Section 11.10 Taylor and Maclaurin Series 769
Table 1. Thus, for all values of x,
` s2x 2 dn ` x2n x2 x4 x6
n−0 n! n! 1! 2! 3!
−o oe2x 2 − s21dn − 1 2 1 2 1 ∙ ∙ ∙
n−0
Now we integrate term by term:
S Dy ye2x2 dx −x2 x4 x6 x2n
12 1! 1 2! 2 3! 1 ∙ ∙ ∙ 1 s21dn n! 1 ∙∙∙ dx
−C1x2 x3 1 x5 2 x7 1 ∙ ∙ ∙ 1 s21dn x 2n11 1 ∙∙∙
3 ? 1! 5 ? 2! 7 ? 3! s2n 1 1dn!
This series converges for all x because the original series for e2x 2 converges for all x.
(b) The Fundamental Theorem of Calculus gives
F Gy1 e2x2 dx − x 2 x3 1 x5 2 x7 1 x9 2 ∙ ∙ ∙ 1
0 3 ? 1! 5 ? 2! 7 ? 3! 9 ? 4! 0
We can take C − 0 in the antid erivative − 1 2 1 1 1 2 1 1 1 2 ∙ ∙ ∙
in part (a). 3 10 42 216
< 1 2 1 1 1 2 1 1 1 < 0.7475
3 10 42 216
The Alternating Series Estimation Theorem shows that the error involved in this
approximation is less than
1 5! − 1 , 0.001 n
11 ? 1320
Another use of Taylor series is illustrated in the next example. The limit could be
found with l’Hospital’s Rule, but instead we use a series.
Example 12 Evaluate lim ex 21 2 x.
x2
xl0
SOLUTION Using the Maclaurin series for e x, we have
S Dlimex212 x 1 1 x 1 x2 1 x3 1 ∙ ∙ ∙ 212x
x2 1! 2! 3!
xl0 − lim
x2
xl0
Some computer algebra systems x2 1 x3 x4 1 ∙∙∙
compute limits in this way. 2! 3! 1 4!
− lim x2
xl0
S D− lim
xl0 1 1 x 1 x2 1 x3 1 ∙ ∙ ∙ 1
2 3! 4! 5! −2
because power series are continuous functions. n
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770 Chapter 11 Infinite Sequences and Series
Multiplication and Division of Power Series
If power series are added or subtracted, they behave like polynomials (Theorem 11.2.8
shows this). In fact, as the following example illustrates, they can also be multiplied and
divided like polynomials. We find only the first few terms because the calculations for
the later terms become tedious and the initial terms are the most important ones.
Example 13 Find the first three nonzero terms in the Maclaurin series for (a) ex sin x
and (b) tan x.
SOLUTION
(a) Using the Maclaurin series for ex and sin x in Table 1, we have
S DS Dex sin x −
1 1 x 1 x2 1 x3 1 ∙ ∙ ∙ x 2 x3 1 ∙ ∙ ∙
1! 2! 3! 3!
We multiply these expressions, collecting like terms just as for polynomials:
1 1 x 1 1 x2 1 1 x3 1 ∙ ∙ ∙
2 6
3 x 2 1 x3 1 ∙ ∙ ∙
6
x 1 x2 1 1 x3 1 1 x4 1 ∙ ∙ ∙
2 6
1 2 1 x3 2 1 x4 2 ∙ ∙ ∙
6 6
x 1 x2 1 1 x3 1 ∙ ∙ ∙
3
Thus ex sin x − x 1 x 2 1 1 x3 1 ∙ ∙ ∙
3
(b) Using the Maclaurin series in Table 1, we have
tan x − sin x − x2 x3 1 x5 2∙∙∙
cos x 12 3! 1 5! 2∙∙∙
x2 x4
2! 4!
We use a procedure like long division:
x 1 31 x 3 1 125 x 5 1 ∙ ∙ ∙
2d11 x 2 1 1 x4 2 ∙ ∙ ∙ x 2 1 x3 1 1 x5 2 ∙ ∙ ∙
2 24 6 120
x 2 21 x 3 1 214 x 5 2 ∙ ∙ ∙
13 x 3 2 310 x 5 1 ∙ ∙ ∙
13 x 3 2 61 x 5 1 ∙ ∙ ∙
125 x 5 1 ∙ ∙ ∙
Thus tan x − x 1 1 x3 1 2 x5 1 ∙ ∙ ∙ n
3 15
Although we have not attempted to justify the formal manipulations used in Exam-
| | | | | |ple 13, they are legitimate. There is a theorem which states that if both f sxd − o cnxn
and tsxd − o bn xn converge for x , R and the series are multiplied as if they were
polynom ials, then the resulting series also converges for x , R and represents f sxdtsxd.
For division we require b0 ± 0; the resulting series converges for sufficiently small x .
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Section 11.10 Taylor and Maclaurin Series 771
1. I f f sxd − o ` bnsx 2 5dn for all x, write a formula for b8. 20. f sxd − x 6 2 x 4 1 2, a − 22
n−0
2. T he graph of f is shown. 21. f sxd − ln x, a − 2 22. f sxd − 1yx, a − 23
y 23. f sxd − e 2x, a − 3 24. f sxd − cos x, a − y2
f 25. f sxd − sin x, a − 26. f sxd − sx , a − 16
1 27. Prove that the series obtained in Exercise 13 represents cos x
for all x.
01 x
28. Prove that the series obtained in Exercise 25 represents sin x
(a) Explain why the series for all x.
1.6 2 0.8sx 2 1d 1 0.4sx 2 1d2 2 0.1sx 2 1d3 1 ∙ ∙ ∙
29. P rove that the series obtained in Exercise 17 represents sinh x
is not the Taylor series of f centered at 1. for all x.
(b) Explain why the series
30. P rove that the series obtained in Exercise 18 represents cosh x
2.8 1 0.5sx 2 2d 1 1.5sx 2 2d2 2 0.1sx 2 2d3 1 ∙ ∙ ∙ for all x.
is not the Taylor series of f centered at 2. 31–34 Use the binomial series to expand the function as a power
series. State the radius of convergence.
3. I f f snds0d − sn 1 1d! for n − 0, 1, 2, . . . , find the Maclaurin
series for f and its radius of convergence. 31. s4 1 2 x 32. s3 8 1 x
4. F ind the Taylor series for f centered at 4 if 33. s2 1 xd3 34. s1 2 xd3y4
1
s21dn n!
f snds4d − 3nsn 1 1d
W hat is the radius of convergence of the Taylor series? 35–44 Use a Maclaurin series in Table 1 to obtain the Maclaurin
series for the given function.
5–10 Use the definition of a Taylor series to find the first four 35. f sxd − arctans x 2d 36. f sxd − sinsxy4d
nonzero terms of the series for f sxd centered at the given value of a.
37. f sxd − x cos 2 x 38. f sxd − e3x 2 e2x
1
5. f sxd − xe x, a − 0 6. f sxd − 1 1 x , a − 2 39. f sxd − x coss 1 x 2d 40. f sxd − x 2 lns1 1 x3d
2
7. f sxd − s3 x , a − 8 8. f sxd − ln x, a − 1 41. f sxd − x 42. f sxd − s2x12
1
9. f sxd − sin x, a − y6 10. f sxd − cos2 x, a − 0 s4 x2 x
H f g 43. fsxd− sin2x Hint: Use sin2x − 1 s1 2 cos 2xd.
2
11–18 Find the Maclaurin series for f sxd using the definition of 44. f sxd − x 2 sin x if x ± 0
a Maclaurin series. [Assume that f has a power series expansion. x3
Do not show that Rnsxd l 0.] Also find the associated radius of
convergence. 1 if x − 0
6
11. f sxd − s1 2 xd22 12. f sxd − lns1 1 xd
; 45–48 Find the Maclaurin series of f (by any method) and its
13. f sxd − cos x 14. f sxd − e22x radius of convergence. Graph f and its first few Taylor polynomials
on the same screen. What do you notice about the relations hip
15. f sxd − 2x 16. f sxd − x cos x between these polynomials and f ?
17. f sxd − sinh x 18. f sxd − cosh x
45. f sxd − cossx 2 d 46. f sxd − lns1 1 x 2 d
19–26 Find the Taylor series for f sxd centered at the given value 47. f sxd − xe2x 48. f sxd − tan21sx 3d
of a. [Assume that f has a power series expansion. Do not show
that Rnsxd l 0.] Also find the associated radius of convergence. 49. Use the Maclaurin series for cos x to compute cos 58 correct
to five decimal places.
19. f sxd − x 5 1 2x 3 1 x, a − 2
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772 Chapter 11 Infinite Sequences and Series
50. Use the Maclaurin series for e x to calculate 1y1s0 e correct to 73–80 Find the sum of the series.
five decimal places.
` x 4n 74. n−`0 s6221nsd2n nd!2n
51. ( a) Use the binomial series to expand 1ys1 2 x 2 . n!
(b) Use part (a) to find the Maclaurin series for sin21x. s21dn
52. (a) Expand 1ys4 1 1 x as a power series. n−0
(b) Use part (a) to estimate 1ys4 1.1 correct to three decimal o o 73.
places. ` 3n 76. n−`0 53n nn !
n5
s21dn21
n−1
o o 75. n
o 77. ` s21dn 2n11
n−0 42n11s2n 1 1d!
53–56 Evaluate the indefinite integral as an infinite series. 78. 1 2 ln 2 1 sln 2d2 2 sln 2d3 1 ∙ ∙ ∙
2! 3!
y y 53. s1 1 x 3 dx 54. x 2 sinsx 2 d dx
9 27 81
55. y cos x 2 1 dx 56. y arctansx 2d dx 79. 3 1 2! 1 3! 1 4! 1 ∙ ∙ ∙
x
1 1 1 1
80. 1?2 2 3 ? 23 1 ?5 ? 25 2 7 ? 27 1 ∙∙∙
57–60 Use series to approximate the definite integral to within 81. Show that if p is an nth-degree polynomial, then
the indicated accuracy.
opsx 1 1d − n p sidsxd
y 57. 1y2 x 3 arctan x dx (four decimal places) i−0 i!
