sin#,
r = cos26>
In a rectangular coordinate system the graph of an equation in x and y consists of all points whose coordinates (x, y) satisfy the equation. However, in a polar coordinate system, points have infinitely many different pairs of polar coordinates, so that a given point may have some polar coordinates that satisfy an equation and others that do not. Given an equation
708
Chapter 10 / Parametric and Polar Curves; Conic Sections
in r and 8, we define its graph in polar coordinates to consist of all points with at least one pair of coordinates (r, 9) that satisfy the equation. > Example 3
Sketch the graphs of (a) r = 1
in polar coordinates. Solution (a). For all values of 9, the point ( 1 , 6 ) is 1 unit away from the pole. Since 9 is arbitrary, the graph is the circle of radius 1 centered at the pole (Figure 10.2.8a). Solution (b). For all values of r, the point (r, n/4) lies on a line that makes an angle of ji/4 with the polar axis (Figure 10.2.8&). Positive values of r correspond to points on the line in the first quadrant and negative values of r to points on the line in the third quadrant. Thus, in absence of any restriction on r, the graph is the entire line. Observe, however, that had we imposed the restriction r > 0, the graph would have been just the ray in the first quadrant. A
> Figure 10.2.8
Equations r = f ( 9 ) that express r as a function of 9 are especially important. One way to graph such an equation is to choose some typical values of 9, calculate the corresponding values of r, and then plot the resulting pairs (r, 9) in a polar coordinate system. The next two examples illustrate this process. Example 4
A Figure 10.2.9
Graph the spiral r — 6 (0 < 0). Compare your graph to that in Figure 10.2.9.
Sketch the graph of r
> 0) in polar coordinates by plotting points.
Solution. Observe that as 9 increases, so does r; thus, the graph is a curve that spirals out from the pole as 9 increases. A reasonably accurate sketch of the spiral can be obtained by plotting the points that correspond to values of 9 that are integer multiples of n/2, keeping in mind that the value of r is always equal to the value of 9 (Figure 10.2.9). -^ > Example 5 points.
Sketch the graph of the equation r = sin 9 in polar coordinates by plotting
Solution. Table 10.2.1 shows the coordinates of points on the graph at increments of 71/6.
These points are plotted in Figure 10.2.10. Note, however, that there are 13 points listed in the table but only 6 distinct plotted points. This is because the pairs from 9 = n on yield
10.2
Polar Coordinates
709
duplicates of the preceding points. For example, (—1/2, ln/6) and (1/2, n/6) represent the same point. -^
Table 10.2.1
e
n 6
JT 3
I
Vs"
V5"
i
2
2
2
(~"f)
(l'f)
(RADIANS)
0
r = sin d
o
2
<,.«
(0,0)
f|'|)
(2 ' 3 )
JT 2
('' f )
2^ 3
5s 6
1
7w "6~
o
-1 1 2
47C
17C
5)r
3
2
3
UJT 6
V3 2
2
V3~ 2
6/
/ V3~ 4;r\ ("T' T)
Cr1i' 32M) /r ^2
1
2n
Q
57rl
y)i
(-!' T)
(0, 2s)
Observe that the points in Figure 10.2.10 appear to lie on a circle. We can confirm that this is so by expressing the polar equation r = sin 6 in terms of x and y. To do this, we multiply the equation through by r to obtain
r 2 = r sin 9 which now allows us to apply Formulas (1) and (2) to rewrite the equation as
x2 + y2 = y Rewriting this equation as x2 + y2 — y = 0 and then completing the square yields A Figure 10.2.10
r = sinfl
which is a circle of radius ^ centered at the point (0, {) in the xy-plane. It is often useful to view the equation r — f(9) as an equation in rectangular coordinates (rather than polar coordinates) and graphed in a rectangular 0r-coordinate system. For example, Figure 10.2.11 shows the graph of r = sin 6* displayed using rectangular 9rcoordinates. This graph can actually help to visualize how the polar graph in Figure 10.2. 10 is generated: • At 9 — 0 we have r = 0, which corresponds to the pole (0, 0) on the polar graph. • As 9 varies from 0 to n/2, the value of r increases from 0 to 1, so the point (r, 9) moves along the circle from the pole to the high point at (1 , n/2). • As 9 varies from jt/2 to n, the value of r decreases from 1 back to 0, so the point (r, 9) moves along the circle from the high point back to the pole. • As 9 varies from n to 3jr/2, the values of r are negative, varying from 0 to — 1 . Thus, the point (r, 9) moves along the circle from the pole to the high point at (1, n/2), which is the same as the point (— 1, 3n/2). This duplicates the motion that occurred for 0 < 9 < n/2. •
As 9 varies from 3n/2 to 2n, the value of r varies from — 1 to 0. Thus, the point (r, 9) moves along the circle from the high point back to the pole, duplicating the motion that occurred for n/2 <9
> Examples
Sketch the graph of r — cos 20 in polar coordinates.
Solution. Instead of plotting points, we will use the graph of r = cos 26 in rectangular coordinates (Figure 10.2.12) to visualize how the polar graph of this equation is generated. The analysis and the resulting polar graph are shown in Figure 10.2.13. This curve is called a four-petal rose. ^
710
Chapter 10 / Parametric and Polar Curves; Conic Sections
r varies from 1 to 0 as e varies from 0 to Ti/4.
r varies from 0 to -1 as 6 varies from
r varies from 0 to 1 as G varies from 3rc/4 to n.
71/4 tO n!2.
r varies from ; i r varies from
1 to 0 as 0
0 to -1 as 0
varies from j ] varies from : n to 571/4. 5n/4 to 37T/2. i
r varies from ; -1 to 0 as 0 j varies from ; 37T/2 to 7/T/4. j
1
r varies from '• 0 to 1 as varies from
A Figure 10.2.13
SYMMETRY TESTS
Observe that the polar graph of r — cos 28 in Figure 10.2.13 is symmetric about the x-axis and the y-axis. This symmetry could have been predicted from the following theorem, which is suggested by Figure 10.2.14 (we omit the proof).
10.2.1
THEOREM (Symmetry Tests)
(a) A curve in polar coordinates is symmetric about the x-axis if replacing 6 by —6 in its equation produces an equivalent equation (Figure 10.2.14a). The converse of each part of Theorem 10.2.1 is false. See Exercise 79.
(b) A curve in polar coordinates is symmetric about the y-axis if replacing 9 by n — 9 in its equation produces an equivalent equation (Figure 10.2.14b). (c) A curve in polar coordinates is symmetric about the origin if replacing 6 by 0 + 7i, or replacing r by ~r in its equation produces an equivalent equation (Figure 10.2.14c).
nil
nil
nil
0,0)
(r,B)
' (r,
(r, -e
(-r, 0) (a)
> Figure 10.2.14
(b)
(c)
> Example 7 Use Theorem 10.2.1 to confirm that the graph of r = cos 29 in Figure 10.2.13 is symmetric about the x-axis and y-axis. Solution, A graph that is symmetric about both the z-axis and the y-axis is also symmetric about the origin. Use Theorem 10.2.1(c) to verify that the curve in Example 7 is symmetric about the origin.
To test for symmetry about the x-axis, we replace 9 by -0. This yields r = cos(-20) = cos 29
Thus, replacing 9 by —9 does not alter the equation. To test for symmetry about the y-axis, we replace 9 by n — 9. This yields r - cos2(;r -9) = cos(2n - 29) = cos(-2#) = cos 29 Thus, replacing 9 by n — 9 does not alter the equation. •«!
10.2 Polar Coordinates
711
*• Example 8 Sketch the graph of r — a(l — cos 9) in polar coordinates, assuming a to be a positive constant. Solution. Observe first that replacing 9 by -9 does not alter the equation, so we know in advance that the graph is symmetric about the polar axis. Thus, if we graph the upper half of the curve, then we can obtain the lower half by reflection about the polar axis. As in our previous examples, we will first graph the equation in rectangular 8 r -coordinates. This graph, which is shown in Figure 10.2.15a, can be obtained by rewriting the given equation as r = a - a cos 9, from which we see that the graph in rectangular 9r -coordinates can be obtained by first reflecting the graph of r — a cos 9 about the jc-axis to obtain the graph of r = —a cos 9, and then translating that graph up a units to obtain the graph of r — a — a cos 9. Now we can see the following: • As 0 varies from 0 to n/3, r increases from 0 to a/2. • As 9 varies from n/3 to n/2, r increases from a/2 to a. • As 9 varies from n/2 to 2n/3, r increases from a to 3a/2. • As 9 varies from 2n/3 to n, r increases from 3a/2 to 2a. This produces the polar curve shown in Figure 10.2.15b. The rest of the curve can be obtained by continuing the preceding analysis from n to 2n or, as noted above, by reflecting the portion already graphed about the x-axis (Figure 10.2.15c). This heart-shaped curve is called a cardioid (from the Greek word kardia meaning "heart" ). -4
la 3a 2 a a 2
j
>"
A
2
I
71 n2n 3 2 3
\. \
I
__™™™™__\™__________™.
71
V, ?
(la, n)
In
(a)
(c)
A Figure 10.2.15
*• Example 9
Sketch the graph of r2 = 4 cos 29 in polar coordinates.
Solution. This equation does not express r as a function of 9, since solving for r in terms of 9 yields two functions: r = 2vcos 29 2
and
r — -2Vcos20
Thus, to graph the equation r = 4 cos 29 we will have to graph the two functions separately and then combine those graphs. We will start with the graph of r = 2\/cos29. Observe first that this equation is not changed if we replace 9 by —9 or if we replace 9 by n — 9. Thus, the graph is symmetric about the ;t-axis and the y-axis. This means that the entire graph can be obtained by graphing the portion in the first quadrant, reflecting that portion about the y-axis to obtain the portion in the second quadrant, and then reflecting those two portions about the Jt-axis to obtain the portions in the third and fourth quadrants.
712
Chapter 10 / Parametric and Polar Curves; Conic Sections
To begin the analysis, we will graph the equation r = 2\/cos 20 in rectangular Orcoordinates (see Figure 10.2.16a). Note that there are gaps in that graph over the intervals jr/4 < 6 < 3n/4 and 5n/4 < 9 < In I'4 because cos 29 is negative for those values of 6. From this graph we can see the following: • As 9 varies from 0 to n/4, r decreases from 2 to 0. • As 0 varies from n/4 to jr/2, no points are generated on the polar graph. This produces the portion of the graph shown in Figure 10.2.l6b. As noted above, we can complete the graph by a reflection about the y-axis followed by a reflection about the x-axis (Figure 10.2.16c). The resulting propeller-shaped graph is called a lemniscate (from the Greek word lemniscos for a looped ribbon resembling the number 8). We leave it for you to verify that the equation r = 2T/cos29 has the same graph as r = —2Vcos26>, but traced in a diagonally opposite manner. Thus, the graph of the equation r2 = 4 cos 29 consists of two identical superimposed lemniscates. •<
nil
4
4
4
4
(a)
(b)
A Figure 10.2.16
FAMILIES OF LINES AND RAYS THROUGH THE POLE
If #o is a fixed angle, then for all values of r the point (r, OQ) lies on the line that makes an angle of 9 = 9o with the polar axis; and, conversely, every point on this line has a pair of polar coordinates of the form (r, 00)- Thus, the equation 9 = 90 represents the line that passes through the pole and makes an angle of #o with the polar axis (Figure 10.2.17
We will consider three families of circles in which a is assumed to be a positive constant: r = 2a cos 6
(*) A Figure 10.2.17
r = 2a sin 9
(3-5)
The equation r = a represents a circle of radius a centered at the pole (Figure 10.2.18a). Thus, as a varies, this equation produces a family of circles centered at the pole. For families (4) and (5), recall from plane geometry that a triangle that is inscribed in a circle with a diameter of the circle for a side must be a right triangle. Thus, as indicated in Figures 10.2.18& and 10.2.18c, the equation r = 2a cos 9 represents a circle of radius a, centered on the jc-axis and tangent to the y-axis at the origin; similarly, the equation r — 2a sin 9 represents a circle of radius a, centered on the y-axis and tangent to the ;c-axis at the origin. Thus, as a varies, Equations (4) and (5) produce the families illustrated in Figures 10.2.18^ and 10.2.18e.
10.2 Polar Coordinates
713
nil
7T/2
(a)
(e)
A Figure 10.2.18 •
FAMILIES OF ROSE CURVES
In polar coordinates, equations of the form Observe that replacing 9 by —9 does not change the equation r = la cos 0
r = a sin n6
(6-7)
r = a cos nO
and that replacing 9 by JT — 9 does not change the equation r = la sin 9. This explains why the circles in Figure 10.2.18d are symmetric about the x-axis and those in Figure 10.2.18e are symmetric about the j>-axis.
in which a > 0 and n is a positive integer represent families of flower-shaped curves called roses (Figure 10.2.19). The rose consists of n equally spaced petals of radius a if n is odd and 2n equally spaced petals of radius a if n is even. It can be shown that a rose with an even number of petals is traced out exactly once as 9 varies over the interval 0 < 9 < 2n and a rose with an odd number of petals is traced out exactly once as 9 varies over the interval 0 < 9 < n (Exercise 78). A four-petal rose of radius 1 was graphed in Example 6.
What do the graphs of one-petal roses look like? ROSE CURVES
n =2
r = a sin nd
n =6
0
o
r = a cos n9
M A Figure 10.2.19
FAMILIES OF CARDIOIDS AND LIMACONS
Equations with any of the four forms
r = a ± b sin 9
r = a±b cos 9
(8-9)
in which a > 0 and b > 0 represent polar curves called limagons (from the Latin word Umax for a snail-like creature that is commonly called a "slug"). There are four possible shapes for a limagon that are determined by the ratio a/b (Figure 10.2.20). If a — b (the case a/b = 1), then the lima§on is called a cardioid because of its heart-shaped appearance, as noted in Example 8.
714
Chapter 10 / Parametric and Polar Curves; Conic Sections
alb< 1
alb= 1
1 < alb < 2
alb > 2
Dimpled limagon
Convex limagon
• Figure 10.2.20
K Example 10 Figure 10.2.21 shows the family of limagons r = a +cos0 with the constant a varying from 0.25 to 2.50 in steps of 0.25. In keeping with Figure 10.2.20, the limagons evolve from the loop type to the convex type. As a increases from the starting value of 0.25, the loops get smaller and smaller until the cardioid is reached at a = 1. As a increases further, the lima§ons evolve through the dimpled type into the convex type. -4
a = 0.25
a = 0.5
a = 0.75
a=l
a = 1.25
a = 1.5
a = 1.75
a=2
a = 2.25
a = 2.5
A Figure 10.2.21
FAMILIES OF SPIRALS
A spiral is a curve that coils around a central point. Spirals generally have "left-hand" and "right-hand" versions that coil in opposite directions, depending on the restrictions on the polar angle and the signs of constants that appear in their equations. Some of the more common types of spirals are shown in Figure 10.2.22fornonnegative values of 6, a, andb. nil
Lituus spiral A Figure 10.2.22
SPIRALS IN NATURE
Spirals of many kinds occur in nature. For example, the shell of the chambered nautilus (below) forms a logarithmic spiral, and a coiled sailor's rope forms an Archimedean spiral. Spirals also occur in flowers, the tusks of certain animals, and in the shapes of galaxies.
10.2 Polar Coordinates
715
Thomas Taylor/Photo Researchers
Rex Ziak/Stone/Getty Images
Courtesy NASA & The Hubble Heritage Team
The shell of the chambered nautilus reveals a logarithmic spiral. The animal lives in the outermost chamber.
A sailor's coiled rope forms an Archimedean spiral,
A spiral galaxy.
GENERATING POLAR CURVES WITH GRAPHING UTILITIES
For polar curves that are too complicated for hand computation, graphing utilities can be used. Although many graphing utilities are capable of graphing polar curves directly, some are not. However, if a graphing utility is capable of graphing parametric equations, then it can be used to graph a polar curve r = f ( 6 ) by converting this equation to parametric form. This can be done by substituting /(0) for r in (1). This yields jr; = /(0)cos0,
y = /(0)sin0
(10)
which is a pair of parametric equations for the polar curve in terms of the parameter 9.
> Example 11
Express the polar equation
se
r = 2 + cos — parametrically, and generate the polar graph from the parametric equations using a graphing utility. Solution. Substituting the given expression for r in x = r cos 6 and y = r sin 6 yields the parametric equations 50' 501 y — 2 + cos — sin 6
501 2 + cos — cos 0,
TECHNOLOGY MASTERY Use a graphing utility to duplicate the curve in Figure 10.2.23. If your graphing utility requires that t be used as the parameter, then you will have to replace 0 by ( in (10) to generate the graph.
T
Next, we need to find an interval over which to vary 0 to produce the entire graph. To find such an interval, we will look for the smallest number of complete revolutions that must occur until the value of r begins to repeat. Algebraically, this amounts to finding the smallest positive integer n such that 2 +cos
5(0 + 2«7T)
= 2 + cos
or
50
r
\
50
cos — + 5nn\ = cos —
50
716
Chapter 10 / Parametric and Polar Curves; Conic Sections
For this equality to hold, the quantity 5nn must be an even multiple of n; the smallest n for which this occurs is n = 2. Thus, the entire graph will be traced in two revolutions, which means it can be generated from the parametric equations 5(91 f 5(9"1 56» I cos9, y = \ 2 + ccos o s — | sin# (0 < 0 < 4n) sm 9 I 2 +cos— 2 J This yields the graph in Figure 10.2.23. -4
A Figure 10.2.23
I/QUICK CHECK EXERCISES 10.2
(See page 719 for answers.)
1. (a) Rectangular coordinates of a point (x, y) may be recovered from its polar coordinates (r, 9) by means of the equations x = and y = (b) Polar coordinates (r, 0) may be recovered from rectangular coordinates (x, y) by means of the equations r2 = and tan 9 = 2. Find the rectangular coordinates of the points whose polar coordinates are given. (a) (4.7T/3) (b) (2,-7T/6) (c) (6, -2;r/3) EXERCISE SET 10.2
(d) (4,5jr/4)
3. In each part, find polar coordinates satisfying the stated conditions for the point whose rectangular coordinates are (a) r > 0 and 0 < 6 < 2rc (b) r < 0 and 0 < 9 < 2n 4. In each part, state the name that describes the polar curve most precisely: a rose, a line, a circle, a Iima9on, a cardioid, a spiral, a lemniscate, or none of these. (a) r = I - d (b) r = 1 + 2 sin 0 (c) r = sin 20 (d) r = cos2 9 (e) r = csc# (f) r = 2 + 2cos6> (g) r = -2sin<9
Graphing Utility
1 -1 Plot the points in polar coordinates. 1. (a) (3.7T/4) (b) (5,27r/3) (d) (4,7^/6) (e) (-6,-jr) 2. (a) (2, -jr/3) (b) (3/2, -7;r/4) (d) (-5,-7T/6) (e) (2.47T/3)
(c) (f) (c) (f)
(l,jr/2) (-l,9;r/4) (-3, 3;r/2) (0, TT)
3-4 Find the rectangular coordinates of the points whose polar coordinates are given. 3. (a) (6,71/6) (c) (-6,-5;r/6) (b) (7.27T/3) (d) (0, -TT) (e) (7, 177T/6) (f) (-5,0) 4. (a) (-2,7r/4) (b) (6, -jr/4) (c) (4,9jr/4) (e) (-4, -37T/2) (f) (0, 3ir) (d) (3,0) 5. In each part, a point is given in rectangular coordinates. Find two pairs of polar coordinates for the point, one pair satisfying r > 0 and 0 < 9 < 2n, and the second pair satisfying r > 0 and -2n < 9 < 0. (a) (-5,0) (b) (2V3, -2) (c) (0, -2) (d) (-8,-8) (e) (-3,373) (f) (1,1) 6. In each part, find polar coordinates satisfying the stated conditions for the point whose rectangular coordinates are (-V3, I).
(a) r > 0 (b) r < 0 (c) r > 0 (d) r < 0
and 0 < 9 < 2n and 0 < 0 < 2n and -2;r < 9 < 0 and -n < 9 < n
7-8 Use a calculating utility, where needed, to approximate the polar coordinates of the points whose rectangular coordinates are given. 7. (a) (3,4) (b) (6,-8) (c) (-1, tan"1 1) 8. (a) (-3,4) (b) (-3, 1.7) (c) (2, siiT'i) 9-10 Identify the curve by transforming the given polar equation to rectangular coordinates. 9. (a) r = 2 (b) r sin 0 = 4 f, (c) r = 3 cos e (d) r = 3 cos 9 + 2 sin 9 10. (a) r = 5 sec 6» (b) r = 2 sin 9 (c) r = 4 cos 0 + 4 sin 9 (d) r = sec Q tan 9 11-12 Express the given equations in polar coordinates. 11. (a) x = 3 (b) x2 + y2 = 7 2 2 (c) x + y + 6y = 0 (d) 9xy = 4
10.2 Polar Coordinates
12. (a) y = -3 (c) x2 + y2 + 4x = 0
(b) x2 + y1 = 5 (d) X2(x2 + y2)
27. r = 3(1+ sinfl)
28. r = 5-5sin0
29. r = 4 - 4 c o s 0
30. r = 1 + 2 sin 0
31. r = -1 -cos0
32. r = 4 + 3 cos 0
FOCUS ON CONCEPTS
33. r = 3 — sin 0
34. r = 3 + 4cos0
13-16 A graph is given in a rectangular 0r-coordinate system. Sketch the corresponding graph in polar coordinates.
35. r - 5 = 3 sin 9
36. r = 5 -2cos0
37. r = -3 - 4 sin 9
38. r2 = cos 29
13.
39. r 2 = 16 sin 29
40. r = 40
41. r = 40
42. r = 40
3 -/•*-
(0 < 0)
717
(0 > 0)
43. r = -2 cos 20
44. r = 3 sin 20
45. r = 9 sin 40
46. r = 2 cos 30
47-50 True-False Determine whether the statement is true or false. Explain your answer. 47. The polar coordinate pairs (—1, :r/3) and (1, —2jr/3) describe the same point.
2
4
48. If the graph of r = /(0) drawn in rectangular 0rcoordinates is symmetric about the r-axis, then the graph of r = f ( 9 ) drawn in polar coordinates is symmetric about the jc-axis.
6
17-20 Find an equation for the given polar graph. [Note: Numeric labels on these graphs represent distances to the origin.]
49. The portion of the polar graph of r = sin 20 for values of 0 between jr/2 and JT is contained in the second quadrant.
17. (a)
50. The graph of a dimpled limac,on passes through the polar origin. | 51-55 Determine a shortest parameter interval on which a complete graph of the polar equation can be generated, and then use a graphing utility to generate the polar graph. Circle
Circle
18. (a)
Cardioid
(b)
(c)
CN
^ (\
19(
Limacon
"
Three-petal rose
Four-petal rose 20. (a)
Limagon
x—4-~%- (b)
Lemniscate
„
(c)
Five-petal rose
r\
r\
51. r = cos — 2
52. r = sin -
0 53. r = 1 - 2 sin 4 0 55. r = cos -
0 54. r = 0.5 + cos -
56. The accompanying figure shows the graph of the "butterfly curve" a r = e c o s e -2cos40 + sin 3 4 Determine a shortest parameter interval on which the complete butterfly can be generated, and then check your answer using a graphing utility.
4
Circle
21-46 Sketch the curve in polar coordinates.
21.0 = -
22. 0 = ~
23. r = 3
24. r = 4cos0
25. r = 6sin0
26. r - 2 = 2cos0
•< Figure Ex-56
718
Chapter 10 / Parametric and Polar Curves; Conic Sections
57. The accompanying figure shows the Archimedean spiral r = 9/2 produced with a graphing calculator. (a) What interval of values for 9 do you think was used to generate the graph? (b) Duplicate the graph with your own graphing utility.
61-62 A polar graph of r = f ( 9 ) is given over the stated interval. Sketch the graph of (a) r = f(-6) (b) r = (d) r = -f(0).
(c) r = f(e +
61. 0 < 9 < 7T/2
62. 71/2 < 0 < 7T
7C/2
(1, 37T/4)
[-9, 9] x [-6, 6] *Scl = l.yScI = 1
•4 Figure Ex-57
58. Find equations for the two families of circles in the accompanying figure.
A Figure Ex-61
A Figure Ex-62
63-64 Use the polar graph from the indicated exercise to sketch the graph of
(a) r = f(9) + 1 63. Exercise 61
A Figure Ex-58
59. (a) Show that if a varies, then the polar equation r = a sec 9
(-?r/2 < 9 < jt/2)
describes a family of lines perpendicular to the polar axis, (b) Show that if b varies, then the polar equation r = bcsc9
(0 < 9 < Tt)
describes a family of lines parallel to the polar axis. FOCUS ON CONCEPTS
60. The accompanying figure shows graphs of the Archimedean spiral r = 6 and the parabolic spiral r = *J(j. Which is which? Explain your reasoning.
(b) r = 2f(9) - 1. 64. Exercise 62
65. Show that if the polar graph of r = f ( 9 ) is rotated counterclockwise around the origin through an angle a, then r = f(9 — a) is an equation for the rotated curve. [Hint: If (/o, 60) is any point on the original graph, then (ro, $o + a ) is a point on the rotated graph.] 66. Use the result in Exercise 65 to find an equation for the lemniscate that results when the lemniscate in Example 9 is rotated counterclockwise through an angle of n/2. 67. Use the result in Exercise 65 to find an equation for the cardioid r = 1 + cos 9 after it has been rotated through the given angle, and check your answer with a graphing utility. (a) n-
(b) |
(c) ji
(d) y
68. (a) Show that if A and B are not both zero, then the graph of the polar equation
r = A sin 0 + B cos 9 is a circle. Find its radius. (b) Derive Formulas (4) and (5) from the formula given in part (a). 69. Find the highest point on the cardioid r = 1 + cos 9. 70. Find the leftmost point on the upper half of the cardioid r = 1 +cos6>. 71. Show that in a polar coordinate system the distance d between the points (r\, 9\) and fa, $2) is - 2r,r 2 cos(<9] - 02) 72-74 Use the formula obtained in Exercise 71 to find the distance between the two points indicated in polar coordinates.
A Figure Ex-60
72. (3,7T/6) and (2,7T/3)
10.3 Tangent Lines, Arc Length, and Area for Polar Curves
73. Successive tips of the four-petal rose r = cos 29. Check your answer using geometry. 74. Successive tips of the three-petal rose r = sin 39. Check your answer using trigonometry. 75. In the late seventeenth century the Italian astronomer Giovanni Domenico Cassini (1625-1712) introduced the family of curves (x1 + y2 + a2)2 -b4 - 4a2x2 = 0
(a > 0, b > 0)
in his studies of the relative motions of the Earth and the Sun. These curves, which are called Cassini ovals, have one of the three basic shapes shown in the accompanying figure. (a) Show that if a = b, then the polar equation of the Cassini oval is r2 = 2a2 cos 29, which is a lemniscate. (b) Use the formula in Exercise 71 to show that the lemniscate in part (a) is the curve traced by a point that moves in such a way that the product of its distances from the polar points (a, 0) and (a, n) is a2.
719
76. Show that the hyperbolic spiral r = 1 /9 (9 > 0) has a horizontal asymptote at y = 1 by showing that y —>• 1 and x —>• +co as 9 —>• 0 + . Confirm this result by generating the spiral with a graphing utility. 77. Show that the spiral r = I/O2 does not have any horizontal asymptotes. 78. Prove that a rose with an even number of petals is traced out exactly once as 9 varies over the interval 0 < 9 < 2n and a rose with an odd number of petals is traced out exactly once as 9 varies over the interval 0 < 9 < n. 79. (a) Use a graphing utility to confirm that the graph of r = 2 - sin (61/2) (0 < 9 < 4n) is symmetric about the x -axis. (b) Show that replacing 9 by —9 in the polar equation r = 2 — sin($/2) does not produce an equivalent equation. Why does this not contradict the symmetry demonstrated in part (a)? 80. Writing Use a graphing utility to investigate how the family of polar curves r = 1 + a cos n6 is affected by changing the values of a and n, where a is a positive real number and n is a positive integer. Write a brief paragraph to explain your conclusions. 81. Writing Why do you think the adjective "polar" was chosen in the name "polar coordinates"?
a
A Figure Ex-75
76-77 Vertical and horizontal asymptotes of polar curves can sometimes be detected by investigating the behavior of x = r cos 9 and y = r sin 9 as 9 varies. This idea is used in these exercises.
I/QUICK CHECK ANSWERS 10.2 1. (a) rcos0; rsin6» (b) x2 + y2; y/x 2. (a) (2,273) (b) (V3,-1) (c) (-3, -3
TANGENT LINES, ARC LENGTH, AND AREA FOR POLAR CURVES In this section we will derive the formulas required to find slopes, tangent lines, and arc lengths of polar curves. We will then show how to find areas of regions that are bounded by polar curves.
TANGENT LINES TO POLAR CURVES
Our first objective in this section is to find a method for obtaining slopes of tangent lines to polar curves of the form r = f ( 9 ) in which r is a differentiable function of 6. We showed in the last section that a curve of this form can be expressed parametrically in terms of the parameter 9 by substituting f ( 0 ) for r in the equations x — r cos 9 and y = r sin 9. This ylddS
x = f ( 9 ) cos 9,
y = f ( 9 ) sin 9
720
Chapter 10 / Parametric and Polar Curves; Conic Sections
from which we obtain — = -/(0) sin 6 + f ' ( 6 ) cos 6 = -r sin 6 + — cos 9 d6 dO
(1) — do Thus, if dx/dO and dy/d9 are continuous and if dx/d0 ^ 0, then y is a differentiable function of x, and Formula (4) in Section 10.1 with 6 in place of t yields
- = f ( 0 ) cos 9 + f ' ( 9 ) sin^ =
dy
dylde
dx
dx/dO
.dr_ r cos 9+sine ~d9 . Q , adr —r sin 9 + cos 0 — d9
(2)
> Example 1 Find the slope of the tangent line to the circle r = 4 cos 0 at the point where 0 = n/4. Solution. From (2) with r = 4 cos 0, so that dr/dO = —4 sin 0, we obtain
Tangent
dy _ 4cos 2 0-4sin 2 0 _ cos20 - sin29 dx —8 sin 0 cos 0 2 sin 0 cos 0 Using the double-angle formulas for sine and cosine, dy cos 20 — = =-cot20 dx sm 20 Thus, at the point where 0 = n/4 the slope of the tangent line is
m—
A Figure 10.3.1
dy dx
= -cot^=0
which implies that the circle has a horizontal tangent line at the point where 9 = 7T/4 (Figure 10.3.1). + > Example 2 Find the points on the cardioid r — 1 — cos 0 at which there is a horizontal tangent line, a vertical tangent line, or a singular point. Solution. A horizontal tangent line will occur where dy/dO = 0 and dx/d9 ^ 0, a vertical tangent line where dy/d& ^ 0 and dx/d9 = 0, and a singular point where dy/d9 = 0 and dxIdO = 0. We could find these derivatives from the formulas in (1). However, an alternative approach is to go back to basic principles and express the cardioid parametrically by substituting r — 1 — cos 0 in the conversion formulas x = r cos 0 and y = r sin 0. This yie
S
x = (1 -cos0)cos0,
y = (l -cos0)sin0
(0 < 0 < 2n)
Differentiating these equations with respect to 0 and then simplifying yields (verify) — = sin 0(2 cos 0-1), dO
— = (l-cos0)(l+2cos0) dO
Thus,dx/d9 — Oif sin0 = Oorcos0 = \, anddy/d9 = Oif cos0 = 1 orcos0 = — i. We leave it for you to solve these equations and show that the solutions of dx/dO = 0 on the interval 0 < 0 < 2n are re 5n -, d9
10.3 Tangent Lines, Arc Length, and Area for Polar Curves
721
and the solutions of dy/dO = 0 on the interval 0 < 9 < 2n are 2n 3
de
4* 3
Thus, horizontal tangent lines occur at 6 = 2n/3 and 0 = 4n/3; vertical tangent lines occur at 0 = jt/3, n, and 5n/3; and singular points occur at 9 — 0 and 0 — 2jt (Figure 10.3.2). Note, however, that r = 0 at both singular points, so there is really only one singular point on the cardioid—the pole. -^
TANGENT LINES TO POLAR CURVES AT THE ORIGIN Formula (2) reveals some useful information about the behavior of a polar curve r = f ( 9 ) that passes through the origin. If we assume that r = 0 and dr/d6 ^ 0 when 9 — OQ, then it follows from Formula (2) that the slope of the tangent line to the curve at 9 = OQ is
A Figure 10.3.2
dy_ dx
0 + sin 6»o 0 + cos 6>o
d0__ dr
cos
= tan 90
(Figure 10.3.3). However, tan 0$ is also the slope of the line 9 = OQ, so we can conclude that this line is tangent to the curve at the origin. Thus, we have established the following result.
71/2
10.3.1 THEOREM If the polar curve r — f ( 9 ) passes through the origin at 0 — 6>o and ifdr/d9 ^ 0 at 9 = 0Q, then the line 9 = 0Q is tangent to the curve at the origin. Slope = tan 00
This theorem tells us that equations of the tangent lines at the origin to the curve r = f ( 9 ) can be obtained by solving the equation /(<9) = 0. It is important to keep in mind, however, that r = f ( 9 ) may be zero for more than one value of 9, so there may be more than one tangent line at the origin. This is illustrated in the next example.
A Figure 10.3.3
6 = 27T/3
nil
8 = Ti/3
> Example 3 The three-petal rose r = sin 39 in Figure 10.3.4 has three tangent lines at the origin, which can be found by solving the equation sin 39 = 0
It was shown in Exercise 78 of Section 10.2 that the complete rose is traced once as 9 varies over the interval 0 < 9 < TC, so we need only look for solutions in this interval. We leave it for you to confirm that these solutions are = 0, 0--.
and
9=
2;r
Since dr/d9 — 3 cos 39 ^ 0 for these values of 9, these three lines are tangent to the rose at the origin, which is consistent with the figure. < A Figure 10.3.4
ARC LENGTH OF A POLAR CURVE A formula for the arc length of a polar curve r — f ( 9 ) can be derived by expressing the curve in parametric form and applying Formula (9) of Section 10.1 for the arc length of a parametric curve. We leave it as an exercise to show the following.
722
Chapter 10 / Parametric and Polar Curves; Conic Sections
10.3.2 ARC LENGTH FORMULA FOR POLAR CURVES If no segment of the polar curve r = f ( 6 ) is traced more than once as 9 increases from a to ft, and if dr/d6 is continuous for a < 0 < ft, then the arc length L from 6 = a to 8 = ft is
-jf
ff>
Ja
Ja
nil
> Example 4 and 6 = it.
/ dr\
= I
p + ( — ) dO V
(3)
\UU /
Find the arc length of the spiral r = ee in Figure 10.3.5 between
=0
Solution. n
(n, e ) A Figure 10.3.5 n
= I Jo > Example 5
- 1) * 31.3
Find the total arc length of the cardioid r — 1 + cos 9.
Solution. The cardioid is traced out once as 9 varies from 9 = 0 to 9 = 2n. Thus, P
/
f2
/dr\
Jr2+( — \ d9= I V \dv J Jo
+cos6dd Id e ntity( 45) A H BR off Appendix
=2 /*27T
r- 1 + cos 0
=2/ Jo
\cos\e\d6
Since cos ^9 changes sign at n, we must split the last integral into the sum of two integrals: the integral from 0 to n plus the integral from it to 2jt. However, the integral from it to lit is equal to the integral from 0 to it, since the cardioid is symmetric about the polar axis (Figure 10.3.6). Thus, L=2 I Jo
A Figure 10.3.6
cos \9 d9 = 4 I cos ± 9 d6 = 8 sin ±0 | = 8 Jo o
AREA IN POLAR COORDINATES
We begin our investigation of area in polar coordinates with a simple case. r=f(6)
10.3.3 AREA PROBLEM IN POLAR COORDINATES Suppose that a and ft are angles that satisfy the condition a < p < a + 2n j and suppose that f ( 9 ) is continuous and nonnegative for a < 9 < ft. Find the area of the region R enclosed by the polar curve r — f ( 9 ) and the rays 9 = a and 9 = ft (Figure 10.3.7).
A Figure 10.3.7
10.3 Tangent Lines, Arc Length, and Area for Polar Curves
723
In rectangular coordinates we obtained areas under curves by dividing the region into an increasing number of vertical strips, approximating the strips by rectangles, and taking a limit. In polar coordinates rectangles are clumsy to work with, and it is better to partition the region into wedges by using rays
A0, A Figure 10.3.8
such that
a < 9\ < 6*2 < • • • < dn-\ < ft
(Figure 10.3.8). As shown in that figure, the rays divide the region R into n wedges with areas A\, A2,..., An and central angles A0], A02, - . . , A0n. The area of the entire region can be written as n A = Al+A2 + - - - + An = J^Ak (4) k=\ /•=/«?)
If A0* is small, then we can approximate the area Ak of the A:th wedge by the area of a sector with central angle A0* and radius /(#£)> where 0 = 0£ is any ray that lies in the A:th wedge (Figure 10.3.9). Thus, from (4) and Formula (5) of Appendix B for the area of a sector, we obtain „ n A = £ Ak « £ £[/(0t*)]2 A0t (5)
A Figure 10.3.9
k=\
k=l
If we now increase n in such a way that max A#t —>• 0, then the sectors will become better and better approximations of the wedges and it is reasonable to expect that (5) will approach the exact value of the area A (Figure 10.3.10); that is,
A —
A Figure 10.3.10
A
lim
max A$i -
Ja
Note that the discussion above can easily be adapted to the case where /(#) is nonpositive for a < 9 < ft. We summarize this result below.
10.3.4 AREA IN POLAR COORDINATES
If a and ft are angles that satisfy the condition
a < ft < a + In and if /(#) is continuous and either nonnegative or nonpositive for a < 0 < ft, then the area A of the region R enclosed by the polar curve r = f ( 9 ) (a < 9 < ft) and the lines 9 — a and 9 = 6 is rP =
I Ja
rfi £[/(0)] 2 <»= / Ja
2
5r
dO
(6)
The hardest part of applying (6) is determining the limits of integration. This can be done as follows: Area in Polar Coordinates: Limits of Integration Step 1. Sketch the region R whose area is to be determined. Step 2. Draw an arbitrary "radial line" from the pole to the boundary curve r — /(#). Step 3. Ask, "Over what interval of values must 0 vary in order for the radial line to sweep out the region /??" Step 4. Your answer in Step 3 will determine the lower and upper limits of integration.
724
Chapter 10 / Parametric and Polar Curves; Conic Sections
> Example 6 r = 1 — cos#. r= 1 - cos 6
Find the area of the region in the first quadrant that is within the cardioid
Solution. The region and a typical radial line are shown in Figure 10.3.11. Forthe radial line to sweep out the region, 9 must vary from 0 to nil. Thus, from (6) with a = 0 and ft = n/2, we obtain fTtll j
A=
Jo
i
i-TC/2
-r2 d8 = 2 2 J0
i
fir/2
(1 -cos6)2 d9 = - / 2 JQ
(1 - 2cos6 + cos2 9) dO
With the help of the identity cos2 0 = |(1 + cos 20), this can be rewritten as
A Figure 10.3.11
2 -1JT/2 1 \ 1 [3 1 Y' 3 2cos6» + -cos29 \dQ = - \ -9 -2sin9 + -sin29 = -jr-l < 2 / 2 2 4 Jo
3
The shaded region is swept out by the radial line as 0 varies from 0 to nil.
2 > Example 7
Find the entire area within the cardioid of Example 6.
Solution. For the radial line to sweep out the entire cardioid, 9 must vary from 0 to 2n. Thus, from (6) with a = 0 and ft = 2n,
A= I Jo
-r2dO = ~l 2 2 JQ
(l-cos0fd9
If we proceed as in Example 6, this reduces to 3n 2
A= -
Alternative Solution. Since the cardioid is symmetric about the jc-axis, we can calculate the portion of the area above the Jt-axis and double the result. In the portion of the cardioid above the jc-axis, 9 ranges from 0 to n, so that
A = 2 I \r2d9= I ( l Jo 2 JQ
=T^
USING SYMMETRY
Although Formula (6) is applicable if r = f ( 9 ) is negative, area computations can sometimes be simplified by using symmetry to restrict the limits of integration to intervals where r > 0. This is illustrated in the next example. > Example 8
Find the area of the region enclosed by the rose curve r = cos 29.
Solution. Referring to Figure 10.2.13 and using symmetry, the area in the first quadrant that is swept out for 0 < 9 < n/4 is one-eighth of the total area inside the rose. Thus, from Formula (6) r-JT/4 i /•7T/4 1
rn/4 /"3T/4
-r2dO =4 I 2 J0
= 8/ Jo
n/4 i
/
cos220d9 /-ir/4
-(l+cos46»)J6> = 2 / 2 JQ /4 r =-sin49 \
i
~
o
( +cos49)dO
10.3 Tangent Lines, Arc Length, and Area for Polar Curves
725
Sometimes the most natural way to satisfy the restriction a < /6 < a + 2n required by Formula (6) is to use a negative value for a. For example, suppose that we are interested in finding the area of the shaded region in Figure 10.3.12a. The first step would be to determine the intersections of the cardioid r = 4 + 4 cos 9 and the circle r = 6, since this information is needed for the limits of integration. To find the points of intersection, we can equate the two expressions for r. This yields 4 + 4 cos 0—6
or
cos 6 = -
which is satisfied by the positive angles 5?r T However, there is a problem here because the radial lines to the circle and cardioid do not sweep through the shaded region shown in Figure 10.3.12b as 9 varies over the interval 7T/3 < 9 < 5:r/3. There are two ways to circumvent this problem—one is to take advantage of the symmetry by integrating over the interval 0 < 0 < n/3 and doubling the result, and the second is to use a negative lower limit of integration and integrate over the interval —7T/3 < 0 < jr/3 (Figure 10.3.12c). The two methods are illustrated in the next example. = - and
nil
nil
nil
7C/2
(a)
A Figure 10.3.12
*• Example 9 Find the area of the region that is inside of the cardioid r = 4 + 4 cos 6 and outside of the circle r — 6. Solution Using a Negative Angle. The area of the region can be obtained by subtracting the areas in Figures 10.3. YLd and 10.3.12e: A=
— (f>Ydd 2
2
I-,13 2
'*irea 'ns't'e cardioid minus area inside circle.
n/3
•£
/32
1
-36]d6
/
(16cos6» + 8cos 2 6»- 10) dO •JT/3
= [16 sin 9 + (49 + 2 sin 29) - 10 <9]*^/3 = 18^3 - 4n Solution Using Symmetry. Using symmetry, we can calculate the area above the polar axis and double it. This yields (verify) fTT/3
A =2 I Jo
-[(4 + 4cos<9) 2 -36]d<9 = 2(9^/3-2:r) = 18^3-4;r 2
which agrees with the preceding result. -4
10.3 Tangent Lines, Arc Length, and Area for Polar Curves
725
Sometimes the most natural way to satisfy the restriction a < ft < a + In required by Formula (6) is to use a negative value for a. For example, suppose that we are interested in finding the area of the shaded region in Figure 10.3.12a. The first step would be to determine the intersections of the cardioid r = 4 + 4 cos 9 and the circle r = 6, since this information is needed for the limits of integration. To find the points of intersection, we can equate the two expressions for r. This yields 4 + 4 cos ( 9 = 6 or
cos6> = -
2
which is satisfied by the positive angles — —
57T
and
T
However, there is a problem here because the radial lines to the circle and cardioid do not sweep through the shaded region shown in Figure W.3.l2b as 9 varies over the interval 7i/3 <9< 5n/3. There are two ways to circumvent this problem—one is to take advantage of the symmetry by integrating over the interval 0 < 6 < n/3 and doubling the result, and the second is to use a negative lower limit of integration and integrate over the interval —n/3 < 9 < n/3 (Figure 10.3.12c). The two methods are illustrated in the next example. nil
(a)
(c)
(d)
(e)
A Figure 10.3.12
> Example 9 Find the area of the region that is inside of the cardioid r = 4 + 4 cos 0 and outside of the circle r — 6. Solution Using a Negative Angle. The area of the region can be obtained by subtracting the areas in Figures 10.3.12d and 10.3.12e:
r i2 = 7-W3
Area inside cardioid minus area inside circle. 1
JT/3 1
/
-.732 2
/•w/3 LV.
f 4cos0r-36]d0 = J-W3 J-jt/3
= [16 sin 0 + (40 + 2 sin 20) - 10 6»]^3/3 = 18V3 - 4n Solution Using Symmetry. Using symmetry, we can calculate the area above the polar axis and double it. This yields (verify) /"fir/3 1
A =2 I Jo
-[(4 + 4cos6») 2 -36]Jc9 = 2(9^3 - 2:r) = 18\/3-47T 2
which agrees with the preceding result. •*
726
Chapter 10 / Parametric and Polar Curves; Conic Sections INTERSECTIONS OF POLAR GRAPHS
In the last example we found the intersections of the cardioid and circle by equating their expressions for r and solving for 9. However, because a point can be represented in different ways in polar coordinates, this procedure will not always produce all of the intersections. For example, the cardioids
r = 1 +cos 8
r = 1 — cos 9
and r = 1 + cos 9
(7)
intersect at three points: the pole, the point (1, ;r/2), and the point (1, 3:r/2) (Figure 10.3.13). Equating the right-hand sides of the equations in (7) yields 1 - cos 9 — 1 + cos 8 or cos 9 = 0, so n 0 = -+kjr, k = 0, ±1,±2, ...
A Figure 10.3.13
Substituting any of these values in (7) yields r = 1, so that we have found only two distinct points of intersection, (1,7T/2) and (1, 3jt/2); the pole has been missed. This problem occurs because the two cardioids pass through the pole at different values of 9—the cardioid r — 1 — cos9 passes through the pole at 0 = 0, and the cardioid r = 1 + cos9 passes through the pole at 9 — n. The situation with the cardioids is analogous to two satellites circling the Earth in intersecting orbits (Figure 10.3.14). The satellites will not collide unless they reach the same point at the same time. In general, when looking for intersections of polar curves, it is a good idea to graph the curves to determine how many intersections there should be.
The orbits intersect, but the satellites do not collide.
A Figure 10.3.14
•QUICK CHECK EXERCISES 10.3
(See page 729 for answers.)
1. (a) To obtain dy/dx directly from the polar equation r = f ( 9 ) , we can use the formula dy dx
dy/d9 dx/dO
(b) Use the formula in part (a) to find dy/dx directly from the polar equation r = esc 6. 2. (a) What conditions on f(0o) and /'(#o) guarantee that the line 9 = OQ is tangent to the polar curve r = f ( 9 ) at the origin?
EXERCISE SET 10.3
B Graphing Utility
(b) What are the values of OQ in [0, 2jt] at which the lines 0 = QQ are tangent at the origin to the four-petal rose r = cos 26>? 3. (a) To find the arc length L of the polar curve r = f ( 9 ) (a < 9 < ft), we can use the formula L = (b) The polar curve r = sec 9 (0 < 6 < jr/4) has arc length L= 4. The area of the region enclosed by a nonnegati ve polar curve r = f(0) (a < 9 < ft) and the lines 9 = a and 9 = ft is given by the definite integral 5. Find the area of the circle r = a by integration.
@ CAS
1 -6 Find the slope of the tangent line to the polar curve for the given value of 9. 1. r = 2 sin 9; 9 = n/6 3. r = \/9; 9 = 2
5. r = sin30; 0 = n/4
2. r = 1 + co&9; 9 = n/2 4. r = a sec 20; 9 = x/6 6. r = 4 - 3 s i n 0 ; 9 = n
7-8 Calculate the slopes of the tangent lines indicated in the accompanying figures. 7. r = 2 + 2 sin 9
8. r = 1 - 2 sin 0
A Figure Ex-7
A Figure Ex-8
9-10 Find polar coordinates of all points at which the polar curve has a horizontal or a vertical tangent line. 9. r = a(l + cos 9) 10. r = a sin 9
10.3 Tangent Lines, Arc Length, and Area for Polar Curves 11-12 Use a graphing utility to make a conjecture about the number of points on the polar curve at which there is a horizontal tangent line, and confirm your conjecture by finding appropriate derivatives. 11. r = sin 9 cos2 9
12. r = 1 - 2 sin 6
13-18 Sketch the polar curve and find polar equations of the tangent lines to the curve at the pole. 13. r = 2cos30
14. r = 4 s i n 6 »
15. r = 4Vcos20
16. r = sin26>
17. r = 1 - 2cos6>
18. r = 29
27. In each part, find the area of the circle by integration. (a) r = 2a sin 9 (b) r = la cos 9 28. (a) Show that r = 1 sin 9 + 2 cos 9 is a circle. (b) Find the area of the circle using a geometric formula and then by integration. 29-34 Find the area of the region described. 29. The region that is enclosed by the cardioid r = 1 + 2 sin &. 30. The region in the first quadrant within the cardioid r = 1 +cos0.
19-22 Use Formula (3) to calculate the arc length of the polar curve.
31. The region enclosed by the rose r = 4 cos 39.
19. The entire circle r = a
33. The region enclosed by the inner loop of the limac,on r — 1 + 2 cos 9. [Hint: r < 0 over the interval of integration.]
20. The entire circle r = 2a cos 0 21. The entire cardioid r = a(l — cos 9) 22. r = e30 from 9 = 0 to 9 = 2 23. (a) Show that the arc length of one petal of the rose r = cos n9 is given by
727
32. The region enclosed by the rose r = 2 sin 29.
34. The region swept out by a radial line from the pole to the curve r = 2/9 as 9 varies over the interval 1 < 9 < 3. 35-38 Find the area of the shaded region.
/* TCI (2n
'•I
(b) Use the numerical integration capability of a calculating utility to approximate the arc length of one petal of the four-petal rose r = cos 29. (c) Use the numerical integration capability of a calculating utility to approximate the arc length of one petal of the n-petal rose r = cosn9 for n = 2, 3, 4, . . . , 20; then make a conjecture about the limit of these arc lengths as n —> +<». 24. (a) Sketch the spiral r = e~0/g (0 < 9 < +00). (b) Find an improper integral for the total arc length of the spiral. (c) Show that the integral converges and find the total arc length of the spiral.
39-46 Find the area of the region described.
25. Write down, but do not evaluate, an integral for the area of each shaded region.
39. The region inside the circle r = 3 sin 9 and outside the cardioid r = 1 + sin6>. 40. The region outside the cardioid r = 2 — 2 cos 9 and inside the circle r = 4. 41. The region inside the cardioid r = 2 + 2 cos 9 and outside the circle r = 3.
r= 1 — cos 9
r = 2 cos 6
• = sin 20
42. The region that is common to the circles r = 2 cos 6 and r = 2 sinfl. 43. The region between the loops of the limac,on r = i + cos 9 . 44. The region inside the cardioid r = 2 + 2 cos 9 and to the right of the line r cos 9 = \ . 45. The region inside the circle r = 2 and to the right of the line
r = 1 - sin 0
r = cos 26
26. Find the area of the shaded region in Exercise 25(d).
46. The region inside the rose r = 2a cos 29 and outside the circle r = a-\/2.
728
Chapter 10 / Parametric and Polar Curves; Conic Sections
47-50 True-False Determine whether the statement is true or false. Explain your answer. 47. The *-axis is tangent to the polar curve r = cos(0/2) at 48. The arc length of the polar curve r = -JO for 0 < 9 < nil is given by fit/2
L
-fJo
49. The area of a sector with central angle 9 taken from a circle of radius r is 9r2.
•4 Figure Ex-54
[c] 55. (a) Show that the Folium of Descartes x3 — 3xy + y3 = 0 can be expressed in polar coordinates as 3 sin 9 cos 0
50. The expression
computes the area enclosed by the inner loop of the Iima9on r =1-
FOCUS ON CONCEPTS
51. (a) Find the error: The area that is inside the lemniscate r2 = a2 cos 29 is
A=
t= 5s
/
1 2J
Jo l
= \a
_
I2* sin 29
/
Jo
=0
Jo (b) Find the correct area. (c) Find the area inside the lemniscate r2 = 4 cos 20 and outside the circle r = \/2.
cos3 9 + sin3 0 (b) Use a CAS to show that the area inside of the loop is | (Figure 2.13d). 56. (a) What is the area that is enclosed by one petal of the rose r = a cos nO if n is an even integer? (b) What is the area that is enclosed by one petal of the rose r = a cos n9 if n is an odd integer? (c) Use a CAS to show that the total area enclosed by the rose r = a cos nO is jta2/2 if the number of petals is even. [Hint: See Exercise 78 of Section 10.2.] (d) Use a CAS to show that the total area enclosed by the rose r = a cos nO is na2/4 if the number of petals is odd. 57. One of the most famous problems in Greek antiquity was "squaring the circle," that is, using a straightedge and compass to construct a square whose area is equal to that of a given circle. It was proved in the nineteenth century that no such construction is possible. However, show that the shaded areas in the accompanying figure are equal, thereby "squaring the crescent."
52. Find the area inside the curve r2 = sin 20. 53. A radial line is drawn from the origin to the spiral r = aO (a > 0 and 9 > 0). Find the area swept out during the second revolution of the radial line that was not swept out during the first revolution. 54. As illustrated in the accompanying figure, suppose that a rod with one end fixed at the pole of a polar coordinate system rotates counterclockwise at the constant rate of 1 rad/s. At time / = 0 a bug on the rod is 10 mm from the pole and is moving outward along the rod at the constant speed of 2 mm/s. (a) Find an equation of the form r = f ( 0 ) for the path of motion of the bug, assuming that 0 = 0 when f = 0. (b) Find the distance the bug travels along the path in part (a) during the first 5 s. Round your answer to the nearest tenth of a millimeter.
•^ Figure Ex-57
E3 58. Use a graphing utility to generate the polar graph of the equation r = cos 30 + 2, and find the area that it encloses. E3 59. Use a graphing utility to generate the graph of the bifolium r = 2 cos 9 sin2 6, and find the area of the upper loop. 60. Use Formula (9) of Section 10.1 to derive the arc length formula for polar curves, Formula (3). 61. As illustrated in the accompanying figure, let P(r, 0) be a point on the polar curve r = f(0), let \jr be the smallest counterclockwise angle from the extended radius OP to the
10.3 Tangent Lines, Arc Length, and Area for Polar Curves
tangent line at P, and let be the angle of inclination of the tangent line. Derive the formula r tan \lr = —;— dr/de by substituting tan 4> for dy/dx in Formula (2) and applying the trigonometric identity tan 0 — tan 9 tan(4> -9) = 1 + tan
729
surface of revolution that is generated by revolving a parametric curve about the x-axis or y-axis. Use those formulas to derive the following formulas for the areas of the surfaces of revolution that are generated by revolving the portion of the polar curve r = f ( 9 ) from 9 = a to 9 = ft about the polar axis and about the line 9 = Tt/2: A
S=
N
— } d8 d9 .
r
S=
Ja
fdrY 2m cos 9 r2 + [— d9 V
About e = 0
About 0 = jr/
\«0/
(b) State conditions under which these formulas hold.
•^ Figure Ex-61
62-63 Use the formula for -ty obtained in Exercise 61. 8 62. (a) Use the trigonometric identity 9 l-cosfl tan - = —: to show that if (r, 9) is a point on the cardioid r = 1 - cos 6 (0 < 9 < 2jr) then ir =9/2. (b) Sketch the cardioid and show the angle ^ at the points where the cardioid crosses the y-axis. (c) Find the angle fy at the points where the cardioid crosses the y-axis. 63. Show that for a logarithmic spiral r = aebe, the angle from the radial line to the tangent line is constant along the spiral (see the accompanying figure). [Note: For this reason, logarithmic spirals are sometimes called equiangular spirals.]