0
82. If f sxd − s1 1 x 3d30, what is f s58ds0d?
y 58. 1 sinsx 4d dx (four decimal places)
0 | | | | 83. Prove Taylor’s Inequality for n − 2, that is, prove that if
f -sxd < M for x 2 a < d, then
y | | 59. s0.4 s1 1 x 4 dx error , 5 3 d1026
0 | R2sxd | M |x |a 3 for | x 2 a | < d
6
y | | 60. s d0.5 x 2e2x 2 dx error , 0.001
0
< 2
61–65 Use series to evaluate the limit. 84. (a) Show that the function defined by
61. lim x 2 lns1 1 xd 62. xlilm0 1112xc2os x Hf sxd − e21yx 2 if x ± 0
x2 e 0 if x − 0
xl0 x
sin x 2 x 1 1 x 3
6
63. lim
xl0 x5 is not equal to its Maclaurin series.
; (b) Graph the function in part (a) and comment on its behav-
64. lim s1 1 x 212 1 x
x2 2 ior near the origin.
xl0
65. lim x 3 2 3x 1 3 tan21x 85. Use the following steps to prove (17) .
xl0 x5
(a) Let tsxd − o ` ( )k x n. Differentiate this series to show
n−0
n
that
66. Use the series in Example 13(b) to evaluate t9sxd ktsxd
11x
tan x 2 x − 21 , x , 1
x3
lim (b) Let hsxd − s1 1 xd2ktsxd and show that h9sxd − 0.
(c) Deduce that tsxd − s1 1 xdk.
xl0
We found this limit in Example 4.4.4 using l’Hospital’s Rule 86. In Exercise 10.2.53 it was shown that the length of the ellipse
three times. Which method do you prefer? x − a sin , y − b cos , where a . b . 0, is
67–72 Use multiplication or division of power series to find the
first three nonzero terms in the Maclaurin series for each function.
67. y − e2x2 cos x 68. y − sec x yL − 4a y2 s1 2 e 2 sin2 d
0
69. y − x x 70. y − e x lns1 1 xd
sin where e − sa 2 2 b 2ya is the eccentricity of the ellipse.
Expand the integrand as a binomial series and use the
71. y − sarctan xd2 72. y − e x sin2 x
result of Exercise 7.1.50 to express L as a series in powers of
the eccentricity up to the term in e 6.
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writing project How Newton Discovered the Binomial Series 773
laboratory Project CAS an elusive limit
This project deals with the function
f sxd − sinstan xd 2 tanssin xd
arcsinsarctan xd 2 arctansarcsin xd
1. U se your computer algebra system to evaluate f sxd for x − 1, 0.1, 0.01, 0.001, and 0.0001.
Does it appear that f has a limit as x l 0?
2. U se the CAS to graph f near x − 0. Does it appear that f has a limit as x l 0?
3. T ry to evaluate limx l 0 f sxd with l’Hospital’s Rule, using the CAS to find derivatives of the
numerator and denominator. What do you discover? How many applications of l’Hospital’s
Rule are required?
4. E valuate limx l 0 f sxd by using the CAS to find sufficiently many terms in the Taylor series
of the numerator and denominator. (Use the command taylor in Maple or Series in
Mathematica.)
5. U se the limit command on your CAS to find limx l 0 f sxd directly. (Most computer algebra
systems use the method of Problem 4 to compute limits.)
6. I n view of the answers to Problems 4 and 5, how do you explain the results of Problems 1
and 2?
writing Project how newton discovered the binomial series
The Binomial Theorem, which gives the expansion of sa 1 bdk, was known to Chinese mathe-
maticians many centuries before the time of Newton for the case where the exponent k is a
positive integer. In 1665, when he was 22, Newton was the first to discover the infinite series
expansion of sa 1 bdk when k is a fractional exponent (positive or negative). He didn’t publish
his discovery, but he stated it and gave examples of how to use it in a letter (now called the epis-
tola prior) dated June 13, 1676, that he sent to Henry Oldenburg, secretary of the Royal Society
of London, to transmit to Leibniz. When Leibniz replied, he asked how Newton had discovered
the binomial series. Newton wrote a second letter, the epistola posterior of October 24, 1676,
in which he explained in great detail how he arrived at his discovery by a very indirect route.
He was investigating the areas under the curves y − s1 2 x 2 dny2 from 0 to x for n − 0, 1, 2, 3,
4, . . . . These are easy to calculate if n is even. By observing patterns and interpolating, Newton
was able to guess the answers for odd values of n. Then he realized he could get the same
answers by expressing s1 2 x 2 dny2 as an infinite series.
Write a report on Newton’s discovery of the binomial series. Start by giving the statement
of the binomial series in Newton’s notation (see the epistola prior on page 285 of [4] or page
402 of [2]). Explain why Newton’s version is equivalent to Theorem 17 on page 767. Then
read Newton’s epistola posterior (page 287 in [4] or page 404 in [2]) and explain the patterns
that Newton discovered in the areas under the curves y − s1 2 x 2 dny2. Show how he was able
to guess the areas under the remaining curves and how he verified his answers. Finally, explain
how these discoveries led to the binomial series. The books by Edwards [1] and Katz [3] contain
commentaries on Newton’s letters.
1. C . H. Edwards, The Historical Development of the Calculus (New York: Springer-Verlag,
1979), pp. 178–187.
2. J ohn Fauvel and Jeremy Gray, eds., The History of Mathematics: A Reader (London:
MacMillan Press, 1987).
3. V ictor Katz, A History of Mathematics: An Introduction (New York: HarperCollins, 1993),
pp. 463–466.
4. D . J. Struik, ed., A Sourcebook in Mathematics, 1200–1800 (Princeton, NJ: Princeton
University Press, 1969).
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774 Chapter 11 Infinite Sequences and Series
In this section we explore two types of applications of Taylor polynomials. First we look
at how they are used to approximate functions––computer scientists like them because
polynomials are the simplest of functions. Then we investigate how physicists and engi-
neers use them in such fields as relativity, optics, blackbody radiation, electric dipoles,
the velocity of water waves, and building highways across a desert.
Approximating Functions by Polynomials
Suppose that f sxd is equal to the sum of its Taylor series at a:
o` f sndsad sx 2 adn
n!
f sxd −
n−0
In Section 11.10 we introduced the notation Tnsxd for the nth partial sum of this series
and called it the nth-degree Taylor polynomial of f at a. Thus
oTnsxd − n f sidsad sx 2 adi
i−0 i!
− f sad 1 f 9sad sx 2 ad 1 f 0sad sx 2 ad2 1 ∙ ∙∙ 1 f sndsad sx 2 adn
1! 2! n!
y Since f is the sum of its Taylor series, we know that Tnsxd l f sxd as n l ` and so Tn
y=´ y=T£(x) can be used as an approximation to f : f sxd < Tnsxd.
y=T™(x)
y=T¡(x) Notice that the first-degree Taylor polynomial
(0, 1) T1sxd − f sad 1 f 9sadsx 2 ad
0x is the same as the linearization of f at a that we discussed in Section 3.10. Notice also
that T1 and its derivative have the same values at a that f and f 9 have. In general, it can
FIGURE 1 be shown that the derivatives of Tn at a agree with those of f up to and including deriva-
tives of order n.
T2sxd x − 0.2 x − 3.0
T4sxd To illustrate these ideas let’s take another look at the graphs of y − ex and its first
T6sxd 1.220000 8.500000 few Taylor polynomials, as shown in Figure 1. The graph of T1 is the tangent line to
T8sxd 1.221400 16.375000 y − ex at s0, 1d; this tangent line is the best linear approximation to ex near s0, 1d. The
T10sxd 1.221403 19.412500 graph of T2 is the parabola y − 1 1 x 1 x2y2, and the graph of T3 is the cubic curve
1.221403 20.009152 y − 1 1 x 1 x2y2 1 x 3y6, which is a closer fit to the exponential curve y − ex than T2.
ex 1.221403 20.079665 The next Taylor polynomial T4 would be an even better approximation, and so on.
1.221403 20.085537 The values in the table give a numerical demonstration of the convergence of the Tay-
lor polynomials Tnsxd to the function y − ex. We see that when x − 0.2 the convergence
is very rapid, but when x − 3 it is somewhat slower. In fact, the farther x is from 0, the
more slowly Tnsxd converges to e x.
When using a Taylor polynomial Tn to approximate a function f, we have to ask the
questions: How good an approximation is it? How large should we take n to be in order
to achieve a desired accuracy? To answer these questions we need to look at the absolute
value of the remainder:
| Rnsxd | − | f sxd 2 Tnsxd |
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Section 11.11 Applications of Taylor Polynomials 775
There are three possible methods for estimating the size of the error:
| |1. I f a graphing device is available, we can use it to graph Rnsxd and thereby
estimate the error.
2. I f the series happens to be an alternating series, we can use the Alternating
Series Estimation Theorem.
3. I n all cases we can use Taylor’s Inequality (Theorem 11.10.9), which says that if
| |f sn11dsxd < M, then
| | | |Rnsxd M
< sn 1 1d! x 2 a n11
Example 1
(a) Approximate the function f sxd − s3 x by a Taylor polynomial of degree 2 at a − 8.
(b) How accurate is this approximation when 7 < x < 9?
SOLUTION f sxd − s3 x − x 1y3 f s8d − 2
(a)
f 9sxd − 1 x22y3 f 9s8d − 1
3 12
f 0sxd − 292 x25y3 f 0s8d − 21414
f -sxd − 10 x28y3
27
Thus the second-degree Taylor polynomial is
T2sxd − f s8d 1 f 9s8d sx 2 8d 1 f 0s8d sx 2 8d2
1! 2!
− 2 1 1 sx 2 8d 2 1 sx 2 8d2
12 288
The desired approximation is
s3 x < T2sxd − 2 1 1 sx 2 8d 2 1 sx 2 8d2
12 288
(b) The Taylor series is not alternating when x , 8, so we can’t use the Alternating
Series Estimation Theorem in this example. But we can use Taylor’s Inequality with
n − 2 and a − 8:
| R2sxd | < M |x 2 8 |3
3!
| |where f -sxd < M. Because x > 7, we have x 8y3 > 78y3 and so
f -sxd − 10 ? 1 < 10 ? 1 , 0.0021
27 x 8y3 27 7 8y3
| |Therefore we can take M − 0.0021. Also 7 < x < 9, so 21 < x 2 8 < 1 and
x 2 8 < 1. Then Taylor’s Inequality gives
| |R2sxd < 0.0021 ? 13 − 0.0021 , 0.0004
3! 6
Thus, if 7 < x < 9, the approximation in part (a) is accurate to within 0.0004. n
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776 Chapter 11 Infinite Sequences and Series
2.5 Let’s use a graphing device to check the calculation in Example 1. Figure 2 shows
T™ | |that the graphs of y − s3 x and y − T2sxd are very close to each other when x is near 8.
y=œ# x„
0 Figure 3 shows the graph of R2sxd computed from the expression
FIGURE 2 | | | |R2sxd − s3 x 2 T2sxd
0.0003 We see from the graph that | |R2sxd , 0.0003
y=| R™(x)| 15
7 when 7 < x < 9. Thus the error estimate from graphical methods is slightly better than
0 the error estimate from Taylor’s Inequality in this case.
FIGURE 3 Example 2
(a) What is the maximum error possible in using the approximation
sin x < x 2 x3 1 x5
3! 5!
9 when 20.3 < x < 0.3? Use this approximation to find sin 12° correct to six decimal
places.
(b) For what values of x is this approximation accurate to within 0.00005?
SOLUTION
(a) Notice that the Maclaurin series
sin x− x 2 x3 1 x5 2 x7 1 ∙ ∙ ∙
3! 5! 7!
| |is alternating for all nonzero values of x, and the successive terms decrease in size
because x , 1, so we can use the Alternating Series Estimation Theorem. The error
in approximating sin x by the first three terms of its Maclaurin series is at most
Z Z | |x7 x 7
7! − 5040
| |If 20.3 < x < 0.3, then x < 0.3, so the error is smaller than
s0.3d7 < 4.3 3 1028
5040
To find sin 12° we first convert to radian measure:
S D S Dsin 12° − sin12
180 − sin 15
S D S D<2 31 51 < 0.20791169
15 15 3! 1 15 5!
Thus, correct to six decimal places, sin 12° < 0.207912.
(b) The error will be smaller than 0.00005 if
| |x 7 , 0.00005
5040
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Section 11.11 Applications of Taylor Polynomials 777
Solving this inequality for x, we get n
| | | |x 7 , 0.252 or x , s0.252d1y7 < 0.821
| |So the given approximation is accurate to within 0.00005 when x , 0.82.
TEC Module 11.10/11.11 graphically | |What if we use Taylor’s Inequality to solve Example 2? Since f s7dsxd − 2cos x, we
shows the remainders in Taylor poly-
nomial approximations. have f s7dsxd < 1 and so
1
4.3 ϫ 10–* | R6sxd | < 7! | x |7
y=| Rß(x)| So we get the same estimates as with the Alternating Series Estimation Theorem.
What about graphical methods? Figure 4 shows the graph of
| | | |( )R6sxd 1 x3 1 x5
− sin x 2 x 2 6 1 120
_0.3 0 | | | |0.3 and we see from it that R6sxd , 4.3 3 1028 when x < 0.3. This is the same estimate
| |that we obtained in Example 2. For part (b) we want R6sxd , 0.00005, so we graph
FIGURE 4 | |both y − R6sxd and y − 0.00005 in Figure 5. By placing the cursor on the right inter-
| |section point we find that the inequality is satisfied when x , 0.82. Again this is the
0.00006
y=0.00005 same estimate that we obtained in the solution to Example 2.
y=| Rß(x)| If we had been asked to approximate sin 72° instead of sin 12° in Example 2, it would
have been wise to use the Taylor polynomials at a − y3 (instead of a − 0) because
they are better approximations to sin x for values of x close to y3. Notice that 72° is
close to 60° (or y3 radians) and the derivatives of sin x are easy to compute at y3.
Figure 6 shows the graphs of the Maclaurin polynomial approximations
_1 0 1 x3
3!
FIGURE 5 T1sxd − x T3sxd − x 2
T5sxd − x 2 x3 1 x5 T7sxd − x 2 x3 1 x5 2 x7
3! 5! 3! 5! 7!
to the sine curve. You can see that as n increases, Tnsxd is a good approximation to sin x
on a larger and larger interval.
y T¡ T∞
FIGURE 6 0x
y=sin x
T£ T¶
One use of the type of calculation done in Examples 1 and 2 occurs in calculators and
computers. For instance, when you press the sin or ex key on your calculator, or when
a computer programmer uses a subroutine for a trigonometric or exponential or Bessel
function, in many machines a polynomial approximation is calculated. The polynomial
is often a Taylor polynomial that has been modified so that the error is spread more
evenly throughout an interval.
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778 Chapter 11 Infinite Sequences and Series
Applications to Physics
Taylor polynomials are also used frequently in physics. In order to gain insight into an
equation, a physicist often simplifies a function by considering only the first two or three
terms in its Taylor series. In other words, the physicist uses a Taylor polynomial as an
approximation to the function. Taylor’s Inequality can then be used to gauge the accu-
racy of the approximation. The following example shows one way in which this idea is
used in special relativity.
Example 3 In Einstein’s theory of special relativity the mass of an object moving
with velocity v is
m − s1 m0
2 v 2yc 2
where m0 is the mass of the object when at rest and c is the speed of light. The kinetic
energy of the object is the difference between its total energy and its energy at rest:
K − mc2 2 m0c2
(a) Show that when v is very small compared with c, this expression for K agrees
1
with classical Newtonian physics: K − 2 m0v2.
| |(b) Use Taylor’s Inequality to estimate the difference in these expressions for K
when v < 100 mys.
SOLUTION
(a) Using the expressions given for K and m, we get
FS D GK − mc2 2 m0c2 −
m0c2 2 m0c2 − m0c2 1 2 v2 21y2
s1 2 v 2yc 2 c2
21
The upper curve in Figure 7 is the | |With x − 2v2yc2, the Maclaurin series for s1 1 xd21y2 is most easily computed as a
graph of the expression for the kinetic
energy K of an object with velocity v binomial series with k − 212. (Notice that x , 1 because v , c.) Therefore we have
in special relativity. The lower curve
shows the function used for K in classi- s1 1 xd21y2 − 1 2 1 x 1 (221 )(2 3 ) x2 1 (221 )(2 3 )(2 5 ) x3 1 ∙ ∙ ∙
cal Newtonian physics. When v is much 2 2 2 2
smaller than the speed of light, the
curves are practically identical. 2! 3!
K − 1 2 1 x 1 3 x 2 2 5 x 3 1 ∙ ∙ ∙
2 8 16
K=mc@-m¸c@
FS D Gand 1 v2 3 v4 5 v6
K − m0c2 11 2 c2 1 8 c4 1 16 c6 1∙∙∙ 21
S D− m0c2 1 v2 3 v4 5 v6
2 c2 1 8 c4 1 16 c6 1∙∙∙
1 If v is much smaller than c, then all terms after the first are very small when compared
2
K = m ¸ √ @ with the first term. If we omit them, we get
0 c √ S DK < m0c2 1 v2
2 c2
FIGURE 7 − 1 m 0 v 2
2
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Section 11.11 Applications of Taylor Polynomials 779
| |(b) If x − 2v 2yc2, f sxd − m0 c2 fs1 1 xd21y2 2 1g , and M is a number such that
f 0sxd < M, then we can use Taylor’s Inequality to write
| |R1sxd < M x2
2!
| |We 3
have f 0sxd − 4 m 0 c 2s1 1 xd25y2 and we are given that v < 100 mys, so
| |f 0sxd − 4s1 3m0 c2 < 4s1 3m0 c2 s−Md
2 v 2yc2 d5y2 2 1002yc2 d5y2
Thus, with c − 3 3 108 mys,
| |R1sxd < 1 ? 4s1 3m0 c2 ? 1004 , s4.17 3 10210 dm0
2 2 1002yc2 d5y2 c4
| |So when v < 100 mys, the magnitude of the error in using the Newtonian expression
for kinetic energy is at most s4.2 3 10210 dm0. n
Another application to physics occurs in optics. Figure 8 is adapted from Optics,
4th ed., by Eugene Hecht (San Francisco, 2002), page 153. It depicts a wave from the
point source S meeting a spherical interface of radius R centered at C. The ray SA is
refracted toward P.
¨r A
¨i
Lo h R ¨t Li
so V ˙
S C P
FIGURE 8 n¡ si
Refraction at a spherical interface n™
Source: Adapted from E. Hecht, Optics, 4e (Upper
Saddle River, NJ: Pearson Education, 2002).
Using Fermat’s principle that light travels so as to minimize the time taken, Hecht
derives the equation
S D1 n1 1 n2 − 1 n2si 2 n1so
,o ,i R ,i ,o
where n1 and n2 are indexes of refraction and ,o, ,i, so, and si are the distances indicated
in Figure 8. By the Law of Cosines, applied to triangles ACS and ACP, we have
2 ,o − sR 2 1 sso 1 Rd2 2 2Rsso 1 Rd cos
,i − sR 2 1 ssi 2 Rd2 1 2Rssi 2 Rd cos
Here we use the identity
coss 2 d − 2cos
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780 Chapter 11 Infinite Sequences and Series
Because Equation 1 is cumbersome to work with, Gauss, in 1841, simplified it by using
the linear approximation cos < 1 for small values of . (This amounts to using the
Taylor polynomial of degree 1.) Then Equation 1 becomes the following simpler equa-
tion [as you are asked to show in Exercise 34(a)]:
3 n1 1 n2 − n2 2 n1
so si R
The resulting optical theory is known as Gaussian optics, or first-order optics, and has
become the basic theoretical tool used to design lenses.