•^ Figure Ex-63
64. (a) In the discussion associated with Exercises 75-80 of Section 10.1, formulas were given for the area of the
•QUICK CHECK ANSWERS
65-68 Sketch the surface, and use the formulas in Exercise 64 to find the surface area. 65. The surface generated by revolving the circle r = cos 9 about the line 9 = n/2. 66. The surface generated by revolving the spiral r = ee (0 < B < a/2) about the line 9 = n/2. 67. The "apple" generated by revolving the upper half of the cardioid r = 1 — cos 9 (0 < 9 < n) about the polar axis. 68. The sphere of radius a generated by revolving the semicircle r = a in the upper half -plane about the polar axis. 69. Writing
(a) Show that if 0 < 6\ <92
A=i (b) Use the formula obtained in part (a) to describe an approach to answer Area Problem 10.3.3 that uses an approximation of the region R by triangles instead of circular wedges. Reconcile your approach with Formula (6). 70. Writing In order to find the area of a region bounded by two polar curves it is often necessary to determine their points of intersection. Give an example to illustrate that the points of intersection of curves r = f ( 9 ) and r = g(9) may not coincide with solutions to f ( 9 ) = g(9). Discuss some strategies for determining intersection points of polar curves and provide examples to illustrate your strategies.
10.3
dr r cos # + sin $ — j ^ g 7 1. (a) -SSU (b) •+ = 0 2. (a) f(90) = 0, f'(90) / 0 (b) 90 = -, —, — , — -r sin 9 + cos 9 —d9
/dr\2
rP
'.(a) / Jr2+ — ) Ja
V
\**"/
(b)
f2" . / \a2 dB = na Jo
730
Chapter 10 / Parametric and Polar Curves; Conic Sections
CONIC SECTIONS In this section we will discuss some of the basic geometric properties of parabolas, ellipses, and hyperbolas. These curves play an important role in calculus and also arise naturally in a broad range of applications in such fields as planetary motion, design of telescopes and antennas, geodetic positioning, and medicine, to name a few.
CONIC SECTIONS
Circles, ellipses, parabolas, and hyperbolas are called conic sections or conies because they can be obtained as intersections of a plane with a double-napped circular cone (Figure 10.4.1). If the plane passes through the vertex of the double-napped cone, then the intersection is a point, a pair of intersecting lines, or a single line. These are called degenerate conic sections.
i
,v Circle
Ellipse
', Parabola i
Some students may already be familiar with the material in this section, in which case it can be treated as a review. Instructors who want to spend some additional time on precalculus review may want to allocate more than one lecture on this material.
10.4 Conic Sections
731
DEFINITIONS OF THE CONIC SECTIONS
Although we could derive properties of parabolas, ellipses, and hyperbolas by defining them as intersections with a double-napped cone, it will be better suited to calculus if we begin with equivalent definitions that are based on their geometric properties.
10.4.1 DEFINITION A parabola is the set of all points in the plane that are equidistant from a fixed line and a fixed point not on the line.
All points on the parabola are equidistant from the focus and directrix.
The line is called the directrix of the parabola, and the point is called the focus (Figure 10.4.2). A parabola is symmetric about the line that passes through the focus at right angles to the directrix. This line, called the axis or the axis of symmetry of the parabola, intersects the parabola at a point called the vertex.
10.4.2 DEFINITION An ellipse is the set of all points in the plane, the sum of whose distances from two fixed points is a given positive constant that is greater than the distance between the fixed points. Directrix A Figure 10.4.2
The two fixed points are called the foci (plural of "focus") of the ellipse, and the midpoint of the line segment joining the foci is called the center (Figure 10.4.3a). To help visualize Definition 10.4.2, imagine that two ends of a string are tacked to the foci and a pencil traces a curve as it is held tight against the string (Figure 10.4.3&). The resulting curve will be an ellipse since the sum of the distances to the foci is a constant, namely, the total length of the string. Note that if the foci coincide, the ellipse reduces to a circle. For ellipses other than circles, the line segment through the foci and across the ellipse is called the major axis (Figure 10.4.3c), and the line segment across the ellipse, through the center, and perpendicular to the major axis is called the minor axis. The endpoints of the major axis are called vertices. The sum of the distances to the foci is constant.
Vertex
Figure 10.4.3
10.4.3 DEFINITION A hyperbola is the set of all points in the plane, the difference of whose distances from two fixed distinct points is a given positive constant that is less than the distance between the fixed points.
The two fixed points are called the foci of the hyperbola, and the term "difference" that is used in the definition is understood to mean the distance to the farther focus minus the distance to the closer focus. As a result, the points on the hyperbola form two branches, each
732
Chapter 10 / Parametric and Polar Curves; Conic Sections
"wrapping around" the closer focus (Figure 10.4.4a). The midpoint of the line segment joining the foci is called the center of the hyperbola, the line through the foci is called the focal axis, and the line through the center that is perpendicular to the focal axis is called the conjugate axis. The hyperbola intersects the focal axis at two points called the vertices. Associated with every hyperbola is a pair of lines, called the asymptotes of the hyperbola. These lines intersect at the center of the hyperbola and have the property that as a point P moves along the hyperbola away from the center, the vertical distance between P and one of the asymptotes approaches zero (Figure 10.4.4&).
Conjugate
The distance from the • farther focus minus the j distance to the closer focus is constant.
z. / Focal ' ax j s ,i / \ Focus Vertex
Vertex
These distances approach zero as the point moves away f rom t ne center.
These distances approach *• zero as the point moves away from the center.
(a) A Figure 10.4.4
EQUATIONS OF PARABOLAS IN STANDARD POSITION
It is traditional in the study of parabolas to denote the distance between the focus and the vertex by p. The vertex is equidistant from the focus and the directrix, so the distance between the vertex and the directrix is also p; consequently, the distance between the focus and the directrix is 2p (Figure 10.4.5). As illustrated in that figure, the parabola passes through two of the corners of a box that extends from the vertex to the focus along the axis of symmetry and extends 2p units above and 2p units below the axis of symmetry. The equation of a parabola is simplest if the vertex is the origin and the axis of symmetry is along the x-axis or j-axis. The four possible such orientations are shown in Figure 10.4.6. These are called the standard positions of a parabola, and the resulting equations are called the standard equations of a parabola.
Directrix A Figure 10.4.5
PARABOLAS IN STANDARD POSITION
= -p
A Figure 10.4.6
10.4 Conic Sections
733
To illustrate how the equations in Figure 10.4.6 are obtained, we will derive the equation for the parabola with focus (/?, 0) and directrix x — —p. Let P(x, y) be any point on the parabola. Since P is equidistant from the focus and directrix, the distances PF and PD in Figure 10.4.7 are equal; that is,
I
rr = rU
D(-p,y)*
(1)
where D(—p, y) is the foot of the perpendicular from P to the directrix. From the distance formula, the distances PF and PD are
PF = V(* - p)2 + y2
and PD =
(2)
p)2
(3)
Substituting in (1) and squaring yields
(x - p)2 and after simplifying
y12 = 4pX
(4)
The derivations of the other equations in Figure 10.4.6 are similar. A TECHNIQUE FOR SKETCHING PARABOLAS
Parabolas can be sketched from their standard equations using four basic steps:
Sketching a Parabola from Its Standard Equation Step 1. Determine whether the axis of symmetry is along the x-axis or the y-axis. Referring to Figure 10.4.6, the axis of symmetry is along the jc-axis if the equation has a y2-term, and it is along the y-axis if it has an x2-term. Step 2. Determine which way the parabola opens. If the axis of symmetry is along the x-axis, then the parabola opens to the right if the coefficient of x is positive, and it opens to the left if the coefficient is negative. If the axis of symmetry is along the y-axis, then the parabola opens up if the coefficient of y is positive, and it opens down if the coefficient is negative. 2P
X A Figure 10.4.8
Step 3. Determine the value of p and draw a box extending p units from the origin along the axis of symmetry in the direction in which the parabola opens and extending 2p units on each side of the axis of symmetry. Step 4. Using the box as a guide, sketch the parabola so that its vertex is at the origin and it passes through the corners of the box (Figure 10.4.8).
*• Example 1
Sketch the graphs of the parabolas
(a) x2 = 12y
(b) y 2 + 8x = 0
and show the focus and directrix of each.
A Figure 10.4.9
Solution (a). This equation involves x2, so the axis of symmetry is along the y-axis, and the coefficient of y is positive, so the parabola opens upward. From the coefficient of y, we obtain 4p = 12 or p = 3. Drawing a box extending p = 3 units up from the origin and 2p = 6 units to the left and 2p — 6 units to the right of the y-axis, then using corners of the box as a guide, yields the graph in Figure 10.4.9. The focus is p = 3 units from the vertex along the axis of symmetry in the direction in which the parabola opens, so its coordinates are (0, 3). The directrix is perpendicular to the axis of symmetry at a distance of p = 3 units from the vertex on the opposite side from the focus, so its equation is y = —3.
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Chapter TO / Parametric and Polar Curves; Conic Sections
Solution (ft).
We first rewrite the equation in the standard form y2 = -8x
This equation involves y2, so the axis of symmetry is along the x-axis, and the coefficient of x is negative, so the parabola opens to the left. From the coefficient of x we obtain 4p = 8, so p — 2. Drawing a box extending p — 2 units left from the origin and 2p = 4 units above and 2p = 4 units below the ;t-axis, then using corners of the box as a guide, yields the graph in Figure 10.4.10. •« > Example 2 Find an equation of the parabola that is symmetric about the ;y-axis, has its vertex at the origin, and passes through the point (5,2). Solution, Since the parabola is symmetric about the y-axis and has its vertex at the origin, the equation is of the form x2 = 4py
A Figure 10.4.10
or x2 = —4py
where the sign depends on whether the parabola opens up or down. But the parabola must open up since it passes through the point (5,2), which lies in the first quadrant. Thus, the equation is of the form x2 = 4py (5) Since the parabola passes through (5, 2), wemusthaveS 2 = 4p • 2or4p = ~. Therefore, (5) becomes
EQUATIONS OF ELLIPSES IN STANDARD POSITION
\
A Figure 10.4.11
It is traditional in the study of ellipses to denote the length of the major axis by 2a, the length of the minor axis by 2b, and the distance between the foci by 2c (Figure 10.4.11). The number a is called the semimajor axis and the number b the semiminor axis (standard but odd terminology, since a and b are numbers, not geometric axes). There is a basic relationship between the numbers a, b, and c that can be obtained by examining the sum of the distances to the foci from a point P at the end of the major axis and from a point Q at the end of the minor axis (Figure 10.4.12). From Definition 10.4.2, these sums must be equal, so we obtain c2 = (a-c)
from which it follows that
a=
(6)
or, equivalently, (7) A Figure 10.4.12
A Figure 10.4.13
From (6), the distance from a focus to an end of the minor axis is a (Figure 10.4.13), which implies that for all points on the ellipse the sum of the distances to the foci is 2a. It also follows from (6) that a > b with the equality holding only when c = 0. Geometrically, this means that the major axis of an ellipse is at least as large as the minor axis and that the two axes have equal length only when the foci coincide, in which case the ellipse is a circle. The equation of an ellipse is simplest if the center of the ellipse is at the origin and the foci are on the x-axis or y-axis. The two possible such orientations are shown in Figure 10.4.14.
T 0.4 Conic Sections
735
These are called the standard positions of an ellipse, and the resulting equations are called the standard equations of an ellipse.
ELLIPSES IN STANDARD POSITION
(0,C)
> Figure 10.4.14
To illustrate how the equations in Figure 10.4.14 are obtained, we will derive the equation for the ellipse with foci on the ;t-axis. Let P(x, y ) be any point on that ellipse. Since the sum of the distances from P to the foci is 2a, it follows (Figure 10.4.15) that PF' +PF=2a so VGc + c)2 + y2 + v/Ot - c)2 + y2 = 2a
Transposing the second radical to the right side of the equation and squaring yields A Figure 10.4.15
(x + c)2 + y2= 4a2 -
- c)2 + y2 + (x - c)2 + y2
and, on simplifying, - c)2 + y2 = a -- x a
(8)
Squaring again and simplifying yields v-2
aL9
a9L — C9L
which, by virtue of (6), can be written as 2
2
X
y 2-+ -
a
b*
(9)
Conversely, it can be shown that any point whose coordinates satisfy (9) has la as the sum of its distances from the foci, so that such a point is on the ellipse.
736
Chapter 10 / Parametric and Polar Curves; Conic Sections A TECHNIQUE FOR SKETCHING ELLIPSES
Ellipses can be sketched from their standard equations using three basic steps:
Sketching an Ellipse from Its Standard Equation
ty
Step 1. Determine whether the major axis is on the x-axis or the y-axis. This can be ascertained from the sizes of the denominators in the equation. Referring to Figure 10.4.14, and keeping in mind that a2 > b2 (since a > b), the major axis is along the x-axis if x2 has the larger denominator, and it is along the y-axis if y2 has the larger denominator. If the denominators are equal, the ellipse is a circle. Step 2. Determine the values of a and b and draw a box extending a units on each side of the center along the major axis and b units on each side of the center along the minor axis.
Rough sketch
A Figure 10.4.16
Step 3. Using the box as a guide, sketch the ellipse so that its center is at the origin and it touches the sides of the box where the sides intersect the coordinate axes (Figure 10.4.16).
Example 3
Sketch the graphs of the ellipses 7T 9
7^ 16
showing the foci of each. Solution (a). Since y2 has the larger denominator, the major axis is along the y-axis. Moreover, since a2 > b2, we must have a2 = 16 and b2 = 9, so a = 4 and b = 3
Drawing a box extending 4 units on each side of the origin along the y-axis and 3 units on each side of the origin along the x-axis as a guide yields the graph in Figure 10.4.17. The foci lie c units on each side of the center along the major axis, where c is given by (7). From the values of a2 and b2 above, we obtain A Figure 10.4.17
c = Ja2 - b2 = V16-9 = V? RS 2.6
Thus, the coordinates of the foci are (0, V"7) and (0, — \/7), since they lie on the y-axis. Solution (b).
We first rewrite the equation in the standard form
Since x2 has the larger denominator, the major axis lies along the x-axis, and we have a2 = 4 and b2 — 2. Drawing a box extending a = 1 units on each side of the origin along the x-axis and extending b = \/2 « 1 .4 units on each side of the origin along the y-axis as a guide yields the graph in Figure 10.4.18. From (7), we obtain _ c = Ja2-b2 = -fl^ 1.4 A Figure 10.4.18
Thus, the coordinates of the foci are (\/2, 0) and (— \/2, 0), since they lie on the x-axis. •<
10.4 Conic Sections
*• Example 4 points (0, ±4). Solution.
737
Find an equation for the ellipse with foci (0, ±2) and major axis with end-
From Figure 10.4. 14, the equation has the form x2 y2 __ L L- — 1
Va2 ~ and from the given information, a = 4 and c = 2. It follows from (6) that
b2 =a2 -c2 = 16-4= 12 so the equation of the ellipse is
2
2
EQUATIONS OF HYPERBOLAS IN STANDARD POSITION
It is traditional in the study of hyperbolas to denote the distance between the vertices by 2a, the distance between the foci by 2c (Figure 10.4.19), and to define the quantity b as
b=
(10)
This relationship, which can also be expressed as + b2
c= A Figure 10.4.19
is pictured geometrically in Figure 10.4.20. As illustrated in that figure, and as we will show later in this section, the asymptotes pass through the corners of a box extending b units on each side of the center along the conjugate axis and a units on each side of the center along the focal axis. The number a is called the semifocal axis of the hyperbola and the number b the semiconjugate axis. (As with the semimajor and semiminor axes of an ellipse, these are numbers, not geometric axes.) If V is one vertex of a hyperbola, then, as illustrated in Figure 10.4.21, the distance from V to the farther focus minus the distance from V to the closer focus is [(c -
A Figure 10.4.20
2a] -(c-d) =
Thus, for all points on a hyperbola, the distance to the farther focus minus the distance to the closer focus is 2a. The equation of a hyperbola has an especially convenient form if the center of the hyperbola is at the origin and the foci are on the Jt-axis or 3>-axis. The two possible such orientations are shown in Figure 10.4.22. These are called the standard positions of a hyperbola, and the resulting equations are called the standard equations of a hyperbola. The derivations of these equations are similar to those already given for parabolas and ellipses, so we will leave them as exercises. However, to illustrate how the equations of the asymptotes are derived, we will derive those equations for the hyperbola b2
We can rewrite this equation as A Figure 10.4.21
(11)
b2 which is equivalent to the pair of equations and
y =
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Chapter 10 / Parametric and Polar Curves; Conic Sections
HYPERBOLAS IN STANDARD POSITION A?
*• Figure 10.4.22
Thus, in the first quadrant, the vertical distance between the line y — (b/a)x and the hyperbola can be written as
a
a
(Figure 10.4.23). But this distance tends to zero as x —> +00 since i
( -x a
\
2
a
j
\/* — a ) = lim - f x — \A:2 — a2 j 2
(x - Jx2 - a2 ) (j* = lim — *^+» a
x + V x2 - a2 ab
= lim
=0
A Figure 10.4.23
The analysis in the remaining quadrants is similar.
A QUICK WAY TO FIND ASYMPTOTES
There is a trick that can be used to avoid memorizing the equations of the asymptotes of a hyperbola. They can be obtained, when needed, by replacing 1 by 0 on the right side of the hyperbola equation, and then solving for y in terms of x . For example, for the hyperbola
a2
_
b2
we would write x2 y2 2 b2 22 — x2 - — 2 = 0 or / = — aba1
which are the equations for the asymptotes.
b
or y = ±-x a
10.4 Conic Sections
739
A TECHNIQUE FOR SKETCHING HYPERBOLAS
Hyperbolas can be sketched from their standard equations using four basic steps:
Sketching a Hyperbola front Its Standard Equation Step 1. Determine whether the focal axis is on the x-axis or the y-axis. This can be ascertained from the location of the minus sign in the equation. Referring to Figure 10.4.22, the focal axis is along the jc-axis when the minus sign precedes the y2-term, and it is along the y-axis when the minus sign precedes the z2-term. Step 2. Determine the values of a and b and draw a box extending a units on either side of the center along the focal axis and b units on either side of the center along the conjugate axis. (The squares of a and b can be read directly from the equation.) Step 3. Draw the asymptotes along the diagonals of the box. Step 4. Using the box and the asymptotes as a guide, sketch the graph of the hyperbola (Figure 10.4.24).
Example 5
Sketch the graphs of the hyperbolas (a) j - - = 1 (b) y2 - x2 = 1
showing their vertices, foci, and asymptotes. Solution (a). The minus sign precedes the y2-term, so the focal axis is along the .x-axis. From the denominators in the equation we obtain a2 = 4
and b2 = 9
Since a and b are positive, we must have a = 2 and b = 3. Recalling that the vertices lie a units on each side of the center on the focal axis, it follows that their coordinates in this case are (2, 0) and (—2, 0). Drawing a box extending a = 2 units along the x-axis on each side of the origin and b — 3 units on each side of the origin along the y-axis, then drawing the asymptotes along the diagonals of the box as a guide, yields the graph in Figure 10.4.25. To obtain equations for the asymptotes, we replace 1 by 0 in the given equation; this yields 2 2 o A Figure 10.4.25
=0
or
y = ±-jc
4 9 2 The foci lie c units on each side of the center along the focal axis, where c is given by (11). From the values of a1 and b2 above we obtain
c = Va + b2 = V4 + 9 = V13 % 3.6 Since the foci lie on the x-axis in this case, their coordinates are (\/13, 0) and (—\/T3, 0). Solution (b). The minus sign precedes the *2-term, so the focal axis is along the y-axis. From the denominators in the equation we obtain a2 — 1 and b1 — 1, from which it follows that a — 1 and b = 1
A Figure 10.4.26
Thus, the vertices are at (0, -1) and (0, 1). Drawing a box extending a — 1 unit on either side of the origin along the y-axis and b = 1 unit on either side of the origin along the je-axis, then drawing the asymptotes, yields the graph in Figure 10.4.26. Since the box is actually
740
Chapter 10 / Parametric and Polar Curves; Conic Sections
—^^^^^^^—^^^^^^^— A hyperbola in which a = b, as in part (b) of Example 5, is called an equilateral hyperbola. Such hyperbolas always have perpendicular asymptotes.
a square, the asymptotes are perpendicular and have equations y = ±x. This can also be by replacing 1 by 0 in the given equation, which yields y 2 — x2 = 0 or y = ±x. Also,
seen
c = \ a2 + b2 = v l + 1 = v 2 ,— .— so the foci, which lie on the y-axis, are (0, — V2) and (0, V2). •* Example 6
Find the equation of the hyperbola with vertices (0, ±8) and asymptotes
Solution. Since the vertices are on the y-axis, the equation of the hyperbola has the form (y 2 /a 2 ) - (x2/b2) = 1 and the asymptotes are
From the locations of the vertices we have a — 8, so the given equations of the asymptotes yield a 8 4 y = ±-x = ±-x — ±-x b b 3 from which it follows that b = 6. Thus, the hyperbola has the equation
z
64
36
=1 ,
TRANSLATED COMICS
Equations of conies that are translated from their standard positions can be obtained by replacing x by x — h and y by y — k in their standard equations. For a parabola, this translates the vertex from the origin to the point (h,k); and for ellipses and hyperbolas, this translates the center from the origin to the point (h, k). Parabolas with vertex (h, k) and axis parallel to x-axis (y — k)2 =
4p(x — h)
[Opens right]
2
(y — k) — —4p(x — h)
[Opens left]
(12) (13)
Parabolas with vertex (h, k) and axis parallel to y-axis (X — h)2 =
4p(y — k)
2
(X — h) = —4p(y — K)
[Opens up] [Opens down]
(14) (15)
Ellipse with center (h, k) and major axis parallel to x-axis (x-h)2 a 5—
H
(y-fc) 2 z ^— =! \-b
(16)
Ellipse with center (h, k) and major axis parallel to y-axis (x - h)2 b2
(y - k)2 a2
=i
(17)
Hyperbola with center (h, k) and focal axis parallel to x-axis (x - h)2 a2
(y - k)2 b2
(18)
Hyperbola with center (h, k) and focal axis parallel to y-axis (y - k)2 a2
(x - h)2 b2
(19)
10.4 Conic Sections
t* Example 7 at (4, 2).
741
Find an equation for the parabola that has its vertex at (1, 2) and its focus
Solution* Since the focus and vertex are on a horizontal line, and since the focus is to the right of the vertex, the parabola opens to the right and its equation has the form (y - k)2 = 4p(x - h)
Since the vertex and focus are 3 units apart, we have p = 3, and since the vertex is at ( h , k ) — (1,2), we obtain (y - 2)2 = 12(:t - 1)
Sometimes the equations of translated conies occur in expanded form, in which case we are faced with the problem of identifying the graph of a quadratic equation in x and y:
Ax2 + Cy2 + Dx + Ey + F = 0
(20)
The basic procedure for determining the nature of such a graph is to complete the squares of the quadratic terms and then try to match up the resulting equation with one of the forms of a translated conic. > Example 8
Describe the graph of the equation
y2 - BJC - 6y - 23 = 0 Solution. The equation involves quadratic terms in y but none in x, so we first take all of the y -terms to one side: , y2 - 6y = &x + 23 (-4,3)-
Next, we complete the square on the y-terms by adding 9 to both sides:
(y - 3)2 = 8* + 32 Finally, we factor out the coefficient of the jc-term to obtain Directrix
A Figure 10.4.27
This equation is of form (12) with h = —4, k = 3, and p = 2, so the graph is a parabola with vertex (—4, 3) opening to the right. Since p = 2, the focus is 2 units to the right of the vertex, which places it at the point (—2,3); and the directrix is 2 units to the left of the vertex, which means that its equation is x = —6. The parabola is shown in Figure 10.4.27. •<
*• Example 9
Describe the graph of the equation 16jc2 + 9y2 - 64x - 54y + 1 = 0
Solution. This equation involves quadratic terms in both x and y, so we will group the x-terms and the y-terms on one side and put the constant on the other: (16.x2 - 64jc) + (9/ - 54;y) = -1 Next, factor out the coefficients of x2 and y2 and complete the squares: I6(x2 - 4x + 4) + 9(y2 - 6y + 9) = -1 + 64 + 81
16(x - 2)2 + 9(y - 3)2 = 144
742
Chapter 10 / Parametric and Polar Curves; Conic Sections
Finally, divide through by 144 to introduce a 1 on the right side: (*-2) 2 (y-3) 2 9 16 This is an equation of form (17), with h = 2, k = 3, a2 = 16, and b2 = 9. Thus, the graph of the equation is an ellipse with center (2, 3) and major axis parallel to the y-axis. Since a = 4, the major axis extends 4 units above and 4 units below the center, so its endpoints are (2, 7) and (2, —1) (Figure 10.4.28). Since b = 3, the minor axis extends 3 units to the left and 3 units to the right of the center, so its endpoints are (—1, 3) and (5, 3). Since
(-1, 3)
(2,-D
(2,3-V?)
c = Va2 - b2 = Vl6 - 9 = V7 the foci lie \/7 units above and below the center, placing them at the points (2, 3 + \/7) and (2, 3-77). +
Example 10 Describe the graph of the equation 2 X
- y2 - 4X + 8y - 21 = 0
Solution. This equation involves quadratic terms in both x and y, so we will group the x -terms and the y-terms on one side and put the constant on the other:
(x2 - 4x) - (y2 - 8y) = 21 We leave it for you to verify by completing the squares that this equation can be written as
(x - 2)2 (y - 4)2 =1 (21) 9 9 This is an equation of form (18) with h = 2, k = 4, a2 = 9, and b2 = 9. Thus, the equation represents a hyperbola with center (2, 4) and focal axis parallel to the jc-axis. Since a = 3, the vertices are located 3 units to the left and 3 units to the right of the center, or at the points (-1, 4) and (5, 4).From(ll),c = Va2 + b2 = V9 + 9 = 3^2, so the foci are located 3^2 units to the left and right of the center, or at the points (2 - 3^2, 4) and (2 + 3\/2, 4). The equations of the asymptotes may be found using the trick of replacing 1 by 0 in (21) to obtain (^ _ 2)2 (y — 4)2 9 9 This can be written as y — 4 = ±(x — 2), which yields the asymptotes y = x + 2 and y = — x + 6 -2l =0 A Figure 10.4.29
With the aid of a box extending a — 3 units left and right of the center and b — 3 units above and below the center, we obtain the sketch in Figure 10.4.29. «i
REFLECTION PROPERTIES OF THE CONIC SECTIONS
Parabolas, ellipses, and hyperbolas have certain reflection properties that make them extremely valuable in various applications. In the exercises we will ask you to prove the following results. 10.4.4 THEOREM (Reflection Property of Parabolas) The tangent line at a point P on na parabola makes equal angles with the line through P parallel to the axis of symmetry.. and the line through P and the focus (Figure 10.4.30a).
10.4 Conic Sections
743
10.4.5 THEOREM (Reflection Property of Ellipses) A line tangent to an ellipse at a point P makes equal angles with the lines joining P to the foci (Figure 10.4.30b).
10.4.6 THEOREM (Reflection Property of Hyperbolas) A line tangent to a hyperbola at a point P makes equal angles with the lines joining P to the foci (Figure 10.4.30c).
• Tangent line at P
(a)
Figure 10.4.30
John MeadyScience Photo Library/Photo Researchers Incoming signals are reflected by the parabolic antenna to the receiver at the focus.
Tangent line at P
(b)
(c)
APPLICATIONS OF THE CONIC SECTIONS Fermat's principle in optics implies that light reflects off of a surface at an angle equal to its angle of incidence. (See Exercise 62 in Section 3.5.) In particular, if a reflecting surface is generated by revolving a parabola about its axis of symmetry, it follows from Theorem 10.4.4 that all light rays entering parallel to the axis will be reflected to the focus (Figure 10.4.3 la); conversely, if a light source is located at the focus, then the reflected rays will all be parallel to the axis (Figure 10.4.31ft). This principle is used in certain telescopes to reflect the approximately parallel rays of light from the stars and planets off of a parabolic mirror to an eyepiece at the focus; and the parabolic reflectors in flashlights and automobile headlights utilize this principle to form a parallel beam of light rays from a bulb placed at the focus. The same optical principles apply to radar signals and sound waves, which explains the parabolic shape of many antennas.
Figure 10.4.31
(a)
Visitors to various rooms in the United States Capitol Building and in St. Paul's Cathedral in London are often astonished by the "whispering gallery" effect in which two people at opposite ends of the room can hear one another's whispers very clearly. Such rooms have ceilings with elliptical cross sections and common foci. Thus, when the two people stand at the foci, their whispers are reflected directly to one another off of the elliptical ceiling. Hyperbolic navigation systems, which were developed in World War II as navigational aids to ships, are based on the definition of a hyperbola. With these systems the ship receives
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Chapter 10 / Parametric and Polar Curves; Conic Sections
synchronized radio signals from two widely spaced transmitters with known positions. The ship's electronic receiver measures the difference in reception times between the signals and then uses that difference to compute the difference 2a between its distances from the two transmitters. This information places the ship somewhere on the hyperbola whose foci are at the transmitters and whose points have 2a as the difference in their distances from the foci. By repeating the process with a second set of transmitters, the position of the ship can be approximated as the intersection of two hyperbolas (Figure 10.4.32). (The modern global positioning system (GPS) is based on the same principle.)
A Figure 10.4.32
UICK CHECK EXERCISES 10.4
(See page 748 for answers.)
1. Identify the conic. (a) The set of points in the plane, the sum of whose distances to two fixed points is a positive constant greater than the distance between the fixed points is (b) The set of points in the plane, the difference of whose distances to two fixed points is a positive constant less than the distance between the fixed points is (c) The set of points in the plane that are equidistant from a fixed line and a fixed point not on the line is 2. (a) The equation of the parabola with focus (p, 0) and directrix x = —p is (b) The equation of the parabola with focus (0, p) and directrix y = — p is 3. (a) Suppose that an ellipse has semimajor axis a and semiminor axis b. Then for all points on the ellipse, the sum of the distances to the foci is equal to (b) The two standard equations of an ellipse with semimajor axis a and semiminor axis b are and
EXERCISE SET 10.4
(c) Suppose that an ellipse has semimajor axis a, semiminor axis b, and foci (±c, 0). Then c may be obtained from a and b by the equation c = 4. (a) Suppose that a hyperbola has semifocal axis a and semiconjugate axis b. Then for all points on the hyperbola, the difference of the distance to the farther focus minus the distance to the closer focus is equal to (b) The two standard equations of a hyperbola with semifocal axis a and semiconjugate axis b are and (c) Suppose that a hyperbola in standard position has semifocal axis a, semiconjugate axis b, and foci (±c, 0). Then c may be obtained from a and b by the equation c= The equations of the asymptotes of this hyperbola are y = ±
Graphing Utility
FOCUS ON CONCEPTS
1. In parts (a)-(f), find the equation of the conic.
(C)
-3
-3-2-1 0
1
2
3
4
1 2 3
-3 -2 -1
1
2 3
10.4 Conic Sections
745
18. Vertex (5,—3); axis parallel to the y-axis; passes through (9,5).
-3-2-1
0
-3-2-1 0
1 2 3
1 2 3
2. (a) Find the focus and directrix for each parabola in Exercise 1. (b) Find the foci of the ellipses in Exercise 1. (c) Find the foci and the equations of the asymptotes of the hyperbolas in Exercise 1. 3-6 Sketch the parabola, and label the focus, vertex, and directrix. 8 3. (a) y2 = 4x (b) x2 = -8y 4. (a) / = -10jc (b) x2=4y 2 5. (a) (y - I) = -12(x + 4) (b) (x -l)2 = 2(y~ \] 6. (a) y2 - 6y - 2x + 1 = 0
(b) y = 4x2 + &x + 5
7-10 Sketch the ellipse, and label the foci, vertices, and ends of the minor axis. x2 v2 (b) 9x2 + y2 = 9 X
-
(b) 4x2 ± y2 = 36
9. (a) U + 3) 2 +4(y-5) 2 = 16 (b) ijc2 + i(y + 2 ) 2 - l = 0 10. (a) 9x2 + 4y2 - 18* + 24y + 9 = 0 (b) 5x2 + 9y2 + 2(k - 54y = -56
11-14 Sketch the hyperbola, and label the vertices, foci, and asymptotes. x y (b) 9y2 ~x2 = 36 12- (a)
-- =
(b) 16*2 - 25y2 = 400
13. (a) (b) 16(* + I)2 - 8(y - 3)2 = 16 14. (a) x2 - 4y2 + 2x + 8y - 7 = 0 (b) I6x2 - y2 - 32;c - 6y = 57 15-18 Find an equation for the parabola that satisfies the given conditions, a 15. (a) Vertex (0, 0); focus (3,0). (b) Vertex (0, 0); directrix y = ~. 16. (a) Focus (6,0); directrix x = —6. (b) Focus (1, 1); directrix y = —2. 17. Axis y = 0; passes through (3, 2) and (2, -\/2).
19-22 Find an equation for the ellipse that satisfies the given conditions. 19. (a) Ends of major axis (±3, 0); ends of minor axis (0, ±2). (b) Length of minor axis 8; foci (0, ±3). 20. (a) Foci (±1,0); b = VI (b) c = 2\/3; a = 4; center at the origin; foci on a coordinate axis (two answers). 21. (a) Ends of major axis (0, ±6); passes through (—3, 2). (b) Foci (—1, 1) and (—1, 3); minor axis of length 4. 22. (a) Center at (0,0); major and minor axes along the coordinate axes; passes through (3, 2) and (1,6). (b) Foci (2,1) and (2, —3); major axis of length 6. 23-26 Find an equation for a hyperbola that satisfies the given conditions. [Note: In some cases there may be more than one hyperbola.] 23. (a) Vertices (±2,0); foci (±3,0). (b) Vertices (0, ±2); asymptotes y = ±|*. 24. (a) (b) 25. (a) (b) 26. (a) (b)
Asymptotes y = ±\x; b = 4. Foci (0, ±5); asymptotes y = ±2x. Asymptotes y = ±|x; c = 5. Foci (±3, 0); asymptotes y = ±2x. Vertices (0, 6) and (6, 6); foci 10 units apart, Asymptotes y = x — 2 and y = — x + 4; passes through the origin.
27-30 True-False Determine whether the statement is true or false. Explain your answer. 27. A hyperbola is the set of all points in the plane that are equidistant from a fixed line and a fixed point not on the line. 28. If an ellipse is not a circle, then the foci of an ellipse lie on the major axis of the ellipse. 29. If a parabola has equation y2 = 4px, where p is a positive constant, then the perpendicular distance from the parabola's focus to its directrix is p. 30. The hyperbola (y2/a2) — x2 = 1 has asymptotes the lines y = ±x/a. 31. (a) As illustrated in the accompanying figure, a parabolic arch spans a road 40 ft wide. How high is the arch if a center section of the road 20 ft wide has a minimum clearance of 12 ft? (b) How high would the center be if the arch were the upper half of an ellipse?
-20ft—H -40ft
•* Figure Ex-31
746
Chapter 10 / Parametric and Polar Curves; Conic Sections
32. (a) Find an equation for the parabolic arch with base b and height h, shown in the accompanying figure, (b) Find the area under the arch.
•^ Figure Ex-32
39. Find an equation of the ellipse traced by a point that moves so that the sum of its distances to (4, 1) and (4, 5) is 12. 40. Find the equation of the hyperbola traced by a point that moves so that the difference between its distances to (0, 0) and (1,1) is 1. 41. Show that an ellipse with semimajor axis a and semiminor axis b has area A = nab. FOCUS ON CONCEPTS
33. Show that the vertex is the closest point on a parabola to the focus. [Suggestion: Introduce a convenient coordinate system and use Definition 10.4.1.] 34. As illustrated in the accompanying figure, suppose that a comet moves in a parabolic orbit with the Sun at its focus and that the line from the Sun to the comet makes an angle of 60° with the axis of the parabola when the comet is 40 million miles from the center of the Sun. Use the result in Exercise 33 to determine how close the comet will come to the center of the Sun. 35. For the parabolic reflector in the accompanying figure, how far from the vertex should the light source be placed to produce a beam of parallel rays?
42. Show that if a plane is not parallel to the axis of a right circular cylinder, then the intersection of the plane and cylinder is an ellipse (possibly a circle). [Hint: Let 9 be the angle shown in the accompanying figure, introduce coordinate axes as shown, and express x' and y' in terms of x and y.]
•4 Figure Ex-42
A Figure Ex-34
A Figure Ex-35
36. (a) Show that the right and left branches of the hyperbola 2
b2 can be represented parametrically as
x— a cosh t, x = —acosht,
y = bsinht y = bsinht
(— oo < t < +°o) (—00 < t < +00)
(b) Use a graphing utility to generate both branches of the hyperbola x2 — y2 = 1 on the same screen. 37. (a) Show that the right and left branches of the hyperbola
43. As illustrated in the accompanying figure, a carpenter needs to cut an elliptical hole in a sloped roof through which a circular vent pipe of diameter D is to be inserted vertically. The carpenter wants to draw the outline of the hole on the roof using a pencil, two tacks, and a piece of string (as in Figure 10.4.3ft). The center point of the ellipse is known, and common sense suggests that its major axis must be perpendicular to the drip line of the roof. The carpenter needs to determine the length L of the string and the distance T between a tack and the center point. The architect's plans show that the pitch of the roof is p (pitch = rise over run; see the accompanying figure). Find T and L in terms of D and p. Source.' This exercise is based on an article by William H. Enos, which appeared in the Mathematics Teacher, Feb. 1991, p. 148.
can be represented parametrically as x= a sect, y = btant x = -a sec t, y = b tan t
(—n/2
(b) Use a graphing utility to generate both branches of the hyperbola x2 — y2 = 1 on the same screen. 38. Find an equation of the parabola traced by a point that moves so that its distance from (2, 4) is the same as its distance to the x -axis.
< Figure Ex-43
44. As illustrated in the accompanying figure on the next page, suppose that two observers are stationed at the points FI(C, 0) and F2(—c, 0) in an xy-coordinate system. Suppose also that the sound of an explosion in the xy-plane is heard by the F\ observer / seconds before it
10.4 Conic Sections 747 is heard by the F2 observer. Assuming that the speed of sound is a constant v, show that the explosion occurred somewhere on the hyperbola x2
y2
2 2
2
c - (v2t2/4) ~
v t /4
F2(-c, 0)
FI(C, 0)
•4 Figure Ex-44
45. As illustrated in the accompanying figure, suppose that two transmitting stations are positioned 100 km apart at points Fi(50, 0) and F2(—50, 0) on a straight shoreline in an jry-coordinate system. Suppose also that a ship is traveling parallel to the shoreline but 200 km at sea. Find the coordinates of the ship if the stations transmit a pulse simultaneously, but the pulse from station F\ is received by the ship 100 microseconds sooner than the pulse from station F2. [Hint: Use the formula obtained in Exercise 44, assuming that the pulses travel at the speed of light (299,792,458 m/s).]
F2(-50, 0)
at the point (XQ, yo) has the equation XXQ +,yyo_, ~# ~& ~ 50. Prove: The line tangent to the hyperbola b2 at the point (XQ, yo) has the equation XXQ yy0 _ a2 " b2 51. Use the results in Exercises 49 and 50 to show that if an ellipse and a hyperbola have the same foci, then at each point of intersection their tangent lines are perpendicular.
52. Consider the second-degree equation Ax2 + Cy2 + Dx + Ey + F = 0 where A and C are not both 0. Show by completing the square: (a) If AC > 0, then the equation represents an ellipse, a circle, a point, or has no graph. (b) If AC < 0, then the equation represents a hyperbola or a pair of intersecting lines. (c) If AC = 0, then the equation represents a parabola, a pair of parallel lines, or has no graph. 53. In each part, use the result in Exercise 52 to make a statement about the graph of the equation, and then check your conclusion by completing the square and identifying the graph. (a) x2 -5y2-2x-\0y-9 = 0 (b) x2 - 3y2 - 6y - 3 = 0 (c) 4x2 + 8y2 + 16* + 16y + 20 = 0 (d) 3x2 + y2 + l2x + 2y + l3 = 0 (e) x2 + 8x + 2y + 14 = 0 (f ) 5x2 + 4Qx + 2y + 94 = 0 54. Derive the equation x2 = 4py in Figure 10.4.6.
•* Figure Ex-45
46. A nuclear cooling tower is to have a height of h feet and the shape of the solid that is generated by revolving the region R enclosed by the right branch of the hyperbola 1521jc2 - 225y2 = 342,225 and the lines x = 0, y = —h/2, and y = h/2 about the y-axis. (a) Find the volume of the tower. (b) Find the lateral surface area of the tower.
47. Let R be the region that is above the x-axis and closed between the curve b2x2 — a2y2 = a2b2 and the line X = Vfl 2 + b2.
(a) Sketch the solid generated by revolving R about the *-axis, and find its volume. (b) Sketch the solid generated by revolving R about the y-axis, and find its volume. 48. Prove: The line tangent to the parabola x2 = 4py at the point (x0, yo) is x0x = 2p(y + y 0 ). 49. Prove: The line tangent to the ellipse x2
y2
55. Derive the equation (x2/b2) + (y2/a2) = 1 given in Figure 10.4.14. 56. Derive the equation (x2/a2) — (y2/b2) = 1 given in Figure 10.4.22. 57. Prove Theorem 10.4.4. [Hint: Choose coordinate axes so that the parabola has the equation x2 = 4py. Show that the tangent line at P(XQ, yo) intersects the y-axis at <2(0, — yo) and that the triangle whose three vertices are at P, Q, and the focus is isosceles.] 58. Given two intersecting lines, let L2 be the line with the larger angle of inclination fa, and let L \ be the line with the smaller angle of inclination 4>\ . We define the angle 0 between L\ and LI by 9 = fa — i- (See the accompanying figure on the next page.) (a) Prove: If LI and L2 are not perpendicular, then m2-ml where L\ and L2 have slopes m\ and m2. (b) Prove Theorem 10.4. 5. [Hint: Introduce coordinates so that the equation (x2/a2) + (y2/b2) = 1 describes the ellipse, and use part (a).] (mm.)
748 Chapter 10 / Parametric and Polar Curves; Conic Sections (c) Prove Theorem 10.4.6. [Hint: Introduce coordinates so that the equation (x2/a2) — (y2/b2) = 1 describes the hyperbola, and use part (a).]
59. Writing Suppose that you want to draw an ellipse that has given values for the lengths of the major and minor axes by using the method shown in Figure 10.4.36. Assuming that the axes are drawn, explain how a compass can be used to locate the positions for the tacks. 60. Writing List the forms for standard equations of parabolas, ellipses, and hyperbolas, and write a summary of techniques for sketching conic sections from their standard equations.
•^ Figure Ex-58
I/QUICK CHECK ANSWERS 10.4 1. (a) an ellipse (b) a hyperbola (c) a parabola Y
2
V
2
2 X
? V
2. (a) y2 = 4px (b) x2 = 4py
3. (a) 2a (b)z - + t? = 1; — + i- = 1 (c) Va2 - b2 a o/ oz a/
X
2
V
2
2 V
9 JC
4. (a) 2a (b) — - yz— = 11; i- _._ = 1 (c) az oz a b
ROTATION OF AXES; SECOND-DEGREE EQUATIONS In the preceding section -we obtained equations of conic sections with axes parallel to the coordinate axes. In this section we will study the equations of conies that are "tilted" relative to the coordinate axes. This will lead us to investigate rotations of coordinate axes.
QUADRATIC EQUATIONS IN x AND y We saw in Examples 8 to 10 of the preceding section that equations of the form
Ax2 + Cy2 + Dx + Ey + F = 0
(1)
can represent conic sections. Equation (1) is a special case of the more general equation
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
(2)
which, if A, B, and C are not all zero, is called a quadratic equation in x and y. It is usually the case that the graph of any second-degree equation is a conic section. If B — 0, then (2) reduces to (1) and the conic section has its axis or axes parallel to the coordinate axes. However, if B ^ 0, then (2) contains a cross-product term Bxy, and the graph of the conic section represented by the equation has its axis or axes "tilted" relative to the coordinate axes. As an illustration, consider the ellipse with foci F\(l, 2) and FI(—\, —2) and such that the sum of the distances from each point P(x, y) on the ellipse to the foci is 6 units. Expressing this condition as an equation, we obtain (Figure 10.5.1)
P(x,y)
V(jc - I) 2 + (y - 2)2 + J(x + I) 2 + (y + 2)2 = 6
Squaring both sides, then isolating the remaining radical, then squaring again ultimately -3 A Figure 10.5.1
as the equation of the ellipse. This is of form (2) with A — 8, B — —4, C = 5, D — 0, E = 0, and F = -36.
10.5 Rotation of Axes; Second-Degree Equations 749 ROTATION OF AXES
To study conies that are tilted relative to the coordinate axes it is frequently helpful to rotate the coordinate axes, so that the rotated coordinate axes are parallel to the axes of the conic. Before we can discuss the details, we need to develop some ideas about rotation of coordinate axes. In Figure 10.5.2a the axes of an .ry-coordinate system have been rotated about the origin through an angle 9 to produce a new jc'y'-coordinate system. As shown in the figure, each point P in the plane has coordinates (x', y') as well as coordinates (x, y). To see how the two are related, let r be the distance from the common origin to the point P, and let a be the angle shown in Figure lQ.5.2b. It follows that x — r cos(0 + a),
y = r sin($ + a)
(3)
and
(4)
Using familiar trigonometric identities, the relationships in (3) can be written as
x = r cos 9 cos a — r sin 9 sin a y = r sin 9 cos a + r cos 9 sin a and on substituting (4) in these equations we obtain the following relationships called the rotation equations: x = x' cos 9 — y' sin 9 y = x' sin 9 + y' cos 9
*• Figure 10.5.2
(b)
(a)
> Example 1 Suppose that the axes of an ry-coordinate system are rotated through an angle of 9 — 45 ° to obtain an jc'y'-coordinate system. Find the equation of the curve x2 - xy + y2 - 6 - 0
in ^'^'-coordinates. Solution. Substituting sin# = sin45° = l/\/2 and cos9 =cos45° = l/\/2 in (5) yields the rotation equations x'
x = —p V2
y'
-p V2
x'
y'
and y = — + — 72 V2
Substituting these into the given equation yields x' y' \2 ( *' y'\( x' y'\ ( x' y' \2 — - ^ = ) -I —--*-)[ — + -?-) + [—+ ^-\ -6 = 0 72 V2/ VV2 >/2/ VV2 x/27 \V2 V2.J
750
Chapter TO / Parametric and Polar Curves; Conic Sections
or x'2 - 2x'y' + y'2 - x'2 + y'2 + x'2 + 2x'y' + y'2
or x'2
y'2
12
4
=1
which is the equation of an ellipse (Figure 10.5.3). <
A Figure 10.5.3
If the rotation equations (5) are solved for x' and y' in terms of x and y, one obtains (Exercise 16): x' = xcos6 + y sind (6) y' = —x sin 6 + y cos 9
*• Example 2 Find the new coordinates of the point (2, 4) if the coordinate axes are rotated through an angle of 9 = 30°. Solution. Using the rotation equations in (6) with x = 2, y — 4, cos 9 = cos 30° = \/3/2, and sin$ — sin 30° — 1/2, we obtain x' = 2(73/2) + 4(1/2) = 73 + 2 y' = -2(1/2) + 4(73/2) = -1 + 2^3 Thus, the new coordinates are (V3 + 2, — 1 + 2-\/3). -^
ELIMINATING THE CROSS-PRODUCT TERM In Example 1 we were able to identify the curve x2 — x y + y2 — 6 — 0 as an ellipse because the rotation of axes eliminated the ry-term, thereby reducing the equation to a familiar form. This occurred because the new x'y'-ax&s were aligned with the axes of the ellipse. The following theorem tells how to determine an appropriate rotation of axes to eliminate the cross-product term of a second-degree equation in x and y.
10.5.1 THEOREM
If the equation
Ax2 + Bxy + Cy2 + Dx + Ey + F — 0
It is always possible to satisfy (8) with an angle 9 in the interval
0 < 9 < ji/2
(7)
is such that B ^ 0, and if an x'y'-coordinate system is obtained by rotating the xy-ax.es ' through an angle 6 satisfying A-C cot 29 = " 7 ^ (8) B , in x' y' -coordinates, Equation (7 ) will have the form A' x'2 + C'y'2 + D'x' + E'y' + F' = 0
We will always choose 6 in this way.
PROOF
Substituting (5) into (7) and simplifying yields A'x'2 + B'x'y' + C'y'2 + D'x' + E'y' + F' = 0
10.5 Rotation of Axes; Second-Degree Equations where
751
A' = A cos2 9 + B cos 9 sin 9 + C sin2 9 B' = B(cos2 9 ~ sin2 0) + 2(C - A) sin 6 cos 9 C" = A sin2 9 - B sin 6 cos 9 + C cos2 9
(9)
D' = D cos 9 + E sin 9
E' - -Dsm9 + Ecos9 F' = F
(Verify.) To complete the proof we must show that B' — 0 if cot 29 =
or, equivalently,
A-C B
A-C B
cos 29
sin 29
(10)
However, by using the trigonometric double-angle formulas, we can rewrite B' in the form 5' = ficos26>-(A-C)sin26>
Thus, B' = 0 if 9 satisfies (10). • *• Example 3
Identify and sketch the curve xy = 1.
Solution. As a first step, we will rotate the coordinate axes to eliminate the cross-product term. Comparing the given equation to (7), we have A = 0,
5 = 1,
C=0
Thus, the desired angle of rotation must satisfy cot2# =
A-C
=
B
0-0 1
=0
This condition can be met by taking 29 = n/2 or 9 = jr/4 = 45 °. Making the substitutions =cos45° - l/v^andsinfl - sin45° = 1/^/2 in (5) yields x' x = —= 72
y' -= 72
and
x' y' y = —= -\—— 72 72
Substituting these in the equation xy — 1 yields ,,'2
A Figure 10.5.4
which is the equation in the *'/ -coordinate system of an equilateral hyperbola with vertices at (\/2, 0) and (— \/2, 0) in that coordinate system (Figure 10.5.4). < In problems where it is inconvenient to solve cot 29 =
A-C B
for 9, the values of sin 9 and cos 0 needed for the rotation equations can be obtained by first calculating cos 29 and then computing sin 9 and cos 9 from the identities /1 - cos 29 sin 9 = , /
and
cos $ =
11 + cos 29
752
Chapter 10 / Parametric and Polar Curves; Conic Sections
Example 4
Identify and sketch the curve 153;t2 - I92xy + 91y2 - 30x - 40y - 200 = 0
Solution. We have A = 153, B =. -192, and C = 97, so 56 ~£~ -~~I92 ~~~'
A
cot 20 =
A Figure 10.5.5
24 Since 0 is to be chosen in the range 0 < 9 < n/2, this relationship is represented by the triangle in Figure 10.5.5. From that triangle we obtain cos 20 = — ^, which implies that / I + cos 29 co& 9 = ,/
sin 9 =
1 - cos 26» 2
V
2
5
Substituting these values in (5) yields the rotation equations
x = \x'- f /
and y = f jc' + |/
and substituting these in turn in the given equation yields
-f (3xr - 4y') - f (4xr + 3y') -200 = 0
which simplifies to A Figure 10.5.6
^+ ^^ _50x, _ ^ = Q
or
x'2 + 9y'2 - 2x' - 8 = 0
Completing the square yields /2
1
+7=1 There is a method for deducing the
kind of curve represented by a seconddegree equation directly from the equation itself without rotating coordinate axes. For a discussion of this topic, see the section on the discriminant that appears in Web Appendix K.
which is the equation in the jc'j'-coordinate system of an ellipse with center (1, 0) in that coordinate system and semiaxes a = 3 and b = 1 (Figure 10.5.6). •*
UICK CHECK EXERCISES 10.5
(See page 754 for answers.)
1. Suppose that an ry-coordinate system is rotated 9 radians to produce a new x'y'-coordinate system. (a) x and y may be obtained from x', y', and 9 using the rotation equations x = and y = (b) x' and y' may be obtained from x, y, and 6 using the equations x' = and y' = 2. If the equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 is such that B ^ 0, then the jry-term in this equation can be
eliminated by a rotation of axes through an angle 6 satisfying cot 29 = __ 3. In each part, determine a rotation angle 9 that will eliminate the xy-term. (a) 2x2 + xy + 2y2 + x - y = 0 (b) x2 + 2V3xy + 3y2-2x + y = (c) 3x2 4. Express 2x2 + xy + 2y2 = 1 in the ^'/-coordinate system obtained by rotating the xy-coordinate system through the angle 9 = jr/4.
10.5 Rotation of Axes; Second-Degree Equations
753
EXERCISE SET 10.5
1. Let an ^'/-coordinate system be obtained by rotating an .ry-coordinate system through an angle of 9 = 60°. (a) Find the ^'^'-coordinates of the point whose xy-coordinates are (—2, 6). (b) Find an equation of the curve V3xy + y2 = 6 in ^'/-coordinates. (c) Sketch the curve in part (b), showing both jry-axes and x'y'-axes. 2. Let an .^'/-coordinate system be obtained by rotating an xy-coordinate system through an angle of 9 = 30°. (a) Find the ^'/-coordinates of the point whose xy-coordinates are (1, —\/3). (b) Find an equation of the curve 2x2 + 2\^3xy = 3 in ^'/-coordinates. (c) Sketch the curve in part (b), showing both xy-axes and x'/-axes. 3-12 Rotate the coordinate axes to remove the xy-term. Then identify the type of conic and sketch its graph. S 2
3. xy = -9
2
4. x - xy + y - 2 = 0
5. x2 + 4xy-2y2-6 = 0 6. 31x2 + lOVsYy + 2ly2 -144 = 0 2
2
7. x + 2^3xy + 3y + 2^3x -2y = 0 8. 34x2 - 24xy + 4ly2 -25 = 0 9. 9x2 - 24xy + 16y2 - 80x - 60y + 100 = 0 10. 5x2 - 6xy + 5y2 - &j2x + &^2y = 8
18. Let an x'/-coordinate system be obtained by rotating an xy-coordinate system through an angle 9. Explain how to find the xy-equation of a line whose ^'/-equation is known.
19-22 Show that the graph of the given equation is a parabola. Find its vertex, focus, and directrix, a 19. x2 + 2xy + y2 + 4^/2x - 4^2y = 0 20. x2 - 2V3xy + 3y2 - 8^3x - 8y = 0 21. 9x2 - 24xy + 16y2 - 80x - 60y + 100 = 0 22. x2 + 2^3xy + 3y2 + I6^3x -I6y-96 = 0 23-26 Show that the graph of the given equation is an ellipse. Find its foci, vertices, and the ends of its minor axis. 23. 288;t2 - 168xy + 337y2 - 3600 = 0 24. 25x2 - I4xy + 25y2 - 288 = 0 25. 31.x2 + lOV3xy + 21y2 - 32* + 32^3y - 80 = 0 26. 43x2 - l4-V3xy + 51 y2 - 36^3x - 36y - 540 = 0 27-30 Show that the graph of the given equation is a hyperbola. Find its foci, vertices, and asymptotes. 27. x2 - \0-V3xy + Uy2 + 64 = 0 28. 17x2 - 3\2xy + 108y2 - 900 = 0 29. 32y2 - 52xy - lx2 + 12^5x - l44^/5y + 900 = 0
11. 52x2 - 72xy + 73y2 + 40x + 30y-15 = 0
30. 2^2y2 + 5^2xy + 2^2x2 + l&x + 18y + 36^2 = 0
12. 6x2 + 24xy - y2 - Ux + 26y + 11 = 0
31. Show that the graph of the equation
13. Let an x'/-coordinate system be obtained by rotating anxycoordinate system through an angle of 45 °. Use (6) to find an equation of the curve 3x'2 + y'2 = 6 in ry-coordinates. 14. Let an x'/-coordinate system be obtained by rotating an xy-coordinate system through an angle of 30°. Use (5) to find an equation in .^/-coordinates of the curve y = x2.
is a portion of a parabola. [Hint: First rationalize the equation and then perform a rotation of axes.] FOCUS ON CONCEPTS
FOCUS ON CONCEPTS
32. Derive the expression for B' in (9).
15. Let an ^'/-coordinate system be obtained by rotating an xy-coordinate system through an angled. Prove: For every value of 9, the equation x2 + y2 = r2 becomes the equation x'2 + y'2 = r2. Give a geometric explanation.