A more accurate theory is obtained by approximating cos by its Taylor polynomial
of degree 3 (which is the same as the Taylor polynomial of degree 2). This takes into
account rays for which is not so small, that is, rays that strike the surface at greater
distances h above the axis. In Exercise 34(b) you are asked to use this approximation to
derive the more accurate equation
F S D S D G4
n1 n2 n2 2 n1 n1 11 2 n2 1 12
so si R 2so so 1 R 2si R 2 si
1 − 1 h2 1
The resulting optical theory is known as third-order optics.
Other applications of Taylor polynomials to physics and engineering are explored in
Exercises 32, 33, 35, 36, 37, and 38, and in the Applied Project on page 783.
; 1. ( a) Find the Taylor polynomials up to degree 5 for 8. f sxd − x cos x, a − 0
f sxd − sin x centered at a − 0. Graph f and these 9. f sxd − xe 22x, a − 0
polynomials on a common screen. 10. f sxd − tan21x, a − 1
(b) Evaluate f and these polynomials at x − y4, y2, CAS 11–12 Use a computer algebra system to find the Taylor poly-
and . nomials Tn centered at a for n − 2, 3, 4, 5. Then graph these
polynomials and f on the same screen.
(c) Comment on how the Taylor polynomials converge 11. f sxd − cot x, a − y4
to f sxd. 12. f sxd − s3 1 1 x 2 , a − 0
; 2. ( a) Find the Taylor polynomials up to degree 3 for 13–22
f sxd − tan x centered at a − 0. Graph f and these (a) Approximate f by a Taylor polynomial with degree n at the
polynomials on a common screen.
number a.
(b) Evaluate f and these polynomials at x − y6, y4, (b) Use Taylor’s Inequality to estimate the accuracy of the approxi-
and y3.
| |mat ion f sxd < Tnsxd when x lies in the given interval.
(c) Comment on how the Taylor polynomials converge
to f sxd. ; (c) Check your result in part (b) by graphing Rnsxd .
13. f sxd − 1yx, a − 1, n − 2, 0.7 < x < 1.3
; 3–10 Find the Taylor polynomial T3sxd for the function f 14. f sxd − x21y2, a − 4, n − 2, 3.5 < x < 4.5
centered at the number a. Graph f and T3 on the same screen.
3. f sxd − e x, a − 1
4. f sxd − sin x, a − y6
5. f sxd − cos x, a − y2
6. f sxd − e2x sin x, a − 0
7. f sxd − ln x, a − 1
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Section 11.11 Applications of Taylor Polynomials 781
15. f sxd − x 2y3, a − 1, n − 3, 0.8 < x < 1.2 32. The resistivity of a conducting wire is the reciprocal of the
16. f sxd − sin x, a − y6, n − 4, 0 < x < y3 conductivity and is measured in units of ohm-meters (V-m).
17. f sxd − sec x, a − 0, n − 2, 20.2 < x < 0.2 The resistivity of a given metal depends on the temperature
18. f sxd − lns1 1 2xd, a − 1, n − 3, 0.5 < x < 1.5 according to the equation
19. f sxd − e x 2, a − 0, n − 3, 0 < x < 0.1
20. f sxd − x ln x, a − 1, n − 3, 0.5 < x < 1.5 std − 20 e ␣st220d
21. f s xd − x sin x, a − 0, n − 4, 21 < x < 1
22. f sxd − sinh 2x, a − 0, n − 5, 21 < x < 1 where t is the temperature in °C. There are tables that list the
values of ␣ (called the temperature coefficient) and 20 (the
23. U se the information from Exercise 5 to estimate cos 80° resistivity at 20°C) for various metals. Except at very low
correct to five decimal places. temperatures, the resistivity varies almost linearly with tem
perature and so it is common to approximate the expression
24. U se the information from Exercise 16 to estimate sin 38° for std by its first- or second-degree Taylor polynomial
correct to five decimal places. at t − 20.
(a) Find expressions for these linear and quadratic
25. U se Taylor’s Inequality to determine the number of terms
of the Maclaurin series for e x that should be used to esti- approximations.
mate e 0.1 to within 0.00001. ; (b) For copper, the tables give ␣ − 0.0039y°C and
26. H ow many terms of the Maclaurin series for lns1 1 xd do 20 − 1.7 3 1028 V-m. Graph the resistivity of copper
you need to use to estimate ln 1.4 to within 0.001? and the linear and quadratic approximations for
2250°C < t < 1000°C.
; (c) For what values of t does the linear approximation agree
with the exponential expression to within one percent?
33. A n electric dipole consists of two electric charges of equal
magnitude and opposite sign. If the charges are q and 2q and
are located at a distance d from each other, then the electric
field E at the point P in the figure is
; 27–29 Use the Alternating Series Estimation Theorem or E − q 2 sD q dd2
Taylor’s Inequality to estimate the range of values of x for D2 1
which the given approximation is accurate to within the stated
error. Check your answer graphically. By expanding this expression for E as a series in powers of
dyD, show that E is approximately proportional to 1yD 3
| | 27. sin x < x 2x3 ( , 0.01) when P is far away from the dipole.
6 error
| | 28. cosx2x4 ( , 0.005) q -q
x<12 2 1 24 error P
Dd
| | 29. a rctan x < x 2x3x5 ( , 0.05)
3 1 5 error 34. (a) Derive Equation 3 for Gaussian optics from Equation 1
by approximating cos in Equation 2 by its first-degree
30. S uppose you know that
Taylor polynomial.
f snds4d − s21dn n!
3nsn 1 1d (b) Show that if cos is replaced by its third-degree Taylor
and the Taylor series of f centered at 4 converges to f sxd polynomial in Equation 2, then Equation 1 becomes
for all x in the interval of convergence. Show that the fifth-
degree Taylor polynomial approximates f s5d with error Equation 4 for third-order optics. [Hint: Use the first two
less than 0.0002. terms in the binomial series for ,o21 and ,i21. Also, use
< sin .]
35. I f a water wave with length L moves with velocity v across a
body of water with depth d, as in the figure on page 782, then
31. A car is moving with speed 20 mys and acceleration v2 − tL tanh 2d
2 mys2 at a given instant. Using a second-degree Taylor 2 L
polynom ial, estimate how far the car moves in the next
(a) If the water is deep, show that v < stLys2d .
second. Would it be reasonable to use this polynomial to (b) If the water is shallow, use the Maclaurin series for tanh
estimate the distance traveled during the next minute? to show that v < std . (Thus in shallow water the
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782 Chapter 11 Infinite Sequences and Series
velocity of a wave tends to be independent of the length of 38. The period of a pendulum with length L that makes a maxi
the wave.)
mum angle 0 with the vertical is
(c) Use the Alternating Series Estimation Theorem to show
that if L . 10d, then the estimate v 2 < td is accurate to Î yT − 4 L y2 dx
within 0.014tL.
t 0 s1 2 k2 sin2x
L
d where k − sins 1 0 d and t is the acceleration due to gravity.
2
(In Exercise 7.7.42 we approximated this integral using
Simpson’s Rule.)
(a) Expand the integrand as a binomial series and use the
result of Exercise 7.1.50 to show that
36. A uniformly charged disk has radius R and surface charge den- Î F GT − 2 L 1 1 12 k2 1 1232 k4 1 123252 k6 1 ∙ ∙ ∙
sity as in the figure. The electric potential V at a point P at a t 22 2242 224262
distance d along the perpendicular central axis of the disk is
V − 2ke (sd 2 1 R2 2 d) If 0 is not too large, the approximation T < 2 sLyt ,
obtained by using only the first term in the series, is often
where ke is a constant (called Coulomb’s constant). Show that used. A better approximation is obtained by using two
terms: ÎT < 2
V < k eR 2 for large d L s1 1 d1k 2
d t
4
(b) Notice that all the terms in the series after the first one have
1
R coefficients that are at most 4 . Use this fact to compare this
d
P series with a geometric series and show that
Î Î2 L s1 1 1 k 2d < T < 2 L 4 2 3k 2
t 4 t 4 2 4k2
37. If a surveyor measures differences in elevation when making (c) Use the inequalities in part (b) to estimate the period of
plans for a highway across a desert, corrections must be made
for the curvature of the earth. a pendulum with L − 1 meter and 0 − 10 °. How does
it compare with the estimate T < 2 sLyt ? What if
(a) If R is the radius of the earth and L is the length of the 0 − 42° ?
highway, show that the correction is
39. I n Section 4.8 we considered Newton’s method for approxi-
C − R secsLyRd 2 R mating a root r of the equation f sxd − 0, and from an initial
approximation x1 we obtained successive approximations
(b) Use a Taylor polynomial to show that
x2, x3, . . . , where
L2 5L 4 xn11 − xn 2 f sxnd
2R 24R 3 f 9sxnd
C < 1
(c) Compare the corrections given by the formulas in parts (a) Use Taylor’s Inequality with n − 1, a − xn, and x − r to show
and (b) for a highway that is 100 km long. (Take the radius
of the earth to be 6370 km.) | | | |that if f 0sxd exists on an interval I containing r, xn, and xn11,
and f 0sxd < M, f 9sxd > K for all x [ I, then
| | | |xn11 2 r M
< 2K xn 2 r 2
LC [This means that if xn is accurate to d decimal places, then xn11
R
is accurate to about 2d decimal places. More precisely, if the
R
error at stage n is at most 102m, then the error at stage n 1 1 is
at most sMy2K d1022m.]
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
applied project Radiation from the Stars 783
applied Project Radiation from the stars
Any object emits radiation when heated. A blackbody is a system that absorbs all the radiation
that falls on it. For instance, a matte black surface or a large cavity with a small hole in its wall
(like a blast furnace) is a blackbody and emits blackbody radiation. Even the radiation from the
sun is close to being blackbody radiation.
Proposed in the late 19th century, the Rayleigh-Jeans Law expresses the energy density of
blackbody radiation of wavelength as
Luke Dodd / Science Source f sd − 8kT
4
where is measured in meters, T is the temperature in kelvins (K), and k is Boltzmann’s con-
stant. The Rayleigh-Jeans Law agrees with experimental measurements for long wavelengths
but disagrees drastically for short wavelengths. [The law predicts that f sd l ` as l 01 but
experiments have shown that f sd l 0.] This fact is known as the ultraviolet catastrophe.
In 1900 Max Planck found a better model (known now as Planck’s Law) for blackbody
radiation:
f sd − 8hc25
e hcyskT d 2 1
where is measured in meters, T is the temperature (in kelvins), and
h − Planck>s constant − 6.6262 3 10234 J ∙ s
c − speed of light − 2.997925 3 108 mys
k − Boltzmann>s constant − 1.3807 3 10223 JyK
1. Use l’Hospital’s Rule to show that
lim f sd − 0 and lim f sd − 0
l 01 l`
fo r Planck’s Law. So this law models blackbody radiation better than the Rayleigh-Jeans
Law for short wavelengths.
2. U se a Taylor polynomial to show that, for large wavelengths, Planck’s Law gives approxi-
mately the same values as the Rayleigh-Jeans Law.
; 3. G raph f as given by both laws on the same screen and comment on the similarities and
differences. Use T − 5700 K (the temperature of the sun). (You may want to change from
meters to the more convenient unit of micrometers: 1 mm − 1026 m.)
4. U se your graph in Problem 3 to estimate the value of for which f sd is a maximum under
Planck’s Law.
; 5. I nvestigate how the graph of f changes as T varies. (Use Planck’s Law.) In particular,
graph f for the stars Betelgeuse (T − 3400 K), Procyon (T − 6400 K), and Sirius
(T − 9200 K), as well as the sun. How does the total radiation emitted (the area under
the curve) vary with T? Use the graph to comment on why Sirius is known as a blue star and
Betelgeuse as a red star.
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784 Chapter 11 Infinite Sequences and Series
11 Review
CONCEPT CHECK Answers to the Concept Check can be found on the back endpapers.
1. ( a) What is a convergent sequence? (b) If a series is convergent by the Comparison Test, how do
you estimate its sum?
(b) What is a convergent series?
(c) If a series is convergent by the Alternating Series Test, how
(c) What does limnl` an − 3 mean? do you estimate its sum?
o
(d) What does ` an − 3 mean? 8. ( a) Write the general form of a power series.
n−1 (b) What is the radius of convergence of a power series?
(c) What is the interval of convergence of a power series?
2. ( a) What is a bounded sequence?
(b) What is a monotonic sequence? 9. S uppose f sxd is the sum of a power series with radius of
(c) What can you say about a bounded monotonic sequence? convergence R.
3. ( a) What is a geometric series? Under what circumstances is (a) How do you differentiate f ? What is the radius of conver-
it convergent? What is its sum? gence of the series for f 9?
(b) What is a p-series? Under what circumstances is it (b) How do you integrate f ? What is the radius of convergence
convergent? of the series for y f sxd dx?
4. S uppose o an − 3 and sn is the nth partial sum of the series. 10. (a) Write an expression for the nth-degree Taylor polynomial of
What is limn l ` an? What is limn l ` sn? f centered at a.
5. S tate the following. (b) Write an expression for the Taylor series of f centered at a.
(a) The Test for Divergence (c) Write an expression for the Maclaurin series of f .
(b) The Integral Test (d) How do you show that f sxd is equal to the sum of its
(c) The Comparison Test
(d) The Limit Comparison Test Taylor series?
(e) The Alternating Series Test (e) State Taylor’s Inequality.
(f) The Ratio Test
(g) The Root Test 11. Write the Maclaurin series and the interval of convergence for
each of the following functions.
6. ( a) What is an absolutely convergent series?
(b) What can you say about such a series? (a) 1ys1 2 xd (b) e x (c) sin x
(c) What is a conditionally convergent series? (d) cos x (e) tan21x (f) lns1 1 xd
7. ( a) If a series is convergent by the Integral Test, how do you 12. Write the binomial series expansion of s1 1 xdk. What is the
estimate its sum? radius of convergence of this series?
TRUE-FALSE QUIZ
Determine whether the statement is true or false. If it is true, o` s21dn − 1
explain why. If it is false, explain why or give an example that n! e
disproves the statement. 10.
n−0
1. I f limnl` an − 0, then o an is convergent. 11. If 21 , ␣ , 1, then limnl` ␣ n − 0.
2. T he series o ` n 2sin 1 is convergent. | | 12. If o an is divergent, then o an is divergent.
n−1
3. I f limn l ` an − L, then limn l ` a2n11 − L. 13. If f sxd − 2x 2 x2 1 1 x3 2 ∙ ∙ ∙ converges for all x,
4. I f o cn6n is convergent, then o cns22dn is convergent. 3
then f -s0d − 2.
5. I f o cn6n is convergent, then o cns26dn is convergent. 14. If han j and hbn j are divergent, then han 1 bn j is divergent.
6. I f o cn x n diverges when x − 6, then it diverges when x − 10. 15. If han j and hbn j are divergent, then han bn j is divergent.
7. T he Ratio Test can be used to determine whether o 1yn 3 16. If han j is decreasing and an . 0 for all n, then han j is
converges. convergent.
8. T he Ratio Test can be used to determine whether o 1yn!
converges.
9. I f 0 < an < bn and o bn diverges, then o an diverges. 17. If an . 0 and o an converges, then o s21dnan converges.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 11 Review 785
18. If an . 0 and limn l` san11yand , 1, then limn l` an − 0. 21. If a finite number of terms are added to a convergent series,
then the new series is still convergent.
19. 0.99999 . . . − 1
o o o` ` `
20. If lim an − 2, then lim san13 2 and − 0. 22. If an − A and bn − B, then an bn − AB.
nl` nl` n−1 n−1 n−1
EXERCISES
1–8 Determine whether the sequence is convergent or divergent. 23–26 Determine whether the series is conditionally conver-
If it is convergent, find its limit. gent, absolutely convergent, or divergent.
1. an − 2 1 n3 2. an − 91n01n1 o o` `
1 1 2n3
23. s21dn21n 21y3 24. s21dn21n 23
n−1 n−1
n3 ` s21dnsn 1 1d3n 26. n−`2 s2l1ndnnsn
1 n−1 2 2 n11
3. an − 1 n 2 4. an − cossny2d o o 25.
5. an − n sin n 6. an − lsnnn 27–31 Find the sum of the series.
n 21 1
` s23dn21 28. n−`1 nsn 11 3d
n−1 23n
7. hs1 1 3ynd4n j 8. hs210dnyn!j o o 27.
` ` s21dn n
n−1 32ns2nd!
tan21ng 30.
n−0
o o 29. 1d
9. A sequence is defined recursively by the equations a1 − 1, ftan21sn 1 2
a n11 − 1 san 1 4d. Show that han j is increasing and an , 2
3
for all n. Deduce that han j is convergent and find its limit.
31. 1 2 e 1 e2 2 e3 1 e4 2 ∙∙∙
2! 3! 4!
; 10. S how that lim n l ` n 4e 2n − 0 and use a graph to find the
smallest value of N that corresponds to « − 0.1 in the pre- 32. Express the repeating decimal 4.17326326326 . . . as a
cise definition of a limit. fraction.
11–22 Determine whether the series is convergent or divergent. 33. Show that cosh x > 1 1 1 x 2 for all x.
2
` n 12. n−`1 nn32 11 11 o sln xdn
o o 11. n−1 1 34. For what values of x does the series ` converge?
n3 1 n−1
` n 3 14. ` s21dn o 35. F ind the sum of the series ` s21dn11 correct to four decimal
n−1 5 n n−1 sn 1 1 n−1 n5
o o 13. places.
S Do o 15. 1 36. (a) Find the partial sum s5 of the series o `n−1 1yn6 and esti-
` nsln n 16. n−`1 ln 3n n1 1 mate the error in using it as an approximation to the sum
n−2
of the series.
(b) Find the sum of this series correct to five decimal places.
` cos 3n 18. n−`1 s1 1n22nn2dn 37. Use the sum of the first eight terms to approximate the sum of
n−1 1 s1.2dn
o o 17. 1 the series o ` s2 1 5nd21. Estimate the error involved in this
n−1
` 1 3 5 ? ∙∙∙ ? s2n 1d approximation.
n−1 5n n!
o 19. ? ? 2 o` nn
s2nd!
38. (a) Show that the series n−1 is convergent.
o 20. ` s25d2n (b) Deduce that lim nn − 0.
n−1 n2 9n nl` s2nd!
o 21. ` sn 39. P rove that if the series o ` an is absolutely convergent, then
n11 n−1
s21dn21
the series S Do`
n−1
n−1
o 22. ` sn 1 1 2 sn 2 1 n11 an
n−1 n n
is also absolutely convergent.
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786 Chapter 11 Infinite Sequences and Series
40–43 Find the radius of convergence and interval of conver- 57–58
gence of the series. (a) Approximate f by a Taylor polynomial with degree n at the
` xn 41. n−`1 sx n14n2dn number a.
n25 ; (b) Graph f and Tn on a common screen.
s21dn
(c) Use Taylor’s Inequality to estimate the accuracy of the
n−1
o o 40. n | |approximation f sxd < Tnsxd when x lies in the given interval.