33. Use (9) to prove that B2 - 4AC = B'2 - 4A'C' for all values of 9. 34. Use (9) to prove that A + C = A' + C' for all values of 9.
16. Derive (6) by solving the rotation equations in (5) for x' and y' in terms of x and y.
35. Prove: If A = C in (7), then the cross-product term can be eliminated by rotating through 45 °.
17. Let an ^'/-coordinate system be obtained by rotating an xy-coordinate system through an angle 9. Explain how to find the ry-coordinates of a point whose x'y'coordinates are known.
36. Prove: If B ^ 0, then the graph of x2 + Bxy + F = 0 is a hyperbola if F / 0 and two intersecting lines if F = 0.
754
Chapter TO / Parametric and Polar Curves; Conic Sections
•QUICK CHECK ANSWERS 10.5 1. (a) x'cosO — y ' s i n O ; x' sinO + y'cosO (b) xcos9 + y sin (3; — x sin 9 + y cos 9 2
2
4. 5x' + 3/ = 2
2.
B
3. (a) — (b) — (c) — 4 3 6
CONIC SECTIONS IN POLAR COORDINATES It will be shown later in the text that if an object moves in a gravitational field that is directed toward a fixed point (such as the center of the Sun), then the path of that object must be a conic section with the fixed point at a focus. For example, planets in our solar system move along elliptical paths with the Sun at a focus, and the comets move along parabolic, elliptical, or hyperbolic paths with the Sun at a focus, depending on the conditions under which they were born. For applications of this type it is usually desirable to express the equations of the conic sections in polar coordinates with the pole at a focus. In this section we will show how to do this. THE FOCUS-DIRECTRIX CHARACTERIZATION OF CONICS To obtain polar equations for the conic sections we will need the following theorem.
It is an unfortunate historical accident that the letter e is used for the base of the natural logarithm as well as for the eccentricity of conic sections. However, as a practical matter the appropriate in-
10.6.1 THEOREM (Focus-Directrix Property of Conies) Suppose that a point P moves in the plane determined by a fixed point (called the focus) and a fixed line (called the directrix), where the focus does not lie on the directrix. If the point moves in such a way that its distance to the focus divided by its distance to the directrix is some constant e (called the eccentricity), then the curve traced by the point is a conic section. Moreover, the conic is
terpretation will usually be clear from the context in which the letter is used.
(a) a parabola if e = 1
(b) an ellipse ifO < e < 1
(c) a hyperbola ife > 1.
We will not give a formal proof of this theorem; rather, we will use the specific cases in Figure 10.6.1 to illustrate the basic ideas. For the parabola, we will take the directrix to be x = —p, as usual; and for the ellipse and the hyperbola we will take the directrix to be x = a2/c. We want to show in all three cases that if P is a point on the graph, F is the focus, and D is the directrix, then the ratio PF/PD is some constant e, where e — 1 for the parabola, 0 < e < 1 for the ellipse, and e > 1 for the hyperbola. We will give the arguments for the parabola and ellipse and leave the argument for the hyperbola as an exercise.
10.6 Conic Sections in Polar Coordinates 755 For the parabola, the distance PF to the focus is equal to the distance PD to the directrix, so that PF/PD — 1, which is what we wanted to show. For the ellipse, we rewrite Equation (8) of Section 10.4 as
c Ia j(x - c}2 + y2 =a- -x = x a a\ c But the expression on the left side is the distance PF, and the expression in the parentheses on the right side is the distance PD, so we have shown that PF = -PD a
Thus, PF/PD is constant, and the eccentricity is (1)
If we rule out the degenerate case where a — 0 or c = 0, then it follows from Formula (7) of Section 10.4 that 0 < c < a , s o O < < ? < 1, which is what we wanted to show. We will leave it as an exercise to show that the eccentricity of the hyperbola in Figure 10.6.1 is also given by Formula (1), but in this case it follows from Formula (11) of Section 10.4 that c > a, so e > 1. ECCENTRICITY OF AN ELLIPSE AS A MEASURE OF FLATNESS
The eccentricity of an ellipse can be viewed as a measure of its flatness—as e approaches 0 the ellipses become more and more circular, and as e approaches 1 they become more and more flat (Figure 10.6.2). Table 10.6.1 shows the orbital eccentricities of various celestial objects. Note that most of the planets actually have fairly circular orbits. Table 10.6.1 CELESTIAL BODY
= 0.9
Ellipses with a common focus ; and equal semimajor axes
A Figure 10.6.2
Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto Halley's comet
ECCENTRICITY
0.206 0.007 0.017 0.093 0.048 0.056 0.046 0.010 0.249 0.970
POLAR EQUATIONS OF CONICS
Directrix
A Figure 10.6.3
Our next objective is to derive polar equations for the conic sections from their focusdirectrix characterizations. We will assume that the focus is at the pole and the directrix is either parallel or perpendicular to the polar axis. If the directrix is parallel to the polar axis, then it can be above or below the pole; and if the directrix is perpendicular to the polar axis, then it can be to the left or right of the pole. Thus, there are four cases to consider. We will derive the formulas for the case in which the directrix is perpendicular to the polar axis and to the right of the pole. As illustrated in Figure 10.6.3, let us assume that the directrix is perpendicular to the polar axis and d units to the right of the pole, where the constant d is known. If P is a point
756
Chapter 10 / Parametric and Polar Curves; Conic Sections
on the conic and if the eccentricity of the conic is e, then it follows from Theorem 10.6.1 that PF/PD = e or, equivalently, that PF = ePD
(2)
However, it is evident from Figure 10.6.3 that PF = r and PD = d — r cos 9. Thus, (2) can be written as r = e(d — r cos9) which can be solved for r and expressed as ed 1 + e cos 6
(verify). Observe that this single polar equation can represent a parabola, an ellipse, or a hyperbola, depending on the value of e. In contrast, the rectangular equations for these conies all have different forms. The derivations in the other three cases are similar.
10.6.2 THEOREM If a conic section with eccentric ity e is positioned in a polar coordinate system so that its focus is at the pole and the c orresponding directrix is d units from the pole and is either paral lei or perpendicular to the polar axis, then the equation of the conic has one offourpos sible forms, depending on its orientation: Directrix
Directrix
Focus \ /
/ Focus
1
ed 1 + ecosO Directrix ri ght of pole
ed 1 - e cos 9 Directrix left of pole
v°<
Directrix ^«- Foe us\
\
\
US/ ^^
Directrix
ed
ed
1 + e sin 0 Directrix above pole
(3-4)
1 — e sin 9
(5-6)
Directrix below pole
SKETCHING CONICS IN POLAR COORDINATES
Precise graphs of conic sections in polar coordinates can be generated with graphing utilities. However, it is often useful to be able to make quick sketches of these graphs that show their orientations and give some sense of their dimensions. The orientation of a conic relative to the polar axis can be deduced by matching its equation with one of the four forms in Theorem 10.6.2. The key dimensions of a parabola are determined by the constant p (Figure 10.4.5) and those of ellipses and hyperbolas by the constants a, b, and c (Figures 10.4.11 and 10.4.20). Thus, we need to show how these constants can be obtained from the polar equations.
10.6 Conic Sections in Polar Coordinates Directrix
Tr/2
Example 1 Sketch the graph of r =
I-cos9
757
in polar coordinates.
Solution. The equation is an exact match to (4) with d = 2 and e = 1. Thus, the graph is a parabola with the focus at the pole and the directrix 2 units to the left of the pole. This tells us that the parabola opens to the right along the polar axis and p = 1. Thus, the parabola looks roughly like that sketched in Figure 10.6.4. < All of the important geometric information about an ellipse can be obtained from the values o f a , b , and c in Figure 10.6.5. One way to find these values from the polar equation of an ellipse is based on finding the distances from the focus to the vertices. As shown in the figure, let ro be the distance from the focus to the closest vertex and r\ the distance to the farthest vertex. Thus,
A Figure 10.6.4
r0 = a — c
and
r\ = a
(7)
from which it follows that
(8-9) Moreover, it also follows from (7) that
A Figure 10.6.5
r0r, =a2-c2=b2 In words, Formula (8) states that a is the arithmetic average (also called the arithmetic mean) of ro and r\, and Formula (10) states that b is the geometric mean of ro and r\.
Thus,
(10)
b—
Example 2
Find the constants a, b, and c for the ellipse r —
2 + cos 9
Directrix
Solution. This equation does not match any of the forms in Theorem 10.6.2 because they all require a constant term of 1 in the denominator. However, we can put the equation into one of these forms by dividing the numerator and denominator by 2 to obtain
1 + \ cos 9 A Figure 10.6.6
This is an exact match to (3) with d — 6 and e — \, so the graph is an ellipse with the directrix 6 units to the right of the pole. The distance ro from the focus to the closest vertex can be obtained by setting 9 = 0 in this equation, and the distance r\ to the farthest vertex can be obtained by setting 9 = n. This yields 1 + icosO
\
1
= cos n
i
Thus, from Formulas (8), (10), and (9), respectively, we obtain
a - \(r\ + r0) = 4,
b — ^7^ — 2-$,
c — \(r\ - r0) — 2
The ellipse looks roughly like that sketched in Figure 10.6.6. -4
A Figure 10.6.7
All of the important information about a hyperbola can be obtained from the values of a, b, and c in Figure 10.6.7. As with the ellipse, one way to find these values from the polar equation of a hyperbola is based on finding the distances from the focus to the vertices. As
758
Chapter TO / Parametric and Polar Curves; Conic Sections
shown in the figure, let ro be the distance from the focus to the closest vertex and r\ the distance to the farthest vertex. Thus, — c —a
r\ = c + a
and
(11)
from which it follows that (12-13)
c= Un +r 0 )
))
Moreover, it also follows from (11) that In words, Formula (13) states that c is the arithmetic mean of rg and n, and Formula (14) states that b \s the geometric mean of rg and r\.
r(>r\ = c2 — a 2 = bj 2
from which it follows that
> Example 3
(14)
Sketch the graph of r =
2
:— in polar coordinates. 1 + 2 sin 9
Solution. This equation is an exact match to (5) with d — 1 and e — 2. Thus, the graph is a hyperbola with its directrix 1 unit above the pole. However, it is not so straightforward to compute the values of ro and r\, since hyperbolas in polar coordinates are generated in a strange way as 9 varies from 0 to 2n. This can be seen from Figure 10.6.8a, which is the graph of the given equation in rectangular Or -coordinates. It follows from this graph that the corresponding polar graph is generated in pieces (see Figure 10.6.8&): • As 9 varies over the interval 0 < 9 < In/6, the value of r is positive and varies from 2 down to 2/3 and then to +00, which generates part of the lower branch. • As 9 varies over the interval In/6 < 9 < 37T/2, the value of r is negative and varies from —oo to —2, which generates the right part of the upper branch. • As 9 varies over the interval 3jr/2 < 9 < 1 ljr/6, the value of r is negative and varies from —2 to —oo, which generates the left part of the upper branch. • As 9 varies over the interval llTT/6 < 9 < 2n, the value of r is positive and varies from +oc to 2, which fills in the missing piece of the lower right branch.
7k 3* 1 to
2 "T
-3
1 + 2 sin 0
Rough sketch
(a)
(b)
A Figure 10.6.8
To obtain a rough sketch of a hyperbola, it is generally sufficient to locate the center, the asymptotes, and the points where e = 0,8 = nil, 6 — n, and 6 = 3jr/2.
It is now clear that we can obtain A-Q by setting 6 = n/2 and r\ by setting 9 = 3n/2. Keeping in mind that ro and r\ are positive, this yields l+2sin(7r/2)
2 3'
2
+2sin(37T/2)
=2
10.6 Conic Sections in Polar Coordinates
759
Thus, from Formulas (12), (14), and (13), respectively, we obtain
a = ~(r\ -r0) = -,
b=
1
c=
4
_(ri+ro) = -
Thus, the hyperbola looks roughly like that sketched in Figure 10.6.8c. •<
APPLICATIONS IN ASTRONOMY In 1609 Johannes Kepler published a book known as Astronomia Nova (or sometimes Commentaries on the Motions of Mars) in which he succeeded in distilling thousands of years of observational astronomy into three beautiful laws of planetary motion (Figure 10.6.9).
10.6.3
Equal areas are swept out in equal times, and the square of the period T is proportional to a 3
Apogee
A Figure 10.6.10
—• Perigee
KEPLER'S LAWS
•
First law (Law of Orbits). Each planet moves in an elliptical orbit with the Sun at a focus.
•
Second law (Law of Areas). The radial line from the center of the Sun to the center of a planet sweeps out equal areas in equal times.
•
Third law (Law of Periods). The square of a planet's period (the time it takes the planet to complete one orbit about the Sun) is proportional to the cube of the semimajor axis of its orbit.
Kepler's laws, although stated for planetary motion around the Sun, apply to all orbiting celestial bodies that are subjected to a single central gravitational force—artificial satellites subjected only to the central force of Earth's gravity and moons subjected only to the central gravitational force of a planet, for example. Later in the text we will derive Kepler's laws from basic principles, but for now we will show how they can be used in basic astronomical computations. In an elliptical orbit, the closest point to the focus is called the perigee and the farthest point the apogee (Figure 10.6.10). The distances from the focus to the perigee and apogee
Johannes Kepler (1571-1630) German astronomer and physicist. Kepler, whose work provided our contemporary view of planetary motion, led a fascinating but ill-starred life. His alcoholic father made him work in a family-owned tavern as a child, later withdrawing him from elementary school and hiring him out as a field laborer, where the boy contracted smallpox, permanently crippling his hands and impairing his eyesight. In later years, Kepler's first wife and several children died, his mother was accused of witchcraft, and being a Protestant he was often subjected to persecution by Catholic authorities. He was often impoverished, eking out a living as an astrologer and prognosticator. Looking back on his unhappy childhood, Kepler described his father as "criminally inclined" and "quarrelsome" and his mother as "garrulous" and "bad-tempered." However, it was his mother who left an indelible mark on the sixyear-old Kepler by showing him the comet of 1577; and in later life he personally prepared her defense against the witchcraft charges. Kepler became acquainted with the work of Copernicus as a student at the University of Tubingen, where he received his master's de-
gree in 1591. He continued on as a theological student, but at the urging of the university officials he abandoned his clerical studies and accepted a position as a mathematician and teacher in Graz, Austria. However, he was expelled from the city when it came under Catholic control, and in 1600 he finally moved on to Prague, where he became an assistant at the observatory of the famous Danish astronomer Tycho Brahe. Brahe was a brilliant and meticulous astronomical observer who amassed the most accurate astronomical data known at that time; and when Brahe died in 1601 Kepler inherited the treasure-trove of data. After eight years of intense labor, Kepler deciphered the underlying principles buried in the data and in 1609 published his monumental work, Astronomia Nova, in which he stated his first two laws of planetary motion. Commenting on his discovery of elliptical orbits, Kepler wrote, "I was almost driven to madness in considering and calculating this matter. I could not find out why the planet would rather go on an elliptical orbit (rather than a circle). Oh ridiculous me!" It ultimately remained for Isaac Newton to discover the laws of gravitation that explained the reason for elliptical orbits.
760
Chapter 10 / Parametric and Polar Curves; Conic Sections
are called the perigee distance and apogee distance, respectively. For orbits around the Sun, it is more common to use the terms perihelion and aphelion, rather than perigee and apogee, and to measure time in Earth years and distances in astronomical units (AU), where 1 AU is the semimajor axis a of the Earth's orbit (approximately 150 x 106 km or 92.9 x 106 mi). With this choice of units, the constant of proportionality in Kepler's third law is 1, since a = 1 AU produces a period of T — 1 Earth year. In this case Kepler's third law can be expressed as T = a372 (15)
Directrix
Shapes of elliptical orbits are often specified by giving the eccentricity e and the semimajor axis a, so it is useful to express the polar equations of an ellipse in terms of these constants. Figure 10.6.11, which can be obtained from the ellipse in Figure 10.6.1 and the relationship c = ea, implies that the distance d between the focus and the directrix is d =
a e
c=
a e
ea —
06)
from which it follows that ed = a(l — e 2 ). Thus, depending on the orientation of the ellipse, the formulas in Theorem 10.6.2 can be expressed in terms of a and e as r^
A Figure 10.6.11
a(l-e2) l i e cos 9
+: Directrix right of pole —: Directrix left of pole
a(\-e2) ~ 1 ± e sin 6
r=
(17-18)
+: Directrix above pole —: Directrix below pole
Moreover, it is evident from Figure 10.6.11 that the distances from the focus to the closest and farthest vertices can be expressed in terms of a and e as = a — ea = a(l — e) and r\ = a + ea = a(\ + e)
Halley's comet
A Figure 10.6.12
(19-20)
> Example 4 Halley's comet (last seen in 1986) has an eccentricity of 0.97 and a semimajor axis of a — 18.1 AU. (a) Find the equation of its orbit in the polar coordinate system shown in Figure 10.6.12. (b) Find the period of its orbit. (c) Find its perihelion and aphelion distances. Solution (a).
From (17), the polar equation of the orbit has the form
a(l-e2) 1 + e cos 9 1 2 But a(\ - e ) = 18.1[1 - (0.97) ] ~ 1.07. Thus, the equation of the orbit is 1.07 r= 1 +0.97 cos 9 Science Photo Library/Photo Researchers Halley's comet photographed April 21, 1910 in Peru.
Solution (b).
From (15), with a = 18.1, the period of the orbit is T = (18.1)3/2?»77years
Solution (c). Since the perihelion and aphelion distances are the distances to the closest and farthest vertices, respectively, it follows from (19) and (20) that r0=a-ea=a(l-e) = 18.1(1 - 0.97) % 0.543 AU n =a + ea = a(l+e) = 18.1(1 + 0.97) % 35.7 AU
10.6 Conic Sections in Polar Coordinates
761
or since 1 AU & 150 x 106 km, the perihelion and aphelion distances in kilometers are r0 = 18.1(1 - 0.97)(150 x 106) % 81,500,000 km n = 18.1(1 + 0.97)(150 x 106) « 5,350,000,000 km + Minimum distance
*• Example 5 An Apollo lunar lander orbits the Moon in an elliptic orbit with eccentricity e = 0.12 and semimajor axis a = 2015 km. Assuming the Moon to be a sphere of radius 1740 km, find the minimum and maximum heights of the lander above the lunar surface (Figure 10.6.13). Solution. If we let rn and rl denote the minimum and maximum distances from the center of the Moon, then the minimum and maximum distances from the surface of the Moon will be
4nin = r0 - 1740 dmax = ri - 1740
or from Formulas (19) and (20) Maximum distance A Figure 10.6.13
dmin = r0 - 1740 = a(l - e) - 1740 = 2015(0.88) - 1740 = 33.2 km rfmax = n - 1740 = a(l + e) - 1740 = 2015(1.12) - 1740 = 516.8 km +
QUICK CHECK EXERCISES 10.6
(See page 763 for answers.)
1. In each part, name the conic section described. (a) The set of points whose distance to the point (2, 3) is half the distance to the line x + y = 1 is (b) The set of points whose distance to the point (2, 3) is equal to the distance to the line x + y = I is (c) The set of points whose distance to the point (2, 3) is twice the distance to the line x + y = 1 is 2. In each part: (i) Identify the polar graph as a parabola, an ellipse, or a hyperbola; (ii) state whether the directrix is above, below, to the left, or to the right of the pole; and (iii) find the distance from the pole to the directrix. (a) r = —J—4 + cos 9
EXERCISE SET 10.6
(b) r =
-4cos6>
B Graphing Utility
1 -2 Find the eccentricity and the distance from the pole to the directrix, and sketch the graph in polar coordinates. 1. (a) r =
2-2cos6> 4 2. (a) r = 2 + 3 cos 9
(b) r =
2 + sin 9 5 (b) r = 3+ 3 sin0
| 3-4 Use Formulas (3)-(6) to identify the type of conic and its orientation. Check your answer by generating the graph with a graphing utility. • 3. (a) r =
1 - sin 9
1 (d) r = 4 + 4 sin 9 1 - sin 9 3. If the distance from a vertex of an ellipse to the nearest focus is r0, and if the distance from that vertex to the farthest focus is n, then the semimajor axis is a = and the semiminor axis is b = 4. If the distance from a vertex of a hyperbola to the nearest focus is rn, and if the distance from that vertex to the farthest focus is r\, then the semifocal axis is a = and the semiconjugate axis is b = (c) r =
(b) r =
4 + 3 sin 9
4. (a) r =
2-3sin0
(b) r =
12 4 + cos 9
5-6 Find a polar equation for the conic that has its focus at the pole and satisfies the stated conditions. Points are in polar coordinates and directrices in rectangular coordinates for simplicity. (In some cases there may be more than one conic that satisfies the conditions.) 5. (a) Ellipse; e = |; directrix x = 2. (b) Parabola; directrix x = 1. (c) Hyperbola; e = |; directrix y = 3.
762
Chapter 10 / Parametric and Polar Curves; Conic Sections
6. (a) Ellipse; ends of major axis (2, jr/2) and (6, 3;r/2). (b) Parabola; vertex (2, JT). (c) Hyperbola; e = A/2; vertex (2, 0). 7-8 Find the distances from the pole to the vertices, and then apply Formulas (8)-(10) to find the equation of the ellipse in rectangular coordinates. 6 1 7. (a) r = ——(b) r = 2 + sin 9 2-cos9 6 8. (a) r = (b) r = 5 + 2 cos 9 4 - 3 sin 6 9-10 Find the distances from the pole to the vertices, and then apply Formulas (12)-(14) to find the equation of the hyperbola in rectangular coordinates.
9. (a) r = , , ^ . „ 1 + 2 sin 9 4 10. (a) r = -2sm8
(b) r =
2-3cos6» 15 (b) r = 2 + 8 cos 0
11-12 Find a polar equation for the ellipse that has its focus at the pole and satisfies the stated conditions. • 11. (a) Directrix to the right of the pole; a = 8; e=\. (b) Directrix below the pole; a = 4; e = |. 12. (a) Directrix to the left of the pole; ft = 4; e = f . (b) Directrix above the pole; c = 5; e = |. 13. Find the polar equation of an equilateral hyperbola with a focus at the pole and vertex (5, 0). FOCUS ON CONCEPTS
14. Prove that a hyperbola is an equilateral hyperbola if and only if e = \/2. 15. (a) Show that the coordinates of the point P on the hyperbola in Figure 10.6.1 satisfy the equation
J(x - c)2 + y2 = -x -a a (b) Use the result obtained in part (a) to show that PF/PD = da. 16. (a) Show that the eccentricity of an ellipse can be expressed in terms of r$ and r\ as r\ -ro r\ +r0
(b) Show that
l-e 17. (a) Show that the eccentricity of a hyperbola can be expressed in terms of ro and r\ as
r\ +r0 r\ (b) Show that
e+1 e- 1
18. (a) Sketch the curves
1 +cos6»
and
r =
1
1 - cos 0
(b) Find polar coordinates of the intersections of the curves in part (a). (c) Show that the curves are orthogonal, that is, their tangent lines are perpendicular at the points of intersection. 19-22 True-False Determine whether the statement is true or false. Explain your answer. S; 19. If an ellipse is not a circle, then the eccentricity of the ellipse is less than one. 20. A parabola has eccentricity greater than one. 21. If one ellipse has foci that are farther apart than those of a second ellipse, then the eccentricity of the first is greater than that of the second. 22. If d is a positive constant, then the conic section with polar equation ^
1 + cos 0 is a parabola. 23-28 Use the following values, where needed: radius of the Earth = 4000 mi = 6440 km 1 year (Earth year) = 365 days (Earth days) 1 AU = 92.9 x 106 mi = 150 x 106 km 23. The dwarf planet Pluto has eccentricity e = 0.249 and semimajor axis a = 39.5 AU. (a) Find the period T in years. (b) Find the perihelion and aphelion distances. (c) Choose a polar coordinate system with the center of the Sun at the pole, and find a polar equation of Pluto's orbit in that coordinate system. (d) Make a sketch of the orbit with reasonably accurate proportions. 24. (a) Let a be the semimajor axis of a planet's orbit around the Sun, and let T be its period. Show that if T is measured in days and a is measured in kilometers, then T = (365 x I(r 9 )(a/150) 3/2 . (b) Use the result in part (a) to find the period of the planet Mercury in days, given that its semimajor axis is a = 57.95 x 106 km. (c) Choose a polar coordinate system with the Sun at the pole, and find an equation for the orbit of Mercury in that coordinate system given that the eccentricity of the orbit is e = 0.206. (d) Use a graphing utility to generate the orbit of Mercury from the equation obtained in part (c). 25. The Hale-Bopp comet, discovered independently on July 23, 1995 by Alan Hale and Thomas Bopp, has an orbital eccentricity of e = 0.9951 and a period of 2380 years. (a) Find its semimajor axis in astronomical units (AU). (b) Find its perihelion and aphelion distances. (««<.)
Chapter 10 Review Exercises (c) Choose a polar coordinate system with the center of the Sun at the pole, and find an equation for the Hale-Bopp orbit in that coordinate system. (d) Make a sketch of the Hale-Bopp orbit with reasonably accurate proportions.
763
On November 6, 1997 the spacecraft Galileo was placed in a Jovian orbit to study the moon Europa. The orbit had eccentricity 0.814580 and semimajor axis 3,514,918.9 km. Find Galileo's minimum and maximum heights above the molecular hydrogen layer (see the accompanying figure).
26. Mars has a perihelion distance of 204,520,000 km and an aphelion distance of 246,280,000 km. (a) Use these data to calculate the eccentricity, and compare your answer to the value given in Table 10.6.1. (b) Find the period of Mars. (c) Choose a polar coordinate system with the center of the Sun at the pole, and find an equation for the orbit of Mars in that coordinate system. (d) Use a graphing utility to generate the orbit of Mars from the equation obtained in part (c). 27. Vanguard 1 was launched in March 1958 into an orbit around the Earth with eccentricity e = 0.21 and semimajor axis 8864.5 km. Find the minimum and maximum heights of Vanguard 1 above the surface of the Earth. 28. The planet Jupiter is believed to have a rocky core of radius 10,000 km surrounded by two layers of hydrogen— a 40,000 km thick layer of compressed metallic-like hydrogen and a 20,000 km thick layer of ordinary molecular hydrogen. The visible features, such as the Great Red Spot, are at the outer surface of the molecular hydrogen layer.
Not to scale A Figure Ex-28
29. Writing Discuss how a hyperbola's eccentricity e affects the shape of the hyperbola. How is the shape affected as e approaches 1 ? As e approaches +00? Draw some pictures to illustrate your conclusions. 30. Writing Discuss the relationship between the eccentricity e of an ellipse and the distance z between the directrix and center of the ellipse. For example, if the foci remain fixed, what happens to z as e approaches 0?
•QIUICK CHECK ANSWERS 10.6 1. (a) an ellipse (b) a parabola (c) a hyperbola 2. (a) (i) ellipse (ii) to the right of the pole (iii) distance = 1 (b) (i) hyperbola (ii) to the left of the pole (iii) distance = j (c) (i) parabola (ii) above the pole (iii) distance = i (d) (i) parabola (ii) below the pole (iii) distance = 4 3. \(r\ + ro); Von 4. \(r\ —TO); «JrQr\
CHAPTER 10 REVIEW EXERCISES
Graphing Utility
0 1. Find parametric equations for the portion of the circle x2 + y2 = 2 that lies outside the first quadrant, oriented clockwise. Check your work by generating the curve with a graphing utility. 02. (a) Suppose that the equations * = f ( t ) , y = g(t) describe a curve C as t increases from 0 to 1. Find parametric equations that describe the same curve C but traced in the opposite direction as t increases from 0 to 1. (b) Check your work using the parametric graphing feature of a graphing utility by generating the line segment between (1,2) and (4, 0) in both possible directions as t increases from 0 to 1. 3. (a) Find the slope of the tangent line to the parametric curve x = t2 + 1, y = t/2 at t = -1 and t = 1 without eliminating the parameter. (b) Check your answers in part (a) by eliminating the parameter and differentiating a function of x.
4. Find dy/dx and d2y/dx2 at t = 2 for the parametric curve x
= \t\y = \t\
5. Find all values of t at which a tangent line to the parametric curve x = 2 cos t, y = 4 sin t is (a) horizontal (b) vertical. 6. Find the exact arc length of the curve
= l- 5t\ y = 4r5 -
(0 < t < 1)
7. In each part, find the rectangular coordinates of the point whose polar coordinates are given. (a) (-8,jr/4) (b) (7, -7T/4) (c) (8, 9ir/4) (d) (5,0) (e) (-2,-3;r/2) (f) (Q,n) 8. Express the point whose ^-coordinates are (— 1, 1) in polar coordinates with (a) r > 0, 0 < e < 2n (b) r < 0, 0 < 0 < 2n (c) r > 0, -TT < 6 < 7t (d) r < 0, -n < 9 < n.
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Chapter 10 / Parametric and Polar Curves; Conic Sections
9. In each part, use a calculating utility to approximate the polar coordinates of the point whose rectangular coordinates are given. (a) (4,3) (b) (2,-5) (c) (l.tair 1 !)
(d) Find a polar equation r = f(9) for the conchoid in part (a), and then find polar equations for the tangent lines to the conchoid at the pole.
10. In each part, state the name that describes the polar curve most precisely: a rose, a line, a circle, a lima§on, a cardioid, a spiral, a lemniscate, or none of these, (a) r = 3cos0 (b) r = cos36>
(d) r = 3 — cos 6 cos 9 (e) r = 1 - 3 cos 9 (f) r2 = 3cos9 2 (g) r = (3cos6>) (h) r = 1+39 11. In each part, identify the curve by converting the polar equation to rectangular coordinates. Assume that a > 0.
^ Figure Ex-21
(c) r =
(a) r = a sec2 -
(c) r = 4 esc
j
(b) r2 cos 29= a1
23. Find the area of the region that is enclosed by the cardioid r = 2 + 2cos9.
(d) r = 4 cos 6 + 8 sin 9
24. Find the area of the region in the first quadrant within the cardioid r = 1 + sin 9.
12. In each part, express the given equation in polar coordinates. (a) x = 1 (b) x2 + y2 = 9 2 2 (c) x + y - 6y = 0 (d) 4xy = 9 13-17 Sketch the curve in polar coordinates. *
13. 0 =
6 15. r = 3(1 - sin 9)
22. (a) Find the arc length of the polar curve r = 1/9 for 7T/4 < 9 < 7i/2. (b) What can you say about the arc length of the portion of the curve that lies inside the circle r = 1 ?
14. r = 6 cos 9
25. Find the area of the region that is common to the circles r = 1, r = 2 cos 9, and r = 2sin0. 26. Find the area of the region that is inside the cardioid r = a(\ + sin 9) and outside the circle r = a sin 9. 27-30 Sketch the parabola, and label the focus, vertex, and directrix, H
16. r2 = sin 29
17. r = 3 - cos 9 18. (a) Show that the maximum value of the y-coordinate of points on the curve r = 1 /\/0 for 9 in the interval (0, ji] occurs when tan 9 = 29. (b) Use a calculating utility to solve the equation in part (a) to at least four decimal-place accuracy. (c) Use the result of part (b) to approximate the maximum value of y for 0 < 9 < it. 19. (a) Find the minimum and maximum x -coordinates of points on the cardioid r = 1 — cos 9. (b) Find the minimum and maximum ^-coordinates of points on the cardioid in part (a).
27. y2 = 6x
28. x2 = -9y
29. (y + I) 2 = -l(x - 4)
30. (x - ±) 2 = 2(y - 1)
31-34 Sketch the ellipse, and label the foci, the vertices, and the ends of the minor axis. • x2 v2 31. — + — = 1 32. 4x2 + 9y2 = 36
33. 9(x - I) 2 + 16(y - 3)2 = 144 34. 3(x + 2)2 + 4(y + I) 2 = 12 35-37 Sketch the hyperbola, and label the vertices, foci, and asymptotes. B v-2
20. Determine the slope of the tangent line to the polar curve r = I + sin 9 at 9 = n/4.
(Q
is called a conchoid ofNicomedes (see the accompanying figure for the case 0 < a < b). (a) Describe how the conchoid
x = cot t + 4 cos t,
16
,,2
4
— i
36. 9y 2 - 4x2 = 36
37.
21. A parametric curve of the form x = acotf +bcost,
— —
y = 1 + 4 sin t
is generated as t varies over the interval 0 < t < lit. (b) Find the horizontal asymptote of the conchoid given in part (a). (c) For what values of / does the conchoid in part (a) have a horizontal tangent line? A vertical tangent line?
38. In each part, sketch the graph of the conic section with reasonably accurate proportions. (a) x2 - 4x + 8y + 36 = 0 (b) 3x2 + 4y2 - 30x - 8y + 67 = 0 (c) 4x2 - 5y2 - 8x - 30y - 21 = 0 39-41 Find an equation for the conic described.
•
39. A parabola with vertex (0, 0) and focus (0, —4). 40. An ellipse with the ends of the major axis (0, ±V5) and the ends of the minor axis (±1,0). 41. A hyperbola with vertices (0, ±3) and asymptotes y = ±x.
Chapter 10 Review Exercises 42. It can be shown in the accompanying figure that hanging cables form parabolic arcs rather than catenaries if they are subjected to uniformly distributed downward forces along their length. For example, if the weight of the roadway in a suspension bridge is assumed to be uniformly distributed along the supporting cables, then the cables can be modeled by parabolas. (a) Assuming a parabolic model, find an equation for the cable in the accompanying figure, taking the y-axis to be vertical and the origin at the low point of the cable. (b) Find the length of the cable between the supports.
46. 7;t2 + 2Vlxy + 5y2 - 4 = 0 47. 4V5.*:2 + 475*3; + ^5y2 + 5x - \0y = 0 48. Rotate the coordinate axes to show that the graph of \7x2 - 3l2xy + 108.V2 + 1080* - 1440^+4500 = 0 is a hyperbola. Then find its vertices, foci, and asymptotes. 49. In each part: (i) Identify the polar graph as a parabola, an ellipse, or a hyperbola; (ii) state whether the directrix is above, below, to the left, or to the right of the pole; and (iii) find the distance from the pole to the directrix.
(a) r = (c) r =
3 + cos 6
1 3(1+sin 61)
(b) r = (d) r =
1
- 3 cos 6» 3 -sin6>
50-51 Find an equation in ^-coordinates for the conic section that satisfies the given conditions. B
- 4200 ft A Figure Ex-42
43. It will be shown later in this text that if a projectile is launched with speed VQ at an angle a with the horizontal and at a height yo above ground level, then the resulting trajectory relative to the coordinate system in the accompanying figure will have parametric equations
x = (DO cos a)?,
765
(VQ sin a)? — ^
where g is the acceleration due to gravity. (a) Show that the trajectory is a parabola. (b) Find the coordinates of the vertex.
50. (a) Ellipse with eccentricity e = | and ends of the minor axis at the points (0, ±3). (b) Parabola with vertex at the origin, focus on the y-axis, and directrix passing through the point (7,4). (c) Hyperbola that has the same foci as the ellipse 3*2 + 16>>2 = 48 and asymptotes y = ±2x/3. 51. (a) Ellipse with center (—3, 2), vertex (2, 2), and eccentricity e = |. (b) Parabola with focus (—2, —2) and vertex (—2, 0). (c) Hyperbola with vertex (—1,7) and asymptotes y - 5 = ±8(* + 1). 52. Use the parametric equations x = a cost, y = bsint to show that the circumference C of an ellipse with semimajor axis a and eccentricity e is fir/2 V 1 — e2s'm2 udu C = 4a I
•« Figure Ex-43
44. Mickey Mantle is recognized as baseball's unofficial king of long home runs. OnApril 17, 1953 Mantle blasted a pitch by Chuck Stobbs of the hapless Washington Senators out of Griffith Stadium, just clearing the 50 ft wall at the 391 ft marker in left center. Assuming that the ball left the bat at a height of 3 ft above the ground and at an angle of 45 °, use the parametric equations in Exercise 43 with g = 32 ft/s 2 to find (a) the speed of the ball as it left the bat (b) the maximum height of the ball (c) the distance along the ground from home plate to where the ball struck the ground. 45-47 Rotate the coordinate axes to remove the xy-term, and then name the conic. •
45. x2 + y2 - 3xy - 3 = 0
Jo 53. Use Simpson's rule or the numerical integration capability of a graphing utility to approximate the circumference of the ellipse 4;c2 + 9y2 = 36 from the integral obtained in Exercise 52. 54. (a) Calculate the eccentricity of the Earth's orbit, given that the ratio of the distance between the center of the Earth and the center of the Sun at perihelion to the distance between the centers at aphelion is |y. (b) Find the distance between the center of the Earth and the center of the Sun at perihelion, given that the average value of the perihelion and aphelion distances between the centers is 93 million miles. (c) Use the result in Exercise 52 and Simpson's rule or the numerical integration capability of a graphing utility to approximate the distance that the Earth travels in 1 year (one revolution around the Sun).
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Chapter 10 / Parametric and Polar Curves; Conic Sections
CHAPTER 10 MAKING CONNECTIONS
0 CAS
[c] 1. Recall from Section 6.6 that the Fresnel sine and cosine functions are defined as
S(x) =
f sin• ('I -
Jo
\ 2 /
dt
and C(x) =
cos cosI|— -^- I| rff V 2/
/./o
triangle and the other end to a fixed point F. A pencil holds the string taut against the base of the triangle as the edge opposite Q slides along a horizontal line L below F. Show that the pencil traces an arc of a parabola with focus F and directrix L.
The following parametric curve, which is used to study amplitudes of light waves in optics, is called a clothoid or Cornu spiral in honor of the French scientist Marie Alfred Cornu (1841-1902):
/"
x = C(t) = I cos Jo
(nu2\
du
\ 2 )
y = S(t) =
( — 00 < t <
+00)
•4 Figure Ex-3
du
(a) Use a CAS to graph the Cornu spiral. (b) Describe the behavior of the spiral as t -> +co and as t-+ —oo. (c) Find the arc length of the spiral for — 1 < t < 1. 2. (a) The accompanying figure shows an ellipse with semimajor axis a and semiminor axis b. Express the coordinates of the points P, Q, and R in terms of t. (b) How does the geometric interpretation of the parameter t differ between a circle x = a cost,
y = a smf
x=acost,
y=
4. The accompanying figure shows a method for constructing a hyperbola. A corner of a ruler is pinned to a fixed point FI and the ruler is free to rotate about that point. A piece of string whose length is less than that of the ruler is tacked to a point F2 and to the free corner Q of the ruler on the same edge as FI . A pencil holds the string taut against the top edge of the ruler as the ruler rotates about the point FI . Show that the pencil traces an arc of a hyperbola with foci FI and ¥^.
and an ellipse
•4 Figure Ex-4
5. Consider an ellipse E with semimajor axis a and semiminor axis b, and set c = ^/a2 — b2. (a) Show that the ellipsoid that results when E is revolved about its major axis has volume V = \jtab1 and surface S = 2jtab I - + - sin"1 a c a •4 Figure Ex-2
3. The accompanying figure shows Kepler's method for constructing a parabola. A piece of string the length of the left edge of the drafting triangle is tacked to the vertex Q of the
(b) Show that the ellipsoid that results when E is revolved about its minor axis has volume V = ^7ta2b and surface area / a b a+c S = Inab - + - In \b c b
EXPANDING THE CALCULUS HORIZON To learn how polar coordinates and conic sections can be used to analyze the possibility of a collision between a comet and Earth, see the module entitled Comet Collision at: www.wiley.com/college/anton
GRAPHING FUNCTIONS USING CALCULATORS AND COMPUTER ALGEBRA SYSTEMS GRAPHING CALCULATORS AND COMPUTER ALGEBRA SYSTEMS
The development of new technology has significantly changed how and where mathematicians, engineers, and scientists perform their work, as well as their approach to problem solving. Among the most significant of these developments are programs called Computer Algebra Systems (abbreviated CAS), the most common being Mathematica and Maple* Computer algebra systems not only have graphing capabilities, but, as their name suggests, they can perform many of the symbolic computations that occur in algebra, calculus, and branches of higher mathematics. For example, it is a trivial task for a CAS to perform the factorization 23x5 + I41x4 - I39x3 - 3464x2 - 2112* + 23040 = (x
5)(x - 3)2(x + 8)3
or the exact numerical computation 63456 V 3 177295
43907 22854377 J
2251912457164208291259320230122866923 382895955819369204449565945369203764688375
Technology has also made it possible to generate graphs of equations and functions in seconds that in the past might have taken hours. Figure A. 1 shows the graphs of the function f ( x ) = x4 — x3 — 2x2 produced with various graphing utilities; the first two were generated with the CAS programs, Mathematica and Maple, and the third with a graphing calculator. Graphing calculators produce coarser graphs than most computer programs but have the advantage of being compact and portable.
Generated by Mathematica
Generated by Maple
Generated by a graphing calculator
A Figure A.I Mathematica is a product of Wolfram Research, Inc.; Maple is a product of Waterloo Maple Software, Inc.
Al
A2
Appendix A: Graphing Functions Using Calculators and Computer Algebra Systems
(b,d)
(a,d)
[c,d]
(b,c)
(a,c)
The window [a, b] x [c, d}
A Figure A.2
VIEWING WINDOWS Graphing utilities can only show a portion of the xy-plane in the viewing screen, so the first step in graphing an equation is to determine which rectangular portion of the ry-plane you want to display. This region is called the viewing window (or viewing rectangle). For example, in Figure A.I the viewing window extends over the interval [—3, 3] in the redirection and over the interval [—4,4] in the y-direction, so we denote the viewing window by [—3, 3] x [—4,4] (read "[—3, 3] by [—4, 4]"). In general, if the viewing window is [a, b] x [c, d], then the window extends between x = a and x = b in the ;t-direction and between y = c and y — d in the y-direction. We will call [a, b] the x-interval for the window and [c, d] the y-interval for the window (Figure A.2). Different graphing utilities designate viewing windows in different ways. For example, the first two graphs in Figure A. 1 were produced by the commands Plot[x"4 - x~3 -2*x~2, {x, -3, (Mathematica)
TECHNOLOGY MASTERY Use your graphing utility to generate the graph of the function 4
f(x) = x
- x^ - 2x2
in the window [—3, 3] x [—4,4].
-2 -4L
- 3 - 2 - 1 0 1 2 : Generated by Mathematica
plot(x~4 - x~3
3}, PlotRange->{-4, 4}]
-2*x~2, x = - 3 . . 3 , y = - 4 . . 4 ) ;
(Maple) and the last graph was produced on a graphing calculator by pressing the GRAPH button after setting the values for the variables that determine the x -interval and y-interval to be xMax = 3,
yMin = —4,
yMax = 4
B TICK MARKS AND GRID LINES To help locate points in a viewing window, graphing utilities provide methods for drawing tick marks (also called scale marks). With computer programs such as Mathematica and Maple, there are specific commands for designating the spacing between tick marks, but if the user does not specify the spacing, then the programs make certain default choices. For example, in the first two parts of Figure A. 1 , the tick marks shown were the default choices. On some graphing calculators the spacing between tick marks is determined by two scale variables (also called scale factors), which we will denote by xScl
and
yScl
(The notation varies among calculators.) These variables specify the spacing between the tick marks in the x- and y-directions, respectively. For example, in the third part of Figure A. 1 the window and tick marks were designated by the settings
Generated by a graphing calculator A Figure A.3
[-5, 5] x [-5, 5] *Scl = 0.5, yScl = 10
A Figure A.4
= —3
jcMax = 3
yMin = -4
yMax = 4
jcScl = 1
yScl = 1
Most graphing utilities allow for variations in the design and positioning of tick marks. For example, Figure A.3 shows two variations of the graphs in Figure A.I; the first was generated on a computer using an option for placing the ticks and numbers on the edges of a box, and the second was generated on a graphing calculator using an option for drawing grid lines to simulate graph paper.
> Example 1 Figure A.4 shows the window [—5, 5] x [—5, 5] with the tick marks spaced 0.5 unit apart in the x-direction and 10 units apart in the y-direction. No tick marks are actually visible in the y-direction because the tick mark at the origin is covered by the jc-axis, and all other tick marks in that direction fall outside of the viewing window.
Appendix A: Graphing Functions Using Calculators and Computer Algebra Systems
A3
*• Example 2 Figure A.5 shows the window [-10, 10] x [-10, 10] with the tick marks spaced 0.1 unit apart in the x- and y-directions. In this case the tick marks are so close together that they create thick lines on the coordinate axes. When this occurs you will usually want to increase the scale factors to reduce the number of tick marks to make them legible. -4
TECHNOLOGY MASTERY
Graphing calculators provide a way of clearing all settings and returning them to default values. For example, on one calculator the default window is [-10, 10] x [—10, 10] and the default scale factors arexScI = landyScI = 1. Read your documentation to determine the default values for your calculator and how to restore the default settings. If you are using a CAS, read your documentation to determine the commands for specifying the spacing between tick marks.
CHOOSING A VIEWING WINDOW [-10, 10] x [-10, 10] xSc\ = 0.1,yScl = 0.1
A Figure A.5
When the graph of a function extends indefinitely in some direction, no single viewing window can show it all. In such cases the choice of the viewing window can affect one's perception of how the graph looks. For example, Figure A.6 shows a computer-generated graph of y — 9 — x2, and Figure A.7 shows four views of this graph generated on a calculator. • In part (a) the graph falls completely outside of the window, so the window is blank (except for the ticks and axes). • In part (b) the graph is broken into two pieces because it passes in and out of the window. • In part (c) the graph appears to be a straight line because we have zoomed in on a very small segment of the curve. • In part (d) we have a more revealing picture of the graph shape because the window encompasses the high point on the graph and the intersections with the x-axis.
A Figure A.6
[-2, 2] x [-2, 2] *Scl = \,ySc\ = 1
[-4,4] x [-2,5] *Scl = l.yScI = 1
(a)
(*)
[2.5, 3.5] x [-1, 1] *Scl =0.1,yScl= 1
[-4, 4] x [-3, 10] *Scl = l , y S c l = 1
(c) A Figure A.7 Four views of y = 9 — x2
A4
Appendix A: Graphing Functions Using Calculators and Computer Algebra Systems
The following example illustrates how the domain and range of a function can be used to find a good viewing window when the graph of the function does not extend indefinitely in both the x- and y-directions. > Example 3 Use the domain and range of the function f(x) = V l 2 — 3x2 to determine a viewing window that contains the entire graph. Solution. The natural domain of / is [—2, 2] and the range is [0, +/12] (verify), so the entire graph will be contained in the viewing window [—2, 2] x [0, V^2]- For clarity, it is desirable to use a slightly larger window to avoid having the graph too close to the edges of the screen. For example, taking the viewing window to be [—3, 3] x [—1,4] yields the graph in Figure A.8. -4 [-3, 3] x [-1,4] *Scl = \,ySc\ = 1 A Figure A.8
Sometimes it will be impossible to find a single window that shows all important features of a graph, in which case you will need to decide what is most important for the problem at hand and choose the window appropriately. > Example 4 Graph the equation y = x3 - ]2x2 + 18 in the following windows and discuss the advantages and disadvantages of each window. (a) [-10, 10] x [-10, 10] with *Scl = 1, yScl = 1 (b) [-20, 20] x [-20, 20] with *Scl = 1, yScl = 1 (c) [-20, 20] x [-300, 20] with jcScl = 1, yScl = 20 (d) [-5, 15] x [-300, 20] with zScl = 1, yScl = 20 (e) [1,2] x [-1, 1] with jtScl = 0.1, yScl = 0.1 Solution (a). The window in Figure A.9a has chopped off the portion of the graph that intersects the y-axis, and it shows only two of three possible real roots for the given cubic polynomial. To remedy these problems we need to widen the window in both the x- and y-directions. Solution (b). The window in Figure A.9b shows the intersection of the graph with the y-axis and the three real roots, but it has chopped off the portion of the graph between the two positive roots. Moreover, the ticks in the y-direction are nearly illegible because they are so close together. We need to extend the window in the negative y-direction and increase yScl. We do not know how far to extend the window, so some experimentation will be required to obtain what we want. Solution (c). The window in Figure A.9c shows all of the main features of the graph. However, we have some wasted space in the x-direction. We can improve the picture by shortening the window in the jc-direction appropriately. Solution (d). The window in Figure A.9d shows all of the main features of the graph without a lot of wasted space. However, the window does not provide a clear view of the roots. To get a closer view of the roots we must forget about showing all of the main features of the graph and choose windows that zoom in on the roots themselves. Solution (e). The window in Figure A.9e displays very little of the graph, but it clearly shows that the root in the interval [1, 2] is approximately 1.3. •<
Appendix A: Graphing Functions Using Calculators and Computer Algebra Systems
AS
[-10, 10] x [-10, 10] jtScI = \,ySc\ = 1
[-20, 20] x [-20, 20] jtScI = \,ySc\ = 1
[-20, 20] x [-300, 20] *Scl = l,ySc\ = 20
(a)
(ft)
(C)
[-5, 15] x [-300, 20] xScI = l.ySd = 20
l = 0.1,jScl =0.1
(d)
Figure A.9
(e)
Sometimes you will want to determine the viewing window by choosing the ^-interval and allowing the graphing utility to determine a y-interval that encompasses the maximum and minimum values of the function over the ^-interval. Most graphing utilities provide some method for doing this, so read your documentation to determine how to use this feature. Allowing the graphing utility to determine the ^-interval of the window takes some of the guesswork out of problems like that in part (b) of the preceding example.
TECHNOLOGY MASTERY
• ZOOMING
The process of enlarging or reducing the size of a viewing window is called zooming. If you reduce the size of the window, you see less of the graph as a whole, but more detail of the part shown; this is called zooming in. In contrast, if you enlarge the size of the window, you see more of the graph as a whole, but less detail of the part shown; this is called zooming out. Most graphing calculators provide menu items for zooming in or zooming out by fixed factors. For example, on one calculator the amount of enlargement or reduction is controlled by setting values for two zoom factors, denoted by xFact and yFact. If jcFact — 10 and
yFact = 5
then each time a zoom command is executed the viewing window is enlarged or reduced by a factor of 10 in the jc-direction and a factor of 5 in the y-direction. With computer programs such as Mathematica and Maple, zooming is controlled by adjusting the ;e-interval and y-interval directly; however, there are ways to automate this by programming.
[-5, 5] x [-1000, 1000] *Scl = l,}-Scl = 500
(a)
•
COMPRESSION
Enlarging the viewing window for a graph has the geometric effect of compressing the graph, since more of the graph is packed into the calculator screen. If the compression is sufficiently great, then some of the detail in the graph may be lost. Thus, the choice of the viewing window frequently depends on whether you want to see more of the graph or more of the detail. Figure A. 10 shows two views of the equation [-5, 5]x[-10, 10] *Scl = l.vScI = 1 A Figure A.10
y = x5(x-2) In part (a) of the figure the y-interval is very large, resulting in a vertical compression that obscures the detail in the vicinity of the jc-axis. In part (b) the y-interval is smaller, and
A6
Appendix A: Graphing Functions Using Calculators and Computer Algebra Systems
consequently we see more of the detail in the vicinity of the jc-axis but less of the graph in the ^-direction. > Example 5 The function f ( x ) = x + 0.01 sin(507T.r) is the sum of f\ (x) — x, whose graph is the line y = x, and /aOc) — 0.01 sin(507r;c), whose graph is a sinusoidal curve with amplitude 0.01 and period 2jr/50jt — 0.04. This suggests that the graph of f ( x ) will follow the general path of the line y — x but will have bumps resulting from the contributions of the sinusoidal oscillations, as shown in part (c) of Figure A. 11. Generate the four graphs shown in Figure A. 11 and explain why the oscillations are visible only in part (c). Solution. To generate the four graphs, you first need to put your utility in radian mode.* Because the windows in successive parts of the example are decreasing in size by a factor of 10, calculator users can generate successive graphs by using the zoom feature with the zoom factors set to 10 in both the x- and v-directions. (a) In Figure A. 1 la the graph appears to be a straight line because the vertical compression has hidden the small sinusoidal oscillations (their amplitude is only 0.01). (b) In Figure A.life small bumps begin to appear on the line because there is less vertical compression. (c) In Figure A.lie the oscillations have become clear because the vertical scale is more in keeping with the amplitude of the oscillations. (d) In Figure A.I la" the graph appears to be a straight line because we have zoomed in on such a small portion of the curve. -^
[-10, 10] x [-10, 10] xSc\ = \,ySc\ = 1
(a)
, l ] x [-1,1] l = 0.1,;yScl = 0.1 (b)
[-0.1, 0.1] x [-0.1, 0.1] ;tScl = 0.01,yScl =0.01
[-0.01, 0.01] x [-0.01, 0.01] jcScI = 0.001,yScl = 0.001
(c)
(d)
A Figure A.ll ASPECT RATIO DISTORTION
Figure A. 12a shows a circle of radius 5 and two perpendicular lines graphed in the window [-10, 10] x [-10, 10] with *Scl = 1 and yScl = 1. However, the circle is distorted and the lines do not appear perpendicular because the calculator has not used the same length for 1 unit on the x-axis and 1 unit on the y-axis. (Compare the spacing between the ticks on the axes.) This is called aspect ratio distortion. Many calculators provide a menu item for automatically correcting the distortion by adjusting the viewing window appropriately. For example, one calculator makes this correction to the viewing window [-10, 10] x [-10, 10] by changing it to [-16.9970674487,16.9970674487] x [-10, 10] (Figure A.\2b). With computer programs such as Mathematica and Maple, aspect ratio distortion is controlled with adjustments to the physical dimensions of the viewing window on the computer screen, rather than altering the x- and y-intervals of the viewing window. In this text we follow the convention that angles are measured in radians unless degree measure is specified.
Appendix A: Graphing Functions Using Calculators and Computer Algebra Systems
[-10, 10] x [-10, 10] >• Figure A. 12
A7
ArScI = l,ySc\ = 1
[-16.9970674487, 16.9970674487] x [-10, 10] xSc\ = l,ySc\ = 1
(a)
(*)
SAMPLING ERROR
63 Pixels
127 Pixels A viewing window with 63 rows of 127 pixels A Figure A.13
TECHNOLOGY MASTERY If you are using a graphing calculator, read the documentation to determine its resolution.
The viewing window of a graphing utility is composed of a rectangular grid of small rectangular blocks called pixels. For black-and-white displays each pixel has two states— an activated (or dark) state and a deactivated (or light state). A graph is formed by activating appropriate pixels to produce the curve shape. In one popular calculator the grid of pixels consists of 63 rows of 127 pixels each (Figure A. 13), in which case we say that the screen has a resolution of 127 x 63 (pixels in each row x number of rows). A typical resolution for a computer screen is 1280 x 1024. The greater the resolution, the smoother the graphs tend to appear on the screen. The procedure that a graphing utility uses to generate a graph is similar to plotting points by hand: When an equation is entered and a window is chosen, the utility selects the xcoordinates of certain pixels (the choice of which depends on the window being used) and computes the corresponding ^-coordinates. It then activates the pixels whose coordinates most closely match those of the calculated points and uses a built-in algorithm to activate additional intermediate pixels to create the curve shape. This process is not perfect, and it is possible that a particular window will produce a false impression about the graph shape because important characteristics of the graph occur between the computed points. This is called sampling error. For example, Figure A.14 shows the graph of y = cos(lO^jc) produced by a popular calculator in four different windows. (Your calculator may produce different results.) The graph in part (a) has the correct shape, but the other three do not because of sampling error: • In part (b) the plotted pixels happened to fall at the peaks of the cosine curve, giving the false impression that the graph is a horizontal line. • In part (c) the plotted pixels fell at successively higher points along the graph. • In part (d) the plotted points fell in some regular pattern that created yet another misleading impression of the graph shape.