` 2nsx 2 2d n 43. ` 2nsx 2 3dn ; (d) Check your result in part (c) by graphing Rnsxd .
n−1 sn 1 2d! n−0 sn 1 3
o o 42. 57. f sxd − sx , a − 1, n − 3, 0.9 < x < 1.1
58. f sxd − sec x, a − 0, n − 2, 0 < x < y6
44. Find the radius of convergence of the series
o` s2nd! xn 59. Use series to evaluate the following limit.
sn!d2
n−1 lim sin x2 x
x3
xl 0
45. Find the Taylor series of f sxd − sin x at a − y6. 60. The force due to gravity on an object with mass m at a
46. Find the Taylor series of f sxd − cos x at a − y3. height h above the surface of the earth is
47–54 Find the Maclaurin series for f and its radius of conver- F − mtR2
gence. You may use either the direct method (definition of a sR 1 hd2
Maclaurin series) or known series such as geometric series, where R is the radius of the earth and t is the acceleration due
binomial series, or the Maclaurin series for e x, sin x, tan21x,
and lns1 1 xd. to gravity for an object on the surface of the earth.
(a) Express F as a series in powers of hyR.
47. f sxd − 1 x2 x 48. f sxd − tan21sx 2 d ; (b) Observe that if we approximate F by the first term in the
1
series, we get the expression F < mt that is usually used
49. f sxd − lns4 2 xd 50. f sxd − xe 2x when h is much smaller than R. Use the Altern ating
Series Estimation Theorem to estimate the range of val-
ues of h for which the approximation F < mt is accurate
to within one percent. (Use R − 6400 km.)
51. f sxd − sinsx 4 d 52. f sxd − 10 x 61. Suppose that f sxd − o ` cn x n for all x.
n−0
53. f sxd − 1ys4 16 2 x 54. f sxd − s1 2 3xd25 (a) If f is an odd function, show that
c0 − c2 − c4 − ∙ ∙ ∙ − 0
ex (b) If f is an even function, show that
x
55. Evaluate y dx as an infinite series. c1 − c3 − c5 − ∙ ∙ ∙ − 0
56. Use series to approximate y01 s1 1 x 4 dx correct to two deci- 62. If f sxd − e x 2, show that f s2nds0d − s2nd! .
mal places. n!
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Problems Plus Example Find the sum of the series o` sx 1 2dn .
sn 1 3d!
Before you look at the solution of the n−0
example, cover it up and first try to
solve the problem yourself. Solution The problem-solving principle that is relevant here is recognizing some-
thing familiar. Does the given series look anything like a series that we already know?
Well, it does have some ingredients in common with the Maclaurin series for the expo-
nential function:
o` xn −11x1 x2 1 x3 1 ∙∙∙
n! 2! 3!
ex −
n−0
We can make this series look more like our given series by replacing x by x 1 2:
oe x12− ` sx 1 2dn − 1 1 sx 1 2d 1 sx 1 2d2 1 sx 1 2d3 1 ∙∙∙
n−0 n! 2! 3!
But here the exponent in the numerator matches the number in the denominator
whose factorial is taken. To make that happen in the given series, let’s multiply and
divide by sx 1 2d3:
o o` sx 1 2dn − sx 1 2d3 ` sx 1 2dn13
sn 1 3d! 1 n−0 sn 1 3d!
n−0
F G− sx 1 2d23 sx 1 2d3 1 sx 1 2d4 1 ∙∙∙
3! 4!
We see that the series between brackets is just the series for ex12 with the first three
terms missing. So
F Go ` sx 1 2dn − sx 1 2d23 e x12 2 1 2 sx 1 2d 2 sx 1 2d2 n
n−0 sn 1 3d! 2!
Problems 1. If f sxd − sinsx 3 d, find f s15ds0d.
2. A function f is defined by
f sxd − lim x 2n 2 1
x 2n 1 1
P¢ nl`
4 P£ W here is f continuous?
2
8 P™ 3. (a) Show that tan 1 x − cot 1 x 2 2 cot x.
1 2 2
A 1 P¡ (b) Find the sum of the series
o` 1 tan x
2n 2n
n−1
P∞ | | | | 4. L et hPn j be a sequence of points determined as in the figure. Thus AP1 − 1,
Pn Pn11 − 2n21, and angle APn Pn11 is a right angle. Find limn l ` /Pn APn11.
FIGURE for problem 4
787
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1 5. T o construct the snowflake curve, start with an equilateral triangle with sides of length 1.
2 Step 1 in the construction is to divide each side into three equal parts, construct an equi-
3 lateral triangle on the middle part, and then delete the middle part (see the figure). Step 2
is to repeat step 1 for each side of the resulting polygon. This process is repeated at each
FIGURE for problem 5 succeeding step. The snowflake curve is the curve that results from repeating this process
indefinitely.
788
(a) Let sn, ln, and pn represent the number of sides, the length of a side, and the total length
of the nth approximating curve (the curve obtained after step n of the construction),
respectively. Find formulas for sn, ln, and pn.
(b) Show that pn l ` as n l `.
(c) Sum an infinite series to find the area enclosed by the snowflake curve.
Note: Parts (b) and (c) show that the snowflake curve is infinitely long but encloses only a
finite area.
6. Find the sum of the series
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ∙∙∙
2 3 4 6 8 9 12
where the terms are the reciprocals of the positive integers whose only prime factors are 2s
and 3s.
7. (a) Show that for xy ± 21,
arctan x 2 arctan y − arctan x2y
1 1 xy
if the left side lies between 2y2 and y2.
(b) Show that arctan 120 2 arctan 1 − y4.
119 239
(c) Deduce the following formula of John Machin (1680–1751):
4 arctan 1 2 arctan 1 −
5 239 4
(d) Use the Maclaurin series for arctan to show that
0.1973955597 , arctan 1 , 0.1973955616
5
(e) Show that
0.004184075 , arctan 1 , 0.004184077
239
(f ) Deduce that, correct to seven decimal places, < 3.1415927.
Machin used this method in 1706 to find correct to 100 decimal places. Recently, with
the aid of computers, the value of has been computed to increasingly greater accuracy.
In 2013 Shigeru Kondo and Alexander Yee computed the value of to more than 12 trillion
decimal places!
8. (a) Prove a formula similar to the one in Problem 7(a) but involving arccot instead of
arctan.
(b) Find the sum of the series o `n−0 arccotsn 2 1 n 1 1d.
9. Use the result of Problem 7(a) to find the sum of the series g n`−1 arctans2yn2d.
10. If a0 1 a1 1 a2 1 ∙ ∙ ∙ 1 ak − 0, show that
lim sa0 sn 1 a1 sn 1 1 1 a2 sn 1 2 1 ∙∙∙ 1 ak sn 1 kd − 0
nl`
If you don’t see how to prove this, try the problem-solving strategy of using analogy (see
page 71). Try the special cases k − 1 and k − 2 first. If you can see how to prove the asser-
tion for these cases, then you will probably see how to prove it in general.
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11. Find the interval of convergence of o ` n3x n and find its sum.
n−1
12. S uppose you have a large supply of books, all the same size, and you stack them at the edge
1 1 of a table, with each book extending farther beyond the edge of the table than the one
4 2
1 beneath it. Show that it is possible to do this so that the top book extends entirely beyond
16
8 the table. In fact, show that the top book can extend any distance at all beyond the edge of
the table if the stack is high enough. Use the following method of stacking: The top book
extends half its length beyond the second book. The second book extends a quarter of its
length beyond the third. The third extends one-sixth of its length beyond the fourth, and so
FIGURE for problem 12 on. (Try it yourself with a deck of cards.) Consider centers of mass.
S Do` 1 2 1 .
n2
13. Find the sum of the series ln
n−2
14. If p . 1, evaluate the expression
1 1 1 1 1 1 1 1 ∙∙∙
2p 3p 4p
1 2 1 1 1 2 1 1 ∙∙∙
2p 3p 4p
15. S uppose that circles of equal diameter are packed tightly in n rows inside an equilateral tri-
angle. (The figure illustrates the case n − 4.) If A is the area of the triangle and An is the
total area occupied by the n rows of circles, show that
lim An −
A 2 s3
nl`
16. A sequence han j is defined recursively by the equations
a0 − a1 − 1 nsn 2 1dan − sn 2 1dsn 2 2dan21 2 sn 2 3dan22
FIGURE for problem 15 Find the sum of the series o ` an.
n−0
P¡ P∞ P™
P˜ 17. If the curve y − e 2xy10 sin x, x > 0, is rotated about the x-axis, the resulting solid looks like
an infinite decreasing string of beads.
Pˆ
Pß (a) Find the exact volume of the nth bead. (Use either a table of integrals or a computer
P¡¸ algebra system.)
P¢ P¶ P£
(b) Find the total volume of the beads.
FIGURE for problem 18
18. S tarting with the vertices P1s0, 1d, P2s1, 1d, P3s1, 0d, P4s0, 0d of a square, we construct
further points as shown in the figure: P5 is the midpoint of P1P2, P6 is the midpoint of
P2P3, P7 is the midpoint of P3P4, and so on. The polygonal spiral path P1P2P3P4 P5P6 P7 . . .
approaches a point P inside the square. 1
2
(a) If the coordinates of Pn are sxn, yn d, show that xn 1 x n11 1 x n12 1 x n13 − 2 and find a
similar equation for the y-coordinates.
(b) Find the coordinates of P.
o 19. ` s21dn .
Find the sum of the series n−1 s2n 1 1d 3n
20. Carry out the following steps to show that
1 1 1 1 1 1 1 1 ∙∙∙ − ln 2
1?2 3?4 5?6 7?8
(a) Use the formula for the sum of a finite geometric series (11.2.3) to get an expression for
1 2 x 1 x 2 2 x 3 1 ∙ ∙ ∙ 1 x 2n22 2 x 2n21
789
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(b) Integrate the result of part (a) from 0 to 1 to get an expression for
1 2 1 1 1 2 1 1 ∙∙∙ 1 1 1 2 1
2 3 4 2n 2 2n
as an integral.
(c) Deduce from part (b) that
Z Zy y11
1?23?4
1 1 1 1 ∙∙∙ 1 s2n 1 2 1 dx , 1 x2n dx
5?6 2 1ds2nd 0 11x 0
(d) Use part (c) to show that the sum of the given series is ln 2.
21. Find all the solutions of the equation
1 1 x 1 x2 1 x3 1 x4 1 ∙∙∙ − 0
2! 4! 6! 8!
[Hint: Consider the cases x > 0 and x , 0 separately.]
1 1 22. R ight-angled triangles are constructed as in the figure. Each triangle has height 1 and its
1 1 base is the hypotenuse of the preceding triangle. Show that this sequence of triangles makes
indefinitely many turns around P by showing that g n is a divergent series.
¨£ ¨™ 1
¨¡ 23. Consider the series whose terms are the reciprocals of the positive integers that can be writ-
ten in base 10 notation without using the digit 0. Show that this series is convergent and the
P1 sum is less than 90.
FIGURE for problem 22 24. (a) Show that the Maclaurin series of the function
x `
of sxd x2
− 1 2 x 2 is fn x n
n−1
where fn is the nth Fibonacci number, that is, f1 − 1, f2 − 1, and fn − fn21 1 fn22
for n > 3. [Hint: Write xys1 2 x 2 x 2d − c0 1 c1x 1 c2 x 2 1 . . . and multiply both
sides of this equation by 1 2 x 2 x 2.]
(b) By writing f sxd as a sum of partial fractions and thereby obtaining the Maclaurin series
in a different way, find an explicit formula for the nth Fibonacci number.
25. Let u−11 x3 1 x6 1 x9 1∙∙∙
3! 6! 9!
v − x 1 x4 1 x7 1 x 10 1 ∙∙∙
4! 7! 10!
w − x2 1 x5 1 x8 1 ∙∙∙
2! 5! 8!
Show that u 3 1 v3 1 w3 2 3uvw − 1.
26. Prove that if n . 1, the nth partial sum of the harmonic series is not an integer.
Hint: Let 2k be the largest power of 2 that is less than or equal to n and let M be the product
of all odd integers that are less than or equal to n. Suppose that sn − m, an integer. Then
M2ksn − M2km. The right side of this equation is even. Prove that the left side is odd by
showing that each of its terms is an even integer, except for the last one.
790
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12 Vectors and the
Geometry of Space
Each of these gears has the
shape of a hyperboloid, a type
of surface we will study in
Section 12.6. The shape allows
the gears to transmit motion
between skew (neither
parallel nor intersecting) axes.
In this chapter we introduce vectors and coordinate systems for three-dimensional space.
This will be the setting for our study of the calculus of functions of two variables in Chapter 14
7et1206un03because the graph of such a function is a surface in space. In this chapter we will see that vectors
04/21/10provide particularly simple descriptions of lines and planes in space.
MasterID: 01462
791
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792 Chapter 12 Vectors and the Geometry of Space
z 3D Space
O To locate a point in a plane, we need two numbers. We know that any point in the plane
can be represented as an ordered pair sa, bd of real numbers, where a is the x-coordinate
x and b is the y-coordinate. For this reason, a plane is called two-dimensional. To locate a
point in space, three numbers are required. We represent any point in space by an ordered
FIGURE 1 triple sa, b, cd of real numbers.
Coordinate axes
In order to represent points in space, we first choose a fixed point O (the origin) and
z y three directed lines through O that are perpendicular to each other, called the coordinate
x axes and labeled the x-axis, y-axis, and z-axis. Usually we think of the x- and y-axes as
being horizontal and the z-axis as being vertical, and we draw the orientation of the axes
FIGURE 2 as in Figure 1. The direction of the z-axis is determined by the right-hand rule as illus-
Right-hand rule trated in Figure 2: If you curl the fingers of your right hand around the z-axis in the direc-
tion of a 90° counterclockwise rotation from the positive x-axis to the positive y-axis,
then your thumb points in the positive direction of the z-axis.
The three coordinate axes determine the three coordinate planes illustrated in Fig-
ure 3(a). The xy-plane is the plane that contains the x- and y-axes; the yz-plane contains
the y- and z-axes; the xz-plane contains the x- and z-axes. These three coordinate planes
divide space into eight parts, called octants. The first octant, in the foreground, is deter-
mined by the positive axes.
y
z
z
xz-plane O yz-plane left wall O right wall
xy-plane x floor y
x y
FIGURE 3 (a) Coordinate planes (b)
z Because many people have some difficulty visualizing diagrams of three-dimensional
aO P(a, b, c) figures, you may find it helpful to do the following [see Figure 3(b)]. Look at any bottom
c
x corner of a room and call the corner the origin. The wall on your left is in the xz-plane,
y
FIGURE 4 the wall on your right is in the yz-plane, and the floor is in the xy-plane. The x-axis runs
b
along the intersection of the floor and the left wall. The y-axis runs along the intersection
of the floor and the right wall. The z-axis runs up from the floor toward the ceiling along
the intersection of the two walls. You are situated in the first octant, and you can now
imagine seven other rooms situated in the other seven octants (three on the same floor
and four on the floor below), all connected by the common corner point O.
Now if P is any point in space, let a be the (directed) distance from the yz-plane to P,
let b be the distance from the xz-plane to P, and let c be the distance from the xy-plane to
P. We represent the point P by the ordered triple sa, b, cd of real numbers and we call
a, b, and c the coordinates of P; a is the x-coordinate, b is the y-coordinate, and c is the
z-coordinate. Thus, to locate the point sa, b, cd, we can start at the origin O and move
a units along the x-axis, then b units parallel to the y-axis, and then c units parallel to the
z-axis as in Figure 4.
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Section 12.1 Three-Dimensional Coordinate Systems 793
The point Psa, b, cd determines a rectangular box as in Figure 5. If we drop a perpen-
dicular from P to the xy-plane, we get a point Q with coordinates sa, b, 0d called the pro-
jection of P onto the xy-plane. Similarly, Rs0, b, cd and Ssa, 0, cd are the projections of
P onto the yz-plane and xz-plane, respectively.
As numerical illustrations, the points s24, 3, 25d and s3, 22, 26d are plotted in Fig-
ure 6.
z z 3 z
(0, 0, c) 0
_4 _5 _2 3
S(a, 0, c) R(0, b, c) 0 y y
P(a, b, c) x
0 (0, b, 0) (_4, 3, _5) x _6
(a, 0, 0)
y
x
Q(a, b, 0) (3, _2, _6)
FIGURE 5
FIGURE 6
|The Cartesian product R 3 R 3 R − hsx, y, zd x, y, z [ Rj is the set of all ordered
triples of real numbers and is denoted by R 3. We have given a one-to-one correspon-
dence between points P in space and ordered triples sa, b, cd in R 3. It is called a three-
dimensional rectangular coordinate system. Notice that, in terms of coordinates, the
first octant can be described as the set of points whose coordinates are all positive.
Surfaces
In two-dimensional analytic geometry, the graph of an equation involving x and y is a
curve in R 2. In three-dimensional analytic geometry, an equation in x, y, and z represents
a surface in R 3.
Example 1 What surfaces in R3 are represented by the following equations?
(a) z − 3 (b) y − 5
|SOLUTION
(a) The equation z − 3 represents the set hsx, y, zd z − 3j, which is the set of all
points in R 3 whose z-coordinate is 3 (x and y can each be any value). This is the
horizontal plane that is parallel to the xy-plane and three units above it as in Figure 7(a).
z z y
3 5
x0 0
x 5y 0x
y
(a) z=3, a plane in R# (b) y=5, a plane in R# (c) y=5, a line in R@
FIGURE 7 (b) The equation y − 5 represents the set of all points in R 3 whose y-coordinate is 5.
This is the vertical plane that is parallel to the xz-plane and five units to the right of it as
in Figure 7(b). ■
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794 Chapter 12 Vectors and the Geometry of Space
NOTE When an equation is given, we must understand from the context whether it rep-
resents a curve in R 2 or a surface in R 3. In Example 1, y − 5 represents a plane in R 3, but
of course y − 5 can also represent a line in R 2 if we are dealing with two-dimensional
analytic geometry. See Figure 7(b) and (c).
In general, if k is a constant, then x − k represents a plane parallel to the yz-plane,
y − k is a plane parallel to the xz-plane, and z − k is a plane parallel to the xy-plane. In
Figure 5, the faces of the rectangular box are formed by the three coordinate planes
x − 0 (the yz-plane), y − 0 (the xz-plane), and z − 0 (the xy-plane), and the planes
x − a, y − b, and z − c.
Example 2
(a) Which points sx, y, zd satisfy the equations
x 2 1 y 2 − 1 and z − 3
(b) What does the equation x2 1 y2 − 1 represent as a surface in R 3?
SOLUTION
(a) Because z − 3, the points lie in the horizontal plane z − 3 from Example 1(a).
Because x2 1 y2 − 1, the points lie on the circle with radius 1 and center on the z-axis.
See Figure 8.
(b) Given that x2 1 y2 − 1, with no restrictions on z, we see that the point sx, y, zd
could lie on a circle in any horizontal plane z − k. So the surface x2 1 y2 − 1 in R 3
consists of all possible horizontal circles x2 1 y2 − 1, z − k, and is therefore the circu-
lar cylinder with radius 1 whose axis is the z-axis. See Figure 9.
zz
3
00
x yx y
FIGURE 8 FIGURE 9
z The circle x 2 1 y 2 − 1, z − 3 The cylinder x 2 1 y 2 − 1 ■
Example 3 Describe and sketch the surface in R 3 represented by the equation y − x.
|SOLUTION The equation represents the set of all points in R 3 whose x- and y-coordi-
nates are equal, that is, hsx, x, zd x [ R, z [ Rj. This is a vertical plane that intersects
y the xy-plane in the line y − x, z − 0. The portion of this plane that lies in the first
0 octant is sketched in Figure 10.
■
x Distance and Spheres
FIGURE 10 The familiar formula for the distance between two points in a plane is easily extended to
The plane y − x the following three-dimensional formula.