[-1,1] x [-1,1] *Scl=0.5,;yScl = 0
(a)
[-12.6, 12.6] x[-l, 1] ;tScl = l.yScI =0.5
[-12.5, 12.6] x[-l, 1] *Scl = l,>Scl = 0.5
[-6, 6] x [-1,1] *Scl = l.yScI = 0.5
(b)
(c)
(d)
A Figure A.14 REMARK | For trigonometric graphs with rapid oscillations, Figure A.14 suggests that restricting the ^-interval to a few periods is likely to produce a more accurate representation about the graph shape.
A8
Appendix A: Graphing Functions Using Calculators and Computer Algebra Systems FALSE GAPS
Sometimes graphs that are continuous appear to have gaps when they are generated on a calculator. These false gaps typically occur where the graph rises so rapidly that vertical space is opened up between successive pixels. >• Example 6 Figure A. 15 shows the graph of the semicircle y = \/9 — x1 in two viewing windows. Although this semicircle has x -intercepts at the points x = ±3, part (a) of the figure shows false gaps at those points because there are no pixels with jc-coordinates ±3 in the window selected. In part (b) no gaps occur because there are pixels with x -coordinates x = ±3 in the window being used. •*
[-5, 5] x [-5, 5] ;tScl = 1,3-Scl = 1
[-6.3, 6.3] x [-5, 5] A:Scl = !,>>Scl = 1
(a)
(b)
> Figure A.15
[-10,10] x [-10,10] *Scl = l.yScI = 1
FALSE LINE SEGMENTS
y = I/O:- 1) with false line segments
In addition to creating false gaps in continuous graphs, calculators can err in the opposite direction by placing/a/se line segments in the gaps of discontinuous curves. *• Example 7 Figure A.16a shows the graph of y = 1 / ( x — 1) in the default window on a calculator. Although the graph appears to contain vertical line segments near x = 1, they should not be there. There is actually a gap in the curve at x — 1, since a division by zero occurs at that point (Figure A.I6b). •«
ERRORS OF OMISSION
| Actual curve shape of y = \l(x-
1)
A Figure A.16
Most graphing utilities use logarithms to evaluate functions with fractional exponents such as f ( x ) — x2/3 — \fx^. However, because logarithms are only defined for positive numbers, many (but not all) graphing utilities will omit portions of the graphs of functions with fractional exponents. For example, one calculator graphs y — x2^ as in Figure A.17a, whereas the actual graph is as in Figure A.I Ib. (For a way to circumvent this problem, see the discussion preceding Exercise 23.)
TECHNOLOGY MASTERY Determine whether your graphing utility produces the graph of the equation y = x1^ for both positive and negative values of x.
[-4, 4] x [-1, 4]
1
-4 -3 -2 -I
xSd = l , y S c l = 1 > Figure A.17
(a)
(b)
2
3
4
Appendix A: Graphing Functions Using Calculators and Computer Algebra Systems
[-10, 10] x [-10, 10] ;tScl = l,yScl = 1
A7
[-16.9970674487, 16.9970674487] x [-10, 10] *Scl = l,ySc\ = 1
> Figure A.12
(*)
SAMPLING ERROR
The viewing window of a graphing utility is composed of a rectangular grid of small rectangular blocks called pixels. For black-and-white displays each pixel has two states— an activated (or dark) state and a deactivated (or light state). A graph is formed by activating appropriate pixels to produce the curve shape. In one popular calculator the grid of pixels consists of 63 rows of 127 pixels each (Figure A. 13), in which case we say that the screen has a resolution of 127 x 63 (pixels in each row x number of rows). Atypical resolution for a computer screen is 1280 x 1024. The greater the resolution, the smoother the graphs tend to appear on the screen. The procedure that a graphing utility uses to generate a graph is similar to plotting points by hand: When an equation is entered and a window is chosen, the utility selects the xcoordinates of certain pixels (the choice of which depends on the window being used) and computes the corresponding y-coordinates. It then activates the pixels whose coordinates most closely match those of the calculated points and uses a built-in algorithm to activate additional intermediate pixels to create the curve shape. This process is not perfect, and it is possible that a particular window will produce a false impression about the graph shape because important characteristics of the graph occur between the computed points. This is called sampling error. For example, Figure A. 14 shows the graph of y — cos(107r;c) produced by a popular calculator in four different windows. (Your calculator may produce different results.) The graph in part (a) has the correct shape, but the other three do not because of sampling error:
63 Pixels
-127 PixelsA viewing window with 63 rows of 127 pixels A Figure A.13
TECHNOLOGY MASTERY If you are using a graphing calculator, read the documentation to determine its resolution.
• In part (b) the plotted pixels happened to fall at the peaks of the cosine curve, giving the false impression that the graph is a horizontal line. • In part (c) the plotted pixels fell at successively higher points along the graph. • In part (d) the plotted points fell in some regular pattern that created yet another misleading impression of the graph shape. '.
i A
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[-12.6, 12.6] X [-1, 1] jcScI = l,yScl = 0.5
[-12.5, 12.6] x [-1, 1] xScI = l.yScI = 0.5
[-6,6] x [-1,1]
*Scl = 0.5, yScI = 0.5
(a)
00
(C)
(d)
= l,.yScl = 0.5
A Figure A.14 REMARK I For trigonometric graphs with rapid oscillations, Figure A.14 suggests that restricting the ^-interval to I a few periods is likely to produce a more accurate representation about the graph shape.
A8
Appendix A: Graphing Functions Using Calculators and Computer Algebra Systems FALSE GAPS
Sometimes graphs that are continuous appear to have gaps when they are generated on a calculator. These false gaps typically occur where the graph rises so rapidly that vertical space is opened up between successive pixels. ^ Example 6 Figure A. 15 shows the graph of the semicircle y = \/9 — x2 in two viewing windows. Although this semicircle has *-intercepts at the points x — ±3, part (a) of the figure shows false gaps at those points because there are no pixels with x -coordinates ±3 in the window selected. In part (b) no gaps occur because there are pixels with .^-coordinates x = ±3 in the window being used. •*
[-5,5] x [-5,5] ;tScl = l,}>Scl = 1
[-6.3, 6.3] x [-5, 5] *Scl = l,yScl = 1
(a)
(b)
>• Figure A.15
[-10, 10] x [-10, 10] ;tScl = l,ySc\ = 1 : y = I/O- 1) with false line segments !
FALSE LINE SEGMENTS
In addition to creating false gaps in continuous graphs, calculators can err in the opposite direction by placing/a/se line segments in the gaps of discontinuous curves.
(a)
> Example 7 Figure A.16a shows the graph of y = l/(x — 1) in the default window on a calculator. Although the graph appears to contain vertical line segments near x = 1, they should not be there. There is actually a gap in the curve at x = 1, since a division by zero occurs at that point (Figure A. 16b). -4
ERRORS OF OMISSION
Actual curve shape of y = I/O- 1) j
A Figure A.16
Most graphing utilities use logarithms to evaluate functions with fractional exponents such as f ( x ) = x213 — Vx2. However, because logarithms are only defined for positive numbers, many (but not all) graphing utilities will omit portions of the graphs of functions with fractional exponents. For example, one calculator graphs y = x2/3 as in Figure A.17a, whereas the actual graph is as in Figure AAlb. (For a way to circumvent this problem, see the discussion preceding Exercise 23.)
TECHNOLOGY MASTERY Determine whether your graphing utility produces the graph of the equation y = x1'^ for both positive and negative values of x.
[-4, 4] x [-1, 4] ScI = l.yScI = 1
> Figure A.17
(a)
1
-4 -3 -2 -1
(b)
2
3
4
Appendix A: Graphing Functions Using Calculators and Computer Algebra Systems A9 H WHAT IS THE TRUE SHAPE OF A GRAPH?
Although graphing utilities are powerful tools for generating graphs quickly, they can produce misleading graphs as a result of compression, sampling error, false gaps, and false line segments. In short, graphing utilities can suggest graph shapes, but they cannot establish them with certainty. Thus, the more you know about the functions you are graphing, the easier it will be to choose good viewing windows, and the better you will be able to judge the reasonableness of the results produced by your graphing utility. • MORE INFORMATION ON GRAPHING AND CALCULATING UTILITIES
The main source of information about your graphing utility is its own documentation, and from time to time in this book we suggest that you refer to that documentation to learn some particular technique. • GENERATING PARAMETRIC CURVES WITH GRAPHING UTILITIES
Many graphing utilities allow you to graph equations of the form y = f ( x ) but not equations oftheformx — g(y). Sometimes you will be able to rewrite x = g (y) in the form y = f ( x ) ; however, if this is inconvenient or impossible, then you can graph x = g(y) by introducing a parameter t = y and expressing the equation in the parametric form x — g ( t ) , y — t. (You may have to experiment with various intervals for t to produce a complete graph.) >• Example 8 Solution. form as
Use a graphing utility to graph the equation x = 3y5 — 5y3 + 1.
If we let t — y be the parameter, then the equation can be written in parametric , , x = 3r5 -5? 3 + l, y = t
Figure A.18 shows the graph of these equations for —1.5 < t < 1.5. ^ Some parametric curves are so complex that it is virtually impossible to visualize them without using some kind of graphing utility. Figure A. 19 shows three such curves.
A Figure A.18
= 31cosf-7cos(31/7)f >> = 31sinr-7sin(31/7)r (0 < t < I4x)
x= 17cosf + 7cos(17/7)f y= 17sin(-7sin(17/7)<
x = cost + (l/2)cos It + (l/3)sin 17* (l/2)sin7r + (l/3)cos lit y=
( 0 < r < Ux)
A Figure A.19
GRAPHING INVERSE FUNCTIONS WITH GRAPHING UTILITIES
Most graphing utilities cannot graph inverse functions directly. However, there is a way of graphing inverse functions by expressing the graphs parametrically. To see how this can be done, suppose that we are interested in graphing the inverse of a one-to-one function /.
A10
Appendix A: Graphing Functions Using Calculators and Computer Algebra Systems
TECHNOLOGY MASTERY Try your hand at using a graphing utility to generate some parametric curves that you think are interesting or beautiful.
We know that the equation y = f(x) can be expressed parametrically as
y = f(t)
x=t,
(1)
and we know that the graph of /"' can be obtained by interchanging x and y, since this reflects the graph of / about the line y — x. Thus, from (1) the graph of /"' can be represented parametrically as (2) = f(t), y = t For example, Figure A. 20 shows the graph of f(x) = x5 + x + 1 and its inverse generated with a graphing utility. The graph of / was generated from the parametric equations
y = t5 +t + 1
x = t,
and the graph of /~' was generated from the parametric equations x =t
5
y -t
t
TRANSLATION
If a parametric curve C is given by the equations x — f ( t ) , y = g(t), then adding a constant to f ( t ) translates the curve C in the x-direction, and adding a constant to g(t) translates it in the y-direction. Thus, a circle of radius r, centered at (XQ, yo) can be represented parametrically as
x = XQ + r cos t,
(0 < t < 2n)
y = yo + r sin f
(Figure A.21). If desired, we can eliminate the parameter from these equations by noting (x - x0)2 + (y - y0)2 = (r cos t)2 + (r sin t)2 = r2
Thus, we have obtained the familiar equation in rectangular coordinates for a circle of radius r, centered at (XQ, yo):
- y0)2 = r2
(x - x0)2
SCALING
If a parametric curve C is given by the equations x = f ( t ) , y — g(t), then multiplying f ( t ) by a constant stretches or compresses C in the jc-direction, and multiplying g(t) by a constant stretches or compresses C in the y-direction. For example, we would expect the parametric equations
A Figure A.21 T E C H N O L O G Y MASTERY Use the parametric capability of your graphing utility to generate a circle of radius 5 that is centered at (3, -2).
x — 3 cos t,
y = 2 sin t
(0 < t < 2;r)
to represent an ellipse, centered at the origin, since the graph of these equations results from stretching the unit circle x = cos t,
y = sin t
(0 < t < In)
by a factor of 3 in the ;c-direction and a factor of 2 in the y-direction. In general, if a and b are positive constants, then the parametric equations y = bsint
(0 < f < 2jr)
(3)
represent an ellipse, centered at the origin, and extending between —a and a on the x-axis and between — b and b on the y-axis (Figure A.22). The numbers a and b are called the semiaxes of the ellipse. If desired, we can eliminate the parameter t in (3) and rewrite the equations in rectangular coordinates as x = a cos t, y = b sin t (0 < t < 2n)
2
2
-T + -^ = 1
(4)
A Figure A.22 TECHNOLOGY I Use the parametric capability of your graphing utility to generate an ellipse that is centered at the MASTERY I origin and that extends between -4 and 4 in the x-direction and between -3 and 3 in the y-direction. Generate an ellipse with the same dimensions, but translated so that its center is at the point (2, 3).
Appendix A: Graphing Functions Using Calculators and Computer Algebra Systems EXERCISE SET A
Graphing Utility
0 1-4 Use a graphing utility to generate the graph of / in the given viewing windows, and specify the window that you think gives the best view of the graph. 1. /CO =x4-x2 (a) [-50, 50] x [-50, 50] (b) [-5, 5] x [-5, 5] (c) [-2, 2] x [-2, 2] (d) [-2, 2] x [-1, 1] (e) [-1.5,1.5] x [-0.5,0.5] 2. /CO = x5 - x3 (a) [-50, 50] x [-50, 50] (b) [-5, 5] x [-5, 5] (c) [-2, 2] x [-2, 2] (d) [-2,2]x[-l,l] (e) [-1.5,1.5] x [-0.5,0.5] 3. /CO = x2 + l2 (a) [-1, 1] x [13, 15] (b) [-2, 2] x [11, 15] (c) [—4,4] x [10,28] (d) A window of your choice 4. (a) [-1, 1] x [-15, -13] (b) [-2, 2] x [-15, -11] (c) [—4,4] x [—28, —10] (d) A window of your choice 5-6 Use the domain and range of / to determine a viewing window that contains the entire graph, and generate the graph in that window. x -x 5. f(x) = 6. f(x) = 07-14 Generate the graph of / in a viewing window that you think is appropriate. 2
7. f(x) = x -9x-36 9. /C0=2cos(8(k) 11. f(x) = 300 - 10*2 + O.OLc3 12. f(x) = x(30 - 2x)(25 - 2x)
13. f(x) = x2 + -
8. /CO =
x+1
x-9 10. /(*) = 12sin(jc/80)
14. /CO =
15-16 Generate the graph of / and determine whether your graphs contain false line segments. Sketch the actual graph and see if you can make the false line segments disappear by changing the viewing window. 15. /(*) = -r^—
16. /(*) =
*
4-x2 17. The graph of the equation x2 + y2 = 16 is a circle of radius 4 centered at the origin. (a) Find a function whose graph is the upper semicircle and graph it. (b) Find a function whose graph is the lower semicircle and graph it. (c) Graph the upper and lower semicircles together. If the combined graphs do not appear circular, see if you can adjust the viewing window to eliminate the aspect ratio distortion. (d) Graph the portion of the circle in the first quadrant. (e) Is there a function whose graph is the right half of the circle? Explain. X
2
-l
Al 1
18. In each part, graph the equation by solving for y in terms of x and graphing the resulting functions together. (a) x 2 /4 + y 2 / 9 = l (b) y2 - x2 = 1 19. Read the documentation for your graphing utility to determine how to graph functions involving absolute values, and graph the given equation. ( a ) ? = |*l ( b ) y = \x-l\ (c)y = M - l ( d ) y = |sin*| (e) y = sin |*| (f) y = \x\ - \x + l\ 20. Based on your knowledge of the absolute value function, sketch the graph of /CO = \x\/x. Check your result using a graphing utility. 21-22 Most graphing utilities provide some way of graphing functions that are defined piecewise; read the documentation for your graphing utility to find out how to do this. However, if your goal is just to find the general shape of the graph, you can graph each portion of the function separately and combine the pieces with a hand-drawn sketch. Use this method in these exercises. 21. Draw the graph of
7^2,
x <2
- 2x - 4, x > 2
22. Draw the graph of x3-x2,
1
X <1
1 -*'
1 <x <4
*2 cos Jx,
4 <x
0 23-24 We noted in the text that for functions involving fractional exponents (or radicals), graphing utilities sometimes omit portions of the graph. If /CO = xp/q , where p/q is a positive fraction in lowest terms, then you can circumvent this problem as follows: • If p is even and q is odd, then graph g(x) = x\p/q instead • If p is odd and q is odd, then graph gCO = (\x\/x)\x\plq instead of /CO23. (a) Generate the graphs of /CO = x2/5 and g(x) = \x\2/5, and determine whether your graphing utility missed part of the graph of/. (b) Generate the graphs of the functions /(*)=* 1//5 and gCO = (|*|/*)|*|1/s, and determine whether your graphing utility missed part of the graph of /. (c) Generate a graph of the function /CO = (x — l) 4/5 that shows all of its important features. (d) Generate a graph of the function /CO = (x + l) 3/4 that shows all of its important features. 24. The graphs of y = (x2 - 4)2/3 and y = [(x2 - 4)2]1/3 should be the same. Does your graphing utility produce
A12
Appendix A: Graphing Functions Using Calculators and Computer Algebra Systems the same graph for both equations? If not, what do you think is happening?
0 25. In each part, graph the function for various values of c, and write a paragraph or two that describes how changes in c affect the graph in each case. (a) y = ex2 (b) y — x2 + ex (c) y = x2 + x + c 0 26. The graph of an equation of the form y2 = x(x — a)(x — b) (where 0 < a < b) is called a bipartite cubic. The accompanying figure shows a typical graph of this type. (a) Graph the bipartite cubic y2 = x(x — \)(x — 2) by solving for y in terms of x and graphing the two resulting functions. (b) Find the x -intercepts of the bipartite cubic
y2 = x(x -a)(x - b) and make a conjecture about how changes in the values of a and b would affect the graph. Test your conjecture by graphing the bipartite cubic for various values of a wdb.
0 27. Based on your knowledge of the graphs of y = x and y = sin x, make a sketch of the graph of y = x sin x. Check your conclusion using a graphing utility. 0 28. What do you think the graph of y = sin (I/*) looks like? Test your conclusion using a graphing utility. [Suggestion: Examine the graph on a succession of smaller and smaller intervals centered at x = 0.] 0 29-30 Graph the equation using a graphing utility.
29. (a) x = y2 + 2y + I (b) x = sin y, —2n < y < 2n 30. (a) x = y + 2y3 - y5 (b) x = tan y, —n/2 < y < jr/2 0 31-34 Use a graphing utility and parametric equations to display the graphs of / and /"' on the same screen. 31. f ( x ) = x3+0.2x- 1, -\<x<2
32. f ( x ) = Jx2 + 2 + x,
-5 < x < 5
33. f(x) = cos(cos0.5jt),
0 < x <3
34. f ( x ) = x + sinx,
0 < x <6
0 35. (a) Find parametric equations for the ellipse that is centered at the origin and has intercepts (4, 0), (-4, 0), (0, 3), and (0, -3). (b) Find parametric equations for the ellipse that results by translating the ellipse in part (a) so that its center is at (-1,2). (c) Confirm your results in parts (a) and (b) using a graphing utility. •^ Figure Ex-26
TRIGONOMETRY REVIEW
ANGLES
Angles in the plane can be generated by rotating a ray about its endpoint. The starting position of the ray is called the initial side of the angle, the final position is called the terminal side of the angle, and the point at which the initial and terminal sides meet is called the vertex of the angle. We allow for the possibility that the ray may make more than one complete revolution. Angles are considered to bepositive if generated counterclockwise and negative if generated clockwise (Figure B.I).
Vertex
• Figure B.I
1 radian
A Figure B.2
Angles generated by more than one revolution
There are two standard measurement systems for describing the size of an angle: degree measure and radian measure. In degree measure, one degree (written 1 °) is the measure of an angle generated by 1/360 of one revolution. Thus, there are 360° in an angle of one revolution, 180° in an angle of one-half revolution, 90° in an angle of one-quarter revolution (a right angle), and so forth. Degrees are divided into sixty equal parts, called minutes, and minutes are divided into sixty equal parts, called seconds. Thus, one minute (written 1') is 1/60 of a degree, and one second (written 1") is 1/60 of a minute. Smaller subdivisions of a degree are expressed as fractions of a second. In radian measure, angles are measured by the length of the arc that the angle subtends on a circle of radius 1 when the vertex is at the center. One unit of arc on a circle of radius 1 is called one radian (written 1 radian or 1 rad) (Figure B.2), and hence the entire circumference of a circle of radius 1 is In radians. It follows that an angle of 360° subtends an arc of In radians, an angle of 180° subtends an arc of n radians, an angle of 90° subtends an arc ofn/2 radians, and so forth. Figure B.3 and Table B.I show the relationship between degree measure and radian measure for some important positive angles. A13
A14
Appendix B: Trigonometry Review
A Figure B.3
Table B.I Observe that in Table B.I, angles in degrees are designated by the degree symbol, but angles in radians have no units specified. This is standard practice—when no units are specified for an angle, it is understood that the units are radians.
DEGREES RADIANS
30°
45°
60°
90°
7T
K
n
n
6
4
3
2
120°
3
135°
150°
180°
270°
371
5TC
6
n
37T
4
360°
2
From the fact that n radians corresponds to 180°, we obtain the following formulas, which are useful for converting from degrees to radians and conversely.
— rad % 0.01745 rad 180 1 80 \° — I w 57° 17' 44.8" jt I
(
(1) (2)
Example 1 (a) Express 146° in radians.
(b) Express 3 radians in degrees.
Solution (a). From (1), degrees can be converted to radians by multiplying by a conversion factor of re1180. Thus, 146° = | — • 146 | rad = — rad % 2.5482 rad V180 90 Solution (b). From (2), radians can be converted to degrees by multiplying by a conversion factor of 180/jr. Thus,
3 rad = = A I 3 . 1!°Y = (^Y « 171.9° ^ RELATIONSHIPS BETWEEN ARC LENGTH, ANGLE, RADIUS, AND AREA
A Figure B.4
There is a theorem from plane geometry which states that for two concentric circles, the ratio of the arc lengths subtended by a central angle is equal to the ratio of the corresponding radii (Figure B.4). In particular, if s is the arc length subtended on a circle of radius r by a
Appendix B: Trigonometry Review
A15
central angle of 9 radians, then by comparison with the arc length subtended by that angle on a circle of radius 1 we obtain s r 0 ~I from which we obtain the following relationships between the central angle 9, the radius r, and the subtended arc length s when 9 is in radians (Figure B.5):
= s/r
and
s =r9
(3-4)
The shaded region in Figure B.5 is called a sector. It is a theorem from plane geometry that the ratio of the area A of this sector to the area of the entire circle is the same as the ratio of the central angle of the sector to the central angle of the entire circle; thus, if the angles are in radians, we have . „
A Figure B.5
Solving for A yields the following formula for the area of a sector in terms of the radius r and the angle 9 in radians: A = i/- 2 6 (5) TRIGONOMETRIC FUNCTIONS FOR RIGHT TRIANGLES The sine, cosine, tangent, cosecant, secant, and cotangent of a positive acute angle 9 can be defined as ratios of the sides of a right triangle. Using the notation from Figure B.6, these definitions take the following form:
A Figure B.6
sin6> =
side opposite 9 hypotenuse
cos 9 —
side adjacent to 9 hypotenuse
tan 9 =
side opposite 9 side adjacent to 9
esc 9 —
hypotenuse side opposite 9
x r'
seed =
hypotenuse side adjacent to 9
r x
y x'
adjacent to 9 cote = side side opposite 9
x y
y r'
r y (6)
We will call sin, cos, tan, esc, sec, and cot the trigonometric functions. Because similar triangles have proportional sides, the values of the trigonometric functions depend only on the size of 9 and not on the particular right triangle used to compute the ratios. Moreover, in these definitions it does not matter whether 9 is measured in degrees or radians.
> Example 2 Recall from geometry that the two legs of a 45°-45°-90° triangle are of equal size and that the hypotenuse of a 30°-60°-90° triangle is twice the shorter leg, where the shorter leg is opposite the 30° angle. These facts and the Theorem of Pythagoras yield Figure B.7. From that figure we obtain the results in Table B.2. ^
> Figure B.7
A16
Appendix B: Trigonometry Review Table B.2 sin 45° = 1A/2,
cos 45° = l/A/2,
tan 45° = 1
esc 45° = A/2,
sec 45° = \/2,
cot 45° = 1
sin 30° = 1/2,
cos 30° = V3/2,
tan 30° = 1/VJ
esc 30° = 2,
sec 30° = 2A/3,
cot 30° = A/3~
sin 60° = A/3/2,
cos 60° = 1/2,
tan 60° = \/3~
esc 60° = 2A/3,
sec 60° = 2,
cot 60° = l/A/3
ANGLES IN RECTANGULAR COORDINATE SYSTEMS
Because the angles of a right triangle are between 0° and 90°, the formulas in (6) are not directly applicable to negative angles or to angles greater than 90°. To extend the trigonometric functions to include these cases, it will be convenient to consider angles in rectangular coordinate systems. An angle is said to be in standard position in an ry-coordinate system if its vertex is at the origin and its initial side is on the positive x-axis (Figure B.8). y Terminal side. Initial side Initial side Terminal side
> Figure B.8
A Figure B.9
A positive angle in standard position
A negative angle in standard position !
To define the trigonometric functions of an angle 9 in standard position, construct a circle of radius r, centered at the origin, and let P(x, y) be the intersection of the terminal side of 0 with this circle (Figure B.9). We make the following definition.
B.I
DEFINITION V
sin 6 = -, r r esc 9 = —, y
JC
V
cos 9 = -, tan 0 = r x r x seed = —, cotO — — x y
Note that the formulas in this definition agree with those in (6), so there is no conflict with the earlier definition of the trigonometric functions for triangles. However, this definition applies to all angles (except for cases where a zero denominator occurs).
Appendix B: Trigonometry Review
(cos 6, sin 6)
A17
In the special case where r = 1, we have sin 9 = y and cos 9 = x, so the terminal side of the angle 9 intersects the unit circle at the point (cos 9, sin 9) (Figure B. 10). It follows from Definition B.I that the remaining trigonometric functions off? are expressible as (verify) land =
sin 9 cos#'
cot6» =
cos 6 sin 9
1 cos#'
1
tan 9'
csc 9 —
1
sin 6*
(7-10)
These observations suggest the following procedure for evaluating the trigonometric functions of common angles: A Figure B.10
•
Construct the angle 9 in standard position in an xy-coordinate system.
•
Find the coordinates of the intersection of the terminal side of the angle and the unit circle; the x- and y -coordinates of this intersection are the values of cos 9 and sin 9, respectively.
•
Use Formulas (7) through (10) to find the values of the remaining trigonometric functions from the values of cos 9 and sin 9.
>• Example 3
Evaluate the trigonometric functions of 9 = 150°.
Solution, Construct a unit circle and place the angle 0 = 150° in standard position (Figure B.ll). SinceZA0Pis30° and AQAPisa30°-60°-90° triangle, the leg APhas length | (half the hypotenuse) and the leg OA has length \/3/2 by the Theorem of Pythagoras. Thus, the coordinates of P are (—73/2, 1/2), from which we obtain 1 sin 150° = -, 2
73 cos 150°=- — , 2
csc 150° = . = 2, sin 150° A Figure B.ll
cot 150° =
> Example 4
tan 150° =
sec 150° =
sin 150° 1/2 = =—= cos 150° -73/2
1 73
2 cos 150°
"73
1 = -73 tan 150°
Evaluate the trigonometric functions of 9 — 5n/6.
Solution. Since 5n/6 = 150°, this problem is equivalent to that of Example 3. From that example we obtain STT 1 sin — — -,
5?r cos — —
73
SJT csc — —2, 6
5;r sec — — 6
2 —, 73
6
> Example 5
2
6
2
,
STT tan — —
6
1 —
V3
5n r cot — — —V3 •* 6
Evaluate the trigonometric functions of 9 — —n/2.
Solution. As shown in Figure B.12, the terminal side of 9 — —jr/2 intersects the unit circle at the point (0, —1), so A Figure B.12
sin(-jr/2) = -1,
cos(-7r/2) = 0
A18
Appendix B: Trigonometry Review
and from Formulas (7) through (10), tan(-7r/2)
j— = —
COS(—7T/2)
(undefined)
0
cos(-7r/2) 0 cot(-jr/2) = -r— — = —- = 0 sin(-jr/2) -1 I sec(-7r/2) — = - (undefined) COS(-7T/2)
0
1
1
CSC(-7T/2) = —
sin(-7T/2)
-
The reader should be able to obtain all of the results in Table B. 3 by the methods illustrated in the last three examples. The dashes indicate quantities that are undefined. Table B.3
6 =0
7T/6
7C/4
71/3
(0°)
(30°)
(45°)
(60°)
nil (90°)
(120°)
3;r/4 (135°)
5nl6 (150°)
TT (180°)
(270°)
In (360°)
sin 8
0
1/2
1A/2
V3/2
1
A/3/2
l/A/2
1/2
0
-1
0
cos 6
1
V3/2
1A/2
1/2
0
-1/2
-l/A/2
-V3/2
-1
0
1
tanfl
0
1/V3
1
V3
—
-V3
-1
-1/V3
0
—
0
CSC0
—
2
V2~
2/V3
1
2/V3
V2~
2
—
-1
—
sec 6
1
2/V3
V2
2
—
-2
-V2
-2/A/3
-1
—
1
cote
—
V3~
1
1A/3
0
-i/Vs
-1
-A/3
—
0
—
27T/3
37T/2
REMARK I It is only in special cases that exact values for trigonometric functions can be obtained; usually, a I calculating utility or a computer program will be required.
sin csc
All +
tan cot
cos sec
A Figure B.13
The signs of the trigonometric functions of an angle are determined by the quadrant in which the terminal side of the angle falls. For example, if the terminal side falls in the first quadrant, then x and y are positive in Definition B.I, so all of the trigonometric functions have positive values. If the terminal side falls in the second quadrant, then x is negative and y is positive, so sin and csc are positive, but all other trigonometric functions are negative. The diagram in Figure B.13 shows which trigonometric functions are positive in the various quadrants. The reader will find it instructive to check that the results in Table B.3 are consistent with Figure B.13. TRIGONOMETRIC IDENTITIES
A trigonometric identity is an equation involving trigonometric functions that is true for all angles for which both sides of the equation are defined. One of the most important identities in trigonometry can be derived by applying the Theorem of Pythagoras to the triangle in Figure B.9 to obtain 2_ 2 2 Dividing both sides by r2 and using the definitions of sin 9 and cos 9 (Definition B.I), we obtain the following fundamental result: sin2 0 + cos2 9 = 1
(11)
Appendix B: Trigonometry Review
A19
The following identities can be obtained from (1 1) by dividing through by cos2 0 and sin2 9, respectively, then applying Formulas (7) through (10): tan2 6 + 1 = sec2 9
(12)
1 + cot2 9 - esc2 9
(13)
If (x, y~) is a point on the unit circle, then the points (—x, y), (—x, — y), and (x, —y) also lie on the unit circle (why?), and the four points form corners of a rectangle with sides parallel to the coordinate axes (Figure B.14a). The x- and y-coordinates of each corner represent the cosine and sine of an angle in standard position whose terminal side passes through the corner; hence we obtain the identities in parts (b), (c), and (d ) of Figure B. 14 for sine and cosine. Dividing those identities leads to identities for the tangent. In summary: sin(:r — 0) = sin0, sin(jr + 0) = — sin 6*, cos(jr — 9) = — cos6>, cos(n + 0) = — cos$, tan(?r - 0) = - tan 0, tan(yr + 6) = tan 0,
sin(— 9) = — sin 9 cos(— 9) = cos 6* tan(-0) = - tan 0
(17-19) (20-22)
(-x,y\
fx,y)
(-x, -y)
sin (n - 0) - sin 6 cos (it-6) = -cos 6
(a)
sin (n + = -sin 6 cos (jc + 8) = -cos B
I sin (-&) = -sin 8 cos (-0) = cos 6
(d)
(b)
A Figure B.14
Two angles in standard position that have the same terminal side must have the same values for their trigonometric functions since their terminal sides intersect the unit circle at the same point. In particular, two angles whose radian measures differ by a multiple of 2jr have the same terminal side and hence have the same values for their trigonometric functions. This yields the identities sin 9 = sin(9 + 2n) = sin(0 - 2n)
(23)
cos0 = cos(0 + 2n) = cos(9 - lit)
(24)
and more generally, sin(9 = sin(0 ± Inn),
n — 0, 1, 2, . . .
(25)
cos 9 = cos(9 ± 2nn),
n = 0,l,2,...
(26)
Identity (21) implies that tan 9 = tan(0 + n)
and
tan 9 = tan(0 - jr)
(27-28)
Identity (27) is just (21) with the terms in the sum reversed, and identity (28) follows from (21) by substituting 9 - n for 9. These two identities state that adding or subtracting jt
A20
Appendix B: Trigonometry Review
from an angle does not affect the value of the tangent of the angle. It follows that the same is true for any multiple of n; thus, tan 6> = tan(9±nn),
n = 0,1,2,...
(29)
Figure B.15 shows complementary angles 6 and (n/2) — 6 of a right triangle. It follows from (6) that side opposite 9 side adjacent to (jr/2) — 6 = cos (I hypotenuse hypotenuse side adjacent to 9 side opposite (jr/2) — 9 cos 9 = = = sin I hypotenuse hypotenuse sin 6* =
- e) i
')
which yields the identities
A Figure B.15
sin(--0\ = cos 9,
cos (- - 0} = sin 9,
tan (--9\= cot 0
(30-32)
where the third identity results from dividing the first two. These identities are also valid for angles that are not acute and for negative angles as well. THE LAW OF COSINES
The next theorem, called the law of cosines, generalizes the Theorem of Pythagoras. This result is important in its own right and is also the starting point for some important trigonometric identities.
B.2 THEOREM (Law of Cosines) If the sides ofa triangle have lengths a, b, and c, and if 9 is the angle between the sides with lengths a and b, then
c2 — a2 + b2 — 2ab cos 9 PROOF Introduce a coordinate system so that 9 is in standard position and the side of length a falls along the positive z-axis. As shown in Figure B.16, the side of length a extends from the origin to (a , 0) and the side of length b extends from the origin to some point ( x , y ) . From the definition of sin 9 and cos 9 we have sin 9 — y/b and cos 9 = x/b, SO
y = bsin0,
x = bcosO
(33)
From the distance formula in Theorem H. 1 of Web Appendix H, we obtain
= (x- a)2
- O)2
so that, from (33),
(a,0) A Figure B.16
c2 = (b cos 9 - a)2 + b2 sin2 9
= a2
b2(cos2 0 + sin2 9) - lab cos 0 b2 -2abcos9
which completes the proof. We will now show how the law of cosines can be used to obtain the following identities, called the addition formulas for sine and cosine: sin(a + ft) = sin a cos ft + cos a sin ft
(34)
cos(a + ft) = cos a cos ft — sin a sin ft
(35)
Appendix B: Trigonometry Review A21
Pt(cosa, sin a)
sin (a — ft) = sin a cos ft — cos a sin ft
(36)
cos(a — ft) = cos a cos ft + sin a sin ft
(37)
We will derive (37) first. In our derivation we will assume that 0 < ft < a < 2n (Figure B.17). As shown in the figure, the terminal sides of a and ft intersect the unit circle at the points PI (cos a, sin a) and P2(cos ft, sin ft). If we denote the lengths of the sides of triangle OP\ P2 by OP\, P\ P2, and OP2, then OP{ - OP2 - 1 and, from the distance formula in Theorem H. 1 of Web Appendix H, (/>, P2)2 = (cos ft - cos a) 2 + (sin ft - sin a)2
= (sin2 a + cos2 a) + (sin2 ft + cos2 ft) — 2(cos a cos ft + sin a sin ft) = 2 — 2(cos a cos ft + sin a sin ft) A Figure B.17
But angle P2Wi — a - ft, so that the law of cosines yields (Pi P2)2 = (OP,}2 + (OP2)2 - 2(OP])(OP2) cos(a - ft) = 2-2cos(a- ft) Equating the two expressions for (P| P2)2 and simplifying, we obtain cos(a — ft) = cos a cos ft + sin a sin ft which completes the derivation of (37). We can use (31) and (37) to derive (36) as follows: sin(a - ft) = cos [| - (a - ft] = cos [(| - a) - (-ft)] = cos ( -- a j cos(— ft) + sin ( -- a J sin(— ft /7T
\
/7T
\
= cos ( -- a 1 cos ft — sm I -- a\ sm ft
= sin a cos ft — cos a sin ft
Identities (34) and (35) can be obtained from (36) and (37) by substituting — ft for ft and using the identities sin(— ft) = — sin ft, cos(—ft)=cosft We leave it for the reader to derive the identities tan (a + ft =
tan a + tan ft 1 — tan a tan ft
tan(a - ft =
tan a — tan ft I + tan a tan ft
(38-39)
Identity (38) can be obtained by dividing (34) by (35) and then simplifying. Identity (39) can be obtained from (38) by substituting -ft for ft and simplifying. In the special case where a — ft, identities (34), (35), and (38) yield the double-angle formulas sin 2ce = 2 sin a cos a (40) cos 2a = cos2 a — sin2 a 2 tan a tan 2ce = ;— l-tan2a
(41)
(42] ( }
By using the identity sin a + cos a = 1, (41) can be rewritten in the alternative forms cos 2a — 2 cos a — 1
and
cos 2a = 1 - 2 sin a
(43-44)
A22
Appendix B: Trigonometry Review
If we replace a by a/2 in (43) and (44) and use some algebra, we obtain the half-angle formulas 1 + cos a 2 oc cos — —
2
and
2
1 — cos a 7 a sin — =
2
2
(45-46)
We leave it for the exercises to derive the following product-to-sum formulas from (34) through (37): sin a cos ft = -[sin(a - ft) + sin(a + ft)] sin a sin ft — - [cos(a — ft) — cos(a + ft)] cos a cos ft — - [cos(a - ft) + cos(a + ft)] We also leave it for the exercises to derive the following sum-to-product formulas:
a + ft a- ft sin a + smft= 2 sin - cos a + ft . a-ft sm a — sin p = 2 cos - sin cos a + cosft— 2 cos - cos . . cos a — cos p = —2 sin - sm
(50) (51) (52) (53)
FINDING AN ANGLE FROM THE VALUE OF ITS TRIGONOMETRIC FUNCTIONS
A- v
There are numerous situations in which it is necessary to find an unknown angle from a known value of one of its trigonometric functions. The following example illustrates a method for doing this.
> Example 6 Unit circle
Find 0 if sin 9 =
2'
Solution. We begin by looking for positive angles that satisfy the equation. Because sin 6 is positive, the angle 9 must terminate in the first or second quadrant. If it terminates in the first quadrant, then the hypotenuse of AOAP in Figure B. 18a is double the leg AP, so 0 = 30° = - radians 6 If 9 terminates in the second quadrant (Figure B.18&), then the hypotenuse of AOAP is double the leg AP, so LAOP — 30°, which implies that 9 = 180° - 30° = 150° = — radians 6
Unit circle
A Figure B.18
Now that we have found these two solutions, all other solutions are obtained by adding or subtracting multiples of 360° (2n radians) to or from them. Thus, the entire set of solutions is given by the formulas G = 30° ± n - 3 6 0 ° ,
n = 0, 1,2, .
Appendix B: Trigonometry Review
A23
and
= 150° ±«-360°,
n = 0,1,2,
or in radian measure, n
= -±n-2n, 6
n = 0,1,2,...
and
~±n-2n, 6
n= 0 , 1 , 2 , . . .
ANGLE OF INCLINATION The slope of a nonvertical line L is related to the angle that L makes with the positive .x-axis. If is the smallest positive angle measured counterclockwise from the jc-axis to L, then the slope of the line can be expressed as
m = tan
(54)
(Figure B.19a). The angle <j>, which is called the angle of inclination of the line, satisfies 0° < 0 < 180° in degree measure (or, equivalently, 0 < <j> < n in radian measure). If 0 is an acute angle, then m = tan 4> is positive and the line slopes up to the right, and if 4> is an obtuse angle, then m = tan <j> is negative and the line slopes down to the right. For example, a line whose angle of inclination is 45 ° has slope m = tan 45 ° — 1, and a line whose angle of inclination is 135° has a slope of m = tan 135° - -1 (Figure B.19fc). Figure B.20 shows a convenient way of using the line x = 1 as a "ruler" for visualizing the relationship between lines of various slopes.
Rise
Run
Negative slope
(a) A Figure B.19
A Figure B.20
(b)
EXERCISE SET B 1-2 Express the angles in radians. 1. (a) 75°
(b) 390°
(c) 20°
(d) 138°
2. (a) 420°
(b) 15°
(c) 225°
(d) 165°
3-4 Express the angles in degrees. 3. (a) Ti/15 (b) 1.5 (c) 8jr/5
4. (a) jr/10
(b) 2
(c) 2;r/5
(d) 37T
(d) ln/6
5-6 Find the exact values of all six trigonometric functions ofO. 5. (a) A (b) (c)
A24
Appendix B: Trigonometry Review
6. (a)
(b)
18. (a)
(c)
(b)
20°
1 7-12 The angle 9 is an acute angle of a right triangle. Solve the problems by drawing an appropriate right triangle. Do not use a calculator, si 7. Find sin 9 and cos 9 given that tan 9 = 3.
19. In each part, let 9 be an acute angle of a right triangle. Express the remaining five trigonometric functions in terms of a. (a) sin0 = a/3 (b) tane = a/5 (c) sec6» = a
8. Find sin 9 and tan 9 given that cos 9 = |.
20 -27 Find all values of 9 (in radians) that satisfy the given equation. Do not use a calculator, a
9. Find tan 9 and esc 9 given that sec 9 = |.
20. (a) cos 6 =-l/>/2
(b) sine = -l/v/2
10. Find cot 9 and sec 9 given that esc 9 = 4.
21. (a) tan 6> = -1
(b) cos e = \
11. Find the length of the side adjacent to 9 given that the hypotenuse has length 6 and cos 9 = 0.3.
22. (a) sine = -i
(b) tan e = 73
23. (a) tan 9 = lA/3
(b) sin6» = -V3/2
24. (a) sing = -1
(b) cose = -1
25. (a) cot e = -1
(b) cote = V3
26. (a) sece = -2
(b) csce = -2
27. (a) csce = 2/73
(b) sectf =2/^3
12. Find the length of the hypotenuse given that the side opposite 9 has length 2.4 and sin 9 =0.8. 13-14 The value of an angle 9 is given. Find the values of all six trigonometric functions of 9 without using a calculator. : 13. (a) 225°
(b) -210°
(c) 5n/3
(d) -3jr/2
14. (a) 330°
(b) -120°
(c) 9n/4
(d) -3n
15-16 Use the information to find the exact values of the remaining five trigonometric functions of e. II
28-29 Find the values of all six trigonometric functions of 9.
28.
29. (-2A/2T, 4)
15. (a) cose = f, 0 < e
tane tane esc e csce
= -l/\/3, n/2 < 9 < n = -lA/3, -n/2 < 9 < 0 = V2, 0 < e < n/2 = V2, n/2 <9
16. (a) sine = |, 0 <9 < n/2 (b) sine -\, n/2 <9
n/2 <9 < n
(f) sece = —|, n < 9 < 3n/2 17-18 Use a calculating utility to find x to four decimal places. • 17. (a) (b)
25°
\2n/9
(-4, -3)
30. Find all values of 9 (in radians) such that (a) sin 9 = 1 (b) cos 9 = 1 (c) tan 9 = 1 (d)csc# = l (e)sece = l (f) cote = 1. 31. Find all values of 9 (in radians) such that (a) sin e = 0 (b) cos 9 = 0 (c) tan 9 = 0 (d) esc e is undefined (e) sec 9 is undefined (f) cote is undefined. 32. How could you use a ruler and protractor to approximate sin 17° and cos 17°? 33. Find the length of the circular arc on a circle of radius 4 cm subtended by an angle of (a) 7T/6 (b) 150°. 34. Find the radius of a circular sector that has an angle of jr/3 and a circular arc length of 7 units. 35. A point P moving counterclockwise on a circle of radius 5 cm traverses an arc length of 2 cm. What is the angle swept out by a radius from the center to P ?
Appendix B: Trigonometry Review 36. Find a formula for the area A of a circular sector in terms of its radius r and arc length s. 37. As shown in the accompanying figure, a right circular cone is made from a circular piece of paper of radius R by cutting out a sector of angle 9 radians and gluing the cut edges of the remaining piece together. Find (a) the radius r of the base of the cone in terms of R and 9 (b) the height h of the cone in terms of R and 8.
A25
45-46 Do not use a calculator in these exercises.
45. If cos 6 = | and 0 < 9 < n/2, find (a) sin 26 (b) cos 29. 46. If tan a = | and tanft= 2, where 0 < a < jt/2 and 0 < ft < n/2, find (a) sin(a ft) (b) cos(a + ft). 47. Express sin 36 and cos 39 in terms of sin 9 and cos 6. 48-58 Derive the given identities. cos 9 sec 6 , 48. = cos2 0 2 T1 + tan 6 cos 9 tan 9 + sin 9 49. = 2cos0
A Figure Ex-37
38. As shown in the accompanying figure, let r and L be the radius of the base and the slant height of a right circular cone. Show that the lateral surface area, S, of the cone is S = jirL. [Hint: As shown in the figure in Exercise 37, the lateral surface of the cone becomes a circular sector when cut along a line from the vertex to the base and flattened.]
•^ Figure Ex-38
50. 2 esc 29 = sec 9 esc 9 51. tan 9 + cot 9 = 2 esc 29 sin 26 cos 26 52. = sec6» sin 9 cos 9 sin 6 + cos 29 - 1 53. = tang cos 9 — sin 29 54. sin 39 + sin 9 = 2 sin 29 cos 9
55. sin 39 - sin 9 = 2 cos 29 sin 9 9 I-cos9 9 sm9 56. tan - = —— 57. tan - = 2 sm9 2 1 + cos 9 58. cos (-
+ cos
cos6>
59-60 In these exercises, refer to an arbitrary triangle ABC in which the side of length a is opposite angle A, the side of length b is opposite angle B, and the side of length c is opposite angle C. 59. Prove: The area of a triangle ABC can be written as
39. Two sides of a triangle have lengths of 3 cm and 7 cm and meet at an angle of 60°. Find the area of the triangle. 40. Let ABC be a triangle whose angles at A and B are 30° and 45°. If the side opposite the angle B has length 9, find the lengths of the remaining sides and the size of the angle C. 41. A 10-foot ladder leans against a house and makes an angle of 67° with level ground. How far is the top of the ladder above the ground? Express your answer to the nearest tenth of a foot. 42. From a point 120 feet on level ground from a building, the angle of elevation to the top of the building is 76°. Find the height of the building. Express your answer to the nearest foot. 43. An observer on level ground is at a distance d from a building. The angles of elevation to the bottoms of the windows on the second and third floors are a and ft, respectively. Find the distance h between the bottoms of the windows in terms of a, ft, and d. 44. From a point on level ground, the angle of elevation to the top of a tower is a. From a point that is d units closer to the tower, the angle of elevation is ft. Find the height h of the tower in terms of a, ft, and d.
area = ^ be sin A Find two other similar formulas for the area. 60. Prove the law of sines: In any triangle, the ratios of the sides to the sines of the opposite angles are equal; that is, a b c sin A sin B sin C 61. Use identities (34) through (37) to express each of the following in terms of sin 9 or cos 9.
(a) sin (^ + (c) sin (
(d) cos I — + i
-i
62. Derive identities (38) and (39). 63. Derive identity (a) (47)
(b) (48)
(c) (49).
64. If A = a + ft and B = a - ft, then a = |(A + B) and ft = ~(A - B) (verify). Use this result and identities (47) through (49) to derive identity (a) (50) (b) (52) (c) (53). 65. Substitute -ft for ft in identity (50) to derive identity (51).
A26
Appendix B: Trigonometry Review
66. (a) Express 3 sin a + 5 cos a in the form C sin (a + 4>) (b) Show that a sum of the form A sin a + B cos a can be rewritten in the form C sin(a + 0). 67. Show that the length of the diagonal of the parallelogram in the accompanying figure is
•4 Figure Ex-67
68-69 Find the angle of inclination of the line with slope m to the nearest degree. Use a calculating utility, where needed. Hi 68. (a) m = \ (b) m = -1 (c) m = 2 (d) m = -57 69. (a) m = -\ (b) m = 1 (d) m = 57 (c) m = -2 70-71 Find the angle of inclination of the line to the nearest degree. Use a calculating utility, where needed, f 70. (a) 3y = 2 - V^x (b) y - 4x + 1 = 0 71. (a) y - Vlx + 2 (b) y + 2x + 5 = 0
SOLVING POLYNOMIAL EQUATIONS We will assume in this appendix that you know how to divide polynomials using long division and synthetic division. If you need to review those techniques, refer to an algebra book. A BRIEF REVIEW OF POLYNOMIALS
Recall that if n is a nonnegative integer, then a polynomial of degree n is a function that can be written in the following forms, depending on whether you want the powers of x in ascending or descending order: Cfl + C\X + C2X2 + n
CnX
1
+ Cn^x"-
h CnXn
(cn ^ 0)
+ • • • + C}X + C0
(cB?feO)
The numbers CQ,C\, ... ,cn are called the coefficients of the polynomial. The coefficient cn (which multiplies the highest power of x) is called the leading coefficient, the term cnx" is called the leading term, and the coefficient CQ is called the constant term. Polynomials of degree 1,2,3,4, and 5 are called linear, quadratic, cubic, quartic, andquintic, respectively. For simplicity, general polynomials of low degree are often written without subscripts on the coefficients: i Constant polynomial p(x) - a p(x) = ax -4-b
(a ^ 0)
2
p ( x ) = ax 4- bx + c 3
2
Linear polynomial
(a ^ 0)
p(x) = ax + for + cjc + d
(a 76 0)
Quadratic polynomial Cubic polynomial
When you attempt to factor a polynomial completely, one of three things can happen: • You may be able to decompose the polynomial into distinct linear factors using only real numbers; for example, x3 + x2 - 2x = x(x2 + x - 2) = x(x - l)(jc + 2)
• You may be able to decompose the polynomial into linear factors using only real numbers, but some of the factors may be repeated; for example,
x6 -
•2x3=.
2) = x\x -
2)
(1)
You may be able to decompose the polynomial into linear and quadratic factors using only real numbers, but you may not be able to decompose the quadratic factors into linear factors using only real numbers (such quadratic factors are said to be irreducible over the real numbers); for example, jc4 - 1 = (x2 - l)(jc2 + 1) = (x - l)(x + l)(x2 + 1)
= (x - l)(x + \)(x Here, the factor x2 + 1 is irreducible over the real numbers. A27
A28
Appendix C: Solving Polynomial Equations
In general, if p(x) is a polynomial of degree n with leading coefficient a, and if complex numbers are allowed, then p(x) can be factored as
p(x) = a(x - r\)(x - r2) • • • (x - rn)
(2)
where r\, r^, • •., rn are called the zeros of p(x) or the roots of the equation p(x) — 0, and (2) is called the complete linear factorization of p(x). If some of the factors in (2) are repeated, then they can be combined; for example, if the first k factors are distinct and the rest are repetitions of the first k, then (2) can be expressed in the form
p(x) = a(x -
(3)
where T], r2,..., r^ are the distinct roots of p(x) = 0. The exponents m \, m2,..., m^ tell us how many times the various factors occur in the complete linear factorization; for example, in (3) the factor (x — r\) occurs m\ times, the factor (x — r2) occurs m2 times, and so forth. Some techniques for factoring polynomials are discussed later in this appendix. In general, if a factor (x — r) occurs m times in the complete linear factorization of a polynomial, then we say that r is a root or zero of multiplicity m, and if (x — r) has no repetitions (i.e., r has multiplicity 1), then we say that r is a simple root or zero. For example, it follows from (1) that the equation x6 — 3x4 + 2x3 — 0 can be expressed as
so this equation has three distinct roots—arootx = Oof multiplicity 3, a root x = 1 of multiplicity 2, and a simple root x = —2. Note that in (3) the multiplicities of the roots must add up to n, since p(x) has degree n; that is, m\ + m2 + • • • + m/i — n For example, in (4) the multiplicities add up to 6, which is the same as the degree of the polynomial. It follows from (2) that a polynomial of degree n can have at most n distinct roots; if all of the roots are simple, then there will be exactly n, but if some are repeated, then there will be fewer than «. However, when counting the roots of a polynomial, it is standard practice to count multiplicities, since that convention allows us to say that a polynomial of degree n has n roots. For example, from (1) the six roots of the polynomial p(x) = x6 — 3x4 + 2x3 are r = 0 , 0, 0, 1, 1, -2 In summary, we have the following important theorem.
C.I THEOREM If complex roots are allowed, and if roots are counted according to their multiplicities, then a polynomial of degree n has exactly n roots.
THE REMAINDER THEOREM
When two positive integers are divided, the numerator can be expressed as the quotient plus the remainder over the divisor, where the remainder is less than the divisor. For example,
t
If we multiply this equation through by 5, we obtain 17 = 5 - 3 + 2
which states that the numerator is the divisor times the quotient plus the remainder.
Appendix C: Solving Polynomial Equations
A29
The following theorem, which we state without proof, is an analogous result for division of polynomials.
C.2 THEOREM If p(x) and s(x) are polynomials, and if s ( x ) is not the zero polynomial, then p(x) can be expressed as p(x) =
s(x)q(x)+r(x)
where q(x) andr(x) are the quotient and remainder that result when p(x) is divided by s ( x ) , and either r (x) is the zero polynomial or the degree ofr(x) is less than the degree ofs(x).
In the special case where p(x) is divided by a first-degree polynomial of the form x — c, the remainder must be some constant r, since it is either zero or has degree less than 1. Thus, Theorem C.2 implies that p(x) = (x -c)q(x) + r and this in turn implies that p(c) — r. In summary, we have the following theorem.
C.3 THEOREM (Remainder Theorem) remainder is p(c).
If a polynomial p ( x ) isdividedbyx — c, thenthe
> Example 1 According to the Remainder Theorem, the remainder on dividing
p ( x ) = 2x3 + 3x2 - 4x - 3 by x + 4 should be p(-4) = 2(-4)3 + 3(-4)2 - 4(-4) - 3 = -67 Show that this is so. Solution.
By long division 2x2-5x+ 16 x + 4\2x + 3x2 - 4x - 3 2x3 + 8;c2 -5x2 - 4x -5x2 - 20x 3
I6x - 3 16* + 64 which shows that the remainder is —67. Alternative Solution. Because we are dividing by an expression of the form x — c (where c — —4), we can use synthetic division rather than long division. The computations are
=4| 2 2
3 -8 -5
-4 20 16
which again shows that the remainder is —67. <«l
-3 -64 -67
A30
Appendix C: Solving Polynomial Equations I THE FACTOR THEOREM To factor a polynomial p(x) is to write it as a product of lower-degree polynomials, called factors of p(x). For s(x) to be a factor of p(x) there must be no remainder when p(x) is divided by s(x). For example, if p(x) can be factored as p(x) = s(x)q(x)
(5)
then
(6) so dividing p(x) by s(x) produces a quotient q(x) with no remainder. Conversely, (6) implies (5), so ,y(*) is a factor of /?(*) if there is no remainder when p(x) is divided by *(*). In the special case where x — c is a factor of p(x), the polynomial p(x) can be expressed as
p(x) = (x -c)q(x)
which implies that p(c) = 0. Conversely, if p(c) = 0, then the Remainder Theorem implies that* - c is a factor of p(x), since the remainder is 0 when p(x) is divided by x - c. These results are summarized in the following theorem.
C.4 THEOREM (Factor Theorem) p(c} = 0.
A polynomial p(x) has a factor x — c if and only if
It follows from this theorem that the statements below say the same thing in different ways: • x — c is a factor of p(x).
• p(c) = 0. • c is a zero of /?(*). • c is a root of the equation p(x) = 0. • c is a solution of the equation p(x) = 0. • c is an x -intercept of y = p(x). > Example 2
Confirm that * - 1 is a factor of
/?(*) = *3 - 3*2 - 13* + 15 by dividing * — 1 into p(x) and checking that the remainder is zero. Solution.
By long division
- 2x - 15 x - lx 3 - 3x2 - I3x + 15
-2x2 - 13* -2*2 + 2* - 15* + 15 - 15* + 15 0 which shows that the remainder is zero.
Appendix C: Solving Polynomial Equations
A31
Alternative Solution. Because we are dividing by an expression of the form x — c, we can use synthetic division rather than long division. The computations are
jj 1 -3 -13 1 -2 1 -2 -15
15 -15 0
which again confirms that the remainder is zero.
USING ONE FACTOR TO FIND OTHER FACTORS
If x — c is a factor of p ( x ) , and if q(x) — p ( x ) / ( x — c), then (7)
so that additional linear factors of p(x) can be obtained by factoring the quotient q(x).
Example 3
Factor
p ( x ) =x3 -3x2 - 13*+ 15
(8)
completely into linear factors. Solution. We showed in Example 2 that x — 1 is a factor of p(x) and we also showed that p(x)/(x - 1) = x2 - 2x - 15. Thus,
x3 - 3x2 - I3x + 15 = (x - l)(x2 -2x- 15) Factoring x2 — 2x — 15 by inspection yields
x3 - 3x2 - 13* + 15 = (x - \)(x - 5)(x + 3) which is the complete linear factorization of p(x). -4
METHODS FOR FINDING ROOTS
A general quadratic equation ax2 + bx + c = 0 can be solved by using the quadratic formula to express the solutions of the equation in terms of the coefficients. Versions of this formula were known since Babylonian times, and by the seventeenth century formulas had been obtained for solving general cubic and quartic equations. However, attempts to find formulas for the solutions of general fifth-degree equations and higher proved fruitless. The reason for this became clear in 1829 when the French mathematician Evariste Galois (1811-1832) proved that it is impossible to express the solutions of a general fifth-degree equation or higher in terms of its coefficients using algebraic operations. Today, we have powerful computer programs for finding the zeros of specific polynomials. For example, it takes only seconds for a computer algebra system, such as Mathematica or Maple, to show that the zeros of the polynomial
p(x) = are v
X — —1
- 23x3 -
29x + 6 and
x —2
(9)
(10)
The algorithms that these programs use to find the integer and rational zeros of a polynomial, if any, are based on the following theorem, which is proved in advanced algebra courses.
A32
Appendix C: Solving Polynomial Equations
C.5
THEOREM
Suppose that — cnx
c\x
is a polynomial with integer coefficients. (a) Ifr is an integer zero of p(x), then r must be a divisor of the constant term CQ. (b)
Ifr = alb is a rational zero of p ( x ) in which all common factors of a and b have been canceled, then a must be a divisor of the constant term CQ, and b must be a divisor of the leading coefficient cn.
For example, in (9) the constant term is 6 (which has divisors ±1, ±2, ±3, and ±6) and the leading coefficient is 10 (which has divisors ±1, ±2, ±5, and ±10). Thus, the only possible integer zeros of p(x) are ±1, ±2, ±3, ±6
and the only possible noninteger rational zeros are i 1 ==2'
i 1 5'
±
_i_ 1 70'
±
±
i 2 5'
±
_i_3 2'
3 5'
±
, 3 TO'
Using a computer, it is a simple matter to evaluate p ( x ) at each of the numbers in these lists to show that its only rational zeros are the numbers in (10). > Example 4 Solve the equation x3 + 3x2 - Ix - 21 = 0. Solution.
The solutions of the equation are the zeros of the polynomial
p ( x ) = x3
3x2 -lx-2l
We will look for integer zeros first. All such zeros must divide the constant term, so the only possibilities are ±1, ±3, ±7, and ±21. Substituting these values into p ( x ) (or using the method of Exercise 6) shows that x — — 3 is an integer zero. This tells us that x + 3 is a factor of p(x) and that p ( x ) can be written as x3 + 3x2 - Ix - 21 = (x + 3)q(x) where q(x) is the quotient that results when x3 + 3x2 - 7* — 21 is divided by x + 3. We leave it for you to perform the division and show that q(x) = x2 — 7; hence, x3 + 3x2 - Ix - 21 = (x + 3)(x2 - 7) = (x + 3)(x + Vl)(x - V?)
which tells us that the solutions of the given equation are x = 3, x = -Jl « 2.65, and x = -77 xi -2.65. *
EXERCISE SET C
CAS
1-2 Find the quotient q(x) and the remainder r(x) that result when p(x) is divided by s(x). 1. (a) p(x) =x4 + 3x3 -5x + 10; s(x) =x2-x + 2 (b) p(x) = 6x4 + IQx2 + 5; s(x) = 3x2 - 1 (c) p(x) =xs+x3 + l; s(x) =x2+x 2. (a) p(x)=2x4 -3x3+5x2 + 2x + l; s(x)=x2-x + { (b) p(x) = 2x5 + 5x4 - 4x3 + 8jc2 + 1; s(x)=2x2-x + (c) P(x) = 5x6 + 4x2 + 5; s(x) - x3 + 1
3-4 Use synthetic division to find the quotient q(x) and the remainder r(x) that result when p(x) is divided by s(x). 3. (a) p(x) = 3x3 - 4x - 1; s(x) =x -2 (b) p(x) =x4 - 5x2 + 4; s(x) =x + 5 (c) p(x)=x5 - 1; s(x) =x-l 4. (a) p(x) = 2x3 - x2 - 2x + 1; s(x) =x-\ (b) p(x) = 2x4 + 3x3 - llx2 - 21x - 9; s(x) = x + 4 (c) p(x) = x7 + 1; s(x)=x-\
Appendix C: Solving Polynomial Equations
A33
5. Let p(x) = 2x4 + x 3 — 3x2 + x — 4. Use synthetic division and the Remainder Theorem to find p(0), p(\), p(— 3), and p(7). 6. Let p(x) be the polynomial in Example 4. Use synthetic division and the Remainder Theorem to evaluate p(x) at jc = ±1, ±3, ±7, and ±21. 7. Let p(x) = x3 + 4x2 + x — 6. Find a polynomial q(x) and a constant r such that (a) P(x) = (x-2)q(x)+r (b) p(je) = (* + !)«(*) +r.
[c] 16. For each of the factorizations that you obtained in Exercises 11-15, check your answer using a CAS.
8. Let p(x) = x5 — 1. Find a polynomial gCx) and a constant r such that (a) p(x) = (x + \)q(x) + r
22. For each of the equations you solved in Exercises 17-21, check your answer using a CAS. 23. Find all values of k for which x — 1 is a factor of the polynomial p(x) = k2x3 - Ikx + 10.
(b) p(jc) = (x-l)q(x)
+ r.
9. In each part, make a list of all possible candidates for the rational zeros of p ( x ) . (a) p(x) = x1 + 3x3 - x + 24 (b) p(x) = 3x4 - 2x2 + lx - 10 (c) p(x)=x35 - 17 10. Find all integer zeros of
p(x) = x6 + 5x5 - 16*4 - 15*3 - 12*2 - 38.x - 21 11-15 Factor the polynomials completely. 11. p(x) = x3 - 2x2 - x + 2
12. 13. 14. 15.
p(x) p(jc) p(;t) p(x)
= 3x3 + x2 - 12* - 4 = x4 + l(k3 + 36x2 + 54x + 27 = 2*4 + x3 + 3x2 + 3x - 9 = x5 + 4x4 - 4x3 - 34^:2 - 45x - 1 8
17-21 Find all real solutions of the equations. 17. x3 + 3x2 + 4x + 12 = 0
18. 2x3 - 5x2 - \0x + 3 = 0
19. lx4 + I4x3 + ]4x2 - &x - 8 = 0 20. 2x4 -x3 - I4x2 ~5x+6 = 0 21. x5 - 2;c4 - 6x3 + 5x2 x + \2 = 0
24. Is x + 3 a factor of x1 + 2187? Justify your answer. 25. A 3 cm thick slice is cut from a cube, leaving a volume of 196 cm3. Use a CAS to find the length of a side of the original cube. 26. (a) Show that there is no positive rational number that exceeds its cube by 1. (b) Does there exist a real number that exceeds its cube by 1? Justify your answer. 27. Use the Factor Theorem to show each of the following. (a) x — y is a factor of x" — yn for all positive integer values of ». (b) x + y is a factor of x" — y" for all positive even integer values of n. (c) x + y is a factor of x" + y" for all positive odd integer values of n.
SELECTED PROOFS
PROOFS OF BASIC LIMIT THEOREMS
An extensive excursion into proofs of limit theorems would be too time consuming to undertake, so we have selected a few proofs of results from Section 1.2 that illustrate some of the basic ideas. D.I THEOREM Let a be any real number, let k be a constant, and suppose that lim f ( x ) = L\ and that lim g(x) = L2. Then:
(a)
lim k = k
(b) Jim [/(*) + g(x)] = Urn /(*) + Urn g(x) = LI + L2
(c)
lim [f(x)g(x)]
x^-a
= (lim /(*)) (lim g(x)} = L,L 2 vc->a
/ vc->a
/
PROOF (a) We will apply Definition 1.4.1 with f ( x ) = k and L = k. Thus, given e > 0, we must find a number S > 0 such that \k-k\<€
if 0<\x-a\<8
or, equivalently,
0 < € if 0 < \x -a\ <S But the condition on the left side of this statement is always true, no matter how S is chosen. Thus, any positive value for S will suffice. PROOF (b)
We must show that given € > 0 we can find a number S > 0 such that
\(f(x)+g(x))-(L}+L2)\<e
if 0<\x-a\<8
(1)
However, from the limits of / and g in the hypothesis of the theorem we can find numbers S\ and 82 such that I / O O - L , ! < e / 2 if 0<\x-a\<8} \g(x)-L2\<€/2
if 0<\x-a < 82
Moreover, the inequalities on the left sides of these statements both hold if we replace Si and 82 by any positive number S that is less than both Si and <52. Thus, for any such 8 it follows that
A34
i / \ r i \i f/•/-( x\) - L ri +\g(x)-L 2\ <e if
r n
0<x-a\<8c
/<-i\ (2)
Appendix D: Selected Proofs
A35
However, it follows from the triangle inequality [Theorem F.5 of Web Appendix F] that
K/GO + gGO) - (L, + L 2 )| = |(/GO - L,) + (gGO - L 2 )| < |/GO-Li| + |gGO-L 2 | so that (1) follows from (2). PROOF (c) We must show that given e > 0 we can find a number 8 > 0 such that |/G<)gGO-LiL2|<e
if 0<\x-a\<8
(3)
To find 8 it will be helpful to express (3) in a different form. If we rewrite /GO and g(x) as /GO = Li + (/GO-Li)
and g(x) = L2 + (g(*) - L2)
then the inequality on the left side of (3) can be expressed as (verify) |Li(gGO - L2) + L2(/(*) - L,) + (/GO - L,)(g(x) - L 2 )| < €
(4)
Since lim /GO = LI
x-»a
lim g(jc) = L 2
and
x->a
we can find positive numbers 81,82,83, and <54 such that
I/GO-L!I < I/GO-L!|<
if
0 < \x — a\ < 8\
if
0<\x-a\<82
IgGO - L2| <
if 0 < | j t - a | < S 3
lgGO-L 2 |<
if
(5) 0 < \x-a\ <84
Moreover, the inequalities on the left sides of these four statements all hold if we replace 8\, 82, 83, and 54 by any positive number S that is smaller than 5i, 52, 63, and 84. Thus, for any such 8 it follows with the help of the triangle inequality that |L,(gGO - L2) + L2(/GO - Lj) + (/GO - LO(gGO - L2)| < |L!(g(jc) - L 2 )| + |L2(/(jc) - L,)| + \(f(x) - L,)(g(x) - L 2 )| = |L,||gW - L 2 | + \ L 2 \ \ f ( x ) - Li| + \f(x) - L,\\g(x) - L^\
Do not be alarmed if the proof of part (c) seems difficult; it takes some experience with proofs of this type to develop a feel for choosing a valid S. Your initial goal should be to understand the ideas and the computations.
+ r-
6 |L |L2|+3 Since
lill
1 and
which shows that (4) holds for the 5 selected. • PROOF OF A BASIC CONTINUITY PROPERTY
Next we will prove Theorem 1.5.5 for two-sided limits.
D.2 THEOREM (Theorem 1.5.5) If limx _^ c g (x) = L and if the function f is continuous at L, then lim^e /(gGO) = /(£)• That is, lim
= / ( lim g(x)) \JT->C
/
A36
Appendix D: Selected Proofs
PROOF We must show that given e > 0, we can find a number S > 0 such that \f(g(x))-f(L)\<e
if 0<\x-c <S
(6)
Since / is continuous at L, we have lim /(«) = /(L)
it—*L
and hence we can find a number <5i > 0 such that !/(«)-/(I)| < e
if | « - L | < « ]
In particular, if u = g(x), then \f(gW) - f ( L ) \ < e
if \g(x)-L\<8J
(7)
But limj^c g(x) = L, and hence there is a number 5 > 0 such that \g(x)-L\<8{
if 0 < x - c | « 5
(8)
Thus, if x satisfies the condition on the right side of statement (8), then it follows that g(x) satisfies the condition on the right side of statement (7), and this implies that the condition on the left side of statement (6) is satisfied, completing the proof. • • PROOF OF THE CHAIN RULE Next we will prove the chain rule (Theorem 2.6.1), but first we need a preliminary result.
D.3
THEOREM Iff
is differentiable at x and ify = f ( x ) , then Ay = /'(x)Ax
where e -> 0 as Ax ->• 0 and € — 0 if AJC = 0.
PROOF Define
f(x + Ax) - f(x) - /'(x) if Ax / 0 e=\ Ax
(9)
if Ax = 0
0
If Ax ^ 0, it follows from (9) that eAx = [/(x + Ax) - /(x)] - /'(x)Ax
(10)
Ay = /(x + Ax) - /(x)
(11)
But
so (10) can be written as eAx = Ay - /'(x)Ax or
(12)
Ay = f ' ( x ) Ax + e Ax
If Ax = 0, then (12) still holds (why?), so (12) is valid for all values of Ax. It remains to show that e ->• 0 as Ax -> 0. But this follows from the assumption that / is differentiable at x, since lim , = lim [/(* + ^)-/(*) _ , A*^0
Ax->0 [_
AjC
We are now ready to prove the chain rule.
1= /(JC) _ f(x)
=Q
„
Appendix D: Selected Proofs
A37
D.4 THEOREM (Theorem 2.6.1) If g is differentiable at the point x and f is differentiable at the point g ( x ) , then the composition fog is differentiable at the point x. Moreover, ify = f ( g ( x ) ) andu — g(x), then dy dx
PROOF
dy du
du dx
Since g is differentiable at x and u = g ( x ) , it follows from Theorem D.3 that AM = g'(jt)A;c + eiAjc
(13)
where e\ -> 0 as AJC -> 0. And since y = /(«) is differentiable at u = g(x), it follows from Theorem D.3 that -,, , Ay = /(M)AM + e 2 AM (14) where €2 -> 0 as AM -> 0. Factoring out the Aw in (14) and then substituting (13) yields Ay = [/'(«) + e 2 ][g'(jc) AJC + €, AJC] or
Ay = [/'(«) or if AJC ^ 0,
But (13) implies that AM ->• 0 as AJC -> 0, and hence ej -> 0 and e2 -> 0 as Ax -> 0. Thus, A rhm — ^- - ffV(u)g ^ V(x)A
or
dy
du
T" aw ' T" dx PROOF THAT RELATIVE EXTREMA OCCUR AT CRITICAL POINTS
In this subsection we will prove Theorem 3.2.2, which states that the relative extrema of a function occur at critical points. D.5 THEOREM (Theorem 3.2.2) Suppose that f is afunction defined on an open interval containing the point XQ. If f has a relative extremum atx = XQ, then x = XQ is a critical point of f; that is, either f'(xo) = 0 or f is not differentiable at XQ.
PROOF Suppose that / has a relative maximum at XQ. There are two possibilities — either / is differentiable at XQ or it is not. If it is not, then JCQ is a critical point for / and we are done. If / is differentiable at XQ, then we must show that /'(XQ) = 0. We will do this by showing that /'(*o) > 0 and /'(XQ) < 0, from which it follows that /'(XQ) — 0. From the definition of a derivative we have . /(*o + h) - /(.ro) so that r = hm and
,ie.. (16)
A38
Appendix D: Selected Proofs
Because / has a relative maximum at XQ, there is an open interval (a, b) containing JCQ in which f ( x ) < f(xo) for all x in (a,b). Assume that h is sufficiently small so that XQ + h lies in the interval (a,b). Thus, f(xo + h) < /(XQ) Thus, if h is negative,
or equivalently
f(xo + h)- /(x0)
f(x0 + h) - f ( x 0 ) < 0 >0
(18)
and if h is positive,
- f(x0) <0 (19) h But an expression that never assumes negative values cannot approach a negative limit and an expression that never assumes positive values cannot approach a positive limit, so that -h)-/(x0) . From (17) and (18) >0 / (*o) = h h and h) - /(x0) ,/, , ,f(xp
J (XQ) = lim
From (16) and (19)
<0
h
Since /'(XQ) > 0 and /'(XQ) < 0, it must be that f(x0) = 0. A similar argument applies if / has a relative minimum at XQ. • PROOFS OF TWO SUMMATION FORMULAS
We will prove parts (a) and (b) of Theorem 4.4.2. The proof of part (c) is similar to that of part (b) and is omitted. D.6
THEOREM (Theorem 4.4.2)
n
n(n + 1) 2
(a) n
2
(b) k=\
n
r s
(c)
L
Jt=l
PROOF (a)
n(n + ])(2n + l) 6
~|2 2
J
Writing
two ways, with summands in increasing order and in decreasing order, and then adding, we obtain =
1
+
2
3
+ • • • + (« -2) + ( « - ! ) +
-2) + - - . +
3
+
2
+
n 1
k=\
= (n + 1) + (n + 1) + (n + 1) + • • • + (n + 1) + (n + 1) + (« + 1)
Appendix D: Selected Proofs
A39
Thus,
n(n + 1) k=\
PROOF (b)
Note that
(k + if - k3 = k3 + 3k2 + 3k + 1 - k3 = 3k2 + 3k + 1 So, n
E k=\
n 2 rtif —
(20)
\£-\j)
k=\
Writing out the left side of (20) with the index running down from k = n to k — 1, we have Thesumin(21) i s a n example of atelescoping sum, since the cancellation of
^—\ 3 3 3 3 / J(* + 1) ~k ] = [(«+!) ~ « J H
each of the two parts of an interior summand with parts of its neighboring summands allows the entire sum
k=\
to collapse like a telescope.
/i3 3 3 o3 n3 h [4 - 3 ] +[3 - 2 ] + [2
= (n + I)3 — 1
i3n
(21)
Combining (21) and (20), and expanding the right side of (20) by using Theorem 4.4.1 and part (a) of this theorem yields
So,
= (« + I) 3 - 3(« + 1)
-[2n2 + n] =
- (n + 1)
(» + l)(2n
Thus,
PROOF OF THE LIMIT COMPARISON TEST
D.7 THEOREM (Theorem 9.5.4) suppose that
Let J] ak and J] i
series with positive terms and
ak
p = lim —
If p is finite and p > 0, then the series both converge or both diverge.
A40
Appendix D: Selected Proofs
PROOF We need only show that ]T bk converges when J] ak converges and that J^ bk diverges when J] ak diverges, since the remaining cases are logical implications of these (why?). The idea of the proof is to apply the comparison test to ^ ak and suitable multiples °f Y^ bk- For this purpose let € be any positive number. Since ak it follows that eventually the terms in the sequence {ak/bk} must be within e units of p; that is, there is a positive integer K such that for k > K we have ak
p-e < -~ < p + e bk In particular, if we take 6 = p/2, then for k > K we have 3
1P < a± 2 '~~ bk
3 .
l or
P k
Thus, by the comparison test we can conclude that
2X
converges if
2_] -pbk
diverges if
v^ 1
N
k=K
converges
(22)
diverges
(23)
k=K
Y^ ak
k=K
But the convergence or divergence of a series is not affected by deleting finitely many terms or by multiplying the general term by a nonzero constant, so (22) and (23) imply that CC
Y"^bk
GO
converges if
Y^a k
k=\
k=\
=0
CO
YJ bk
diverges if
Y"] ak
k=\
converges diverges •
k=\
PROOF OF THE RATIO TEST
D.8
THEOREM (Theorem 9.5.5)
that
Let
k
be a series with positive terms and suppose
p = lim
(a) I f p < l , the series converges. (b)
If p > I or p = +00, the series diverges.
(c) If p = 1, the series may converge or diverge, so that another test must be tried.
PROOF (a) The number p must be nonnegative since it is the limit of uk+\/uk, which is positive for all k. In this part of the proof we assume that p < 1, so that 0 < p < 1. We will prove convergence by showing that the terms of the given series are eventually less than the terms of a convergent geometric series. For this purpose, choose any real number r such that 0 < p < r < 1. Since the limit of uk+i/uk is p, and p < r, the terms of the sequence {uk+]/uk} must eventually be less than r. Thus, there is a positive integer K such that for k > K we have Uk+l
uk
or
uk+i < ruk
Appendix D: Selected Proofs This yields the inequalities
A41
UK+\ < ruK uK+2 < ruK+i < r2uK
UK+I < ruK+2 < r^uK uK+4 < r4uK
(24)
ButO < r < 1, so ruK + ruK
is a convergent geometric series. From the inequalities in (24) and the comparison test it follows that ' UK+2 + U r+3 + - Uk + • • • converges by The-
must also be a convergent series. Thus, u\ + «2 + «3 + orem 9.4.3(c).
PROOF (b) In this part we will prove divergence by showing that the limit of the general term is not zero. Since the limit of Uk+i/Uk is p and p > 1, the terms in the sequence {uk+\/uk} must eventually be greater than 1. Thus, there is a positive integer K such that for k > K we have uk+\ > 1 or uk+\ > uk uk This yields the inequalities
UK uK+2 > UK+I > UK
(25) UK
uK+4
Since UK > 0, it follows from the inequalities in (25) that lim^^+a, M* ^ 0, and thus the series u\ + u-i + • • • + uk + • • • diverges by part (a) of Theorem 9.4.1. The proof in the case where p = +00 is omitted. PROOF (c) The divergent harmonic series and the convergent p-series with p — 2 both have p = 1 (verify), so the ratio test does not distinguish between convergence and divergence when p = 1. • PROOF OF THE REMAINDER ESTIMATION THEOREM
D.9 THEOREM (Theorem 9.7.4) If the function f can be differentiated n + 1 times on an interval containing the number XQ, and if M is an upper bound for |/ ( " +1) (x)| on the interval, that is, \f(-n+^(x)\ < M for all x in the interval, then \Rn(x)\<7^-1\X-X0\"+l
(n + 1)!'
for all x in the interval.
PROOF We are assuming that / can be differentiated n + 1 times on an interval containing the number XQ and that i f ("+!)(• vM < Af (26) l/ ( ' for all x in the interval. We want to show that M \X -X0\
n+l
(27)
A42
Appendix D: Selected Proofs
for all x in the interval, where
(28) k=0
In our proof we will need the following two properties of Rn(x): RH(XQ) = R'n(XQ) = • • • = R^(X0) = 0
(29)
+1
(30)
fl(" > (X) = /(«+') (X) for all x in the interval
These properties can be obtained by analyzing what happens if the expression for Rn(x) in Formula (28) is differentiated j times and XQ is then substituted in that derivative. If j < n , then the j th derivative of the summation in Formula (28) consists of a constant term /^(*o) plus terms involving powers of x — XQ (verify). Thus, Rf, (XQ) — 0 for j < n, which proves all but the last equation in (29). For the last equation, observe that the nth derivative of the summation in (28) is the constant fM(x0), so R^(x0) = 0. Formula (30) follows from the observation that the (n + l)-st derivative of the summation in (28) is zero (why?). Now to the main part of the proof. For simplicity we will give the proof for the case where x > x0 and leave the case where x < x0 for the reader. It follows from (26) and (30) that |/^n+1)(jt)| < M, and hence
-M <
<M
Thus, t-x
r-x
I -Mdt< Jxa
R(^+1\t)dt<
Jxo
px
Mdt
(31)
Jxo
However, it follows from (29) that /?
Thus, performing the integrations in (31) we obtain the inequalities -M(x - *o) < R(n\x) < M(x - x0) Now we will integrate again. Replacing x by t in these inequalities, integrating from XQ to x, and using R^~I)(XQ) = 0 yields
If we keep repeating this process, then after n + 1 integrations we will obtain
M
< Rn(x) <
which we can rewrite as
M (X (n + 1)!
- XQ) n+1
M <
(» + !)! This completes the proof of (27), since the absolute value signs can be omitted in that formula when x > XQ (which is the case we are considering). • PROOF OF THE EQUALITY OF MIXED PARTIALS
D.10 THEOREM (Theorem 13.3.2) Let f be a function of two variables. If fxy and fyx are continuous on some open disk, then fxy = fyx on that disk.
Appendix D: Selected Proofs
A43
PROOF Suppose that / is a function of two variables with fxy and fyx both continuous on some open disk. Let (x, y) be a point in that disk and define the function w(Ajc, Ay) = f(x + Ax, y + Ay) - f(x + Ax, y) - f(x, y + Ay) + /(x, y) Now fix y and Ay and let
g(x) = f ( x , y + Ay) - /(x, y) so that
w(Ax, Ay) = g(x + Ax) - g(x)
(32)
Since / is differentiable on an open disk containing ( x , y ) , the function g will be differentiable on some interval containing x and x + AJC for Ax small enough. The Mean-Value Theorem then applies to g on this interval, and thus there is a c between x and Ax with
But
so from Equation (32) w(Ax, Ay) = g(x + Ax) - g(x) = g'(c)Ax = (fx(c, y + Ay) - fx(c, y))Ax
(33)
Now let h(y) = fx(c, y). Since fx is differentiable on an open disk containing (x, y), h will be differentiable on some interval containing y and y + Ay for Ay small enough. Applying the Mean-Value Theorem to h on this interval gives a d between y and y + Ay with But h'(d) = fxy(c, d), so by (33) and the definition of h we have w(Ax, Ay) = (fx(c, y + Ay)- fx(c, y))Ax = (h(y + Ay) - h(y))Ax = h'(d)AyAx =
fxy(c,d)AyAx
and
fxy(c,d) =
w(Ax, Ay) Ay Ax
(34)
Since c lies between x and Ax and d lies between y and Ay, (c, d) approaches (x, y) as (A;c, Ay) approaches (0, 0). It then follows from the continuity of fxy and (34) that fxv(x,y) =
lim
(A*,Ay)-»(0,0)
fxy(c,d)=
lim
(A*,Ay) -> (0,0)
w(Ax, Ay) AyAjc
In similar fashion to the above argument, it can be shown that =
lim
(Ax, Ay) -» (0,0)
w(AJC, Ay) Ay AJC
and the result follows. • PROOF OF THE TWO-VARIABLE CHAIN RULE FOR DERIVATIVES
D.ll THEOREM (Theorem 13.5.1) Ifx = x ( t ) and y = y(t) are differentiable at t, and ifz = f ( x , y) is differentiable at the point (x(t), y ( t ) ) , then z — f ( x ( t ) , y(t)) is differentiable at t and ^z _ dz dx dz dy dt ~ 9jc ~dt
+
dy~dt
A44
Appendix D: Selected Proofs
PROOF Let AJC, Ay, and Az denote the changes in x, y, and z, respectively, that correspond to a change of Af in t . Then dz
dx Ax — — hm — , at AJ^O Af
Az
— — hm — , at AI^O A;
dy
Ay
— = lira — at A<^O At
Since f(x, y) is differentiable at (x(t), y(f)), it follows from (5) in Section 13.4 that
(Ay) 2
+ ~ Ay + €(Ax, dy
Az =
(35)
where the partial derivatives are evaluated at (x(t), y(f)) and where e(Ax, Ay) satisfies e(Ax, Ay) ^ 0 as (Ax, Ay) -> (0, 0) and e(0, 0) = 0. Dividing both sides of (35) by Af yields /\ 7 f17 /\Y ft /\V £{ / \ V / \ V I , / l / \ r i ' " - t - [ / \ V I ' Az_dz_Ax_ 9 7z A y (36) H Af Af ~ ~dx~At 9y7u~ Since
(Ay) 2
lim
, +
= hm
hm
At
Af
AyV hm — r ^ O At J
we have lim
e(Ax, Ay)^(Ax)2 + (Ay)2 At
= lim , Ay)| • lim
= lim
Therefore, 2
+ (Ay) 2
Af
Taking the limit as Af ->• 0 of both sides of (36) then yields the equation dz dt
dz dx dx dt
dz dy dy dt
(Ay)2 |Af|
ANSWERS TO ODD-NUMBERED EXERCISES Exercise Set 0.1 (Page 12) _ (a) -2.9, -2.0, 2.35, 2.9 (b) none (c) y = 0 ( d ) - 1 . 7 5 < j c < 2.15 (e) >>max = 2.8 at* = -2.6; >>min = -2.2 at x = 1.2 (a) yes (b) yes (c) no (d) no (a) 1999, about $47,700 (b)1993, $41,600 (c) first year (a) -2; 10; 10; 25; 4; 27r 2 - 2 (b) 0; 4; -4; 6; 272; /(3f) = l/3r forr > 1 and /(3f) = 6t fort < 1 (a) domain: x / 3; range: y / 0 (b) domain: x / 0; range: {— 1, 1) (c) domain: x < -Vlorx > V3; range: y > 0 (d) domain: — oo < x < +co; range: y > 2 (e) domain: * 7^ (2n + |) JT, n = 0, ±1, ±2, . . . ; range: y > \ (f ) domain: — 2<x<2orx>2; range: 0 < > > < 2 o r ; y > 2 11. (a) no; births and deaths (b) decreases for 8 hours, takes a jump upward, and repeats 13.
33. (a) r % 3.4, h x> 13.7 (b) taller (c) r R* 3.1 cm, h ^ 16.0 cm, C « 4.76 cents 35. (i) * = 1, -2 (ii) g(jt) = * + 1, all * 37. (a) 25° F (b) 13°F (c)5°F 39. 15°F
^ Exercise Set 0.2 (Page 24)
(b)
1. (a)
1
2
1
1
-4
2
3. (a)
M 2
25-* , -5 < x < 0 25-* 2 , 0<x<5 Responses to True-False questions may be abridged to save space. 19. False; for example, the graph of the function f ( x ) = x1 — 1 crosses the x-axis at x = ± 1. 21. False; the range also includes 0. 23. (a) 2,4 (b) none (c) x < 2; 4 < x (d) y m j n = — 1; no maximum 25. h = L(\ -cosd) '1-2*, x<0 <0 27. (a) 0 <1 4;c+
* >0
29. (a) V = (8 - 2*)(15 - 2x)x (b) 0 < * < 4 (c) 0 < V < 90, approximately + 17 -
:-
1,
y
(b)
1
17. function; y =
3
(d)
(c)
IS. function; y = •
2
X\ 1
2
3
1
i
y
(c)
,
(d)
1
I
2
-1
3
X
/I
1
-1
A 2 3,
-1
^ /\/ 1
-1
2
x
3
20 -1 4
-2
X>1
31. (a) L = x + 2y (b) L = x + 2000/x (c) 0 < x < 100 (A)x ^45ft,;y » 2 2 f t
-100
-6
11.
9.
L
lii 12
3.2
1 2 3 4 (d) V appears to be maximal forj: zz 1.7.
20 40 60 80 100
A45
A46 13.
Answers to Odd-Numbered Exercises 10
15.
17.
12
(c)
59. (a) even (b)odd (c) even 63. (a) y-anis 65. (b) origin (c) x-axis, y-axis, origin
19.
(d) neither
(e) odd 2
(f)even
-3
-5
21.
2.5
-1.5
-2 4
0
23.
69. (a)
1
2.3
25. (a)
-2n-n
n
71. yes;/(*) = **,$(*)=*" ^ Exercise Set 0.3 (Page 35) 1. (a) y = 3^: + b (C) (b) y = 3x + 6
(0, \2x,
x<0 x >0
27. 3V* - 1, x > } ; V* - 1, * > 1; 2* - 2, jc > 1; 2, x > 1 29. (a) 3 (b) 9 (c) 2 (d) 2 (e) 72 + ft (f) (3 + ft)3 + 1 31. 1 -x,x < 1; Vl --r 2 , kl < 1 3. (a) y = mx + 2 (c) 1
2
35. (a) g(j;) = VT> ^(Jc) = - + 2 (b) g(;t) = \x\, h(x) = x — 3x +5 37. (a)g(jc) = x2,h(x) — sin^r (b)g(z) = 3/x, h(x) = 5 + cosz 39. (a)g(x)=x3,h(x) = \
Responses to True-False questions may be abridged to save space. 41. True; see Definition 0.2.1. 43. True; see Theorem 0.2.3 and the definition of even function that follows. 45.
,9-x0x 5. y — + /Q
i-2
7.
^-intercepts represent current value of item being depreciated.
y=f(g(x)) -3 •
49. ±1.5, ±2
9. (a) slope: -1 y = -x + 3.5 51. 6x + 3h,3w + 3x 53. -
55. /: neither, g: odd, ft: even 57. (a)
I
1
x(x + h)'
xw
(b) y-intercept: y = — 1
Answers to Odd-Numbered Exercises (c) pass through (—4, 2)
(d) 4 0
10
(c)
-5
1V
I f "
A47
3 0
-50
~D ^
-40
-2 2 21. (a) newton-meters (N-m) (b)20N-m
V(L)
0.25
0.5
1.0
1.5
2.0
]
P (N/m2) 80 x I0 3 40 x 103 20 x 103 13.3 x I03 lOxlO 3 ]
13. (a)
23. (a) * = 0.000045 N-m 2 (b) 0.000005 N (d) The force becomes infinite; the force tends to zero.
(C)
5
10
Responses to True-False questions may be abridged to save space. 25. True; see Figure 0.3.2(6). 27. False; the constant of proportionality is 2 • 6 = 12. 29. (a)II;y = 1,* = -1,2 (b) I; y = 0, x = -2, 3 (c) IV; y = 2 (d) III; y = 0, x = -2 31. (a) y = 3 sin(jr/2) (b) y = 4 cos 2x (c) y = -5 sin 4* 33. (a) y = sin[x + (ir/2)] (b) y = 3 + 3 sin(2;c/9)
35. (a) amplitude = 3, period = jr/2
37. x = 2V2sin(2jrt + jr/3) -80
(b) amplitude = 2, period = 2
A48
1. 3. 5. 7.
Answers to Odd-Numbered Exercises
Exercise Set 0.4 (Page 44) (a) yes (b)no (c)yes (d) no (a) yes (b) yes (c)no (d)yes (e)no (a) yes (b) no (a) 8, -1,0 (b) [-2, 2], [-8, 8] (£)
=t y
.
23. A: (-fjr, 1-V3);B: (f)no
27. (a) i(* + 1)1/3
(b) none
(c) - - 1 (d)
—1
^ Exercise Set 1.1 (Page 59) 1. (a) 3 (b)3 (c)3 (d)3 3. (a)-l (b)3 (c) does not exist (d) 1 5. (a) 0 (b) 0 (c) 0 (d) 3 7. (a) -oo (b) -oo (c) -oo (d) 1 9. (a) 1 (b) -oo (c) does not exist (d) -2 11. (i) ~~ 0.001 0.01 0.1 -0.01 -0.001
13. -V3X* [(5/2)-*, j c > 1/2 0 < x < 1/2 ' \llx. 17. *1/4 -2 for* > 16 19. <0
-2
C: (|jr, 1 + V3");
D: (f;r, 1-V3")
f(x)
Responses to True-False questions may be abridged to save space. 21. False;/-'(2) = 2 23. True; see Theorem 0.4.3
1.9866933 1.9998667 J1.9999987J 1.9999987
The limit appears to be 2. 13. (a) 3 (b)+oo (c)-oo 15. (a) 3 (b) does not exist Responses to True-False questions may be abridged to save space. 17. False; see Example 6. 19. False; the one-sided limits must also be equal.
2a -b-^b2 -4a(c-x) (b)/-'(*) = 2a 104 4 27. (a) y = (6.214 x 10~ );c (b) x = 6.214"1 (c) how many meters in y miles 29. (b) symmetric about the line y = x 31. 10
^- Chapter 0 Review Exercises (Page 46) 3. 70°
5. (a) C = 5x2 + (64/'x) (b) x > 0 7. (a) V = (6-2;t)(5-;c);cft 3 (b) 0 < x < 3 (c) 3.57 ft x 3.79 ft x 1.21ft
31. (a) rest length
-2 11.
JC
m g(x)
(/"*)«
-4 0 o
4
Of°/)M -1
-1
0
1 _3
-3 -2
-1
1
-3
-4
-3 -2 -1 2 2
1
4
1
2
3
-2
-1
—4 0
-2
3
4
-3
4
-4
4
-2
0
-4
2
3
2
0
•3
13. 0,-2 15. l/(2-;t 2 ),;t ^ ± 17. (a) odd (b) even (c) neither (d) even 19. (a) circles of radius 1 centered on the parabola y = x2 (b) parabolas congruent to y = x2 that open up with vertices on the linej> = jc/2 (b) January 11 (c) 122 days 21. (a)
(b) 0. As speed approaches c, length shrinks to zero.
^ Exercise Set 1.2 (Page 69) 1. (a)-6 (b)13 (c)-8 (d) 16 (e)2 ( f ) - ± 3. 6 5. | 7. 4 9. -f -J 11. 11. -3 13. -3\ 15. +00 17. does not exist 19. —oo 21. +00 23. does not exist 25. +00 27. +00 29. 6 31. (a) 2 (b) 2 (c) 2 Responses to True-False questions may be abridged to save space. 33. True; this is Theorem 1.2.2(a). 35. False; see Example 9. 37. \ 39. (a) 3 (b)
-3 1 41. (a) Theorem 1.2.2(a) does not apply. (b) lim (- - -^ = lim (^-^] = -oo 43. a = 2 2 x^O+\X
X J
x^0+ V *
/
45. The left and/or right limits could be ±00; or the limit could exist and equal any preassigned real number. ^ Exercise Set 1.3 (Page 78)
1. (a)-oo (b)+oo 3. (a)0 (b)-l 5. (a)-12 (b)21 (c)-15 (d)25 (e) 2 (f)-f -20 -.
(g)0
Answers to Odd-Numbered Exercises
7. (a)
l/w
7. (a)
'
100,000 1,000,000 | 10 | 100 1000 10,000 0.9534636 0.9950372 0.9995004 0.99995000 0.9999950 1 0.9999995 |
The limit appears to be 1. 9. -co 11. +00 13. \ 15. 0 17. 0 19. -oo 21. -| _ 3/5
23. --— 25. -75 27. lA/6 29. V3 31. 0 Responses to True-False questions may be abridged to save space. 33. False; the limit is +00. 35. True; consider /(*) = (sin x)/x. 37. lim n(t) = +00; lim e(t) = c 39. (a) +00 (b) -5 (->+«
f->+00
43. (a) no (b) yes; tan x and sec x at x = nn + nil, and cot x and esc x at* =nn,n =0, ±1,±2,... 47. +00 49. +00 51. x + 2 53. I-* 2 55. sin* ^ 1. 3. 5.
(b) One second could cost you one dollar.
9. (a) t
Exercise Set 1.4 (Page 87) (a) |*| < 0.1 (b) |* - 3| < 0.0025 (c) \x - 4| < 0.000125 (a) *0 = 3.8025, *i = 4.2025 (b) S = 0.1975 5 = 0.0442 7. 5 = 0.13 2.2 1
11. 1(3* - 5) - 1| = |3* - 6| = 3 • |* - 2| < 3 • S = e 13.5 = 1 15. S = \e 17.5 = e/2 19. 5 = e 21. 5 = e Responses to True-False questions may be abridged to save space. 23. True; |/(*) - /(a)| = \m\\x - a\ 25. True; constant functions 29. (b) 65 (c) e/65; 65; 65; e/65 31. 5 = min (l, £e) 33. 5 = min(l,e/(l + f)) 35. 5 = 2e 39.
(a) _
e
'V
e
Y e
53. N = (1 +2/e)
2
55. (a)|*|< -|L ( b ) | * - l | < ^ 57. 65. 71. 73.
(b)
33. (a) \y
V e
41. 10 43. 999 45. -202 47. -57.5 49. N = -= 51. N =
2
11. none 13. none 15. -1/2,0 17. -1,0,1 19. none 21. none Responses to True-False questions may be abridged to save space. 23. True; the composition of continuous functions is continuous. 25. False; let / and g be the functions in Exercise 6. 27. True; /(*) = vTW • vTW 29. (a) k = 5 (b) k = f 31. k = 4, m = 5/3
5 = lA/M 59. 5 = 1/M 61. S = l/(-M) 1/4 63. S = e 5 = e2 67. 5 = e 69. (a) 5 = -1/M (b) 5 = 1/M (a) N = M - 1 (b) AT = M - 1 (a) 0.4 amps (b) about 0.39474 to 0.40541 amps (c) 3/(7.5 + 5) to 3/(7.5 - 5) (d) 5 ss 0.01870 (e) current approaches +00
^ Exercise Set 1.5 (Page 98) 1. (a) not continuous, x = 2 (b) not continuous, * = 2 (c) not continuous, * = 2 (d) continuous (e) continuous (f ) continuous 3. (a) not continuous, * = 1,3 (b) continuous (c) not continuous, * = 1 (d) continuous (e) not continuous, x = 3 (f) continuous 5. (a) no (b) no (c)no (d)yes (e)yes (f)no (g) yes
35. (a) * = 0, not removable (b) * = —3, removable (c) * = 2, removable; * = —2, not removable 37. (a) * = 5, not removable; at* = —3, removable (b) (2*-!)(* +3) -5
45. /(*) = 1 forO < * < 1, /(*) = -1 for 1 < * < 2 49. x = -1.25,* = 0.75 51. * = 2.24 Exercise Set 1.6 (Page 105) _ 1. none 3. * = njt, n = 0, ±1, ±2, . . . 5. * = nx,n = 0, ±1,±2, ... 7. Irni + (n/6), 2nn + (5jr/6), n = 0, ±1, ±2, ... 9. (a) sin*,* 3 +7* + 1 (b)|*|,sin* (c)* 3 ,cos*,* + 1 11. 1 13. 3 15. -3 17. +00 19. I 21. 0 23. 0 25. 1 27. 2 29. does not exist 31. alb 33. (a) | * |
4
| 4.5
4.9
,1
5.5
6
! f(x) 0.093497 | 0.100932 0.100842 0.098845 0.091319 0.076497 The limit appears to be -^. (b) ^
A49
A50
Answers to Odd-Numbered Exercises
Responses to True-False questions may be abridged to save space. 35. True; use the Squeezing Theorem. 37. True; see the Remark after Exercise 3. 39. (b);r/180 41. 1 43. k=\ 45. (a) 1 (b)0 (c) 1 47. — 7i 49. limx->Qsin(l/x) doesnotexist 51. The limit is 0.
*y
y=f(x)
11. (a) 2
53. (b)
9. Answers may vary.
7. Answers may vary.
(b)0
y=f(x)
(c)4x0
55. (a) Gravity is strongest at the poles and weakest at the equator. Secant
9.83 (b)-2
1
9.82
15. (a) 2x0
9.81
Responses to True-False questions may be abridged to save space. 19. True; set h = x — 1, so x = 1 + h and h -* 0 is equivalent to x ->• 1. 21. False; velocity is a ratio of change in position to change in time. 23. (a) 72° Fat about 4:30 P.M. (b)4°F/h (c)-7°F/h at about 9 P.M. Growth rate (cm/year) 25. (a) first year (d) (b) 6 cm/year 40 (c) 10 cm/year at about age 14 30
(c) 0.739
15 3/45 60 75 90
9.79 9.78
^ Chapter 1 Review Exercises (Page 107) 1. (a) 1 (b) does not exist (c) does not exist (d) 1 (e) 3 ( f ) 0 (g) 0 (h) 2 (i) i 3. 1 5. -3/2 7. 32/3 9. (a) y = 0 (b) none (c) y = 2 11. 1 13. 3-it 15. -i 17. (a) 2x/(x — 1) is one example. 19. (a) lim, _,. 2 /(*) = 5 (b) S = 0.0045 21. (a) S = 0.0025 (b) S = 0.025 (c) S = 1/9000 (Some larger values also work.) 25. ( a ) — 1 , 1 (b)none (c)-3, 0 27. no; not continuous at x = 2 29. Consider/(*) = xfoix £ 0, /(O) = 1, a = -\,b = 1, it = 0.
17. (a) 1 +
20 10 h
5 10 15 20 27. (a) 19,200 ft
(b)480ft/s
(c) 66.94 ft/s
Exercise Set 2.2 (Page 131) 1. 2,0, -2,- 1 3. (b)3 (c)3
(d)1440ft/s
7. y = 5x - 16 9. 4x, y = 4x - 2 11. 3*2; y = 0
^ Chapter 1 Making Connections (Page 109) Where correct answers to a Making Connections exercise may vary, no answer is listed. Sample answers for these questions are available on the Book Companion Site. 4. (a) The circle through the origin with center (0, j) (b) The circle through the origin with center (0, |) (c) The circle does not exist. (d) The circle through the origin with center (0, ^ ) (e) The circle through (0, 1) with center at the origin.
2V* + 1 ' 19. -l/(2jt 3/2 ) 21. 8( + 1 23. (a)D (b)F (c) B (d) C (e)A 25. (a) (b)
(f)E
(f) The circle through the origin with center I 0, (g) The circle does not exist. ^- Exercise Set 2.1 (Page 119) 1. (a)4m/s (b)?? 5
5
10 15 20 Time (s) 3. (a) 0 cm/s (b) t = 0, t = 2, and t = 4.2 (c) maximum: t = I; minimum: t = 3 (d) —7.5 cm/s 5. straight line with slope equal to the velocity
Responses to True-False questions may be abridged to save space. 27. False;/'(a) = 0 29. False; for example, f ( x ) = \x\ 31. (a) v/Jt, 1 (b)* 2 ,3 33. -2
Answers to Odd-Numbered Exercises
17. 2(l+x-')(jc- 3 +7) + (2A:+ IX-*-2)^3 + 7) + (2x + 1)(1 +x~l)(-3x~4) 19. 3(7*6 + 2)(x7 + 2x - 3)2 21. -29 23. 0 25. (a)-f (b)-f| 27. (a) 10 (b)19 (c) 9 (d)-l 29. -2 ± V3 31. none 33. -2 37. F"(x)=xf"(x) + 2 f ' ( x ) 39. /?'(120) = 1800; increasing the price by Ap dollars increases revenue by approximately 1 800 Ap dollars. 41. f ' ( x ) = -nx-"-]
35. y = -2x + 1
37. (b)
-3
1.001 1.0001 1.00001 1.3868 1.3863 1.3863 [/(«>)-/(l)]/(u> - 1) 11.6569 ! 1.4355 1.3911 1.1
1.5
f(2
39. (a) 0.04, 0.22, 0.88 (b)best:
1.01
~
; worst:
41. (a) dollars per foot (b) the price per additional foot (c) positive (d)$1000 43. (a) F & 200 Ib, dF/d6 fa 50 Ib/rad (b) \t, = 0.25 45. (a) T ^ }\5°F,dT/dt sa -3.35°F/min (b)k = -0.084 ^ Exercise Set 2.3 (Page 140) 1. 28*6 3. 24* 7 +2 5.0 7. -i(7*6 + 2) 4 8 9. -3*- -7*- 11. 13. /'(*) = nx"
jOWIo)
15. 3a*2 + 2bx + c 17. 7 19. 2t - 1 21. 15 23. -8 25. 0 27. 0 29. 32f 31. 3jrr2 Responses to True-False questions may be abridged to save space. 33. True; apply the difference and constant multiple rules. 35. False; —[4/(*) + *3] = [4/'(*) + 3*2] =32 dx x=2 s=2 2 37. (a) 4jrr (b) lOOw 39. y = 5* + 17 41. (a) 42* - 10 (b) 24 (c) 2/*3 (d) 700*3 - 96* 43. (a)-210*"8 + 60*2 (b)-6*~4 (c)6a 45. (a)0 49. (1, f ) (2, f ) 15
^ 1. 5. 9.
Exercise Set 2.5 (Page 151) _ — 4sinjc + 2cos* 3. 4x2 sin x - 8* cos x (1 — 5cosx)/(5 + sinx)2 7. sec x tan x — V2 sec2 j CSCJE —4 esc x cot x + esc2* 11. sec3 * + sec * tan2 * 13. — + esc* 1 15. 0 17. 19. -*cos* - 2sin* 2 v . +* tan*) 21. —* sin* +5cos* 23. —4sin* cos* 25. (a) y = x (b) y = 2x~ (jr/2) + 1 (c) y = 2* + (it/2) - 1
29. (a) * = ±n/2, ±3n/2 (b) * = -3n/2, nil (c) no horizontal tangent line (d) * = ±2jr, ±JT, 0 31. 0.087 ft/deg 33. 1.75m/deg Responses to True-False questions may be abridged to save space. 35. False; by the product rule, g'(*) = /(*) cos* + /'(*) sin *. 37. True; /(*) = (sin*)/(cos*) = tan*, so /'(*) = sec2*. 39. (a)-cos* 41. 3,7, 11,... 43. (a) all* (b)all* (c)* ^ (jr/2) +nn, n = 0, ±1, ± 2 , . . . (d)* ^ n 7 T , n = 0 , ± l , ± 2 , . . . (e)* + (n/2) + njr,n =0, ±1, ± 2 , . . . (f)xj=nx,n=Q,±],±2,... (g)* £ (2n + l)jr, n = 0, ± 1 , ± 2 , . . . (h)* ^mt/2,n = 0, ±1, ± 2 , . . . (i) all * ^- Exercise Set 2.6 (Page 157) 1. 6 3. (a) (2* - 3)5, 10(2* - 3)4 (b) 2*5 - 3, 10*4 5. (a) -7 (b) -8 7. 37(*3 + 2*)36(3*2 + 2) 7 \ 24(1 - 3*) -^ H. (3* 2 -2* + I) 4
(*'-;)"
(c) 360
4V*74 + 3V*
15.
-x-2
53. * = i
55. (2 + x/3, -6 - 473), (2 - V3~, 2GmM 57. —2xo 61. -- =— 63. /'(*) > 0 for all x ^ 0
65. yes, 3
\ r^ /
17. -20 cos4* sin*
5 sin(5*) 2Vcos(5*) 25. -3[* +csc(*3 3) cot(*3 + 3)] 3 2 2 27. 10* sin 5* cos 5*+ 3* sin 5* 29. -*3 sec | - I t,n -I _) \+ _L 5* <;».44 sec | -
•e)-a) 2
2(2* + 3) (52* + 96* + 3) 39. (2* + I)4 (4*2 - I) 9 5[* sin 2* + tan4(*7)]4[2* cos2* + sin 2* + 28*6tan3(*7) sec2(*7)] y = -x 45.y = -l 47. y = %^tx - 8?r 49. y = \x - | -25*cos(5*) - 10sin(5*) - 2cos(2*) 53. 4(1 -*)'3 3 cot2 ecsc20 57. n(b - a) sin 2jtco 4 — 2*2 (a) (c) 2 V4 - *2 33(* - 5)
41. 43. 51. 55. 59.
33. — 6cos 2 (sin2*)sin(sin2*)cos2* 3 5
r—pp*+ »)•(! - V*) V* 2 2
35. 35(5*
^ Exercise Set 2.4 (Page 147) 1. 4* + l 3. 4*3 5. 18*2- |* + 12 7. -15*~2 - 14*~3 + 48*~4 + 32*~5 9. 3*2 2 2 -3* - 8* 3* - 8* 10*1/2 + 4 11. 13. 15. 2 (3* - 4) (x + 3)2 IF
(^}
Zj.
31. sin(cos*)sin*
-3-2^1 ' 1 2 3 67. not differentiable at * = 1 69. (a)* = \ (b)* = ±2 71. (b) yes 73. ( a ) n ( n - l ) ( n - 2 ) . - . l (b) 0 (c)ann(n - \)(n - 2) • • • 1 79. -12/(2* + I ) 3 81. -2/(* + l)3
T cos
v-3
21. 28*6 sec2(*7)tan(*7)
19. —— cos(3V*)s 51. y = 3x2
A51
-2
A52
Answers to Odd-Numbered Exercises
125 ft/s 21. 704 ft/s ToT 23. (a) 500 mi, 1716 mi (b) 1354 mi; 27.7 mi/min
i - 7T = -L(r - 11 3
19.
27. 125jrft3/min
25. —- ft/min 20?r 31.
36
ft/min
33. — km/s 35. 600^/7 mi/h
Responses to True-False questions may be abridged to save space.
37. (a) — y units per second
61. False; by the chain rule, — U/y"] = — = LV 2,/y dx 63. False; by the chain rule, dy/dx = (— sin[g(x)]) • g'(x). 65. (c) / = 1/r (d) amplitude = 0.6 cm, T = 2?r/15 seconds per oscillation, / = 15/(2jr) oscillations per second 67.^76 69. (a)101b/in 2 ,-21b/in 2 /mi (b) ^0.61b/in2/s f cosx, 0 < x < jr
41. x = ±. /
(—COSX,
29. 250mi/h
(b) falling
39. —4 units per second
43. 4.5 cm/s; away 47. — cm/s
^- Exercise Set 2.9 (Page 181) 1. (a) /(*) « 1 + 3C* - 1) (b) /(I + Ax) RS 1 + 3Ax 3. (a)l + ^,0.95,1.05 13. \x\ < 1.692
(c) 1.06
(b)
—71 < X < 0
dy
73. (c) — cos —h sin - (d) limit as x goes to 0 does not exist x x x 75. (3)21 (b)-36 77. l/(2x) 79. \x 83. f'(g(h(x)))g'(h(x))h'(x)
Ay
^ Exercise Set 2.7 (Page 166) i
-y-\)lx
I. 3/2
9.
11.
(b)4x-2A
1 - 2xy - 3y3 + 9xy2
3. --
45° (c)0.694765 15. \x\ < 0.3158 17. (a)0.0174533 19. 83.16 21. 8.0625 23. 8.9944 25. 0.1 27. 0.8573
i - 2xy2 cos(x2y2) 2x 2 ycos(x 2 y 2 )
31. (a)4,5 (b)
1 - 3y2 tan 2 (xy 2 + y) sec2(xy2 + y) 3(2xy + 1) tan 2 (xy 2 + y) sec2(xy2 + y)
13. -
9>>
3
17. T19. (1+cosy) 3 Responses to True-False questions may be abridged to save space. 21. False; the graph of / need only coincide with a portion of the graph of the equation in x and y. 23. False; the equation is equivalent to x2 = y2 and y = \x\ satisfies this equation. -0.1312 29. -A
35. points(2, 2), (-2, -2);y' = -1 atbothpoints 37. a = j , f c = | 41. (a) (c) 2 72/3
-1
^ Exercise Set 2.8 (Page 172) 1. (a) 6 (b)-i 3. (a)-2 (b)675
(c) — = 2x — (d) 12 ft 2 / min dt dt 7. (a) — =n(r2 h 2rh — } (b) -20jr in3/s; decreasing 5. (b) A = x 2
dt
\ dt
dx\ y— I
:mi/h
5
(b) — -fg rad/s; decreasing
15. 4860jrcm3/min
3
Responses to True-False questions may be abridged to save space. 41. False; dy = (dy/dx) dx 43. False; consider any linear function. 45. 0.0225 47. 0.0048 49. (a) ±2 ft2 (b) side: ±l%;area: ±2% 51. (a) opposite: ±0.151 in; adjacent: ±0.087 in (b) opposite: ±3.0%; adjacent: ±1.0% 53. ±10% 55. ±0.017 cm2 57. ±6% 59. ±0.5% 61. 15W2cm3 63. (a)a = 1.5 x 10-5/°C (b) 180.1 cmlong ^ Chapter 2 Review Exercises (Page 183) 3. (a) 2x (b) 4 5. 58.75 ft/s 7. (a) 13 mi/h (b) 7 mi/h 9. (a) -2/7§ - 4x (b) l/(x + I) 2 11. (a)x = -2, -1, 1, 3 (b) (-co, -2), (-1, 1), (3, +00) (c) (-2,-!),(!, 3) (d)4 13. (a) 78 million people per year (b) 1.3% per year 15. (a) x2 cosx + 2xsinx (c)4xcosx + (2 — x2) sinx 17. (a) (6x2 + &x- 17)/(3x + 2)2 (c) 118/(3x + 2)3 19. (a) 2000 gal/min (b) 2500 gal/min 21. (a) 3.6 (b)-0.777778 23. /(I) = 0,/'(I) = 5 25. y =-16x,y = -145x/4
29. (a) 8x7 -
dt /
cos2e ( dy
11. — in 2 / min 13.
7/2
33. 3x2dx,3x2&x + 3x(Ax)2 + (Ax) 3 35. (2x - 2) Jx, 2xAx + (Ax) 2 - 2Ax 37. (a)(12x 2 - 14x)rfx (b) (-x sin x + cosx) dx
39.
(c) ;t = -y2 or x = y2 + 1
-2
o.i
-0.1
17. ft/s
31. 33. 37. 41.
*
:-15x-4
(b)(2x+ l)100(1030x2 +
-1414)
(x-l)(15x + l) (a) (b)-3(3x + l) 2 (3x + 2)/x7 273x + 1 x = — |, — j, 2 35. y = ±2x x = «TT ± (a/4), « = 0, ±1, ±2,... 39. y = -3x + (1 + 9jr/4) (a)4073 (b)7500
Answers to Odd-Numbered Exercises 3
(2jt + l) 45. (a)
2 - 3x2 - -
21 y sec(xy) tan(xy) 51. — 53. 21 (2-it) 1 - *sec(;t;y) tan(*y) 16y3 2 57. (1/4/3, 1/2/3) 59. 500jrm /min 61. (a)-0.5, 1,0.5 (b)?r/4, 1.7T/2 (c)3,-1.0 63. (a) between 139.48 m and 144.55 m (c) \d
^ Chapter 2 Making Connections (Page 186) Where correct answers to a Making Connections exercise may vary, no answer is listed. Sample answers for these questions are available on the Book Companion Site. 2. (c)k = 2 (A)h'(x) = 0 3. (b) /' • g • h • k + f • g' • h • k + f • g • h1 • k + f • g • h • k' h
4.(c)f'-f
-f-^h
+
27. increasing: [—jr/4, 3ir/4]; decreasing: [—x, —it/4], [3jr/4, it]; concave up: (—37T/4, n/4); concave down: (—it, —371/4), (n/4, n); inflection points: —3W4, n/4 29. increasing: none; decreasing: (—JT, it); concave up: (—it, 0); concave down: (0, jt); inflection point: 0 31. increasing: [—it, —3it/4], [—it/4, n/4}, [3jr/4, it]; decreasing: [-37T/4, -it/4], [n/4, 3jr/4]; concave up: (-jr/2, 0), (it/2, n); concave down: (—it, —it12), (0, n/2); inflection points: 0, ±jr/2 33. (a) f ? (b)
f-s-" (c)
^- Exercise Set 3.1 (Page 194) 1. (a) /' > 0, /" > 0 (b) /' > 0, /" < 0
i,y
N/"
iy
L
35. 1 + \x - fyl+x > O i f x > 0 37. x > sin* 2.5
(c) /' < 0, /" > 0
(d) /' < 0, /" < 0
V
10
0 0
\ 2
3. A: dyldx < 0,d y/dx > 0; B: dy/dx > Q,d2y/dx2 < 0; C: dy/dx <0,d2y/dx2 <0 5. x = -1,0, 1,2 7. (a) [4, 6] (b) [1,4], [6, 7] (c) (1,2), (3,5) (d) (2, 3), (5, 7) (e)jc = 2,3,5 9. (a) [1,3] (b) (-o=, 1], [3, +00) (c) (-00, 2), (4, +00) (d) (2, 4) (e)x = 2,4 Responses to True-False questions may be abridged to save space. 11. True; see definition of decreasing: f ( x \ ) > /fe) whenever 0 < x\ < X2 < 2. 13. False; for example, f(x) = (x — I) 3 is increasing on [0, 2] and 15. (a) [3/2,+00) (b) (-00, 3/2] (c) (-00,+00) (d)none (e)none 17. (a) (-00, +00) (b) none (c) (-1/2, +00) (d) (-00, -1/2) (e) -1/2 19. (a) [1, +») (b) (-00, 1] (c) (-00,0), ( f , +0=) (d) (0, |) (e) 0, f 3-V5 3 _ V 5 | h+jv/S (b) -oo, 21. (a) 2 '" ,+oo
(d)(-oo,0),
4-V6
4±
(e)0, 23. (a) [-1/2,+00)
(b) (-00, -1/2] (c)(-2, 1)
25. (a)[-l,0],[l,+oo) (b)(-oo, -1],[0, 1] (c) (-00, 0), (0,+00) (d) none (e) none
200
points of inflection at jc = —2, 2; concave up on (—5, —2), (2, 5); concave down on (—2, 2); -5 increasing on [-3.5829, 0.2513] and [3.3316, 5]; decreasing on [-5, -3.5829], -200 [0.2513,3.3316] 41. -2.464202,0.662597,2.701605 45. (a) true (b) false 49. (c) inflection point (1,0); concave up on (1, +00); concave down on (—00, 1)
39.
2
A53
A
55.
57.
_ Inflection point
Exercise Set 3.2 (Page 205) (b) 1. (ah
A54
Answers to Odd-Numbered Exercises 47. intercepts: (-1,0), (0, 0), (2,0); stationary points: (—1,0) (max), (min), 1W3
5. (b) nothing
(c) / has a relative minimum at x = 1, g has no relative extremum at x = 1. 7. critical: 0, ±\/2; stationary: 0, ±-fl 9. critical: -3, 1; stationary: -3,1 11. critical: 0, ±5; stationary: 0 13. critical: n?r/2 for every integers; stationary: nn + n/2 for every integers Responses to True-False questions may be abridged to save space. 15. False; for example, f ( x ) = (x — l ) 2 ( x — 1.5) has a relative maximum six = l,but/(2) = 0.5 > 0 = /(1). 17. False; to apply the second derivative test (Theorem 3.2.4) at x. = 1, /'(I) must equal 0. 19. 21. (a) none (b) x = 1 (c) none (d)
inflection points: (
j=,
V2
49. intercepts: (0,-1), (-1, 0), (1, 0); stationary points: (—1/2, —27/16) (min), (1,0) (neither); inflection points: (0, -1), (1, 0)
51. intercepts: (-1, 0), (0, 0), (1, 0); stationary points: (—1,0) (max), I (min), 25-V/5/
V5'
23. (a) 2 (b)0 ( c ) l , 3
4= - -^7= ) ( max )> (LO) (min); ^5 25 V5/ inflection points: I —J —,
J — \,
1 2 3 4 25. 29. 31. 33. 35. 37. 39. 41. 43.
0 (neither), ^5 (min) 27. -2 (min), 2/3 (max) relative maximum at (4/3, 19/3) relative maximum at (;r/4, 1); relative minimum at (3?r/4, — 1 ) relative maximum at ( 1 , 1 ); relative minima at (0, 0), (2, 0) relative maximum at (— 1,0); relative minimum at (-3/5, -108/3125) relative maximum at (— 1, 1); relative minimum at (0, 0) no relative extrema relative maximum at (3/2, 9/4); relative minima at (0, 0), (3, 0) intercepts: (0, -4), (-1,0), (4, 0); stationary point: (3/2, —25/4) (min); inflection points: none
b 53. (a)
I
6?
1
-<>
(c)
,y
(b)
2
' >
y
AA L
(d)
45. intercepts: (0,5),
-7± V57
stationary points: (—2,49) (max), (3, -76) (min); inflection point: (1/2,-27/2)
2
"
x 2
JC
2
y
0.6
0.6
A
25V5 /
0.3
'
T
16_\
'
-2
/-0.2
55. relative min of 0 at x = jr/2,n, 3jr/2; relative max of 1 at x = jr/4, 3jr/4, Sjr/4, 7;r/4
l\ -0.2
j
, i
2
Answers to Odd-Numbered Exercises 57. relative min of 0 at x = jr/2, 3jr/2; relative max of 1 at x = n
ASS
13. stationary points: none; inflection points: none; asymptotes: x = 1, y — — asymptote crossings: none 2n
59. relative minima at x = -3.58,3.33; relative max at x = 0.25 150
15. (a)
-5
61. relative maximum at x ss —0.272; relative minimum at * « 0.224 63. relative maximum at x = 0; relative minima at JE ~ ±0.618 65. (a) 54 (b)9
-W
17. 10
^ Exercise Set 3.3 (Page 214) 1. stationary points: none; 3. stationary points: none; inflection points: none; inflection point: (0,0); asymptotes: x = 4, y = -2; asymptotes: x = ±2, y = 0; asymptote crossings: none asymptote crossings: (0, 0)
y
'
/% + 3
-A ' 1
\\
^
^
10
X
*
ll
19. stationary point: ( - -J-, f -v/2 ); \ -n i inflection point: (1,0); 2 asymptotes: y = x , x = 0; asymptote crossings: none
I -4 -
5. stationary point: (0,0); inflection points: (±-7?. ?) asymptote: y = 1; asymptote crossings: none
stationary point: (0, — 1 ) ; inflection points: (0, — 1), asymptotes: x = 1, y = 1; asymptote crossings: none
21. stationary points: (-4,-27/2), (2, 0); inflection point: (2,0); asymptotes: x = 0, y = x — 6; asymptote crossing: (2/3,—16/3)
23. stationary points: (-3, 23), (0, -4); inflection point: (0, -4); asymptotes: x = —2, y = x2 — 2;c; asymptote crossings: none 9. stationary point: (4,11/4); 11 stationary point: (—1/3,0); inflection point: (6,25/9); inflection point: ( — 1 , 1 ) ; asymptotes: x = 0, y = 3; asymptotes: x = 1, y = 9; asymptote crossing: (2, 3) asymptote crossing: (1/3,9)
25. (a) VI (b)I (c)III (d)V (e)IV (f)II Responses to True-False questions may be abridged to save space. 27. True; if deg P > deg Q, then f(x) is unbounded as x -» ±00; if deg P < deg Q, then f(x) ->• 0 as jc -> ±00. 29. False; for example, /(*) = (jc — 1)1/3 is continuous (with vertical tangent line) at x = 1, but f ' ( x ) = j- has a vertical asymptote 3(x - I) 2 '- 3 atx = 1.
A56
Answers to Odd-Numbered Exercises
31. critical points: (±1/2,0); 33. critical points: (-1, 1), (0, 0); inflection points: none inflection points: none
49. Graph misses zeros at x = 0, 1 and min at x = 5/6.
47.
100 20
-0.01 - (L
-1.5-0.5 0.5 1.5
\6'
35. critical points: (0, 0), (1, 3); inflection points: (0, 0), (-2, -6 3/2). It's hard to see all the important features in one graph, so two graphs are shown:
625 \ 93,3121
Exercise Set 3.4 (Page 222) 1. relative maxima at x = 2, 6; absolute max at x = 6; relative min at x — 4; absolute minima at x = 0, 4
3. (a)
37. critical points: (0,4), (1,3); inflection points: (0,4), (8,4)
39. extrema: none; inflection points: x = Ttn for integers n
(1,3) -10
30
41. minima: x = lx/6 + 2im for integers n; maxima: x = jr/6 + 2itn for integers n; inflection points: x = 2n/3 + 7m for integers n
(lL 2\ ^ 'J 2.
(-£.-*) BN
43. relative minima: 1 at x = n; -1 at x = 0, 2?r; relative maxima: 5/4 at x = -2?r/3, 2?r/3,4;r/3, 8jr/3; -l±x/33 inflection points where cos* = : (-2.57, 1.13), (-0.94,0.06), (0.94, 0.06), (2.57, 1.13), (3.71, 1.13), (5.35, 0.06), (7.22, 0.06), (8.86, 1.13)
7. max = 2 at * = 1, 2; min = 1 at x = 3/2 9. max = 8 at x = 4; min = — 1 at x = 1 11. maximum value 3/\/5 at x = 1; minimum value — 3/V5 at x = — 1 13. max = 1/2- tt/4 at * = -jr/4; min = 7T/3 - A/3 at x = nil, IS. maximum value 17 at x = —5; minimum value 1 at x = — 3. Responses to True-False questions may be abridged to save space. 17. True; see the Extreme-Value Theorem (3.4.2). 19. True; see Theorem 3.4.3. 21. no maximum; min = —9/4 at x = 1/2 23. maximum value/(I) = 1; no minimum 25. no maximum or minimum 27. max = —2 — 2\/2 at x = — 1 — \/2; no minimum 29. no maximum; min = 0 at x = 0, 2 31. maximum value 48 at x = 8; 33. no maximum or minimum minimum value 0 at x = 0, 20 25 50
(2.57, 1.13)
10 35. max = 2x/2+ 1 at* = 37T/4;min = A/3 at x = jr/3
45. (a) x = 1,2.5,3,4 (b) (-00, 1], [2.5,3] (c) relative max at x = 1,3; relative min at x = 2.5 (d)*^ 0.6, 1.9,4
1.5 37. maximum value sin(l) sa 0.84147; minimum value — sin(l) ss -0.84147
Answers to Odd-Numbered Exercises AST 39. maximum value 2; minimum value — J 41. max = 3 at* = 2nw, min = -3/2 at* = ±2jr/3 + 2mt for any integer n 45. 2, at * = 1 49. maximum y = 4 at t = n, STT; minimum y = 0 at t = 0, 2n
Exercise Set 3.5 (Page 233)
19. (a) v(t) = 37i sin(;rt/3), a(t) = n2 cos(7rt/3) (b) s(\) = 9/2 ft, u(l) = speed = 3->/3jr/2 ft/s, a(l) = jr2/2 ft/s2 (c) t = 0 s, 3 s (d) speeding up: 0 < t < 1.5, 3 < t < 4.5; slowing down: 1.5 < t < 3,4.5 < t < 5 (e) 31.5 ft 21. (a)V5 (b)V5/10 0.25
1. (a) 1 (b) i 5. 9. 15. 17. 19. 25. 27. 31. 35. 37. 39. 41. 43. 49.
3. 500 ft parallel to stream, 250 ft perpendicular 500 ft ($3 fencing) x 750 ft ($2 fencing) 7. 5 in x -^ in 10V2 x lOx/2 11. 80 ft ($1 fencing), 40 ft ($2 fencing) maximum area is 108 when * = 2 maximum area is 144 when * = 2 11,664 in3 21. ™ ft3 23. base 10 cm square, height 20 cm ends -V/3V/4 units square, length | ^3^/4 height = 2R/-V/3, radius = J2/3R height = radius = ^SOO/TTcm 33. L/12by L/12by L/12 height = Ll \/3, radius = V273L height = 2V75/jrcm, radius = \/2V75/jr cm height = 4R, radius = V2R R(x) = 225* -0.25*2;R'(*) = 225 - 0.5*; 450 tons (a) 7000 units (b)yes (c)$15 45. 13,722 Ib 47. 373 height = rA/2 51. (V2, 5) 53. (-1/V3, \\
0.2
v(t) (c) speeding up for V5 < t < >/15; slowing down for 0 < t < \/5 and VI5 < t
°-01 10
-0.15 0(0
23.
t = 3/4
Constant speed
«t=0
2/3 3/2
55. (a)jrmi (b)2sirr'(l/4)mi 57. 4(1 + 2 ) ft 59. 30 cm from the weaker source 61. V24 = 2^6 ft
Speeding up
25. Slowing down
Exercise Set 3.6 (Page 243) 1. (a) positive, negative, slowing down (b) positive, positive, speeding up (c) negative, positive, slowing down s(m) 3. (a) left (b) negative (c) speeding up (d) slowing down
(Stopped) ( = •
2
(Stopped) / 16 18 20 Slowing down
27.
(Stopped Permanently)
Slowing down
t > 2n
-1
0 V 1 Speeding up 29. (a) 12 ft/s (b) t = 2.2 s, s = -24.2 ft 31. (a) t = 0, s = 0, v = 2, t = 7C/2, s = 0, u = -2 (b)r = 7i/4,s = I , a = -4 33. (a) IJ (b)V2
Responses to True-False questions may be abridged to save space. 9. False; a particle has positive velocity when its position versus time graph is increasing; if that positive velocity is decreasing, the particle would be slowing down. 11. False; acceleration is the derivative of velocity (with respect to time); speed is the absolute value of velocity. 13. (a) 7.5 ft/s2 (b) / = 0 s 15. (a) t s v a (b) stopped at t = 2; moving right at t = 1; 1 0.71 0.56 -0.44 moving left at t = 3, 4, 5 0 -0.62 2 1 (c) speeding up at t = 3; -0.44 0.71 -0.56 3 slowing down at t = 1, 5; 4 0 -0.79 0 neither at r = 2, 4 -0.56 0.44 5 -0.71 17. (a) v(t) = 3t2 - 6t,a(t) = 6t - 6 (b)s(l) = -2ft,i;(l) = -3 ft/s, = 3ft/s,a(l) = 0ft/s 2 (c) ( = 0, 2 s (d) speeding up for 0 < t < 1 and 2 < t , slowing down for 1 < t < 2 (e) 58 ft
35. (b) | unit (c) 0 < t < 1 and t > 2 ^ Exercise Set 3.7 (Page 250) 1. 1.414213562 3. 1.817120593 5. x ^ 1.76929 7. * as 1.224439550 9. x « -1.24962 11. x «» 1.02987 13. * a* 4.493409458 5_ 15. * «j 0.68233
T
ASS
Answers to Odd-Numbered Exercises 11. increasing on [0, ?r/4], [3?r/4, TI\; decreasing on [jr/4, 3jr/4]; concave up on (nil, n); concave down on (0, jr/2); inflection point: (ir/2, 0)
17. -0.474626618, 1.395336994 4
0.5
-0.5
-0.5
13. (a) g
Responses to True-False questions may be abridged to save space. 19. True; x = xn+\ is the jc-intercept of the tangent line to y = f(x) at
6
4
21. False; for example, if f(x) = x(x — 3)2, Newton's Method fails (analogous to Figure 3.7.4) with x\ = 1 and approximates the root x = 3 forx\ > 1. 23. (b) 3.162277660 25. -4.098859132
27. (0.589754512,0.347810385) 29. (b)6> SB 2.99156rador 171° 31. -1.220744085,0.724491959 33. i = 0.053362 or 5.33% 35. (a) The values do not converge. ^ Exercise Set 3.8 (Page 257) 1. c = 4 3. C = TT 5. c = 1 9. (a) [-2, 1] .(b)c^ -1.29
7. c=l 6
(c)-1.2885843
2 1 2 3 4
b
15. < 0 17. (a) at an inflection point 2a 21. (a) x = zb-s/2 (stationary points) (b) x = 0 (stationary point) 23. (a) relative max at x = 1, relative min at jc = 7, neither at x = 0 (b) relative max at x = jr/2, 3?r/2; relative min at* = litId, lljr/6 (c) relative max at x = 5 v 25. lim f ( x ) = +00, lim f ( x ) = +00; relative min at x = 0; points of inflection at x = j, 1; no asymptotes
-2 -2
Responses to True-False questions may be abridged to save space. 11. False; Rolle's Theorem requires the additional hypothesis that / is differentiable on (a, b) and /(a) = f ( b ) = 0; see Example 2. 13. False; the Constant Difference Theorem applies to two functions with equal derivatives on an interval to conclude that the functions differ by a constant on the interval. 15. (b) tan x is not continuous on [0, TT]. 33. (b) f(x) = sin jc, g(x) = COSJE 35. | y 39. a = 6, b = -3
y=f(x)
27. lim f ( x ) does not exist; critical point at x = 0; relative min *^±°o
atjc = 0; point of inflection when 1 +4x2tan(x2 + 1) = 0; vertical asymptotes at x = ±Jn (n + ^) — \,n = Q, \,2, ... (0,tan 1)
(-1.42,-0.12)
(1.42,-0.12)
29. critical points at x = —5, 0; relative max at x = —5, relative min at x = 0; points of inflection atz ~ —7.26, —1.44, 1.20; horizontal asymptote y = 1 for x —> ±00 {_$ i\ (-7.26, 1.22
Chapter 3 Review Exercises (Page 259)
i. (a 3. 5. 7. 9.
> /te); f(x\) = f(x2)
(b) /' > 0; /' < 0; /' = 0 (a) [f , +00) (b) (-00, |] (c) (-00, +00) (d) none (e) none (a) [0, +0=) (b) (-00, 0] (c) (-7273, ^273) (d)(-oo, -V273), (%/273, +00) (e) -V273, V2?3 (a)[-l,+oo) (b)(-oo, -1] (c) (-00, 0), (2, +00) (d)(0,2) (e)0,2 increasing on [71, 2n]; decreasing on [0, JT]; concave up on (x/2, 3jr/2); concave down on (0, nil), (3ir/2, In); inflection points: (nil, 0), (3jr/2, 0)
(-1.44, 0.49)
-20 (0,0) 31. lim f ( x ) = +00, lim f ( x ) = -co; x^-cc
20
x -^+cc
critical point at x = 0; no extrema; inflection point at x = 0 (/ changes concavity); no asymptotes
-2 (0,0) -2
33. no relative extrema 35. relative min of 0 at x = 0 37. relative min of 0 at x = 0
Answers to Odd-Numbered Exercises 40
39. (a)
(b) relative max at x = relative min at x =
A59
(b)
-5 -0.4
-40
1
0.0001
(c) The finer details can be seen when graphing over a much smaller ^-window.
0.1
-0.0001
41. (a)
43. -5
2
3
4
5
-l.2\-LJ-LJ
*• 6
LJ-L-LJ
(c)t <* 0.64, £«* 1.2 ( d ) 0 < f < 0 . 6 4 (e) speeding up when 0 < f < 0.36 and 0.64 < r < 1.1, otherwise slowing down (f) maximum speed ~ 1.05 when? « 1.10 63. *!«-2.11491, 0.25410, 1.86081 65. *«-1.165373043 67. 249x10 6 km 2 69. (a) yes, c = 0 (b) no (c) yes, c = \/7r/2 71. use Rolle's Theorem
10 -2
-10
-5
1r
horizontal asymptote y = 1/3 vertical asymptotes at x = (-l±VT3)/6
Chapter 3 Making Connections (Page 263) Where correct answers to a Making Connections exercise may vary, no answer is listed. Sample answers for these questions are available on the Book Companion Site. 1. (a) no zeros
47. (a) true (b) false 49. (a) no max; min = —13/4 at x = 3/2 (b) no max or min (c) max = 0 at x = 2; min = — 3 at x = 3 51. (a) minimum value 0 for x = ±1; (b) max = 1/2 at x = 1; no maximum min = 0 at x = 0 10 0.5
(b) one (c) lim g'(x) = 0
4 -
slope -2/3
3. x = -4, 5 ^ Exercise Set 4.1 (Page 270) 10 0
(c) maximum value 2 at x =0; minimum value -v/3 at x = jr/6 2
7.
(b) minimum: (-2.111985, -0.355116); maximum: (0.372591,2.012931)
55. width = 4-V/2, height = 3\/2 57. 2 in square 59. (a) yes (b)yes ;(r 4 + 2 f 2 - l ) 3(8 + 10(6- 12f 4 -6r 2 61. (a)v = -2——-. =—-,a = 2 ff4 i i \2 4 3 +l)
9. 3 ( x - l ) 11. x(x + 2) 13. Responses to True-False questions may be abridged to save space. 15. False; the limit would be the area of the circle 4n. 17. True; this is the basis of the antiderivative method. 19. area = A(6) - A(3) 23. f ( x ) = 2x;a = 2 Exercise Set 4.2 (Page 278)
x2 cos(l + x3) dx = - sin(l
5. ^[v^ dx Lv J 2V^3 + 5 2V^3 + 5 d . r cos(2v^) COS( ) y ^ = sin(2^) + C. 7. -—[sin(2V^)] = j=— ,so/' ax ^/x j V* 9 nr 9. (a)(^ /9) + C (b) l,x '
A60
11.
Answers to Odd-Numbered Exercises
x
4 + C
12
x
13.
2
4/3
..5/4
47.
-
7/3
+C 15. (x /2) +V/5) + C 17. 3* - ^x + x2 1 1 19. 1 j + C 21. -3cosx -2tan* + C 23. tan * + sec*-)-C 25. tan0 + C 27. secJt + C 29. 8 - cos 9 + C 31. tan* -sec* + C Responses to True-False questions may be abridged to save space. 33. True; this follows from Equations (1) and (2). 35. False; the initial condition is not satisfied since y(0) = 2. 37. (b) y = - cos t + t + 1 - 7T/3 = §* 3/2 +2jc 1/2 - |
Exercise Set 4.4 (Page 297) 10
1. (a) 36 (b)55 (c)40 (d)6 (e) 11 ( f ) 0 6
50
1- E(-l) t+1 (2*:-l)
k=\
-5h
,4
2+
(b)
,
3(n-l)\4 3 n
I
n
n
k=(l
27. (a) 46 (b)52 (c)58 29. (a) f (b)0 (c)-f 31. (a)0.7188,0.7058,0.6982 (b) 0.6688,0.6808,0.6882 (c) 0.6928,0.6931,0.6931 33. (a) 4.8841,5.1156, 5.2488 (b) 5.6841, 5.5156, 5.4088 (c) 5.3471, 5.3384,5.3346 35. £ 37. 18 39. 320 41. ^ 43. 18 45. 16 47. i 49. 0 51.f 53. '• + 2n . (n + I)2 if« is odd 57. 317 - 34 59. 55. if n is even; 4 4 61. (b) \ 65. (a) yes (b) yes
55. (b)
>• Exercise Set 4.5 (Page 307) 1. (a) if
(b)2
3. (a)--^
(b) 3
-3x)dx 61. (a) F'(x) = G'(x) = cosx, x 63. tanx-
9. (a)
Exercise Set 4.3 (Page 286) _
1. (a)
+ C (b) — 24 4 3. (a) i tan(4x + 1) + C (b) |(H- 2j.2)3/2 + C 5. (a)-4cot2;c + C (b) ^(1 + sinf) 10 + C 7. (a) f (1 +*) 7/2 - 4(1 +*) 5/2 + |(1 +^) 3/2 + C (b)-cot(smx) + C 11. i(4jc-3)'° + C 13. -icos7^: + C 15. ^ (7r2 + 12)3/2 + C 4
27. 33. 37. - '
19.
+ C 23.
lim lim
,,nc4 ,
r-2 j_ j)24
17.
11.5050
k=\
13.2870 15.214,365 17. \(n + \) 19. i(n - I) 2 Responses to True-False questions may be abridged to save space. 21. True; by parts (a) and (c) of Theorem 4.4.2. 23. False; consider [a, b] = [-1, 0].
41. s(t) = I6t2 + 20 43. s(t) = 2r3/2 - 15 45. f ( x ) = ±x5/2 + Cix + C2 47. y = x2 + x - 6 49. f ( x ) = cos* + 1 51. y = x3 - 6x +1
53. (a)
5.
50
9. (a) £2*: (b)£(2fc-l)
k=\
I
3. Tk
13. (a) A = |
+C 25. -i
+C
+ 2)2 ) + C 29. -i(2-sin40) 3/2 + C 31. £ sec3 2* + C 1) 3/2 -i(2y + l) 1/2 + C 35. -icos2e + icos 3 2(9 + C
;
b n+1 .&(» + !) 41. (a) 5 sin2 x + Ci; -1 cos2 * + C2 (b) They differ by a constant. n3/2 _ 158 43. Af5v40pjc-l-i; 15
X
TJ —j-i— AA;^ ; a = 0, fc =
Answers to Odd-Numbered Exercises 15. (a) A = 10
41. area = 3/2
(b) AI — A2 = 0 by symmetry
Aj\
A61
n
-
1.5
2.5
m X
(d) 7T/2
:)>ii + A2 = f .y
(«io
X" *N
\ k
-1
17. (a) 2 (b)4 (c) 10
Ay 1
-1
. \ "-2 \ ^Vf (\>*i
r
;
1
(b)-2.6 f£\
I g
frit
m
3 2 2
21. -1 23. 3 25. -4 27. (l+jr)/2 Responses to True-False questions may be abridged to save space. 29. False; see Theorem 4.5.8(o). 31. False; consider f(x) = x - 2 on [0, 3]. 33. (a) negative (b) positive 37. ^JT 39. | 45. (a) integrable (b) integrable (c) not integrable (d) integrable ^ Exercise Set 4.6 (Page 320) 1. (a)/ 0 2 (2-*)<** = 2 (b) x* is any point in [—1, 1], /(**) 3. (a) ** = /(**) = ! Ay *.y = 2-X
i
2
43. (a) /P cosjcd* = sin0.8 (b) degree mode; 0.717 Jo 45. (a) change in height from age 0 to age 10 years; inches (b) change in radius from time t = 1 s to time t = 2 s; centimeters (c) difference between speed of sound at 100°F and at 32°F; feet per second (d) change in position from time t\ to time f 2 ; centimeters 47. (a)3* 2 -3 49. (a)sin(;t2) (b) Vl -cos* 51. -jcsec* 53. (a)0 (b)5 (c) | 55. (a) x = 3 (b) increasing on [3, +00), decreasing on (-co, 3] (c) concave up on (—1, 7), concave down on (-co, —1) and (7, +00) 57. (a) (0, +00) (b) x = 1 59. (a) 120 gal (b) 420 gal (c) 2076.36 gal 61. 1 ^ Exercise Set 4.7 (Page 329) 1. (a) displacement = 3; distance = 3 (b) displacement = —3; distance = 3 (c) displacement = —^', distance = | (d) displacement = |; distance = 2 3. (a)35.3m/s (b)51.4m/s 5. (a)? (b)4t + 3 - ± sin3r 7. (a)|? 2 + ? - 4 (b)t + (\/t) 9. (a) displacement = 1 m; distance = 1 m (b) displacement = — 1 m; distance = 3 m 11. (a) displacement = | m; distance = ^ m (b) displacement = 2\/3 — 6m; distance = 6 — 2\/3 m 13. 4, 13/3 15. 296/27, 296/27 17. (a)s =
(b)s = 21. 180'
o
i
5. f 7. 14 9. (a) \ 17. 0 19. 72 21. -12
(b) -7 11. 48 13. 3 15. ^ 23. (a) 5/2
Responses to True-False questions may be abridged to save space. 23. True; if a(t) = ao> then v(t) = agt + t>o. 25. False; consider v(t) = sin f on [0, 2?r]. 27. (a)
*2
25. (a)£ (b)r; ;
T' "- 1 3 x i T + 6'
Responses to True-False questions may be abridged to save space. 27. False; since |x | is continuous, it has an antiderivative. 29. True; by the Fundamental Theorem of Calculus. 31. 0.6659; f 33. 3.1060; 2tanl 35. 12 37. |
(b) 5/2 - sin(5) + 5 cos(5)
A62
Answers to Odd-Numbered Exercises
29. (a)
51. (a) F(x) i s O i f x = 1, positive if x > 1, and negative if x < 1. (b)F(;c)isOif;c = - 1 , positive if - 1 < x < 2, and negative if —2 < x < —I. 53. (a) \ (b) 1 ± i 73 55. -fg 59. |r4 - f t3 + t + 1 61. / 2 - 3 f 63. 1 2 m , 2 0 m 65. 67. displacement = —6 m; distance = y m 69. (a) 2.2s (b) 387.2 ft 71. u 0 /2 ft/s 73. -^f1 75. § 77. 0
' (b) 4
_2
8
12
-6 -10 (c) 120cm,-20cm 31. 35. 37. 41.
(d) 131.25 cmaU = 6.5 s
2
(a)-^ ft/s (b)|fs ( c ) f f s 33. 50s, 5000 ft (a) 16 ft/s, -48 ft/s (b) 196 ft (c) 112 ft/s (a) I s ( b ) £ s 39. (a) 6.122s (b) 183.7m (c) 6.122s (d)60m/s (a) 5s (b) 272.5m (c)10s (d)-49m/s (e) 12.46 s (f) 73.1 m/s 43. 113.42 ft/s
1. (a) 4 (c) ' (b)2 4
':'/ /
3. 6 5. 2/7T 7. 1/21
*
^ Exercise Set 4.9 (Page 340) (b) f/ 2 5 7Srfw
(c)-/ T ./-jr/
2
3. 15. 25. 35. 43.
5
(d)/! (« + 1)M rfa 10 5. 0 7. ilp 9. 8-(472) 11. -^ 13. 25ir/6 7T/8 17. 2/jrm 19. 6 21. 2 23. §(710-272) 2(77-73) 27. 1 29. 0 31. (73-l)/3 33. ^f (a) ^ 37. (a) f (b) f (c)-i 41. (a) 0.45 (b) 0.461 (a) 2/jr 47. (b) \ (c) jr/4 Chapter 4 Review Exercises (Page 342) 1
,.« 3/2
C
3. -4cosjt+ 2sin*+ C
5. (a) >>(*) = 27? - P72
4 3
(b) y(jc) = sinx - f *2 + 1
.4/3
(c) y(x) = \ + \x
9. 1(;c4 + 2)3/2 -7 C 4 _ | _ 2 1 1 c* 0 T T~1 »3
1. (b)b2 -a2
3. 12
5. (a) they are equal
9. (a) 5.28 (b) 4.305 (c)4 11. (a)-± (b) I Responses to True-False questions may be abridged to save space. 15. False; let g(x) = cos*; f ( x ) = 0on [0, 3jr/2]. 17. True; see Theorem 4.5 A(b). 19. (a) ^ (b)31 21. 1404jrlb 23. 97 cars/min 27.27
1. (a) ± f f u 3 d u
^ Chapter 4 Making Connections (Page 346) Where correct answers to a Making Connections exercise may vary, no answer is listed. Sample answers for these questions are available on the Book Companion Site.
r3
(b) / g(x)dx=
I
f9
f(u)du
^ Exercise Set 5.1 (Page 353) 1. 9/2 3. 1 5. (a) 4/3 (b)4/3 7. 49/192 9. 1/2 11. 72 13. 24 15. 37/12 17. 472 19. \ Responses to True-False questions may be abridged to save space. 21. True; use area Formula (1) with f ( x ) = g(x) + c. 23. True; the integrand must assume both positive and negative values. By the Intermediate-Value Theorem, the integrand must be equal to 0 somewhere in [a, b]. 25.9152/105 27.9/74 29. (a)4/3 (b)m = 2 - 7 4 31. 1.180898334 33. 2.54270 35. racer 1 's lead over racer 2 at time t = 0 37. (a) (area above graph of g and below graph of /) minus (area above graph of / and below graph of g) (b) area between graphs of / and g 39. a1 Id ^ Exercise Set 5.2 (Page 362) 1. »JT 3. 13JT/6 5. (1 - 72/2)1 7. 8* 9. 32/5 11. 256?r/3 13. 2048JT/15 15. 3/5 17. 2n 19. 72jr/5 Responses to True-False questions may be abridged to save space. 21. False; see the solids associated with Exercises 9 and 10. 23. False; see Example 2 where the cross-sectional area is a linear function of*. 25. 4nab2/3 21.71 29. /j" rt[f(x) - k]2 dx 31. (b) 40^/3 33. 6487T/5 35. jr/2 37. ir/15 39. 40,000jrft3 41. 1/30 43. (a)2;r/3 (b)16/3 (c) 473/3 45.0.710172176 49. (b)left R* 11.157; right <* 11.771; V « average = 11.464cm3
11.
55. 16r3/3
3a ax + b 14
19
15. (a) 2^(k + 4)(k + 1) (b) ^(k - l)(k - 4) k=0
17. f
k=S
19. 0.718771,0.668771,0.692835
23. (a) | (b)-f (c)-3! (d)-2 (f) not enough information 25. (a) 2 + (jr/2)
27. 48 29. | 37. area = i
(e) not enough information
(b) i (103/2 - 1) - ^
(c) ;r/8
31. | - sec 1 33. f
35.
Exercise Set 5.3 (Page 370) 1. lSjr/2 3. 7T/3 5. 2?r/5 7. 4jr 9. 20?r/3 11. n/2 13. n/5 Responses to True-False questions may be abridged to save space. 15. True; this is a restatement of Formula ( 1 ). 17. True; see Formula (2). 19. 2n2 21. 1.73680 23. (a) 7jr/30 (b) easier 25. (a) f 2n(l-x)xdx (b) f 2jr(l +y)(l -y)dy Jo Jo 27. 7;r/4 29. 7tr2h/3 31. V = t;7r(L/2) 3 33. b = 1 >• Exercise Set 5.4 (Page 374) 1. L = 75 3. (85785 - 8)/243 5. ^(SOTlO - 137l3) 7. ^ Responses to True-False questions may be abridged to save space. 9. False; /' is undefined at the endpoints ± 1.
Answers to Odd-Numbered Exercises 11. True; if f ( x ) = mx + c over [a, b], then L = Vl + m2(b — a), which is equal to the given sum. 13. (a) (b) dy/dx does not exist at x = 0. 4 ) (c)L = (13713 + 80710-16)/27
15. (a) They are mirror images across the line y = x.
Exercise Set 5.7 (Page 398) 1. (a) positive: ni2 is at the fulcrum, so it can be ignored; masses m \ and m^ are equidistant from position 5, but m\ < 013, so the beam will rotate clockwise, (b) The fulcrum should be placed y units to the right of mi. 3. (1,1) 5. (1,1) 7. ( f , i ) 9. (£,§) 11. (f, I) 13. (-1,4) 15. ( l , f ) 17. (I,, I,) 21. I - (II) 23. 3;(0,§) 25. 8; (f , f ) Responses to True-False questions may be abridged to save space. 27. True; use symmetry. 29. True; use symmetry.
= ^/u transforms the first integral into the second. ,4 / f ,2 i (c) / /H dy, I y 1 + 4y2 dy (d) 4.0724,4.0716 (e) The first: Both are underestimates of the arc length, so the larger one is more accurate. (f) 4.0724, 4.0662 (g) 4.0729
.3 17. (a) They are mirror images across the line y = x.
(b) 1 +
J4/
1
•dx,
~ D4
u = 1 + j transforms thefirstintegral into the second.
37. 2n2abk
39. (a/3,fe/3)
^ Exercise Set 5.8 (Page 405) 1. (a) F = 31,200 Ib; P = 312 lb/ft 2 (b) F = 2,452,500 N; P = 98.1 kPa 3. 499.21b 5. 8.175 x 105 N 7. 1, 098,720 N 9. yes 11. pa 3 /V21b Responses to True-False questions may be abridged to save space. 13. True; this is a consequence of inequalities (4). 15. False; by Equation (7) the force can be arbitrarily large for a fixed volume of water. 17. 61,748 Ib 19. (4.905 x 10 9 )V3N 2 1 . (b) 80p0 Ib/min ^- Chapter 5 Review Exercises (Page 407)
•f ( g ( x ) - f ( x ) ) d x
7. (a) / (/(*)(c)
I (f(x)-g(x))dx 1
(b)ll/4
Jc
9. 4352JT/105
11. 17/4
(d) 2.1459, 2.1463
^ Exercise Set 5.5 (Page 380) 1. 35wV2 3. 8?r 5. 407TV^2 7. 24?r 9. 167T/9 11. 16,91l7r/1024 13. 2jrh/2 + l n ( V 2 + l ) ] Responses to True-False questions may be abridged to save space. 15. True; use Formula (1) with r\ = 0, r^ = r, / = Vr2 + h2. 17. True; the sum telescopes to the surface area of a cylinder.
13. 9 15. - (653/2 - 373/2)
19. ( f , 0 ) '
(e) The second is more accurate since Ax is smaller. (f) 2.1443, 2.1371 (g) 2.1466 21. t=1.83 23. 196.31 yards 25. ( 2 v ^ - l ) / 3 27. ji 29. (b)9.69 (c) 5. 16 cm
19. 14.39
A63
f4
(b) F = I p(\ + x)2x dx lb/ft 2
Chapter 5 Making Connections (Page 408) Where correct answers to a Making Connections exercise may vary, no answer is listed. Sample answers for these questions are available on the Book Companion Site.
(b)a =
1.
r
2A2
3. / Jo
2nrf( r) dr
21. 5 = f Ja
31. -jr(17VT7-1) 3
33. — (17x/l7-1) 24
^ Exercise Set 5.6 (Page 389) 1. 7.5 ft-lb 3. d = l/4 5. 100 ft-lb 7. 160 J 9. 20 lb/ft Responses to True-False questions may be abridged to save space. 11. False; the work done is the same. 13. True;joules 15. 47,385jrft-lb 17. 261,600J 19. (a) 926,640 ft-lb (b) hp of motor = 0.468 21. 75,000 ft-lb 23. 120,000 ft-tons 25. (a)2,400,000,000/x2 Ib (b) (9.6 x 10'°)/(;c + 4000)2 Ib (c) 2.5344 x 1010 ft-lb 27. vf = lOOm/s 29. (a) decreases of 4.5 x 1014 J (b) « 0.107 (c) « 8.24 bombs
Exercise Set 6.1 (Page 4 18) 1. (a) -4 (b)4 (c) \ 3. (a) 2.9690 (b) 0.0341 5. (a) 4 (b)-5 (c)l (d) ± 7. (a) 1.3655 (b) -0.3011 9. (a) 2r + S- + '-
(b) s-3r-t
11. (a) (b)21n|x| 2 5
13. log !
- i ln(;t2
1S
-
ln
l/x(x + I) 2 In 3
D
17. 0.01 In I
19. e2
29. -2
21. 4
A64
Answers to Odd-Numbered Exercises
31. (a) domain: (—00, +00); range: (— 1,+°°)
35.
*^T 1 * U 3(1 2* 2 1 + ^ )J (* 2 --8) V *3 + 1 2* *0 - 7 * -1-5 ~\l
1/3
37.
p
39.
^
.
^ 2 3*
41 v
(b)
£
" 7 " 6*^ — 7
£*
° 4^ v
l
* ^^
45. 47. A (w) = w/2 51. /(*) = l n ( * 4 l ) 53. (a) 3 (b)-5 55. (a)0 (b)V2 57. 21n|*| - 3 c o s * 4 C 54. l n | l n * | + C 61. j; l n | l + * 5 [ + C 63. t + \n\t\ + C 65. f i n 5 67. ; 69. (ln3)/2 71. y = l n | r [ + 5
(b) domain: x 7^ 0; range: (—00, +=c)
Exercise Se 16.3(1'age 431) 3. --2/*2
i. (b) \ 1 7.
v
(b) yes
(c) yes
(d) yes
10;j 4 ~>y 17. *2 e* (* 4 3)
15. lelx
33. (a) domain: x ^= 0; range: (—°o, +0=)
5. (a) no
9 19
/
r
ai
"
23. (l-Se3-1)^^''1'
77 2*2*2 In 2
29. 7r sinjr (lnjr)cos*
21. 25.
e*- ;c
-2x)l"x
31.
(b) domain: (—00, +=o); range: (0, 1]
M
Responses to True-False questions may be abridged to save space. 35. False; exponential functions have constant base and variable exponent. 37. True; In* = log,* 39. 2.8777, -0.3174
fin Y-1
-2
1
3
]
[_*3 — n* _i_
Responses to True-False questions may be abridged to save space. 37. False; consider y = Ae*. 39. True; use the chain rule. 41. (b) 1 - (x/3/3) 43. (b) y = (88* - 89)/7 49. r = 1, K = 12 51.3 53. In 10 55. 2In |* 1 + 3e* + C 57. -\e-^ + C 59. i« 2 *4C
61. e s i n * 4 C
63. -^-^ + C 65. -e~* + C _ -i 67. /[ln(e' : )4ln(e-- t )]d* = C 69. 5e3 - 10 71.^ 2 73. I n f ^ 75. % 48,233,500,000 77. (ln7)/2
^ Exercise Set 6.4 (Page 438)
-3
1. O(min) 3. — l(min), l(max) 5. maximum value ^?e~3 at* = minimum value 64/e8 at * = 4
43. (a) no (b) y = 2*/4 (c) y = 2-" (d) y = (-J5)
45. log i < 0, so 3 log i < 2 log \ 47. 201 days 49. (a) 7.4, basic (b) 4.2, acidic (c) 6.4, acidic (d) 5.9, acidic 51. (a) 140 dB, damage (b) 120 dB, damage (c) 80 dB, no damage (d) 75 dB, no damage 53.^200 55. ( a ) % 5 x 10 16 J (b) «s 0.67
^- Exercise Set 6.2 (Page 425) 1. \/x
3. !/(!+*)
5. 2*/(* 2 -l)
13. 1 + l n *
11.
2*Vln*
0.2
7. max = 5 In 10-9 at* = 3; min = 5 ln(10/9) - 1 at* = 1/3
_ 7.
'~*
+ x2)
9. 2/* 2
2* 15. 2*log 2 (3-2*) (In2)(3-2x)
19. l/(*ln*) 21. 2 csc 2* (1+log*) 2 1 3 8* 23. --sin(ln*) 25. 2cot*/(ln 10) 27. --,--^-2
9. (a)+«>,0
(b)
17.
x
x- 1
29. - tan * H
x +1
T2
4 - 3* Responses to True-False questions may be abridged to save space. 31. True; lira — = +00 33. True; l/x is an odd function. 1^0+ X
(-2, -0.27)),. (-1.-0.37)
, 0.05)
Answers to Odd-Numbered Exercises 13. (a) 0,0 (b) relative max = 1 /e at x = ± 1 ; relative min = 0 at x = 0; , inflection points where x = ±,
about (±0.47, 0.18), (±1.51, 0.23); asymptote: y = 0
27. (a) does not exist, 0 (b) y = ex and y = e" cos* intersect x = 2nn, and y = —ex and y = ex cos* intersect for x = Inn + n, for all integers n.
A65
(c) Graphs for a = 1: for t> = 3 6 -
20
-6 -
(1.51,0.23) -40
h (-0.47,0.18)
1\* (0.47,0.18) 15. (a)-oo, 0 17. (b) relative max = -e2 at AT = 2; no relative min; no inflection points; asymptotes: y = 0, x — 1 10
x=\
(a)0,+00 (b) critical points at A: = 0,2; relative min at x = 0, relative max at A: = 2; points of inflection at x = 2 ± i/2; horizontal asymptote y = 0 as x ->• +00 lim f ( x ) = +oc
Responses to True-False questions may be abridged to save space. 29. False; these graphs are reflections of one another across the line y — x. 31. True; f ( x ) = In x is concave down for x > 0, so the average value of / over [1, e2] exceeds j[/(l) + /(e2)] = 1. LAk \n(a/b) 35. the eighth day 39. / = 33. (a) a —b 1000 (c)lnA
2 3 4
-20
(0,0) 21. (a)+oo,0
19. (a)+oo,0 (b)
(b)f>
51. (a)
23. (a)+oo,0 (b) no relative max; relative min = 3/2
2e
at x = e
2
;
inflection point: (e ,3e/2); no asymptotes. It's hard to see all the important features in one graph, so two graphs are shown:
53. x % 0.58853 or 3.09636
1 2 3 4 5 (b) 3/2 + 6e~5 55. x = -lor x % 0.17951 57. 4jr 59. jrln2 61. L = V2(eW2 - 1) 63. L = l n ( l + V 2 ) 6 5 . 5 ^ 2 2 . 9 4 ^- Exercise Set 6.5 (Page 448) 1. (a) \ (b) § Responses to True-False questions may be abridged to save space. 3. True; the expression (In x)/x is undefined if A: < 0. 5. False; applying L'Hopital's rule repeatedly shows that the limit is 0. 7. 1 9. 1 11. -1 13. 0 15. -oo 17. 0 19. 0 21. n 23. - 1 25.
(b) relative max at A: = \/b; inflection point at x = 2/b 10
1000
19
-1.5
="0.5
A66
Answers to Odd-Numbered Exercises
51. no horizontal asymptote
53. y = \
43. (a)< = 3 (b)f = 1,5 (c)r = 5 (d)r = 3 (e) F is concave up on (0, ^) and (2, 4), concave down on (|, 2) and (4, 5).
1.02
0
10000
-16
55. (a)0 (b)+oo (c)0 (d)-co (e)+cc (f) -o= 57. 1 59. does not exist 61. Vt/L 65. (b) Both limits equal 0. 67. does not exist
45. (a) relative maxima at x = ±^/4k + 1, k = 0, 1,...; relative minima at* = ±V4t- \,k = 1,2,... 47. /(*) = 2e2x, a = In 2 49. 0.06 °^_
^ Exercise Set 6.6 (Page 459) 1. (a)
Exercise Set 6.7 (Page 469) 1
2
3. (a) 7 (b)-5 (c)-3 (d)6 5. 1.603210678; magnitude of error is < 0.0063
(c) t
, 4 3 3 5 5 I- 5 . 5 . 4 . 3 . 4
3. (a) 0 < x < TC (b) -1 < x < 1 (c) -7T/2 < x < ;r/2 (d) -co < X < +00
5.
7- (a) 9. (a)
7. (a)x~\x > 0 (b)* 2 ,jc^0 (c) — x2, — oo < x < +00 (d) -x, -co < x < +00 (e)x*,x>0 (f)\nx +x,x > 0 (g) x — yic, -co < x < +00 (h) e*/x, x > 0 9. (a)e* ln3 (b)e^21"2 11. (a) 4~e (b)e2 13. 3*8 - 3*5 15. (a)3/x (b)l 17. (a)0 (b)0 (c) 1 Responses to True-False questions may be abridged to save space. 19. True; both equal -In a. 21. False; the integrand is unbounded on [-1, e] and thus the integrand is undefined.
23. (a)2*Vl+;t
2
2/,2_Ll)3/2+2(;c2+1)5/2_4^
(b)-f(jc
25. (a) - cos(*3) (b)-tan 2 * 27. -3
:- 1
• +2x-
29. (a) 3x2 sin2(x3) - 2x sin2(;c2) (b) ——^ 31. (a) F(0) = 0, F(3) = 0, F(5) = 6, F(7) = 6, F(10) = 3 (b) increasing on [|, 6] and [^, 10], decreasing on [0, |] and [6, ^] (c) maximum y at j; = 6, minimum — | at x = \ (d)
x<0 35. y(x) = (\+x2)/2, x>0 37. y(x) = tanx + cos* - (V2/2) 39. P(x) = PO + /Q r(t) dt individuals 41. I is the derivative of II.
11. (a) 0.25545, error
(b) \x\ < sin 1 13. 3/Vl - 9;c 17. 3x2/(l+x6)
15. 21.
:
+ ex sec"1 x
19. -
23. 0 25.
27. \ sin""1 x -3 tan"1 A: + C 29. |sin~'(2x) + C 31. tan-'e-' + C 33. sin-'(tanA:) + C 35.71/4 37. jr/12 = 41. jr/6 43. ^(6^) \3v3 1 47. (a) sin' (}*) + C (b) ^ tan"1 +C 39.
45. ?r/9
Responses to True—False questions may be abridged to save space. 49. False; for example, sin" ' (sin ?r) = 0 / 71. 51. True; y = nil is an asymptote for the graph of y — sec"1 x. 53. (a) cor'Ot) csc-'M
33. F(x) =
1 2 3
-2-1
1 2
l
(b)cot x:allx, 0 < v < n cx~l x: \x\ > 1, 0 < \y\ < Till 55. (a) 55.0°
(b)33.6°
65. (a) 21.1 hours
(c)25.8°
(b) 2.9 hours
61. 0 63.
67.29°
69. k * 0.9973
Answers to Odd-Numbered Exercises 71. 7?/4
73. ^7i(7i2 - 8) % 1.46838
75.
79. y = 3 sin- 1 1 - n 81. y = ^ tan" 1 / 3 ^
(b) about 10 years
11. (a)
% 12(73-1)
A67
(c) 220 sheep
*
^ Exercise Set 6.8 (Page 481) 1. (a) ss 10.0179 (b) SB 3.7622 (c) 15/17 ss 0.8824 (d) SB -1.4436 (e) % 1.7627 (f) ^ 0.9730 13. (b) 3.654, 332105.108
5.
x+l sinh ;c0
i
tanh^Q ! coth^Q
COSh XQ
sech XQ
2
V5
2/^5
V5/2
i/Vs
1/2
| (b)
3/4
5/4
3/5
| 5/3
4/5
4/3
I
4/3
5/3
4/5
1
3/5
3/4
9. 4cosh(4*-8)
5/4
x +2
(d) curve must have a horizontal tangent line between x = 1 and x = e
2 + 5 cosh(5x) si 13. -4 csch | - | coth ( - ) 15. x2 \xj \xj cosh2 (5x)
1 l
2
(cosh~ x)Jx
-1
23. -
(tann"1*)"2 1-x 2
s*
jc > 0 x* <<0 0
27. -+ ej:sech~17* Isinh*! (-1, 2*71-* 7 372 29. jsinh A: + C 31. f (tanhx) + C 33. \ In(cosh2z) + C 35. 37/375 37. i sinh"1 3^: + C 39. -sech"1^) + C 41. -csch"112^| + C 43. ^ I n 3 Responses to True-False questions may be abridged to save space. 45. True; see Figure 6.8.1. 47. True; f ( x ) = sinh x. 49. 16/9 51. 5n 53. | 55. (a)+00 (b)-oo ( C )l (d)-l (e)+oo (f)+oo 63. |H| < 1: tanh"1 M + C; |u| > 1: tanh"1 (l/») + C 65. (a) In 2 (b) 1/2 71. 405.9ft 73. (a) 650 (b) 1480.2798 ft (c) ±283.6249 ft (d) 82°
-300 75. (b) 14.44 m (c) 15 In 3 ^ 16.48 m ^ Chapter 6 Review Exercises (Page 484)
1. (a) \ \n(x - 1) (b)
1.6 x 1055 miles 7T
JT
7. 15*+ 2 37T
x +4 29. 2*
11. —csch x
tanh(^x) sech
x +3
csch *o
fw (c)
15. 0
JT
7T
43. e2
45. e1/e
47. No; for example, /(*) = A 3 . 49. (A,e) (b) The population tends to 19. (c) The rate tends to zero. 0 0
20
-100
55. 7i 57. +00, +00: yes; +00, — oo: no; — oo, +00: no; — oo, — oo: yes 59. +00 61. £ 63. (a) (-oo,0] (b)[0,+°o) (c) (-00, -lA/2), (1A/2, +00) (d) (-1/V2, 1/V2) (e) ±lA/2 65. relative min of 0 at x = 0 67. m = e2/4 at x = 2 69. maximum value 1 /(-2 - 73) SB 0.84; minimum value /(-2 + 73) « -0.06
-0.5 71. 3x1/3 -5e* + C 73. tan"1 x + 2 sin"1 x + C 75.0.35122,0.42054,0.38650 77. e - 1 79. e3 - e 81. (a) X*) = sin* - 5e* + 5 (b) X*) = \^ ~ \ 83. 2-2/Ve 85. 3/2 + In 4 ^ Chapter 6 Making Connections (Page 487)
_
Answers are provided in the Student Solutions Manual. ^ Exercise Set 7.1 (Page 490) _ 3. itan(* 2 ) + C 5. -\ ln(2 + cos 3*) + C :
9. e t a n ; c +C 11. -^cos 6 5A: + C + 4) + C 15. 2e^~^ + C 17. 21. icoth2 x 27. -icos(x 2 )
25. 2
31. (a) i sin z + C
+C
-4' 29. --
(b) -| cos2^; + C
In 16
+C
A68
Answers to Odd-Numbered Exercises
(?-£Uc
33. (b) In tan - + C (c) In
+C
Exercise Set 7.2 (Page 498) 1. -e~
15. In |V* - 2*e* + 2ex + C
3.
2
5. — ^jccos3jc + g sin3jt + C
x2
x
9. —\nx
7. x sinjt + 2jtcosjt — 2sinjc + C
II. x(\nx)2 -2x\nx
13. x In (3* - 2) - x - I In (3* - 2) + C 15. x sin"1 * + Vl - x2 + C 17. 19. - COSJE) + C ) - 5 ln(l + 9jt 2 ) + 21. (jt/2)[sin(ln;c) -cos(lnj;)] + C 23. jtrtanjc + In |cosjc| + C
25. ±x2e*
C
-
27. |(
4
31. 31n3-2
33. — -x/3 + 6 37. - (2V3JT - - - 2 + In 2)
+ l)
29. (2e3
35. -jr/2
33. L = V5 - V2
37. tan"'(A: -2)
Responses to True-False questions may be abridged to save space. 39. True; see the subsection "Guidelines for Integration by Parts." 41 . False; ex isn't a factor of the integrand. 43. 2(V* - \)e
51. 55. 61. 65. 69.
- e2jc +C 2 1 . 2 / 3 23. (V3 - V2)/2 19. isin~'(<;*) 10V3 + 18 25. 243 Responses to True-False questions may be abridged to save space. 27. True; with the restriction —n/2 <9< nil, this substitution gives Va 2 - Jc2 = acos# anddx = acosddO. 29. False; use the substitution jc = a sec 6 with 0 < 0 < nil (x. > a) or n/2 < 6 < n (x < -a). 31. | l n U 2 + 4 ) -
eax
-j (a sin bx - b cos bx) + C 53. (a) i sin 2 A: + C a +b2 (a) A = 1 (b) V = jt(e - 2) 57. V = 2n2 59. jr3 - 6jr (a) -| sin 3 jccosjt - | sin jt cos* + |jt + C (b)8/15 (a) I tan 3 x - tan x + x + C (b) j sec2 jc tan x + \ tan jt + C (c) JE V - 3* V + 6xe* - 6ex + C (jc + l)ln(je + l ) - j e + C 71. (x2 + 1) tan' 1 x - x + C 1
= ^[sin 2 A:Vl -sin 4 jc + sin ' (sin2 ;t)] + C +9 x 51. (a)sinh~'(;c/3) + C (b)ln
Exercise Set 7.5 (Page 521)
Exercise Set 7.3 (Page 506) 1. -|cos4jc
3. -- — si
5.
5. — cos3 ad - cos aO + C 1. — sin2 ax + C 3a la n t + C 11. g x — jj sin
9
'
13. --jL cos 5jc + i cos jc + C 15. -| cos(3*/2) - cos(jc/2) + C 17. 2/3 19. 0 21. 7/24 23. { tan(2jc - 1) + C 25. ln|cos(e~ A r )| + C 27. ± In | sec4;t + tan4jt| + C 29. ^tan 3 jc + C 31. isec^jc + C 33. I7 sec7 A: - i sec5 A: + C 16 35. I sec3 jt tan Jt — § sec j; tan jc + | In | sec jc + tan x\ + C 31. 5 sec3 f + C
41. I tan 4* + £ In | cos4*| + C 43. f tan 45.
372
jc + | tan
772
jc + C
47. - i + In 2 49. - j esc5 jc + ± esc3 jc + C
2 8 51. —\ esc2 jc - In | sin jc| + C Responses to True-False questions may be abridged to save space. 53. True; f sin5 x cos8 jt dx = f sin jc (1 — cos2 jc)2 cos8 jc dx = -/(I -u2)2u*du = - f ( u * ~2ul° + ul2)du 55. False; use this identity to help evaluate integrals of the form / sin mjc cos nx dx. 59. L = ln(V2 + 1) 61. V = n/2 1 | Va2 + b2 + a cos jc — b sin x 67. , a sin A"
69. (a) f
(b)3Wl6
(c)-^
(d) 5jr/32
^ Exercise Set 7.4 (Page 513) W16-Z 2 4/ 2 7. Vjc2 - 9 - 3sec-'(;c/3) + C
1. 2 sin~' (jc/2) + i W4 - jc2 + C 3. 8 5. ^tan-'(jc/2)+
n ,^
8(4 + .
+C
17.
19. In \x2 -3x21. x + ^- +ln
39. tan jc + i tan3 jc + C
2
13.
B A B 3. - + ^ jc - 3 jc +4 A B C Ax + B Cx + D - +— + -^J + jc jcz A: ! n + C 11. § In \2x - 11 + 3 In \x + 4| + C 5 2 x (x + 3) + C 15. 3j In \n\x x-3 2 3jc -2| - +C
+C
23. 31n|jc| - l n | j c - 1| +C 2 25. h l n | j c -3| +ln\x + 1| + C A: — 3 27. ^- r ~—-!-- y +ln|A- + l| + C
29. 31. 3 ttan~' A: + i ln(;c2 + 3) + C x2 1 33. y Responses to True-False questions may be abridged to save space. 35. True; partial fractions rewrites proper rational functions P(x)/Q(x) as a sum of terms of the form 7y
_|_
7
9v
^
r
(Bjc + C)* 9
and/or -
GX + HY
i
37. True; ^-±^ = ^ + 4z = 1 + 4. xz Jtz J: JC Jf2 j 43. V = ;r(-y — | In 5) 45. —= tan"
1
+C
47. | In IA-- 11 - 5 In |jc - 2| + -^ In |jc - 3| - -[^ In |jc + 3| + C
Answers to Odd-Numbered Exercises
1.
55. M2= /•VPTT, , ,
Exercise Set 7.6 (Page 531) Formula (60): fjc + ^ln|3*-
2
^3/2
- / » V ^ i ) ^ = -(
3. Formula (65): - li
57. u =
1/3
59 H
1 / 4 /I
5. Formula (102): i( 7. Formula (108):
3
"
+c
2/3
2
c
,1/4 ^ 7 / 1 1
1
' J u(\ -u)
61. H =
9. Formula (69): 8
3
f
J u3 ~ u
A69
_L r1
\\
3
: l / 6 + 61n|A: 1 ' r \ 63. M = Vl +x2, 1 (UL — \)du = -(1 J 3 2 ' d U _ /• ' , 6< /" 65 2n 1 - w 2 1 + w2 J M+l 'J
6 f
11. Formula (73): 13. Formula(95): -Vx 2 + 4-21n(x + \/x2 +4) x 9 x 15. Formula (74): - V9 - x2 + - sin"1 - + C
1+M2
l+»2
= In | tan(x/2) + \\ + C
17. Formula (79): V4 - x2 - 2 In sin Ix
19. Formula (38):
67.
du
I —^— = Jf 4 = -cot(e/2) + c u2
J 1 - cosD9 ( 69. I J 2w
1 1--
x4 16
21. Formula (50): — [4 In * - 11 + C 23. Formula (42):
1 + M2
2 f I -u2 =-2 • ^2 aw = / dw 1+ w 1+ u J 2u
1 2w
+
1 + U2 ' 1 - M 2
= - In | tan(jc/2)l - - tan 2 (x/2) + C
[-2 sin(3x) - 3 cos(3x)] + C
udu 1 Cf udu 1 Tf 25. Formula (62): - / == — 2j (4-3«) 2 18 L' 2 f du 1 _, 27. Formula (68): - / —=2 = - tan 3 J w +4 3 29. Formula (76): - I 2 J
"
VM 2 — 9
4
+ In
+C
73. /I =6+ f sin"1 f
71. * = -^—^
75. ,4 = ^ l n 9
77. V = 7r(jr - 2) 79. V = 27T(1 - 4e~3) 81. i, = V65 + i In(8 + V65) 83. S = 2?r [^2 + ln(l + V2)l 85.
= - In \2x + 2
91. J r (
4-
-
.
93. -i]
31. Formula (81): - / —== du = 2 1 2
33. Formula (26): i): I sin u rfw = \ In A: - | sin(2 In A:) + C x
r
/
37. « = sin 3x, Formula (67): - / r 3 J u(u + I) 2
sin 3* I sin 3x
2
39. u = 4x , Formula (70): - / —^— = — In 8 j u1 - 1 16
+C
41. u =2^, Formula (74): -
45. u = 2x, Formula (44): I usinud = sin 2^: — 2x cos 2;c + C 47. u = -VI, Formula (51): 2 / we" du = -2 49. x2 + 6x - 1 = (x + 3)2 - 16, u = x + 3, Formula (70): +C
w2-16 8 51. x2 - 4x - 5 = (x - 2)2 - 9, u = x - 2, Formula (77): f x -2 +C
d = v
/ift ; -
5/2
+c
+ 1 dx = A4- as 4.66667
(a) MIO = 4.66760; \EM\<* 0.000933996 (b) TIQ = 4.66480; \ET\^ 0.00187099 (c) S20 = 4.66667; \ES\ ^ 9.98365 x 10^7 /•jr/2 / cos x dx = 1 Jo (a) MIO = 1.00103; \EM\ « 0.00102882 (b) TIO = 0.997943; \ET\™ 0.00205701 (c) S20 = 1.00000; \ES\^2.\\541 x 10~7 + £ fi
dx =
43. u = 3x, Formula (112):
du
6 9 12 15
Exercise Set 7.7 (Page 544) 3
35. Formula (51): - / ue" du = -(-2x - \)e^ + C 4J
3 \ sin 3x + 1
3
RS 0.0664283
(a) M I O = 0.0659875; |£ M I ~ 0.000440797 (b) r,0 = 0.0673116; \Er\ ^ 0.000883357 (c) S20 = 0.0664289; |£5| ss 5.87673 x 10'7 9 (»)\EM\ < =0.0028125 ~ 3200 (b) | ET | < -?-= 0.005625 1600 81 (c) |£sl < * 7.91016 x 10-6 10,240,000 9. (»)\EM\ <
JT3
19,200 9600
- 2)3/2 + C (c)|£sl <
- 0.00161491 : 0.00322982
921,600,000
• 3.32053 x
A70
Answers to Odd-Numbered Exercises
11. (a)|£ M |< (b)|£ r |<
75e2 2 TSe2
: 0.00180447 ' 0.00360894
1 • 2.40596 x 1 56,250e2 13. (a) n = 24 (b) n = 34 (c) n = 8 15. ( a ) n = 1 3 (b)n = 18 (c) n = 4 17. (a)n = 43 (b)« = 61 (c)n = 8 Responses to True-False questions may be abridged to save space. 19. False; Tn is the average of Ln and Rn. 21. False; £50 = 5^25 + 5 23. 25. 27.
= 1.49367; = 3.80678
29. S,o = 0.904524; I
x V*2 + 9 91n(|V*2 + 9-M) ^ +C 'in 2 2 5 " x+4 2 29. \x - 2x + 6 In \x + 2| + C 4 2 l /^ 31. h i •*v T^1 •<-O l 11 T T^ *— x + 2 (jc + 2)2 cos 16.x 1 cos 2* 1 f" Fnrmnln (A(\\35. ^UIHlUla ^^rUJ. -p *--
25.
1.49365 3.80554 « 0.904524
31. (a) MIo = 3.14243°; error EM -0.000833331 (b)rio = 3.13993; error £r s 0.00166666 (c)S20 = 3.14159; error E5 % 6.20008 x lO"10 33. 5,4 = 0.693147984, |£5| « 0.000000803 = 8.03 x 10~7 35. n = 116 39. 3.82019 41. 1071ft 43. 37.9mi 45. 9.3 L 47. (a) max \f"(x)\ & 3.844880 (b) n = 18 (c) 0.904741 49. (a) The maximum value of |/(4) (*)| is approximately 12.4282. (b) n = 6 (c) 55 = 0.983347
37. Formula (113): ^(8*2 - 2x -3)7* - x2 + ^sin ~'(2;t - 1) + C 39. Formula (28): itan2j:-jc + C 3
41.
/
A — 9./O - — 4Z V •£ ~ '^ 1 1 .171 1 / 1S7 J/
—
7* + 1 (a)Mio = 1. 17138; \EM\ ™ 0.000190169 (b) 7,0 = 1.17195; \ET\ % 0.000380588 (c) 520 = 1.17157; \ES\ ^ 8.35151 x 10~8
43. (a) \EM\ < -l—p ^ 0.000441942 1600V2 (b) \ET\< -- « 0.000883883 800 V2 7 • 3.22249 x 10" 15,360,000^ 45. (a) n = 22 (b) n = 30 (c) n = 6 47. 1 49. 6 4_ r* ^7 5 51. e-] 53. a 7T/9 5^ 2 3
5
' In 9
sVsT^
59. g sin 2x — i sin 2* + C 61. -^ e2' cos 3^ -|- -fje2* sin 3^: + C 63. | + -rj In |JT + 2| + -^ In * — 3| + C 65. A.
TT
f>n
V^TT+l 12 2 1 / 2 _i_ o v _i - 9 _1_ O I n f - / v 2 r 0,- 1 T 1 .- 1 T\ _L /" "71 71 ^ Exercise Set 7.8 (Page 554) 1. (a) improper; infinite discontinuity at x = 3 (b) not improper (c) improper; infinite discontinuity at x = 0 Chapter 7 Making Connections (Page 559) (d) improper; infinite interval of integration Where correct answers to a Making Connections exercise may vary, no (e) improper; infinite interval of integration and infinite discontinuity answer is listed. Sample answers for these questions are available on the at x = 1 (f) not improper Book Companion Site. 3. 5. In 2 7. i 9. -| 11. | 13. divergent 15. 0 3. (a) r(i) = i (c) r(2) = i, r(3) = 2, r(4) = e 17. divergent 19. divergent 21. n/2 23. 1 25. divergent 5. (b) 1.37078 seconds 27. | 29. divergent 31. n/2 ^ Exercise Set 8.1 (Page 566) Responses to True-False questions may be abridged to save space. 3. (a) first order (b) second order 33. True; see Theorem 7.8.2 with p = | > 1. Responses to True-False questions may be abridged to save space. 35. False; the integrand is continuous on [1, 2]. 5. False; only first-order derivatives appear. x(x - 3) 37. 2 39. 2 41. \ 7. True; it is third order. 43. (a) 2.726585 (b) 2.804364 (c) 0.219384 (d) 0.504067 45. 12 15. y(x) = e-2* - 2ex 17. y(x) = 2e2* - 2xe2x 47. -1 49. i 51. (a)V = 7T/2 (b) S = jrK/2 + ln(l + 72)] 19. y(x) = sin 2x + cos 2* 21. y(x) = -2x2 + 2* + 3 53. (b) \/e (c) It is convergent. 55. V = n 23. y(x) = 2/(3 - 2x) 25. y(x) = 2/x2 59.
kr
61. (b) 2.4 x 107mHb 63. (a) -j 67. (a) 1.047 71. 1.809
1. 9. 13. 15. 19. 21.
27. (a) ^ = ky2, y(0) = y0 (k > 0) at
l
(b) -j
(c)
Chapter 7 Review Exercises (Page 557) 3. -\ J cos3/26> 2. I (a)2sin~'(V;t72) + C;-2sin '( sin"1 (x -xe~x -e~* + C 11. x \n(2x + 3) - x + \ In (2* + 3) + C (4*4 - 12*2 + 6) sin(2;t) + (Sx3 - \2x) cos(2x) + C 50-35 sin 100 + C 17. - g cos3* + ^ cosjc + C -1 sin3 (2*) cos 2*- ^ cos 2* sin 2* + \x + C | sin" 1 (jc/3) - A^\/9 - *2 + C 23. In |z + V*2 - 11 + C
Exercise Set 8.2 (Page 575) -vl+* - 1,C / O 5. 2 In |. \+y = e* C 7. y = \n(secx + C) 1 (j ^ _ ^ 1 — C(cscx — cotjc)' 3 2 2 11. / + siny = x +7t 13. y ~2y = t2 + t + 3 2
Answers to Odd-Numbered Exercises
A71
61. (a) v = cln
15. (a)
- gt (b) 3044 m/s mo — kt 63. (a) hfa(2- 0.003979f )2 (b) 8.4 min 65. (a) v = 128/(4f + 1), x = 321n(4f + 1) ^ Exercise Set 8.3 (Page 584) i 1 \ \2-
- -
1\ X O ) = 2
y(0) = 1
_2
2"*
- —-
/)
,
i
if 5
/ \
X ^--217. y =
2
-
5. y -> 1 as x ->+00 7. 1 1 " I °
2
1
3
4
|
5
6
7
8
0.5
1
! 1.5
2
: 2.5
3
) 3.5
4
1.50
2.07
1 yn [ i.oo
2.71
3.41 [ 4.16 1 4.96 | 5.82 6.72
7 -
-I
Responses to True-False questions may be abridged to save space. 21. True; since - dy = dx. 23. True; since (5) 32 = 1.
29. (a) y'(r) = y(0/50, y(0) = 10,000 (b) y(t) = 10,OOOe'/5° (c) 50 In 2 % 34.66 hr (d) 50 ln(4.5) % 75.20 hr (b) y = 5.0 x 107e-°-18"
31. (a) — = -ky, k «s 0.1810 33. 39. 43. 47. 51.
(c) fa 219,000 atoms (d) 12.72 days 50 ln(100) % 230.26 days 35. 3.30days (b) 70 years (c) 20 years (d) 7% (a) no (b) same, r% 45. (b)ln(2)/ln(5/4) ^ 3.106 hr (a) $1491.82 (b) $4493.29 (c) 8.7 years (a) " (b)
0.5 1 1.5 2 I I . 0.62 Responses to True-False questions may be abridged to save space. 13. True; the derivative is positive. 15. True; y = yo is a solution if yo is a root of p. 17. (b)y (1/2) = 73/2 19. (b) The ^-intercept is In 2. 23. (a) y' =
3*y2 -:
f n + l\n 25. (b) lim yn = lim =, "^+" n^+™\ n } ^ Exercise Set 8.4 (Page 592) _
(d)
1. y = e~3x + Ce~4x
3. y = e~x sin
) + Ce~
7- y = - + — 9. y = 4e* - 1
2
53. 3-0 ~ 2, L fa 8, k fa 0.5493 55. (a)yo = 5 ( b ) L = 1 2 (c) k = 1 (d) t = 0.3365
2 10
°°
57. (a) y = (b)
r
(c)
0 i
y(t) 20 r
2x
Responses to True-False questions may be abridged to save space. 11. False; y = x1 is a solution to dy/dx = 2*,but y + y = 2x2 is not. 13. True; it will approach the concentration of the entering fluid. +oo if yo > 1/4 17. lim y = ifyo < 1/4
8
y(f) 49
2
3
4
5
6|7
22 25 28 31 35 39 '44
9 10 11 54 61
12 13 3 67 75 83 93|
(b) 6.22 min
1
2
3
4
5
n
0
0.2
0.4
0.6
0.8
1.0
yn
1
1.20 1 1.48
1.86
2.35
2.98 |
n
3 59. (a) T = 21 + 74e-*'
0
x
6
9
12
j
A72
Answers to Odd-Numbered Exercises 19. e-',4e" 2 ,9e~ 3 , 16e~ 4 ,25e~ 5 ; converges, lim « 2 e~" = 0 -
x
0
0.2
0.4 ! 0.6
0.8
1.0
y<X)
1
1.24
1.58 : 2.04
2.65
3.44
Absolute error Percentage error
0
0.04
0.10 ; 0.19
0.30
0.46
0
3
11
13
n
6
1
9
21.2,.^ 23.
/25
21. (a) 200- 175e-' oz (b)136oz 23. 25 Ib 27. (a)/(0 = 2-2e' 2 ' A ( b ) / ( f ) - » 2 A
27.
^ Chapter 8 Review Exercises (Page 594) 3. y = tan(^ 3 /3 + C) 5. In |y\ + y2/2 = ex + C and y = 0 7. 9.
1
13.
n
0
i
n
0
0.5
y/i
i
1.50
' 1
1 3 | 1.5
2n In
1n~\
; converges, hm =I . n -»+00 2/1 +c 1 l ° (-!)"+' — } ;converges, lim (-!)"+' 3— = 0 ! 3"L=i "^+t° "
(-
converges, lim (-1)"+' ( --|=0 n->+oo \n » + !/ + 29. \~Jn + 1 - V« + 2} * ; converges,n-^+co lim (V« + 1 - \/» + 2) = 0 '«=]
2
3
4
| 5
4 6
7
8
2
| 2.5
3
3.5
4
6.91
8.23
2.11 i 2.84 3.68 i 4.64 5.72
- 1 ; converges, lim 6/ n -> +o
Responses to True-False questions may be abridged to save space. 31. True; a sequence is a function whose domain is a set of integers. 33. False; for example, ( (- 1 )"+ ' ) diverges with terms that oscillate between 1 and -1. 35. The limit is 0. 37. for example, {(-l)")+=i and jsin(jr«/2) + !/«)+;, n, n odd 39. (a) 1,2, 1,4, 1,6 (b)a n = 1/2", neven \/n, n odd (c)a,,= neven l/(n (d) (a) diverges; (b) diverges; (c) lim an = 0 n—> +co
43. (a)(0.5)2"
(c) lim an = 0 ( d ) - l < a n < l (b) lim (2"+3") 1 / n = 3
45. (a) 30
Ov
0 49. (a)N=4 15.
1.00
0
1
2
3
4
\ <„
0
0.2
0.4
0.6
0.8
I 3-*
1.00
1.20
1.26
1.10
0.94 ] 1.00
I n
2
5 j__
3j:
17. about 2000.6 years 19. y = e" * + Ce~ 21. y = -1 +4
_
Where correct answers to a Making Connections exercise may vary, no answer is listed. Sample answers for these questions are available on the Book Companion Site. 1. (b) v = 2 - 3e~2A 3. (a) du/dx = (/(«) - u)/x
2
(b) .v - 2xy - y = C _
1
(b) 1. (a) 3"-' 3"-' 2n 3. (a) 2, 0,2,0 ( b ) l , - l , l , - l (c) 2(1 + (-I)"); 2 + 2cos«jr 5. (a) The limit doesn't exist due to repeated oscillation between -1 a n d l . (b)-l; 1;-1; 1;-1 (c) no 7. , f ; converges, ^ lirn^ 9. 2, 2, 2, 2, 2; converges, lim 2 = 1
In 1 In 2 In 3 In 4 In 5 In n , , , , ; converges, lim =0 1 2 3 4 5 n^+so n 13. 0, 2, 0, 2, 0; diverges 15. — 1 , — , — — , — , — — ; diverges 11.
- f.f'l'l'l; conve
(c) N = 1000
^- Exercise Set 9.2 (Page 613) 1. strictly decreasing 3. strictly increasing 5. strictly decreasing 7. strictly increasing 9. strictly decreasing 11. strictly increasing Responses to True-False questions may be abridged to save space. 13. True; an+\ — an > 0 for all n is equivalent to a\ < at < aj < • • • < a,, < • • • . 15. False; for example, )(-l)" + 1 ) = (1, -1, 1, -1,...) is bounded but diverges. 17. strictly increasing 19. strictly increasing 21. eventually strictly increasing 23. eventually strictly increasing 25. Yes; the limit lies in the interval [1,2].
27. (a) 72, v/2 + 72, \/2 + \/2 + 72 2
^- Exercise Set 9.1 (Page 605)
17
(b)N=\0
1 / 1W 2\ 1 l i m - ! +- ! +- = ]-*• +«> 2 \ /i/V n7 2
(e) L = 2
Exercise Set 9.3 (Page 621) 1. (a) 2, ^, g, f^f; f (l - (j)") ; n l i m ^ s n = f (converges) (b) \,\,\, f ; - \(\ - 2"); J^Jn = +^ (diverges) (c) -, - , — , - ; -
—; n lim^.s n = - (converges)
3. | 5. 6 7. 5 9. 1 11. diverges 13. ^ 15. (a) Exercise 5 (b) Exercise 3 (c) Exercise 7 (d) Exercise 9 Responses to True-False questions may be abridged to save space. 17. False; an infinite series converges if its sequence (Apartial sums converges. 19. True; the sequence of partial sums [sn] for the harmonic series satisfies n+ 1 sy > , so this series diverges.
21. 1 23. ^
27. 70m
Answers to Odd-Numbered Exercises 29. (a)Sn = - l n ( n + 1); lim Sn = -oo (diverges) »-* +CC
5,, = -In2
33. |error| < 0.125 35. |error|<0.1 37. n = 9999 39. n = 39,999 41. |error| < 0.00074; j, 0 ^ 0.4995; S = 0.5 43. 0.84 45. 0.41 47. (c)n = 50 (-1)* 49. (a) If a^ = —-;=-, converges and J2 af diverges. = then
•Jk
31. (a) converges for \x\ < 1; S = -
(-1)*
, then ^ a* converges and ^ a 2 also converges.
(b) converges for |JT| > 2; S =
If at =
(c) converges for x > 0; S = —
(b) If UK = - , then J] a 2 converges and ^ at. diverges. If a\< = —j , k K. then J^ a2, converges and ^ a^ also converges.
33. «„ = ^rfOl + r—f + 2^2 ^
h
T'nH™-^"
=
'
Exercise Set 9.7 (Page 657) (a)l ~x + \x2, 1 -x (b)l-jj: 2 , 1 (a)l + \(x- ) ) - \(x- I) 2 (b) 1.04875 5.1.80397443 po(x) = 1, pi (x) = 1 - x, P2(x) = 1 - x + i*2, 1 , 1,
^ Exercise Set 9.4 (Page 629) 1. (a) | (b)-| 3. (a) p = 3, converges (b) p = |, diverges (c) p = 1, diverges (d) p = |, diverges 5. (a) diverges (b) diverges (c) diverges (d) no information 7. (a) diverges (b) converges 9. diverges 11. diverges 13. diverges 15. diverges 17. diverges 19. converges 21. diverges 23. converges 25. converges for p > 1 29. (a) diverges (b) diverges Responses to True-False questions may be abridged to save space.
= l,P2(Jt) = 1 - - J : 2 , P 3 ( J : ) = 1 - ~*2, L«/2J
P4(x) = 1 - ~^ 2 + ^r*4;
31. False; for example, Y^ 2"* converges to 2, but VJ —^ = YJ 2*' ""2
A=0
diverges. 33. True; see Theorem 9.4.4. 35. (a) (jr2/2) - (jr4/90) (b) (n2/6) - (5/4) 37. (d)-jV < ^ 2 - . v 1 0 < 4,
11.
Jn
(See Exercise 70 of
Section 0.2.) ) = 0, pi (^) = x, P2(x) = x - {x2, p 3 (jc) = JT - ix
(c) jr4/90
/•+
39. (a)) /
13.
—rdx =
T;
apply Exercise 36(b)
(b)« =
(0^*1.08238 41. converges ^- Exercise Set 9.5 (Page 636) 1. (a) converges (b) diverges 5. converges 7. converges 9. diverges 11. converges 13. inconclusive 15. diverges 17. diverges 19. converges Responses to True-False questions may be abridged to save space. 21. False; the limit comparison test uses a limit of the quotient of corresponding terms taken from two different sequences. 23. True; use the limit comparison test with the convergent series
cise 70 of Section 0.2.) 15. po(x) = 0, pi (x) = 0, P2(x) = x2, p 3 (jc) = x2, -x2k+2 (See Exercise 70 of Sec1)!i tion 0.2.) 17. p0(x) = e, pi (Jc) = e + e(x - 1), P2(x)
=
.£(*-!)* + !(*-1)3 H 25. converges 27. converges 29. converges 31. converges 33. diverges 35. diverges 37. converges 39. converges 41. diverges 43. converges 45. converges 47. converges 49
A73
*!
' "* = 1 . 3 . 5 . . ( 2 * - 1 ) ; P = * ^ WTl 51. (a) converges (b) diverges 53. (a) converges
Exercise Set 9.6 (Page 646) 3. diverges 5. converges 7. converges absolutely 9. diverges 11. converges absolutely 13. conditionally convergent 15. divergent 17. conditionally convergent 19. conditionally convergent 21. divergent 23. conditionally convergent 25. converges absolutely 27. converges absolutely Responses to True-False questions may be abridged to save space. 29. False; an alternating series has terms that alternate between positive and negative. 31. True; if a series converges but diverges absolutely, then it converges conditionally.
4!
19.
p 4 (x) = -1 - (x + 1) - (x + I) 2 - (x + I) 3 - (x + D4
21. PO(J:) =
E
(See Exercise 70 of Section 0.2.)
A74
Answers to Odd-Numbered Exercises
23. po(x) = 0,P l ( x ) = (x- 1), p2(x) = (*-!)- i(x - I) 2 , P4(x)
^ Exercise Set 9.8 (Page 667)
= (*-!)- {(x - I) 2 + i(jc - I) 3 - \(x - I) 4 ;
X—^ ^ —
'
/ -J
1r
k
1 \&
r2*+2
7. 2^
*=l 25. (a) p3(jt) = I + 2* - x2 + x3 (b) p 3 (jt) = I + 2(x - I) - (x - I) 2 + (x - I) 3 27. poOt) = ' n . r r l = 1 - 9r
Ot
k 13. J (-l)(x + l) fc
4
a
p3(x) = 1 -2j + 2^ 2 - |;c3
15.
^Vi)* 19.-i 1
23. (a)-2<x<2 (b) /(O) = 1; /(I) = f Responses to True-False questions may be abridged to save space. 25. True; see Theorem 9.8.2(c). 27. True; the polynomial is the Maclaurin series and converges for all x. 29. /?=!;[-!, 1) 31. * = +«=; (-00, +0.) 33. R = ±; [-i, i] 35. « = 1; [-1, 1] 37. R = l; (-1, 1] 39. .R = +c°; (-00, +00) 41. /? = !;[-!, 1] 43. I? = 1; (-2,0] 45. « = f ; (-f , -^-) 47. R = +°°; (-00, +«,) 49. (-00, +00) 55. radius = R 61. (a) n = 5; ss % 1.1026 (b) f (3.7) % 1.10629 21. 1
-1.25 Responses to True-False questions may be abridged to save space. 31. True; y = /(JCQ) + f'(xo)(x — XQ) is the first-degree Taylor polynomial for / about x = XQ. 33. False; pf(x0) = fm(x0) 35.1.64870 37. IV 39. (a) 3
<3;
^ Exercise Set 9.9 (Page 676) _ 3. 0.069756 5. 0.99500 7. 0.99619 9. 0.5208
11.
2k - 1
(b) 0.223
13. (a) O4635; 0.3218 (b) 3.1412 15. (a) error < 9 x 10~8 (b)
1 *
l/w !/>,«
-1.000 -0.750 -0.500 -0.250 0.( )00 0.250 0.500 1 0.750 1.000 0.506 0.619 0.781 l.( XX) 1.281 1.615 i 1.977 2.320 0.431
JP 2 W
0.000
0.250
0.500
0.500
0.531
0.625
(c) \e * -
| <0.0 for -f .14 < x <0.14 0.01
( i)
^
* ~ (
+
**T)
t=2
23.23.406%
<0.0
^ Exercise Set 9.TO (Page 686) 1. (a) 1 - x + x2 + (-1)*** + • • • ; « = (b) 1 + x2 + x4 + • • • + x2k + • • •; R = 1
for -0.50 < x < 0.50 0.015
(c) 1 + 2x + 4x2 H . . 1 1 1 ^ -0.15
0.15
.
-0.6
(c) no 0.00000005
17. (a)
0.750 l.(XX) 1.250 1.500 1 1.750 2.000 0.781 l.( XX) 1.281 1.625 2.031 2.500
An
11.
*=o
0.6
h 2*** -I
r
;R = 1 ,, ,
_
" ^TTI ~ 25/2 29/2 • 2! ~ 2 1 3 / 2 - 3 ! (b) (1 - x r = \+2x2 + 3*4 + 4x6 + •'• • 23 25 27 5. (a) 2x - —x* + —x5 - —x1 + - - .; 5 = +00 2 2
41. (a) [0,0.137] (b)0-002
(b) 1 - 2x + 2x2 - -x3 + • • •; R = +co (c) 1 + x2 + i*4 + ^x6 + • • •; R = +0=
0.2
43. (a) (-0.569,0.569) 0.0005
45. (-0.311,0.311) 4.0005
-0.2
0.2
2 4! 6! 7. (a)^:2 - 3^;3 + 9x* - 21x5 + • • • ; « = | 25 4 25 6 2? 8 + 3!^ + ST" + 7!X 3 (c) x - -x + -j:5 + —x1 + • • • ; / ? = !
,
Answers to Odd-Numbered Exercises
(b) 12*3 - 6*6 + 4x9 - 3xn + • • • 11. (a) 1 - (x - 1) + (x - I) 2 + (-!)*(* - 1)* + • • • X3
XS
3
X3
30
24
1 ^ (**\ I _L ^ *-2 _i_ 5 X.,4 _i_ 61 V6 i 13. (aj 1 •+• ^-^ i 24 T 720-^ + * * '
19, 2
4vt -h 2jc^
X4
(b) (0,2)
7j
24 ~ 1920*
/u\ v !-2 i 1 .,3 1 ~5 _i_ \u) X — X + ^-^ "~ 30^ * ' '
4jt^ -
25. [-1,1]; [-1,1) 27. (a)I>
(C)/(»)(O) = B!CII = 29. (a)l
(b)-i
!
J»
fa\\^ •
, fneven
31. 0.3103 33. 0.200
35. w2_>-rj-;R = •1C
*=0
37. (a)3/4 (b)ln(4/3)
. D
t=o *' 39. (a) * - k3 + AJ
(-i)*(o.oooi2irr
41.
(c) 0.9998790073TO
64
Chapter 9 Review Exercises (Page 689) 9. (a) true (b) sometimes false (c) sometimes false (d)true (e) sometimes false (f) sometimes false (g) false (h) sometimes false (i) true (j) true (k) sometimes false (1) sometimes false n +2 \+x ,. n+ 2 _ l_ 11. (a) ; converges, lim („ + 1)2 _ B 2 ( B=] l)2-n2 ~ 2 (b) (-!)"+' 2 } ; diverges I n + lJn=i 15. (a) converges (b) converges 17. (a) converges 19. (a) diverges (b) converges
21.
(b) diverges
13. A: = 5 cos r, v = -5 sin t (0 < t < 2;r) 17. x = r2, v = t (-1 < f < 1) 19. (a) 15
„
23. (a) 2 (b) diverges (c) 3/4 (d) n/4 25. p > 1 29. (a) p0(x) = 1, pi (x) = 1 - Ix, p2(x) = l-7x + 5x2, 2
33.(a)e -\ (b)0 (c)cose (d) i 4 7 17 ^ol v 2 r5 4 , 17 /u\ , 2 _3 , 2 ^ 37. (a) ^ — j2A;_3 +, TJ-*~ TTs(b)x— jjr + -^x —315* ^ Chapter 9 Making Connections (Page 691) _ Where correct answers to a Making Connections exercise may vary, no answer is listed. Sample answers for these questions are available on the Book Companion Site. a sin 9 1. (a) - (b)acsc0 (c)acote 1 - cos 8 2.
y
4. (a) 124.58 < d < 124.77
0
l-i y i
X
1 i 2
"j o |T I2 3
(b)f t
0
1
2
X
0
5.5
8
4.5 -8 -32.5
y
1
1.5
3
5.5
(c) t = 0, 2V3 21. (a) 3
3
4 9
5
13.5
(d) 0 < t < 2^2 (e)t = (b) . 16
2 n +
-5
23. (a) IV (b)II (c)V (d)VI (e)III (f)I 27. (b)from(j:o,>'o)to(xi,yi) (c) x = 3 - 2(t - 1), v = -1 + 5(t - 1) 29. _§_ 31. 3
Exercise Set 10.1 (Page 700) 1. (a)y =x + 2(-l <x < 4)
(
_
-35
(b) 1243 < 5 < 1424
6.
(c)
15. x = 2, >> = t
3
4
2
3
4
5
-3
2. -3
25. (b) i
(c)
_
A75
A76
Answers to Odd-Numbered Exercises
Responses to True-False questions may be abridged to save space. 33. False; x = sin t, y = cos21 describe only the portion of the parabola y = 1 - x1 with -1 < x < 1. 35.True;^ = ^ = 1 2 f 3 - f o 2 dx dx/dt x'(t)
37.
|> 2*
5. (a) (5, TT), (5, -n) (b) (4, 1 ljr/6), (4, -jr/6) (c) (2, 3?r/2), (2, -jr/2) (d) (8V2, 57T/4), (8^2, -3W4) (e) (6, 2ir/3), (6, -4^/3) (f) (V2, ?r/4), (x/2, -7?r/4) 7. (a) (5, 0.92730) (b) (10,-0.92730) (c) (1.27155, 2.47582) 9. (a) circle (b) line (c) circle (d) line 11. (a) r = 3 sec 0 (b) r = (d) r 2 cos Q sin 0 = 4/9
13. 3 -
39. (a) x = 4 cos t, y = 3 sin t (b) x = - 1 + 4 cos t, y — 2 + 3 sin t 41. -4,4 43. both are positive 45. 4,4 47. 2A/3, -l/(3-v/3) 49.^3,4 51. y = -e-2x + 2e~< 53. (a)0,jr, 2n (b)n/2,3ji/2 55. (a) (b) y = -2x, y = 2x
-3 -3
17. (a)/-= 5 (b)r = 6cos0 ( c ) r = l - c o s e 19. (a)r = 3sin26> (b) r = 3 + 2sin<9 (c)r 2 21. |^ , 23.
25.
57. y = 2x - 8, y = -2x + 8 59.
63.
65. 73. 75.
(b) 6 ^ -0.4345 dx l-3cosf (a) ellipses with fixed center, varying axes of symmetry (b) ellipses with varying center, fixed shape, size, and orientation (c) circles of radius 1 with centers on line y = x — I i(5V5-8) 67. 3jr 69. ^-(e1 - e~2) (b) x = cos t + cos 2t, y = sin t + sin 2t (c) yes S = 49n 77. S = *j2n
^ Exercise Set 10.2 (Page 716)
1.
nn
3. (a) (3x73,3) (b) (-7/2,7x/3/2)
37.
(c) (3A/3, 3) (d) (0, 0) (e) (-7V3/2, 7/2) (f) (-5, 0) Lemniscate
Answers to Odd-Numbered Exercises
A77
72 67. (a)r = 1 + — (cos 0 + sin (9) (b) r = 1 + s i n e
41.
72
(c) /- = 1 - cos 0 (d) r = 1
69. (3/2,7r/3)
(cos 0 + sin 0)
73. 72
^ Exercise Set 10.3 (Page 726) tan2 2 5. 1/2 7. 1,0, -1 2tan2+ 1 9. horizontal: (3a/2, 7T/3), (0, JT), (3a/2, 5ir/3); vertical: (2a, 0), (a/2, 2jr/3), (a/2, 4?r/3) 11. (0, 0), (72/4, Tt/4), (72/4, 3?r/4) 13. nV15.
,. 73 3.
Four-petal rose
45.
Eight-petal rose
Responses to True-False questions may be abridged to save space. x\ ~K 47. True; (/ -1, — I describes the same point as (I 1, — + it \I , which describes the same point as (1, - + JT - 2n\ = I I , - — I.
9 = ±^
49. False; — 1 < sin 26 < 0 for jr/2 < 9 < IT, so this portion of the graph is in the fourth quadrant. 51. 0 < 6 < 4ir 53. 0 < 6 < STT 1
3
fl = ±s -3
55. 0 < 9 < 57t
19. L = 2;ra 21. L = 8a 23. (b)«2.42
__2_j
i 2 3 4 5 6 I 2.4221 1 2.22748 2.14461 2.10100 2.07501 2.05816 1
11
10
12
13
14
2.04656 2.03821 2.03199 2.02721 2.02346 2.02046 2.01802
18 16 17 19 20 n 1 15 1 2.01600 2.01431 2.01288 2.01167 2.01062 2.00971
\
57. (a) -471 < 9 < 4?r 61. (a) */2
L
1
(b) 25. (a) /
nil b) / 2c Jo
i(l-
Jn/2
rnll
(c) / Jo
^sm22ed6
l*2K
(d)
-2S2
f JO
/•W4
/•IT/2
(e) / J-ir/2 2
i(l -sin0) 2
(f) / Jo
co^20de
27. (a)jra (b)?ra 29. 6;r 31. 4^ 33. ^-373/2 35. nl2-\ 37. lOjr/3-473 39. it 41. 973/2 - TT 43. (w + 373)/4 45. n - 2 Responses to True-False questions may be abridged to save space. 47. True; apply Theorem 10.3.1: cos— = 371
63. (a) t
= 0 and — dO
1 - ^ 0, so the line 0 = 3?r (the *-axis) is tangent to the curve at the origin. 0
7
1
7
49. False; the area of the sector is — • nr~ = -6r . lit 2 51. (b) a 2 (c) 273 - — 53. 8?r3a2
A78
Answers to Odd-Numbered Exercises
59. 7i/Id
65. 7T2
' 0
67. 32ir/5 (-2,-2)4
^i
(0, -2 - V5) (0, -5)
Exercise Set 10.4 (Page 744) 1. (a)x = y2
(b)-3y = x2
(c) — + — = 1 9 4
4
9
11. (a) vertices: (±4, 0); foci: (±5,0); asymptotes: y = ±3x/4
(b) vertices: (0, ±2); foci: (0, ±2vTO); asymptotes: y = ±x/3
(e) y2 - x2 = 1 (f) ~ - y— = 1 4 4 3. (a) focus: (1,0); (b) focus: (0,-2); vertex: (0,0); vertex: (0,0); directrix: x = — 1 directrix: y = 2 V(0, 0) y=2
13. (a) c1 = 3 + 5 = 8, c =
5. (a)
7. (a)
9. (a)
(-7,5)**
.
»*(I,5) 15. (a) y 2 = 12* (b).*2 = -y r2
v2
17. >2 = 2(* - 1)
r2
v2
v2
r2
' -
Answers to Odd-Numbered Exercises A79
Responses to True-False questions may be abridged to save space. 27. False; the description matches a parabola. 29. False; the distance from the parabola's focus to its directrix is 2p; see Figure 10.4.6. 31. (a) 16ft (b)8V3ft 35. ^ ft 39. ±(x-4)2 + ^0> -3) 2 = 1 43. L = D^l + p2, T = \pD 45. (64.612, 200)
21. vertex: (4/5, 3/5); focus: (8/5, 6/5); directrix: 4x + 3y = 0 23. foci: ±(477/5, 377/5); vertices: ±(16/5,12/5); ends: ±(-9/5, 12/5) 25. foci: (1 - 75/2, -73 + 7T5/2), (1 + 71/2, -73 - 7l5/2); vertices: (-1/2, 73/2), (5/2, -573/2); ends: (1 + 73, 1 - 73), (1 - 75, -1 - 73) 27. foci: ±(715, 75); vertices: ±(273,2); 573 ±8 asymptotes: y = x
2 47. (a) V = ^(b - 2<j 2 )Vfl 2 + b2 + -a£2jr 3a2 3
vertices: (2/75, 11/75), (-275, 75); asymptotes: y = 7^/4 + 375, y = -x/% + 375/2
^ Exercise Set 10.6 (Page 761) 1. (»)e= \,d= \ n!2
-2
2
53. (a) (* - I) - 5(y + I) = 5, hyperbola (b)x2 -3(j> + I) 2 = 0,x = ±\/3(y + 1), two lines (c) 4(jc + 2)2 + 8(y + I) 2 = 4, ellipse (d) 3(x + 2)2 + (y + I) 2 = 0, the point (-2, -1) (degenerate case) (e) (jc + 4)2 + 2y = 2, parabola (f) 5(x + 4)2 + 2y = -14, parabola 1. (a) Jt' = -1 + 3 V3, / = 3 + 73 3. y'2 - x'2 = 18, hyperbola (b)3;t' 2 -/ 2 = (c) \\
-15
9. /2 = 4(*' - 1), parabola A)'
/
-20
-10
4±3cos0 1+COS0 7. (a) r0 = 2, r\ = 6; ^~x2 + ±(y + 2)2 = 1
7. / = jc'2, parabola
11. \ (x1 + 1 ) 2 + /2 = 1, ellipse .\
(c)r =
12 2 +cos 9
64 25- 15 sin 6»
572-5 „., — 1 + 72 cos Q 1 + 72 cos 0 Responses to True-False questions may be abridged to save space. 19. True; the eccentricity e of an ellipse satisfies 0 < e < 1 (Theorem 10.6.1). 21. False; eccentricity correlates to the "flatness" of an ellipse, which is independent of the distance between its foci. 23. (a) T sa 248 yr (b) r0 =» 4,449,675,000 km, n % 7,400,325,000 km 13. r =
(c)r
1+ 0.249 cos 0
AU
X
13. x2 + xy + y2 = 3 19. vertex: (0,0); focus: (-1A/2, lA/2); directrix: >• = A: - V2
12
3±4sin6»
9. (a) r0 = 1, n = 3; (y - 2)2 - — = 1
11. (a)r=
5. jjc'2 - ^y'2 = 1, hyperbola
5
5. (a)r =
^ Exercise Set 10.5 (Page 753)
v'
(b) ellipse, directrix above the pole
3. (a) parabola, opens up 10
2
-50 •
A80
Answers to Odd-Numbered Exercises
25. (a) a ^ 178.26 AU (b)r 0 % 0.8735 AU, n % 355.64 AU 1.74 AU 1 +0.9951 cos 6 21. 563 km, 4286 km
n/2
(d)
40
x 7 4 = ~^:2) Y (2 + Vl3,4)X
-180
Chapter 10 Review Exercises (Page 763) 1. x = V2cost,y = -V2sinf(0 < f < 3jr/2) 3. (a) -1/4, 1/4 5. (a)f = :r/2 + njrfor/z = 0 , ± l , . . . (b)f = nxforn = 0, ±1, . 7. (a) (-4^2, -4V2) (b) (7/V2, -7/V2) (c) (472, 4x/2) (d)(5,0) (e)(0,-2) (f)(0,0) 9. (a) (5, 0.6435) (b) (V29, 5.0929) (c) (1.2716, 0.6658) 11. (a) parabola (b) hyperbola (c) line (d) circle 13. 15. 17. 3
Cardioid
(2 - Vn, 4) 39. x2 = -1
41.y2-x2 =
«n sm a 43. (b) * = — sin a cos a; y = yo • g ^s 45. 8 = jr/4; 5(y') 2 - (x')2 = 6; hyperbola 47. & = tan" 1 (1/2); y' = (x')2; parabola 49. (a) (i) ellipse; (ii) right; (iii) 1 (b) (i) hyperbola (ii) left; (iii) 1/3 (c) (i) parabola; (ii) above; (iii) 1/3 (d) (i) parabola; (ii) below; (iii) 3
(C)
6 Line
I"
37.
:
(3i
~ 5 ) - 16(X
I) 2 = 1
53. 15.86543959 Limagon
19. (a) -2, 1/4 (b) -3V3/4, 3x/3/4 21. (a) The top is traced from right to left as t goes from 0 to jr. The bottom is traced from right to left as / goes from n to 2jr, except for the loop, which is traced counterclockwise as t goes from JT + sin"1 (1/4) to 2^-sin" 1 (1/4). (b)y = l (c) horizontal: / = jr/2, 3jr/2; vertical: t = n + sin"1' 2jr - sin
Chapter 10 Making Connections (Page 766) Where correct answers to a Making Connections exercise may vary, no answer is listed. Sample answers for these questions are available on the Book Companion Site. 1. (a)
F 1-
-1 -
(c) L = I
cos2 ( —
2. (a) P: (bcost,bsiut); Q: (a cost, a sin t); R: (a cos t, b sin t) focus: (9/4,-1); vertex: (4, -1); directrix: x = 23/4
(0, -5 foci: (0, ±V2T); vertices: (0, ±5); ends: (±2,0)
^ Appendix A (Page Al) 1. (e) 3. (b), (c) 5. [-3, 3] x [0, 5] 7. [-5, 14] x [-60,40] 9. [-0.1,0.1] x [-3,3] 40 -5
0.1
-60
-3
Answers to Odd-Numbered Exercises 11. [-400, 1050] x [-1500000, 10000] 10000 -400 1050
A81
The graph is translated vertically.
13. [-2, 2] x [-20, 20] 20
-5
-20
-1500000
17. (a) f ( x ) = Vl6-;t 19. (a)
2
(b) f ( x ) = -716-jt
2
(e) no
(b)
-2K
33.
3
-3
-3 35. (a) * = 4 cos f , y = 3 sin /
/3
0 (b) x = - 1 + 4 cos f , y = 2 + 3 sin t
Appendix B (Page A1 3) 1. (a) ^JT (b) f ^ (c) i;r (d) ITT 3. (a) 12° (b)(270/jr)° (c)288° (d) 540° 5.
= 1/VlO 9. tan 0 = V2T/2, esc 6 = 5/V2T
25. (a) /" c=
4
The graph is stretched in the vertical direction, and reflected across the .x-axis if c < 0.
The graph is translated so its vertex is on the parabola
3/5 j -4/3 j -5/4 i 5/3 3/21zlA\i j _ A/3/2 i -1/V3 I -2 I 2/V3 c = -2 c = -3
17. (a)1.2679
(b) 3.5753
A82
Answers to Odd-Numbered Exercises
-R
(b)
39. 2i
41. 9.2ft
45. (a)4>/5/9 (b)-^ sin 36> = 3 sin e cos2 e - sin3 e, cos 30 = cos3 0 - 3 sin2 6> cos 0 (a)cos# (b)-sin0 (c)-cosS (d)sin0 (a) 153° (b)45° (c)117° (d) 89° 71. (a)60° (b) 117° 21. (a)3W4±rut,n = 0, 1,2,... (b) 7T/3 ± 2nit and 5x/3 ± 2mt, n = 0, 1, 2,... 23. (a) jr/6 ± n;r, n = 0, 1,2,... (b) 47T/3 ± 2n;r and 5^/3 ± 2n;r, « = 0,1,2,... 25. (a) 3;r/4 ± n?r, n = 0, 1,2,... (b) jr/6 ± OTT, n = 0 , 1 , 2 , . . . 27. (a) ^/3 ± 2rm and 2^/3 ± 2nn, n = 0, 1, 2 , . . . (b)jr/6±2n7z-andlljr/6±2n7r,/i =0, 1 , 2 , . . . 29. sin6> = 2/5,cos0 = -V2T/5,tan0 = -2/^21, csc0 = 5/2, sec0 = -5A/2T, cot0 = -V21/2 31. (a)0 = ± n j r , n = 0 , 1,2,... (b)0 = jr/2±«jr, n = 0, 1, 2 , . . . (c) 0 = ±n?r, n = 0, 1, 2,... (d) 0 = ±nw, n = 0, 1, 2,... (e) 0 = 7T/2 ± nit, n = 0, 1, 2,... (f) 0 = ±«TT, n = 0, 1, 2 , . . . 33. (a) 2;r/3 cm (b) 10^/3 cm 35. f
Appendix C (Page A27) _ = x2+4x + 2, r(x) = -1 ) = 2x2+4,r(x) =9 (c) q(x) =x3 -x2 + 2x-2, r(x) = 2x + 1 (a) q(x) = 3x2 + 6x + 8, r(x) = 15 db)q(x) = jc3 - 5^2 + 20x - 100, r(x) = 504 (c) q(x) = x4 + x3 + x2 + x + 1, r(x) = 0 0
P(x)
1 ! -3
- 4 - 3 101 5001!
7. (a) q(x) = A:2 + 6x + 13, r = 20 (b) q(x) = x2 + 3x - 2, r = -4 9. (a) ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24 (b)±l,±2, ±5, ±10, ± | , ± f , ± f , ± f (c)±l,±17 11. (x + \)(x-\)(x-2) 13. (* + 3)3(;t + l) 15. (x + 3)(x + 2)(x + l)2(x - 3) 17. -3 19. -2,-|,-1±73 21. -2,2,3 23. 2,5 25. 7cm
INDEX
Abel, Niels Henrik, 247 abscissa, Web-Gl absolute convergence, 641, 642 ratio test for, 643, 644 absolute error, 97 Euler's Method, 583 absolute extrema Extreme-Value Theorem, 217 finding on finite closed intervals, 218 finding on infinite intervals, 219, 220 functions with one relative extrema, 221 on open intervals, 220 absolute extremum, 217 absolute maximum, 217 absolute minimum, 217 absolute value function, 5 continuity, 95 derivative of, 126 properties, 5 absolute values, Web-Fl-Web-F5 and square roots, Web-Fl, Web-F2 acceleration, 240 constant, 325, 326, Web-Gil due to gravity, 107, 327 instantaneous, 240 rectilinear motion, 240 sign of, 240 acceleration function, 240 addition formulas for sine and cosine, A20 of functions, 15, 16 air resistance, 590 Alexeev, Vasili, 383 algebraic functions, 32 integration formulas, 489 algebraic manipulation, of integrand, 488 alternating current, 341 alternating harmonic series, 639 alternating series, 638-641 alternating series test, 638, 645, 671 amplitude alternating current, 341 simple harmonic motion, 159, Web-L5 sin* and cos jc, 33 analytic geometry, 225, Web-E3 Anderson, Paul, 383 angle(s),A13-A15 finding from trigonometric functions, A22 of inclination, A23 polar, 706 rectangular coordinate system, A16-A18 standard position, A16 angular coordinate, 706 angular frequency, simple harmonic motion, 159
annuity, 251 antiderivative, 271 antidifferentiation, 269, 272 aphelion, 760 apogee, 759 artificial Earth satellite, 173 approximation(s) area under a curve, 268 left, right, and midpoint, 533 local linearity, 176-177 local quadratic, 648 Maclaurin and Taylor polynomial, 649, 651,652,655,656 midpoint, 534, 536, 537 Riemann sum, 301 of roots using Intermediate-Value Theorem, 96-97 of roots using Newton's Method, 247-249 of roots by zooming, 98 Simpson's rule, 537, 538, 540 trapezoidal, 534 arc length, 371, 372, 374, A14, A15 finding by numerical methods, 374 and improper integrals, 553 parametric curves, 376, 697 polar curve, 721, 722 Archimedean spiral, 714, 718 Archimedes, 265, 266 palimpsest, 266 area antiderivative method, 269, 270 computing exact value of, 293 definition, 291 left endpoint approximation, 294 midpoint approximation, 294 net signed, 296, 297, 302, 324 polar coordinates, 724, 725 rectangle method, 267, 270 right endpoint approximation, 294 surface of revolution, 377, 379 total, 313, 324 between two curves, 348, 350-352 under a curve, 291 area problem, 267 argument, 3 arithmetic mean (or average), 332, 757 aspect ratio distortion, A6 Astronomia Nova, 759 astronomy, planetary motions, 759 asymptote(s), 80, 207 a curve as an, 80 curvilinear, 211 horizontal, 32, 71 oblique, 211
slant, 211 vertical, 32, 58 asymptotes of a hyperbola, 732 finding, 738 asymptotic curves, 80 autonomous differential equation, 585 auxiliary equation, Web-L2 average rate of change, 117 average value, 332, 333 and average velocity, 334 average velocity, 114, 332, 334 geometric interpretation, 115 axis (axes) coordinate, Web-Gl of an ellipse, 732 of a hyperbola, 732 of a parabola, 732 polar, 705 of revolution, 358 rotation of, 749, 750 Badel, Hannskarl, 482 base b exponential function, 455 Bell, Alexander Graham, 417 Bernoulli equation, 595 Bernoulli, Daniel, 700 Bernoulli, Jakob I, 698-700 Bernoulli, Johann (John), 3, 442 Bernoulli, Johann (John) I, 698-700 Bernoulli, Nikolaus, 700 Bessel, Friedrich Wilhelm, 666 Bessel equation of order one, 689 Bessel equation of order zero, 689 Bessel functions, 666 derivative of, 679 bifolium, 728 binomial coefficient, 677 binomial series, 674, 675 bipartite cubics, A12 Bolzano, Bernhard, 128 Bopp, Thomas, 762 bounded functions, 306 Bowditch curves, 705 Bowditch, Nathaniel, 705 Boyle's law, 30, 36 brachistochrone problem, 699 Brahe, Tycho, 759 branches, of hyperbola, 731 breakpoints, 6 butterfly curve, 717 carbon dating, 574 cardioid, 711 1-1
1-2
Index
area, 724 families of, 713 intersection with circle, 725 carrying capacity, 80, 563 Cartesian coordinate system, Web-Gl Cassini, Giovanni Domenico, 719 Cassini ovals, 719 catenary, 473, 474 Cauchy, Augustin, 649, 651 Cavalieri, Bonaventura, 364 Cavalieri's principle, 364 center circles, Web-H3 ellipse, 731 hyperbola, 732 center of gravity, 391 lamina, 391-395 centimeter-gram-second (COS), 383 centroid, 397 geometric property, 395 of a region, 395-397 chain rule, 153 alternative version, 154 generalized derivative formulas, 155 Leibniz notation, 177 proof, A36 two-variable, A43 chaos and Newton's Method, 249 circle(s), 730, Web-H2-Web-H4 degenerate cases, Web-H4, Web-H5 and discriminant, Web-Kl families, 712, 713 involute, 704 osculating, 109 closed form, sigma notation, 290 closed interval(s), Web-E4 absolute extrema on finite, 217 continuity on, 93 clothoid, 766 coefficient of friction, 133, 159 coefficient of linear expansion, 183 coefficient of volume expansion, 183 coefficients, A27 correlation, Web-J2 leading, A27 polynomials, 31 cofunction, 160 Commentaries on the Motions of Mars, 759 common logarithms, 413 comparison test, 631, 632, 645 complete elliptic integral of the first kind, 560, 685 complete linear factorization, A28 Completeness Axiom, 611 completing the square, 20, Web-H4 complex numbers, Web-El, Web-E2 compliance, Web-Gl2 composition of functions, 17 continuity, 94, 95 derivatives, 153 w-substitutions for, 281-284 compounding interest annually, 251,577 continuously, 577 quarterly, 577 semiannually, 577 computer algebra system(s) (CAS), 530, Al differentiation using, 157 integration using, 285, 488, 528, 529 linear systems, 519
Maple, Al,A2 Mathematica, A1, A2 polynomial roots, A31 concave down, 190 concave up, 190 concavity, 190, 207 conchoid of Nicomedes, 764 conditionally convergent, 643 conic sections, 730-732 applications, 743 degenerate, 730, Web-K2 focus-direction characterization, 754 polar equation, 755, 756 quadratics, 748-752, Web-K2 reflection properties, 742, 743 sketching in polar coordinates, 756-758 translated, 740-742 conjugate axis, 732 Constant Difference Theorem, 256 constant function, 27, 62, 134, 188 constant of integration, 272 constant of proportionality, 29, 36, Web-Gil inverse proportionality, 29, 36 constant polynomial, A27 constant term, polynomials, A27 constant(s) derivative, 134 factor, in a derivative, 136 factor, in a limit, 64 and graph reflections, 22 and graph translations, 20 integration formulas, 489 second-order linear homogeneous equations, Web-L2 continuity, 90, 92-98 applications, 92 compositions, 94-95 and differentiability, 127 exponential function, 454 Intermediate-Value Theorem, 96, 97 on an interval, 93 inverse functions, 96 from the left/right, 92 logarithm function, 451, 456 polynomials and rational functions, 94 proof of basic property, A35, A36 trigonometric functions, 101 continuous compounding, 577 continuous functions average value, 332-334 properties, 93 convergence, 600, 610, 611, 614-616 absolute, 641,642 and algebraic properties of series, 624, 625 conditional, 643 and eventual behavior of sequences, 610 improper integral, 549-551 infinite series, 616 monotone sequences, 610, 611 radius/interval of, 662, 664 sequences, 600 convergence set, 662 convergence tests alternating series, 638-641 comparison test, 631, 632 integral test, 626, 628 limit comparison test, 633, 634 p-series, 627 ratio test, 634, 635 root test, 635 summarized, 645
coordinate axes, Web-Gl coordinate line, Web-E3 coordinates, Web-E3, Web-Gl, Web-G2 Cartesian, Web-Gl polar, 705, 706 Copernicus, 759 corner points, 126 Cornu, Marie Alfred, 766 Cornu spiral, 766 correlation, Web-J 1 correlation coefficient, Web-J2 cosecant, A15 continuity, 102 derivative, 149 hyperbolic, 473 cosine, A15 continuity, 101 derivative of, 148, 152 family, 32, 34 formulas, A20, A22 hyperbolic, 473 integrating powers of, 500, 501, 505 integrating products with sine, 501, 503 rational functions of, 527 trigonometric identities, A18-A20 cost function, 231 cotangent, A15 continuity, 102 derivative, 149 hyperbolic, 473 Coulomb's law, 36 CRC Standard Mathematical Tables and Formulae, 488, 523 Crelle, Leopold, 247 critical point, 198 cross-product term, 750, 751 cross section, 351, 355 cubic polynomial, 31, A27 curve(s) as asymptote, 80 integral, 276, 562 length of, 371-373, 697 orthogonal, 167 parametric, 694, Web-I2 polar, 707 position versus time, 238 smooth, 371 velocity versus time, 324 curve sketching, 207 curvilinear asymptotes, 473 cusp, 212 cycle, 341 cycloid, 698 apple of discord, 699 cycloids prolate, 703 role in mathematics history, 700 cylinder, 355, 356 cylindrical shells, 365-367, 369 da Pisa, Leonardo (Fibonacci), 607 damping constant, Web-L8 decay constant, 571 interpreting, 572 decibels, 417 decimal representations, Web-E2 decreases without bound, 57 decreasing functions, 188 invertibility, 42 decreasing sequence, 607
Index definite integrals defined, 301 and Fundamental Theorem of Calculus, 310,312,313,316,317 integration by parts, 496 properties of, 303-305 substitution, 337-339 degenerate conic sections, 730, Web-K2 degree, polynomials, 31, A27 degree measure, A13 density lamina, 391 mass, 402 weight, 402 dependent variable, 3 derivative function, 122-124 defined, 122 domain, 122 "prime" notation, 122 slope producing, 123 derivative(s) compositions, 153 constant function, 134 constant multiple rule, 136 defined, 122, 129, 130 difference quotient, 122 equation of the tangent line, 123 exponential functions, 429 left-hand, 129 of a linear function, 124 logarithmic functions, 421 notation, 122, 129, 130 rath order, 139 one-sided, 129 of position function, 125 power rule, 135,424 of a product, 142, 143 of a quotient, 144 of a reciprocal, 146 right-hand, 129 trigonometric functions, 148-151 velocity as, 125 Descartes, Rene, 162, 225, 401, 698, Web-E3 deterministic phenomena, Web-Jl difference quotient, 122 differentiability, 125, 127, 129, 130 and continuity, 127 functions defined implicitly, 165 geometric interpretation, 126 lack of, 126 differentiable on (a,b), 125 everywhere, 125 on an interval, 129 atjco, 125 differential equation(s), 561-562 autonomous, 585 Bernoulli, 595 homogeneous, 595 initial-value problems, 277 and integration, 276, 277 order of, 561 and rates of change, 561 separable, 568-571 slope/direction field for, 277, 580 solutions, 561 differentials, 177, 179-180 error propagation, 177-179 formulas, 180
differentiation, 129 computer algebra systems, 157 implicit, 163-165 as inverse property of integration, 318 logarithmic, 423,429 of power series, 678, 679 and related rates, 168-172 rules, 146 techniques of, 134, 135, 137-139, 142-146, 153 Diophantine analysis, 252 Diophantus, 225 direct proportionality, Web-Gil, Web-G12 direction field, 277, 580 direction of increasing parameter, 694, Web-I2 directrix, 731,754 discontinuities, 29, 91, Web-G4 infinite, 91 and integrability, 305 jump, 91 limits at, 99 physical significance, 92 removable, 91, 99 atjc = c, 91
discriminant, 752, Web-Kl, Web-K2 disease, first-order differential equations modeling spread of, 564 displacement, 114 finding by integration, 319, 323 from velocity versus time curve, 324 distance formula, Web-Hi distance traveled finding by integration, 323 rectilinear motion, 324 from velocity versus time curve, 324 divergence absolute, 641 improper integral, 549-551 of infinite series, 616 of sequences, 600 divergence test, 623, 624, 645 division of functions, 15, 16 by zero, Web-E2 domain, 7 effect of algebraic operations on, 8 inverse functions, 40, 44 natural, 7 physical considerations in applications, 9, 10 restricted, 44 restricting to make functions invertible, 44 double-angle formulas, A21 doubling time, 572-573 drag force, 590 dummy variables, 314, 315 dx,177 dy, 177 dy/dx, 129, 177 dyne-centimeters (erg), 383 dynes, 383 e,421,454,455 Earth Mercator projections, 506 Earth-observing satellite, 152 earthquake intensity, 417, 420 eccentricity, 754 as measure of flatness, 755 economics, optimization problems, 231, 232 elementary functions, 456
elements, of sets, Web-E4 ellipses, 730, 731 area, 746 and discriminant, Web-K 1 eccentricity as measure of flatness, 755 focus-directrix property, 755 polar equation, 756 reflection properties, 743 sketching, 736, 737 sketching in polar coordinates, 756-758 standard equations, 734, 735 translated, 740 elliptical orbits, 760 empty set, Web-E4 end behavior, 71 and limits, 73, 76 Endpaper Integral Table, 523-528 energy, 382 kinetic, 382, 388 relationship to work, 387-388 equiangular spirals, 729 equilateral hyperbola, 29, 740 equilibrium, 393 erg, 383 error absolute, 97, 583 in local linear approximations, 177 measurement, 179 percentage, 180,583 propagated, 179 relative, 179 error function [erf(jt)L 457, 530 error propagation, 179 Euler, Leonhard, 3, 622 Euler's equidimensional equation, Web-L7 Euler's Method, 581-583 even function, 23 eventually holding properties, 610 existence theorem, 217 exp, 412 exponential decay, 571-573 exponential functions, 410 analyzing growth with L'Hopital's rule, 445 approximating with power series, 672 with base b, 410 derivatives, 429^130 equation solving, 414, 415 graph, 411 integration formulas, 489 natural, 412 exponential growth, 417, 571-573 Extreme-Value Theorem, 217, 224 extremum (extrema), 197, 217, 218 absolute, 217 relative, 197,207 extremum problem applied, 224, 226-228, 230-232 one-variable, 217 Factor Theorem, A30 factorial («!), 603, 650 factors, polynomials, A30 false gaps, graphing utilities, A8 false line segments, graphing utilities, A8 families of cardioids, 713, 714 of circles, 712, 713 of curves, 27 of functions, 30-34 oflimacons, 713, 714 of rose curves, 713 of spirals, 714
1-3
1-4
Index
families of functions cosine, 32, 34 exponential, 410 sine, 32, 34 Fermat's last theorem, 225 Fermat's principle, 237 Fermat, Pierre de, 224, 698 Fibonacci sequence, 607 finite intervals, Web-E4 first-degree equation * andy,Web-G10 first derivative test, 198, 199 first-order differential equations, 561, 562, 570,586-591 functions of two variables, 579-580 linear, 586 mixing problems, 589-590 modeling with, 571-574, 589-591 separable, 569-570 slope fields, 580 first-order initial-value problem, 562 first-order model, pendulum period, 686 fluid, 400 dynamics, 400 force, vertical surface, 403^104 homogeneous, 402 pressure, 402 statics, 400 focal axis, 732 focus, 731,754 ellipse, 731 hyperbola, 731 parabola, 731 focus-directrix property, of conies, 754 Folium of Descartes, 162, 164, 728 Folsom points, 576 foot-pounds, 383 force(s) constant, 382-384 fluid, 404 pressure, 400^102 variable, 384-386 four-cusped hypocycloid, 26, 166 four-petal rose, 709, 713 fractals, and Newton's Method, 249 free-fall model, 327 integration, 327, 328 with resistance, 590-591 free motion, Web-L8 frequency alternating current, 341 simple harmonic motion, 159, Web-L6 sin x and cos*, 33 Fresnel cosine function, 457,458 Fresnel sine function, 457, 458 Fresnel, Augustin, 457 function(s), 1, 6 adding, 15 approximating with power series, 670-675 arithmetic operations on, 15 bounded, 306 compositions, 17, 18 constant, 27 decreasing, 42 defined by integrals, 456, 458 defined by power series, 665, 666 defined explicitly and implicitly, 161 differentiable, 125 domain, 7 elementary, 456 even, 23, 28, 29
expressing as composition, 18, 19 graphical information analysis, 1 graphing utilities, Al graphs of, 4 increase without bound, 417 increasing, 42 increasing and decreasing, 187-189 interpolating, Web-Jl inverse, 38-40 inversing, 42 invertible, 41 logarithmic, 413 natural domain, 7 odd, 23, 28, 29 one-to-one, 41 piecewise-defined, 6 properties, 4-10 range, 7 relative maxima and minima, 197 restriction, 44 smooth, 371 of two variables, 579-580 vertical line test, 4 ways of describing, 2 Fundamental Theorem of Calculus, 309-311, 316,317 dummy variables, 314, 315 and integrating rates of change, 318 relationship between definite and indefinite integrals, 312, 313 g, acceleration due to gravity, 107 Gabriel's horn, 555 Galileo, 698, 763 Galois, Evariste, A31 Gamma function, 559 gases, 400 Gateway Arch, 482 Gauss, Carl Friedrich, 301 general equation circle, Web-H5 lines, Web-GlO general logarithms, 456 general solution, 562 general term, 597, 623 geometric curves, Web-G2 geometric mean, 757, 758 geometric series, 617, 619 graphing using calculus and technology together, 213 graphing utilities aspect ratio distortion, A6 errors of omission, A8 false gaps and line segments, A8 inverse function graphing, A9, Web-I4 nonelementary functions, 461 parametric curve generation, A9, Web-I3 polar curves, 715 resolution, A7 root approximation by zooming, 98 sampling error, A7 and true shape of graph, A9 graphs, Web-G2 conic sections in polar coordinates, 756-758 functions, 4 hyperbolic functions, 473, 474 inverse functions, 43 logarithmic functions, 413 polar coordinates, 707, 709 power functions, 29
properties of interest, 207 quadratic equation, Web-H5-Web-H7 rational functions, 207 reflections, 21 roots, 4 sequence, 599 stretches and compressions, 22 symmetry, 23 translations, 20 vertical tangents and cusps, 211 ;t-intercepts, 4 zeros, 4 Great Pyramid at Giza, 186 greatest integer function, 27 greatest lower bound, 612 Gregory, James, 673 Gregory of St. Vincent, 450 grid lines, A2 growth constant, 571 interpreting, 572 growth rate, population, 184 Hale, Alan, 762 Hale-Bopp comet, 762 half-angle formulas, A22 half-closed intervals, Web-E4 half-life, 572-573 half-open intervals, Web-E4 Halley's comet, 760 hanging cables, 473, 474, 765 harmonic motion, simple, 159, 567 harmonic series, 620 height right cylinder, 356 Hippasus of Metapontum, Web-El homogeneous differential equation, 595 homogeneous lamina, 391 Hooke, Robert, 385, Web-G12 Hooke's law, 385, Web-G12 horizontal asymptotes, 32, 71 polar curves, 719 horizontal line test, 42 Hubble, Edwin P., Web-J8 Hubble's constant, Web-J8 Hubble's law, Web-J8 Humason, Milton L., Web-J8 hyperbolas, 730, 731 and discriminant, Web-Kl equilateral, 29, 740 polar equation, 756 reflection properties, 743 sketching, 739, 740 sketching in polar coordinates, 756, 757, 759 standard equation, 737, 738 translated, 740 hyperbolic cosecant, 473 hyperbolic cosine, 473 hyperbolic cotangent, 473 hyperbolic functions, 472, 473 applications arising in, 473, 474 derivative formulas, 475 graphs, 473, 474 integration formulas, 476, 489 inverse, 477^79 and unit hyperbola, 475 hyperbolic identities, 474, 475 hyperbolic navigation systems, 743 hyperbolic secant, 473 hyperbolic sine, 473 hyperbolic spiral, 714, 719
Index hyperbolic substitution, 514 hyperbolic tangent, 473 hyperharmonic series, 627 ill posed problems, 230 image of x under /, 3 implicit differentiation, 162, 164, 165 implicit functions, 161 improper integrals, 547, 549-551, 553 arc length and surface area, 553 integrands with infinite discontinuities, 551 over infinite intervals, 548-550 improper rational functions, integrating, 520, 521 increases without bound, 57 increasing functions, 188 invertibility, 42 increasing sequence, 607 increment, 129 Euler's method, 582 indefinite integrals, 272, 273 and Fundamental Theorem of Calculus, 312,313 properties, 274, 275 independent variable, 3 indeterminate form, 67, 441, 444-^47 of type oo/oo, 444 of type oo — oo, 446 of type 0/0, 441 of type 0°, 00°, 1=°, 447 of type 0 • co, 445 index for sequence, 598 of summation, 288 inequalities with absolute values, Web-F3, Web-F4 algebraic properties, Web-E5, Web-E6 solving, Web-E6-Web-E8 triangle, Web-F5 infinite discontinuities, 91 integrands with, 551 infinite intervals, Web-E4 absolute extrema on, 219, 220 integrals over, 548-550 infinite limits, 56, 85, 86 informal view, 57 infinite sequence, 596 infinite series, 596, 614-617, 619, 620 absolute convergence, 641, 642 algebraic properties of, 624—626 alternating harmonic series, 639 alternating series, 638-641 binomial series, 674, 675 comparison test for, 631-633, 645 conditional convergence of, 643 convergence/divergence of, 616 convergence tests for, 623-628, 631-635, 638, 640-643, 645 divergence test for, 624, 645 general term of, 623 geometric series, 617-619 harmonic series, 620 hyperharmonic series, 627 integral test for, 626, 628 limit comparison test for, 633, 634, 645, A40 Maclaurin series, 660-663, 669-675, 681, 683, 684, 686 nth partial sum, 616 power series, 661-666 p-series, 627
ratio test for, 634, 635 root test for, 635 roundoff error with, 672 sums of, 614—616 Taylor series, 659-663, 669, 670, 681, 683-685 terms of, 614 truncation error with, 672 infinite slope, 696, Web-G5 infinity, limits at, 71-77 inflection point at XQ, 191 inflection points, 191-193, 207 applications, 193 inhibited population growth, 563-564 initial conditions, 277, 562, Web-L4 initial side, of angles, A13 initial-value problems, 277 Euler's Method for, 581-583 first-order differential equations, 562 second-order linear homogeneous differential equations, Web-L4 instantaneous acceleration, 240 instantaneous rate of change, 49, 117 instantaneous velocity, 115 definition, 115 geometric interpretation, 115 as a limit of average velocities, 115 rectilinear motion, 115, 125 instantaneous velocity function, 125 integer powers, 409 integers, Web-El integrability and discontinuities, 305 integrable functions, 301 integrable on [a, b], 301 integral(s) complete elliptic, of the first kind, 560, 685 curves, 276, 562 functions defined by, 456, 457 Mean-Value Theorem for, 315, 316, 333 notation, 272 relationship between definite and indefinite, 312, 313 Riemann, 301 sign (/), 272, 273 tables, 489, 490, 523-528 integral test, 626, 628, 645 integrand, 272 integrating factor, 587 integration, 272 constant of, 272 and differential equations, 276, 277 and direction fields, 278 hyperbolic substitution, 514 as inverse property of differentiation, 318 limits of, 301 notation for, 272, 273 of power series, 679, 680 of rates of change, 318 and slope fields, 277 w-substitution, 281-284 integration by parts, 491^93 definite integrals, 496 LIATE, 493, 494 reduction formulas, 497 repeated, 493 tabular, 495 integration formulas, 273, 489,490 inverse trigonometric functions, 468, 490 integration methods, 488, 490 intensity, of sound, 416 intercepts, Web-G4
Intermediate-Value Theorem, 96 International System of Units (SI), 383 interpolating function, Web-Jl intersection, intervals, Web-E5 interval of convergence, 662 finding, 662, 663 power series in x — XQ, 663, 664 intervals, Web-E4, Web-E5 intrinsic error, 540 inverse cosine function, 463 inverse functions(s), 38, 39, 413 cancellation equations, 39 continuity, 96 domain, 40 domain and range, 40 existence, 41 finding, 40 graphing utilities, A9, Web-I4 graphs, 43 horizontal line test, 42 notation, 39 parametric equations, 695 range, 40 inverse hyperbolic functions, 477 derivatives and integrals, 479—480 logarithmic forms, 477, 479 inverse proportionality, 29 inverse secant function, 463 inverse sine function, 462 inverse tangent function, 463 inverse trigonometric functions, 462 derivatives, 467 evaluating, 464 identities, 465 invertible functions, 41 involute of a circle, 704 irrational exponents, 409,454 derivatives, 424 irrational, 409 irrational numbers, Web-El irreducible quadratic factors, A27 joule, 383 jump discontinuity, 91 Kepler, Johannes, 759 Kepler's equation, 262 Kepler's laws, 262, 759 first law, 759 second law, 759 third law, 759 kinetic energy, 382, 388 Kramer, Gerhard (Mercator), 506 L'Hopital, Guillaume Francois Antoine De, 442, 698, 699 L'Hdpital's rule, 441-447 Lame's special quartic, 166 lamina, 391 center of gravity of a, 391-395 centroid of, 395-397 density of, 391 first moment of a, 394 homogeneous/inhomogeneous, 391 mass, 391 moment(s) of, 393-395 Laplace, Pierre-Simon de, 224 Laplace transform, 556 Law of Areas, 759 Law of Cosines, 469, A20-A22 Law of Orbits, 759
1-5
1-6
Index
Law of Periods, 759 Law of Sines, A25 Law of Universal Gravitation, 2, 37 leading coefficient (term), A27 least squares line of best fit, Web-J2 least squares principle, 237 least upper bound, 612 left-endpoint approximation, 533 left-hand derivatives, 129 Leibniz, Gottfried Wilhelm, 49, 268, 442, 698,699 derivative notation, 177 integral notation, 177, 272 lemniscate, 166,712 length arc, 372, 373, A14 lever arm, 393 LIATE stategy for integration by parts, 493 Libby, W. R, 573 limacons, 713, 714 limit comparison test, 633, 634, 645 proof, A39 limit(s), 49-58, 60, 62-68, 71-79, 81-86 area defined as, 291,292 computing, 62, 67, 73, 76 decimals and, 51 at discontinuities, 99 does not exist, 55, 56 end behavior, 73, 76 epsilon-delta definition, 82 fail to exist, 54-56, 77, 104 horizontal asymptotes, 71 indeterminate forms, 441^(47 indeterminate forms of type 0/0, 67 infinite, 56, 85, 86 at infinity, 71-72 informal view of, 53 intuitive approach, 49, 55, 56 nth root, 63 one-sided, 54, 82 piecewise-defined functions, 68 polynomials as x —> ±00, 73, 74 polynomials as x -> a, 64, 66 product, 63 proof of basic theorems, A34, A35 quotient, 63 radicals, 67 rational functions as x. -> ±00, 74 rational functions as x —» a, 64, 66 of Riemann sums, 301 rigorous discussion, 81-83, 85, 86 sampling pitfalls, 53 sequences, 599, 600, 602 simple functions, 62, 63, 73 Squeezing Theorem, 102 sum, 63 of summations, 293 trigonometric functions, 101 two-sided, 54-55, 81-82 x" as * ^ ±00, 73 *^±o=, 84, 85 limits of integration, 301 linear algebra, 519 linear depreciation schedule, 35 linear differential equation, 586, Web-Ll-Web-L4, Web-L6 linear equation, Web-GlO linear factor rule, 516 linear factors, 516,517
linear functions, 31 mathematical models, Web-J3, Web-J4 practical problems, Web-Gl 1 linear polynomial, 31, A27 linearly dependent functions, Web-L 1 linearly independent functions, Web-L 1 lines, Web-G5, Web-G7-Web-G9, Web-GlO, Web-Gil angle of inclination, A23 degenerate conies, 730, Web-K2 determined by point and slope, Web-G8 determined by slope and y-intercept, Web-G9, Web-GlO families of, 712 parallel to the coordinate axes, Web-G8 liquids, 400 Lissajous curve, 703 Lissajous, Jules Antoine, 703 Lituus spiral, 714 local linear approximation, 176, 649 local linearity, 175 differentials, 177, 179-180 local quadratic approximations, 648 locally linear, 175 logarithmic differentiation, 423,429 logarithmic function(s), 413,414 base b, 413 derivatives, 420-422 equation solving, 414, 415 integral, 493 properties, 414 properties of exponential functions compared, 414 logarithmic growth, 417 logarithmic scales, 416, 417 logarithmic spiral, 714, 729 logarithms, 413 approximating with power series, 673 change of base formula, 416 properties, 414 logistic curve, 436, 577 differential equation, 564 growth, 436, 439, 564 lower bound greatest, 612 monotone sequence, 611 sets, 612 lower limit of integration, 301 lower limit of summation, 288 m, slope of line, Web-G5 Machin's formula, 677 Maclaurin, Colin, 649 Maclaurin polynomials, 649-651 sigma notation, 652, 654, 655 Maclaurin series, 660-663, 669, 670, 672-675,681,683-685 binomial series, 674, 675 differentiation, 678 exponential functions, 672 integrating, 680 logarithm approximation, 673 practical ways to find, 683, 684 trigonometric function approximation, 670, 672 various functions, 675 Magellan spacecraft, Web-J3, Web-J4 magnitude, 5, Web-Fl major axis, 731 Mantle, Mickey, 765
manufacturing cost, 231 Maple, Al,A2 marginal analysis, 232 marginal cost, 232 marginal profit, 232 marginal revenue, 232 Mars, 763 mass of a lamina, 391 mass density, 402 Mathematica, A1, A2 mathematical models, Web-J 1 linear functions as, Web-J3, Web-J4 quadratic and trigonometric functions as, Web-J4-Web-J6 maximum absolute, 217 relative, 197 mean value, 333 Mean-Value Theorem, 253-256 consequences, 255 proof, 254 velocity interpretation, 255 Mean-Value Theorem for Integrals, 315, 333 measurement error, 179 mechanic's rule, 251 members, of sets, Web-E4 Mercator (Gerhard Kramer), 506 Mercator projection, 506 mesh size, of partition, 300 method of cylindrical shells, 365-367, 369 of disks, 358 Newton's, 247, 249 of washers, 359 method of integrating factors, 587-589 midpoint approximation, 533, 540, 543 error estimate, 540-542 errors, 535 midpoint formula, Web-H2 Miller, Norman, 236 minimum absolute, 217 relative, 197 minor axis, 731 minutes (angle), A13 mixing problems, first-order differential equations, 589-590 modeling with differential equations, 561, 563-566, 571-574, 589-591 differential equations, 561 doubling time, 572, 573 exponential growth/decay, 571-573 half-life, 572-573 Newton's Law of Cooling, 565 pharmacology, 564 population growth, 563-564 spread of disease, 564-565 vibration of springs, 565-566 moment lamina, 393-395 about a line, 393 about a point, 392 monotone sequences, 607 convergence, 610, 611 monotonicity testing, 608, 610 properties that hold eventually, 610 motion constant acceleration, 325 free-fall, 327, 328, 590, 591 rectilinear, 238, 240-242, 323, 325, 326
Index simple harmonic, 159, 567 m t a n , 111 multiplication, of functions, 15, 16 multiplicity, 201, A28 geometric implications of, 201 natural domain, 7 natural exponential function, 412 formal definition, 454 natural logarithm, 413 approximating with Maclaurin series, 673 derivative, 451 integral, 493 properties, 451 natural numbers, Web-El negative angles, A13 negative direction, Web-E3 net signed area, 296, 297 Newton, Isaac, 3, 49, 268, 698, 699, 759 derivative notation, 177 Maclaurin disciple of, 649 solution of the brachistochrone problem in his own handwriting, 700 Newton's Law of Cooling, 133, 437, 565 Newton's Law of Universal Gravitation, 2, 37 Newton's Method, 248 difficulties with, 249 Newton's Second Law of Motion, 387 Newtonian kinetic energy, 691 newton-meters, 383 newtons, 383 nonnegative number, Web-E5 nth Maclaurin polynomial, 650, 656, 659 nth order derivative, 139 nth partial sum, 616 nth remainder, 655, 668 estimating, 669, 670 nth Taylor polynomial, 653, 655, 659 null set, Web-E4 numbers complex, Web-El, Web-E2 integers, Web-El natural, Web-El rational, Web-E 1 real, Web-El numerical analysis, 540, 672 numerical integration, 533-543 absolute error, 538 absolute errors, 535 error, 535, 538, 540 midpoint approximation, 534 Riemann sum approximation, 533 Simpson's rule, 537-540 tangent line approximation, 536 trapezoidal approximation, 534 oblique asymptote, 211 odd function, 23 Ohm's law, 89 On a Method far the Evaluation of Maxima and Minima, 224 one-sided derivatives, 129 one-sided limits, 54, 82 one-third rule, 540 one-to-one functions, 41, 428 open form, sigma notation, 290 open interval, Web-E4 optimization problems, 187 absolute maxima and minima, 218-221 applied maximum and minimum problems, 224-233
categories, 224 economics applied, 231, 232 five-step procedure for solving, 226 ill posed, 230 involving finite closed intervals, 224-229 involving intervals not both finite and closed, 229-231 order of the derivative, 139 differential equations, 561-562 ordinate, Web-Gl Oresme, Nicole, 622 orientation, of a curve, 694, Web-I2 origin, 705, Web-E3, Web-Gl symmetry about, 710 orthogonal curves, 167 trajectories, 167 osculating circle, 109 output, of function, 2, 3 outside function, 18 overhead, 231 Pappus, Theorem of, 397-398 Pappus of Alexandria, 398 parabolas, 730, Web-H5 defined, 731 and discriminant, Web-Kl focus-directrix property, 755 Kepler's method for constructing, 766 polar equation, 756 reflection properties, 742, 743 semicubical, 697 sketching, 733, 734 sketching in polar coordinates, 756 standard equations, 732, 733 translated, 740 parabolic antenna, 743 parabolic mirror, 743 parabolic spiral, 714, 718 parallel lines, Web-G7 parameters(s), 27, 692, Web-Il parametric curves, 694, Web-I2 arc length, 376, 698 generating with graphing utilities, A9, Web-I3 orientation, 694, Web-12 piecewise-defined, 702, Web-I7 scaling, Web-I5,A10 tangent lines, 695-697 translation, Web-I4, Web-I5, A10 parametric equations, 692, 693, Web-11 expressing ordinary functions parametrically, 694, A9, Web-I3 graphing utilities, 715 orientation, 694, Web-12 partial fraction decomposition, 515 partial fractions, 514, 515 improper rational functions, 520, 521 linear factors, 516, 517 quadratic factors, 518-520 partial sums, 616 partition of the interval [a, ft], 300 regular, 302 Pascal, Blaise, 401 pascal (Pa), 401 Pascal's Principle, 406 peak voltage, 341 pendulum, Taylor series modeling, 685 percentage error, 180
Euler's Method, 583 perigee, 759 artificial Earth satellite, 173 perihelion, 760 period, 183 alternating current, 341 first-order model of pendulum, 686 pendulum, 183 simple harmonic motion, 159 simple pendulum, 560 sinx and cos*, 33 periodicity, 207 perpendicular lines, Web-G7 pH scale, 417, 420 phenomena deterministic, Web-J 1 probabilistic, Web-Jl physical laws, Taylor series modeling, 685 71
approximating, 266, 267, 673, 674 famous approximations, Web-E9 piecewise-defined functions, 6 limits, 68 pixels, A7 planes distance between two points in, Web-H 1, Web-H2 planetary orbits, 262, 705, 759 plot, Web-G2 Pluto, 762 points distance between two in plane, Web-Hi, Web-H2 point-slope form, Web-G9 polar angle, 706 polar axis, 705 polar coordinates, 705, 706 area in, 719,724, 725 graphs, 707 relationship to rectangular, 706 sketching conic sections, 756, 758, 759 symmetry tests, 710-712 polar curves arc length, 721 area bounded by, 719, 724, 725 conic sections, 755, 756 generating with graphing utilities, 715 intersections, 726 tangent lines, 719-721 tangent lines at origin, 721 pole, 705 Polonium-210, 576 polygonal path, 372 polynomial in x, 31 polynomial of degree n, A27 polynomials, A27, A28 coefficients, A27 continuity, 93 degree, A27 Factor Theorem, A30 geometric implication of multiplicity of a root, 201,202 graphing, 201-204 limits as x —> ±00, 73 limits as x —>• a, 64, 66 Maclaurin, 649-651 method for finding roots, A31, A32 properties, 202, 207 quick review, 31 Remainder Theorem, A29 roots, 246
1-7
1-8
Index
Taylor, 653, 654 population growth, 563-564 carrying capacity, 80, 563 first-order differential equations, 563 inhibited, 563-564
rate, 184 the logistic model, 564 uninhibited, 563 position finding by integration, 322 position function, 114, 238 derivative of, 125 position versus time curve, 114, 238 analyzing, 241 positive angles, A13 positive direction, Web-E3 positive number, Web-E5 pounds, 383 power functions, 28 fractional and irrational exponents, 409, 455 noninteger exponents, 30 power rule, 135,424 power series, 661, 664 convergence of, 662, 664 differentiation, 678, 679 exponential function approximation, 672 functions defined by, 665, 666, 675 integrating, 679, 680 interval of convergence, 662, 663 logarithm approximation, 673 JT approximation, 673, 674 and Taylor series, 681 trigonometric function approximation, 670-672 pressure, 401^102 probabilistic phenomena, Web-Jl product, of functions, 15, 16 product rule, 143,491, 493 product-to-sum formulas, A22 profit function, 231 projective geometry, 401 prolate cycloid, 703 propagated error, 179 proper rational function, 515 p-series, 627 quadrants, Web-G2 quadratic approximations, local, 648 quadratic equation(s) discriminant, Web-Kl eliminating cross-product terms, 750, 751 in x, Web-H5 in x and y, 741,748 in y, Web-H7 quadratic factor rule, 518 quadratic factors, 518-520 quadratic formula, Web-E2 quadratic mathematical model, Web-J4, Web-J5 quadratic polynomial, 31, A27 quadratic regression, Web-J4 quadratrix of Hippias, 185 quartic polynomials, 31, A27 quintic polynomials, 31, A27 quotient rule, 144 r, polar coordinate, 706, 708 radial coordinate, 706 radian measure, A13 radians, A13
radicals, limits involving, 67 radioactive decay, 487, 573 radius, A14 circles, Web-H3 radius of convergence, 662, 664 Radon-222, 576 Ramanujan, Srinivasa, 677 Ramanujan's formula, 677 range, 7 inverse functions, 40 physical considerations in applications, 9, 10 rate(s) of change, 116-118 applications, 119 average, 117 differential equations, 561 instantaneous, 117 integrating, 318 related rate, 168-172 ratio, geometric series, 617 ratio test, 634, 635, 645 for absolute convergence, 643, 645 proof, A40 rational functions, 31, 32 continuity, 93 graphing, 207 integrating by partial functions, 514, 516-521 limits as x —> ±00, 74 limits as x -> a, 64, 66 proper, 515 properties of interest, 207 of sin x and cos x, 527 rational numbers, Web-El rays, 712 real number line, Web-E3 real numbers, 611, Web-El Completeness Axiom, 611 decimal representation, Web-E2 real-valued function of a real variable, 3 rectangle method for area, 267 rectangular coordinate systems, Web-Gl angles, A16-A18 rectangular coordinates relationship to polar, 706 rectilinear motion, 113-115, 238-242, 322, 324-327 acceleration, 240 average velocity, 114 constant acceleration, 325, 326 distance traveled, 324 free-fall, 327, 328 instantaneous speed, 239 instantaneous velocity, 115, 125 position function, 114 position versus time, 114
speed, 239 velocity, 125, 239 recursion formulas, 604 recursively defined sequences, 604 reduction formulas, 497 integral tables, matches requiring, 525, 526 integrating powers and products of trigonometric functions, 500, 501 reflections, 21 region, 397 regression line, Web-J2 quadratic, Web-J4 regular partition, 300 related rates, 168-172
strategy for solving, 169 relative decay rate, 572 relative error, 179 relative extrema, 197, 207 and critical points, 198 first derivative test, 198 second derivative test, 200 relative growth rate, 572 relative maximum, 197 relative minimum, 197 relativistic kinetic energy, 691 relativity, theory of, 61, 79 Remainder Estimation Theorem, 655, 656, 669,671 proof, A41, A42 Remainder Theorem, A28, A29 removable discontinuity, 91, 99 repeated integration by parts, 493 repeating decimals, Web-E2 represented by power series, 665 residuals, Web-Jl resistance thermometer, Web-Gl5 resistivity, Web-J7 resolution, in graphing utilities, A7 restriction of a function, 44 revenue function, 231 revolution solids of, 358 surfaces of, 377 Rhind Papyrus, Web-E9 Richter scale, 417, 420 Riemann, Bernhard, 301 Riemann integral, 301 Riemann sum approximations, 533 Riemann sums, 301, 347, 533 Riemann zeta function, 668 right cylinder, 355, 356 height, 356 volume, 356 width, 356 right endpoint approximation, 533 right-hand derivatives, 129 right triangle, trigonometric functions, A15,
A16 rise, Web-G5 RL series electrical circuit, 593 Rolle, Michel, 252 Rolle's Theorem, 252 root test, 635, 645 root-mean-square, 341 roots, A28 approximating by zooming, 98 approximating using Intermediate-Value Theorem, 96-97 approximating using Newton's Method, 247, 248 of functions, 4 multiplicity of, 201 simple, 201 rose curves, 713 rotation equations, 749, 750 roundoff error, 540 in power series approximation, 672 Rule of 70, 576 run, Web-G5 Ryan, Nolan, 328 Saarinan, Eero, 482 sampling error, A7 scale and unit issues, graphs, 10 scale factors, A2
Index scale marks, A2 scale variables, A2 scaling, parametric curves, Web-I5, A10 secant, A15 continuity, 102 derivative, 149 hyperbolic, 473 integrating powers of, 503, 504 integrating products with tangent, 504, 505 second derivative, 138 second derivative test, 200 second-degree equation, 748 second-order initial-value problem, Web-L4 second-order linear differential equation, Web-Li second-order linear homogeneous differential equations, Web-Li complex roots, Web-L3 distinct real roots, Web-L2 equal real roots, Web-L3 initial-value problems, Web-L4 second-order model, pendulum period, 686 seconds (angle), A13 sector, A15 segment, 251 semiaxes, Web-I5, A10 semiconjugate axis, 737 semicubical parabola, 697 semifocal axis, 737 semimajor axis, 734 semiminor axis, 734 separable differential equations, 568-571 separation of variables, 568-569 sequence of partial sums, 616 sequences, 596, 597, 599, 600, 602, 604 convergence, 600 defined recursively, 604 general term, 597 graphs, 599 increasing/decreasing, 607 limit, 599, 600,602 lower bound, 611 monotone, 607, 609-612 of partial sums, 616 properties that hold eventually, 610 Squeezing Theorem, 602, 603 strictly increasing/decreasing, 607 types of, 607 upper bound, 611 sets, Web-E4 shells, cylindrical, 365-367, 369 Shroud of Turin, 574 sigma notation (£), 287, 288 changing limits of, 288 properties, 289 Taylor and Maclaurin polynomials, 652, 654, 655 simple harmonic model, 159 simple harmonic motion, 567 simple pendulum, 559 simple root, 201, A28 Simpson, Thomas, 539 Simpson's rule, 537-540 error estimate, 541, 542 error in, 538 sine, A15 continuity, 101 derivative of, 148, 152 family, 32, 34 formulas, A20, A22 hyperbolic, 473
integrating powers of, 500, 501, 505 integrating products with cosine, 501, 503 rational functions of, 527 trigonometric identities, A18-A20 singular points, 696 slope, Web-G5 slope field, 277, 580 slope of a line, A23 slope-producing function, 123 slope-intercept form, Web-G9 slowing down, 240 small vibrations model, 686 smooth curve, 371 smooth function, 371 Snell, Willebrord van Roijen, 238 Snell's law, 237, 238 solids of revolution, 358 solution, Web-G2 of differential equation, 561-562 inequalities, Web-E6 solution set, Web-E6, Web-G2 sound intensity (level), 416, 417 speed, 113,239 instantaneous, 239 terminal, 591 speeding up, 240 spherical cap, 364 spirals, 714 equiangular, 729 families of, 714 spring constant, 385 spring stiffness, 385 springs, 565, Web-L5, Web-L6 free motion, Web-L8 Sprinz, Joe, 331 square roots, and absolute values, 5, 6, Web-Fl,Web-F2 squaring the circle/crescent, 728 Squeezing Theorem, 103 Squeezing Theorem for Sequences, 603 standard equations circle, Web-H3 ellipse, 735 hyperbola, 737,738 parabola, 732, 733 standard positions angles, A16 ellipse, 735 hyperbola, 737 parabola, 732,733 stationary point, 198 step size, Euler's method, 582 strictly decreasing sequence, 607 strictly increasing sequence, 607 strictly monotone sequence, 607 substitution(s) definite integrals, 337-339 hyperbolic, 514 integral tables, 524, 526-528 trigonometric, 508-512 «-substitution, 281 w-substitution, 282-284 substitution principle, 80 subtraction, of functions, 15, 16 sum absolute value, Web-F4 open/closed form, 290 partial sums, 616 telescoping, 299, A39
summation formulas, 290, A38, A39 index of, 288 notation, 287 sums of infinite series, 614-616 and convergence, 616 sum-to-product formulas, A22 Sun, planetary orbits, 759 superconductors, 90 surface area and improper integrals, 553 surface of revolution, 377-379 surfaces of revolution, 377 symmetry, 23, 207 area in polar coordinates, 724, 725 about the origin, 23 about the x-axis, 23 about the y-axis, 23 symmetry tests, 23 polar coordinates, 710-712 tabular integration by parts, 495 tangent, Al 5 continuity, 102 derivative, 149 double-angle formulas, A21 hyperbolic, 473 integrating powers of, 503—505 integrating products with secant, 504, 505 trigonometric identities, A18-A22 tangent line(s), 110, 112, 113 definition, 111 equation, 111, 123 as a limit of secant lines, 110 parametric curves, 695-697 polar curves, 719-721 polar curves at origin, 721 slope, 111 vertical, 211,696 tangent line approximation, 536 tautochrone problem, 699 Taylor, Brook, 653 Taylor polynomials, 653 sigma notation, 652, 654, 655 Taylor series, 660, 661, 670 convergence, 668, 669 finding by multiplication and division, 684 modeling physical laws with, 685 power series representations as, 681, 682 practical ways to find, 682, 684 Taylor's formula with remainder, 655, 656 telescoping sum, 619, A39 temperature scales, Web-G14 terminal side, of angles, A13 terminal speed, 591 terminal velocity, 484, 591 terminating decimal(s), 622, Web-E2 terms infinite sequences, 596 infinite series, 614 test(s) alternating series, 638-641, 645 comparison, 631-633, 645 divergence, 623, 645 integral, 626, 628, 645 limit comparison, 633, 634, 645, A39 ratio, 634, 635, 645, A40 root, 635,645 symmetry, 23,708,710-712 test values, Web-E7 Theorem of Pappus, 397-398 theory of relativity, 61, 79
1-9
1-10
Index
0, polar coordinate, 706 thin lens equation, 175 third derivative, 139 tick marks, A2 Torricelli's law, 578 tractrix, 483 trajectory, 692, Web-Il transform, 556 transformations, 488 translated conies, 740-742 translation, 20 parametric curves, Web-I4, Web-I5, A10 trapezoidal approximation, 534, 540, 543 error estimate, 540-542 error in, 535 triangle inequality, 5, Web-F4, Web-F5 trigonometric functions approximating with Taylor series, 670-672 continuity, 101 derivatives, 149 finding angles from, A22 hyperbolic, 472^77, 479, 480 integration formulas, 489 inverse, 462^65, 468 limits, 102 mathematical model, Web-J4-Web-J6 right triangles, Al5 trigonometric identities, A18-A20 trigonometric integrals, 500, 501, 503-506 trigonometric substitutions, 508-512 integrals involving ax2 + bx + c, 512 trisectrix, 166 truncation error, 540, 557 power series approximation, 672 two-sided limits, 54-55 undefined slope, Web-G5 uninhibited growth model, 564 uninhibited population growth, 563 union, intervals, Web-E5 unit circle, Web-H3 unit hyperbola, 475 units, graphs, 10 Universal Law of Gravitation, 37
universe, age of, Web-J8 upper bound least, 612 monotone sequence, 611 sets, 612 upper limit of integration, 301 upper limit of summation, 288 ^-substitution, 281-285, 337-339, 488 guidelines, 282 value of / atx, 3 Vanguard 7, 763 variables dependent and independent, 3 dummy, 314, 315 separation of, 569 velocity, 113,239 average, 332, 334 finding by integration, 322 function, 125, 239 instantaneous, 125, 239 rectilinear motion, 125, 324 terminal, 484, 591 versus time curve, 324 vertex (vertices) angles, A13 ellipse, 73) hyperbola, 732 parabola, 731, Web-H5 Vertical asymptotes polar curves, 719 vertical asymptotes, 32, 58 vertical line test, 4 vertical surface, fluid force on, 403^1-04 vertical tangency, points of, 126 vertical tangent line, 211, 696 vibrations of springs, 565-566 vibratory motion, springs, 565, Web-L5, Web-L6 viewing rectangle, A2 viewing window, A2 choosing, A3, A5 graph compression, A5 zooming, A5
volume by cylindrical shells method, 365-367, 369 by disks and washers, 358-360 slicing, 355-357 solids of revolution, 358-360 Wallis cosine formulas, 508 Wallis sine formulas, 508 washers, method of, 359 wedges area in polar coordinates, 723 Weierstrass, Karl, 82, 527 weight, 385 weight density, 402 width, right cylinder, 356 Wiles, Andrew, 225 wind chill index (WCT), 6, 15 work, 382, 384-388 done by constant force, 382-383 done by variable force, 384 work-energy relationship, 382, 387-388 world population, 572 doubling time, 573 Wren, Sir Christopher, 699 A:-axis, Web-Gl *-coordinate, Web-Gl ;t-intercepts, 207, Web-G4 of functions, 4 ^-interval, for viewing window, A2 xy-plane, Web-Gl y-axis, Web-Gl y-coordinate, Web-Gl y-intercepts, 207, Web-G4 y-interval, for viewing window, A2 zeros, A28 of functions, 4 zone, of sphere, 381 zoom factors, A5 zooming, A5 root approximating, 98
RATIONAL FUNCTIONS CONTAINING POWERS OF a + bu IN THE DENOMINATOR
"•Si +2 bu
c u du _ i n (a + bu)
1
61
-2a(a + bu) + ai
r
f J ,. f 67 / 67. J 66.
C u2du \ \ a2 -— -2a\n\a 2a\n\a + bu\\+C bu\ + C 633. / 2 = — 3 U>« J (« + 6») b [_ a+bu
o
i i
J (a + bu)3 ~ b2 \_2(a + bu)2
65
C udu _ 1 \a +bu\ I +C + ln|a J (a + bu)2 b2 \_a + bu
62
r udu _ j_ r
64.
b2
a + bu\
du
+C u(a + bu) a a + bu I du_ ln a + bu 1 ~2 bu) = - au- + a u2(a + i—; du 1 1 +C -— —-j1 = + — In a + bu u(a + bu) a(a + bu) a2
RATIONAL FUNCTIONS CONTAINING a2 ± o2 IN THE DENOMINATOR (a > 0)
du
= -tan ' - + C
+C
70.
• + u1 a a f du 1 u +a 69. / -=2 r- = — In +C J a - u2 2a u —a
u +a , ~u + c b T -, c i u 71. / -;2 ,2 du = - ln(a z + « 2 ) + - tan ' - + C
+H
2
a
a
INTEGRALS OF v/a2Tu2, Va 2 - u2, \/u2 - a2 AND THEIR RECIPROCALS (a > 0) . / v7"2
=
du
= ln(w + Vw 2 +a2) + C
f /
du
r~ = In |» 4- y;/ 2 — nl\ + C.
I
2"
+a2+ —l
y
vV -a - a22du = -Vw 2 -a 2 73.. / VV
In |M + Vw 2 - a 2 | + C
.
2 i—-,-i " •. If va — u*- 9au, = —" VaH>— w H ? -- °sin • — \- C rJ 2 2 a
'
J \lu- — a2
:
2~
1 sin" - +C : a
POWERS OF u MULTIPLYING OR DIVIDING %/a2 - u2 OR ITS RECIPROCAL
/
sin ' - + C M y a 2 — u2 du = = —-(2u - a )Va2 -u2 -\ 8 a 8
79,
'•/^ /" V«2 — W 2 C / M
80
V«2 — W
/
a + Va / — u
= \/a2 — u2 — a In
u2 du u /—r a2 , 2. . =--VQ2-M2 + — 2 Va - w
/• du 1 , a + Va — M . / = --a In J wVa 2 - w 2
-sin" 1 - +C a
POWERS OF u MULTIPLYING OR DIVIDING Vu^i a2 OR THEIR RECIPROCALS i
M-v/w 2 + a2 du = - (u2 + a2)3'2 + C >. / « V M 2 — a2
r
90.
f du *Juu22 ± ± a2 I . . 2 , ,2 2 =Ta j J « VH ±a » _
i = \(u2-a2Y12-
y
C -
a + Vw 2 + a2
/ w 2 V" 2 — a2du = — (2u2 — fl2)V«2 — a 2 J 8
93. / 93. ,/
+c
a '• Jf uvu~ ^—a2 =~
••/ ••/^?-
92.
w
94. / J
• — a2 — a sec = v« 2 + a2 — a\n
" +C a a + •Ju2 + a2
+ a2) + C
4
,.
du
I. / —-== J u^Ju2 + a2
,j4
« 2 V M 2 + a2 du = - (2w 2 + fl2)V«2 + a 2 - — ln(w + 8 8
91.
Jw = d« ==
w
8
In \u + \
2
- a2\ + C
— + ln(« + \/M 2 +a 2 ) + C — + In \u + ^u2 -a2\ + C
M
9S
' J Ju2+a2dU=2 " +" " Y l n ( " + " +a } " U2 a1 u —== du = - \/M 2 - a2 -\ In \u + yV -« 2 | 2 2 / Vw - a 2 2
INTEGRALS CONTAINING (a2 + u 2 ) 3/2 , (a2 - u 2 ) 3/2 , (u2 - a 2 ) 3/2
(a > 0) „
hC
91. j^^ = ^=
i. / («2 + a 2 ) 372 du = -(2u2+5a2)jit2 + J f
98, 99.
/ 2 _i_ a 2 vu ± 2 3/2
- w )
a
:
+C
01 .
^ 4
(u2 - a2)3'2 du = "(2w 2 - 5a 2 )\/« 2 - a 2 + — In |u + \/« 2 - a 2 | + C J 8 8
^ 3a4 , u Jw = --(2w - 5a2)^/a2 - w 2 + -r- sin" 1 - + C 2
-- ln(w
POWERS OF u MULTIPLYING OR DIVIDING Va + bu OR ITS RECIPROCAL 102. 2. /I uja uj~a~+ budu = J
/a + bu - Ja
bu - 2a)(a
1Jtr
4.y«"^
104. / u"Ja + budu =
'•/
^T^
fe«) b(2n + 3)
2w"(« +
-. tan
3/2
2a«
b(2n + 3)
II
+ C (a > 0)
)8
_, la +bu
+ C (a < 0)
/• „_, r—— / M va +fewrf«
du iC J un~Ja +bu
•
3fe2 v"
2 " 2d " (3*V-4«*« /a + bu + C aTto" 15fc3 n n C u"du 2an C u••"~ ldu rfu _2u 2u" ja Ja + bu J Ja + bu ~ ~b(2, b(2n + l) b(2n + 1) J J~a + bu
106.
a + bu + -Ja
, f du '/ wv^T^ =
" r . 103. 3. f u2Ja~+b^du = —2—^(15b2u2- }2abu + Sa2)(a + bu)3/2 + C
=
Ja + bu — b(2n -3) if du a(n — l)u"~l 2a(n - \) J u"~lj~a + bu
J
C -Ja + bu du i f du = 2Ja + bu + a I ——= u J u-Ja + bu
J
C Ja + bu du _ u"
0. I
(a+bu)312 a(n -!)«"-'
b(2n - 5) C Ja + bu du 2a(n - 1) J u"-]
POWERS OF u MULTIPLYING OR DIVIDING \/2au -u2 OR ITS RECIPROCAL M —a . | +C i i f i - / " , du 2 = .in--^Ur 112. / \/2a« — u2 du = \/2a« — w 2 H — ssin"' | J / z J J2au - u \ a / du / p. ^ , 2»2 - au - 3a2 2 11 J. / uy2au — u- du = 2au - « + y sin"' ( - —} + C 117. J 6
/ J2au - w 2 du ,— 114.4. // - = v\J2ait ' 2 a « -—Mu-2 + a s i n - 1 I - — - } + C J u u \ a ) . t J2au — u2 du 2\/2au — u1, ( u — a\ 115. / J
= u-
=
sin'
u
\ a /
f
I
J Il
+C
udu
-^2au - u2 + a sin"1 I
\/2au — u2
r_*d»ss_(jL+ J
\/2au — u2
I+C . _, (u-a\ - sm -
V " /
2
+c
INTEGRALS CONTAINING (2au - u2)3/2 120
f udu_ J (2au - u22)3'2
= +C
aj2au-u2
THE WALLIS FORMULA rn/2
122. / Jo
/-ir/2
sin u du = I Jo
1-3-5 cos udu = —-—;— 2-4-6
I n an even \ (n-1) n I . • — integer and n 2
or
(n - 1)
2-4-6 —-———
n>2
n
3-5-7
/ n an odd integer and n>3
TRIGONOMETRY REVIEW (-L -L) \,lo' -Jl' ^3 i. (cos 0, sin 0) T' 2'
PYTHAGOREAN IDENTITIES
SIGN IDENTITIES
COMPLEMENT IDENTITIES sin ( /n
0 ) = cos 9 \
esc ( — - 9 I = sec 9
cos I
sin(-0) = - s i n 0
cos(-0) = cos0
tan(-0) = - tan 0
csc(-0) = -csc0
sec(-0) = sec0
cot(-0) = - cotO
SUPPLEMENT IDENTITIES (it tan(- - 0 I = cot 9
0 ] = sin 0
lit
sec I — — (
cot (-
— 0) = sin 9 CSC(TT — 9) = esc 0 sin(7T + 0) = - sin 0 csc(jr + 0) =-csc0
cos(^ — (9) = sec(jr — 0) = -sec0 COS(TT + 9) = — cos0 sec(jr + 6») = -sec 9
tan(jr — 9) = — t cot(;r-0) = - c o t 0 tan(7r + 60 = tan0 cot(x + 9) =cotd
ADDITION FORMULAS sin(o! + R) = sin a cos 8 + cos a sin 8
tan a + tan 8 -~ tana tan P
tan (a
sin (a - B) = sin a cos fi - cos a sin 8
DOUBLE-ANGLE FORMULAS sin 2ct = 2 sin a cos a cos 2a = cos a — sin a
cos(a + B) = cosacos/3 - sin a sin /i cos(a-B) =
, . tan a - tan B t3.n (^CT — ajD ) — ' + tan a tan/
HALF-ANGLE FORMULAS 2
cos 2a = 2 cos a — 1 cos 2a = 1 — 2 sin a
a
1 — cos a
a 1 + cos a cos" - =
UIOO
00006 N8SI