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Section 12.1 Three-Dimensional Coordinate Systems 795
| |Distance Formula in Three Dimensions The distance P1P2 between the points
P1sx1, y1, z1d and P2sx2, y2, z2 d is
| |P1P2 − ssx2 2 x1d2 1 s y2 2 y1d2 1 sz2 2 z1d2
z To see why this formula is true, we construct a rectangular box as in Figure 11, where
P¡(⁄, ›, z¡) P™(¤, fi, z™) P1 and P2 are opposite vertices and the faces of the box are parallel to the coordinate
planes. If Asx2, y1, z1d and Bsx2, y2, z1d are the vertices of the box indicated in the figure,
then
| P1A | − | x2 2 x1 | | AB | − | y2 2 y1 | | BP2 | − | z2 2 z1 |
0 B(¤, fi, z¡) y Because triangles P1BP2 and P1AB are both right-angled, two applications of the Pythago
x A(¤, ›, z¡) rean Theorem give
FF IIGGUURREE111 1 | | | | | |P1P2 2 − P1B 2 1 BP2 2
and | P1B |2 − | P1A |2 1 | AB |2
7et120111
09/18/08 Combining these equations, we get
MasterID: 01380
| | | | | | | |P1P2 2 − P1A 2 1 AB 2 1 BP2 2
| | | | | |− x2 2 x1 2 1 y2 2 y1 2 1 z2 2 z1 2
Therefore − sx2 2 x1d2 1 s y2 2 y1d2 1 sz2 2 z1d2
| |P1P2 − ssx2 2 x1d2 1 s y2 2 y1d2 1 sz2 2 z1d2
Example 4 The distance from the point Ps2, 21, 7d to the point Qs1, 23, 5d is ■
| | PQ − ss1 2 2d2 1 s23 1 1d2 1 s5 2 7d2 − s1 1 4 1 4 − 3
z Example 5 Find an equation of a sphere with radius r and center Csh, k, ld.
0 P(x, y, z) | | | |SOLUTION By definition, a sphere is the set of all points Psx, y, zd whose distance from
x r
C(h, k, l ) C is r. (See Figure 12.) Thus P is on the sphere if and only if PC − r. Squaring both
FIGURE 12 sides, we have PC 2 − r 2 or
y
sx 2 hd2 1 s y 2 kd2 1 sz 2 l d2 − r 2 ■
The result of Example 5 is worth remembering.
Equation of a Sphere A n equation of a sphere with center Csh, k, ld and radius r
is
sx 2 hd2 1 s y 2 kd2 1 sz 2 l d2 − r 2
In particular, if the center is the origin O, then an equation of the sphere is
x2 1 y2 1 z2 − r2
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796 Chapter 12 Vectors and the Geometry of Space
Example 6 Show that x2 1 y2 1 z2 1 4x 2 6y 1 2z 1 6 − 0 is the equation of a
sphere, and find its center and radius.
SOLUTION We can rewrite the given equation in the form of an equation of a sphere if
we complete squares:
sx 2 1 4x 1 4d 1 s y 2 2 6y 1 9d 1 sz2 1 2z 1 1d − 26 1 4 1 9 1 1
sx 1 2d2 1 s y 2 3d2 1 sz 1 1d2 − 8
Comparing this equation with the standard form, we see that it is the equation of a
sphere with center s22, 3, 21d and radius s8 − 2s2 . ■
Example 7 What region in R3 is represented by the following inequalities?
1 < x 2 1 y 2 1 z2 < 4 z < 0
z SOLUTION The inequalities 1 < x2 1 y2 1 z2 < 4
can be rewritten as 1 < sx2 1 y2 1 z2 < 2
0
1 so they represent the points sx, y, zd whose distance from the origin is at least 1 and at
2
x y most 2. But we are also given that z < 0, so the points lie on or below the xy-plane.
Thus the given inequalities represent the region that lies between (or on) the spheres
FIGURE 13
x 2 1 y 2 1 z2 − 1 and x 2 1 y 2 1 z2 − 4 and beneath (or on) the xy-plane. It is
sketched in Figure 13. ■
1. S uppose you start at the origin, move along the x-axis a 7. D escribe and sketch the surface in R3 represented by the equa-
distance of 4 units in the positive direction, and then move tion x 1 y − 2.
downward a distance of 3 units. What are the coordinates
of your position? 8. Describe and sketch the surface in R3 represented by the equa-
tion x 2 1 z 2 − 9.
2. S ketch the points s1, 5, 3d, s0, 2, 23d, s23, 0, 2d, and
s2, 22, 21d on a single set of coordinate axes. 9–10 Find the lengths of the sides of the triangle PQR. Is it a right
triangle? Is it an isosceles triangle?
3. W hich of the points As24, 0, 21d, Bs3, 1, 25d, and Cs2, 4, 6d 9. Ps3, 22, 23d, Qs7, 0, 1d, Rs1, 2, 1d
is closest to the yz-plane? Which point lies in the xz-plane?
10. Ps2, 21, 0d, Qs4, 1, 1d, Rs4, 25, 4d
4. W hat are the projections of the point (2, 3, 5) on the xy-, yz-,
and xz-planes? Draw a rectangular box with the origin and 11. D etermine whether the points lie on a straight line.
s2, 3, 5d as opposite vertices and with its faces parallel to the (a) As2, 4, 2d, Bs3, 7, 22d, Cs1, 3, 3d
coordinate planes. Label all vertices of the box. Find the length (b) Ds0, 25, 5d, Es1, 22, 4d, Fs3, 4, 2d
of the diagonal of the box.
12. Find the distance from s4, 22, 6d to each of the following.
5. What does the equation x − 4 represent in R2? What does it
represent in R3? Illustrate with sketches. (a) The xy-plane (b) The yz-plane
6. What does the equation y − 3 represent in R3? What does (c) The xz-plane (d) The x-axis
z − 5 represent? What does the pair of equations y − 3, z − 5
represent? In other words, describe the set of points sx, y, zd (e) The y-axis (f) The z-axis
such that y − 3 and z − 5. Illustrate with a sketch.
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Section 12.1 Three-Dimensional Coordinate Systems 797
13. Find an equation of the sphere with center s23, 2, 5d and 40. T he solid cylinder that lies on or below the plane z − 8 and on
radius 4. What is the intersection of this sphere with the or above the disk in the xy-plane with center the origin and
yz-plane? radius 2
14. F ind an equation of the sphere with center s2, 26, 4d and 41. T he region consisting of all points between (but not on) the
radius 5. Describe its intersection with each of the coordinate spheres of radius r and R centered at the origin, where r , R
planes.
42. T he solid upper hemisphere of the sphere of radius 2 centered
15. F ind an equation of the sphere that passes through the point at the origin
s4, 3, 21d and has center s3, 8, 1d.
43. T he figure shows a line L1 in space and a second line L2,
16. Find an equation of the sphere that passes through the origin which is the projection of L1 onto the xy-plane. (In other words,
and whose center is s1, 2, 3d. the points on L2 are directly beneath, or above, the points
on L1.)
17–20 Show that the equation represents a sphere, and find its
center and radius. (a) Find the coordinates of the point P on the line L1.
(b) Locate on the diagram the points A, B, and C, where
17. x 2 1 y 2 1 z2 2 2x 2 4y 1 8z − 15
the line L1 intersects the xy-plane, the yz-plane, and the
18. x 2 1 y 2 1 z 2 1 8x 2 6y 1 2z 1 17 − 0 xz-plane, respectively.
19. 2x 2 1 2y 2 1 2z 2 − 8x 2 24 z 1 1 z
20. 3x 2 1 3y 2 1 3z 2 − 10 1 6y 1 12z L¡
21. (a) Prove that the midpoint of the line segment from
S DP1sx1, y1, z1d to P2sx2, y2, z2 d is
x1 1 x2 y1 1 y2 z1 1 z2 P
2 2 2
, ,
(b) Find the lengths of the medians of the triangle with ver 1 1 L™
tices As1, 2, 3d, Bs22, 0, 5d, and Cs4, 1, 5d. (A median of a 0 y
triangle is a line segment that joins a vertex to the midpoint 1
of the opposite side.)
x
22. F ind an equation of a sphere if one of its diameters has end-
points s5, 4, 3d and s1, 6, 29d. 44. Consider the points P such that the distance from P to
As21, 5, 3d is twice the distance from P to Bs6, 2, 22d. Show
23. Find equations of the spheres with center s2, 23, 6d that touch that the set of all such points is a sphere, and find its center and
(a) the xy-plane, (b) the yz-plane, (c) the xz-plane.
radius.
24. F ind an equation of the largest sphere with center s5, 4, 9d that
is contained in the first octant.
25–38 Describe in words the region of R 3 represented by the 45. F ind an equation of the set of all points equidistant from the
equation(s) or inequality. points As21, 5, 3d and Bs6, 2, 22d. Describe the set.
25. x − 5 26. y − 22
27. y , 8 28. z > 21 46. Find the volume of the solid that lies inside both of the spheres
29. 0 < z < 6 30. y 2 − 4 x 2 1 y 2 1 z2 1 4x 2 2y 1 4z 1 5 − 0
31. x 2 1 y 2 − 4, z − 21 32. x 2 1 y 2 − 4 and x2 1 y2 1 z2 − 4
33. x 2 1 y 2 1 z 2 − 4 34. x 2 1 y 2 1 z 2 < 4 47. Find the distance between the spheres x 2 1 y 2 1 z 2 − 4 and
x 2 1 y 2 1 z 2 − 4x 1 4y 1 4z 2 11.
35. 1 < x 2 1 y 2 1 z 2 < 5 36. x − z
37. x 2 1 z 2 < 9 38. x 2 1 y 2 1 z 2 . 2z 48. Describe and sketch a solid with the following properties.
When illuminated by rays parallel to the z-axis, its shadow is a
39–42 Write inequalities to describe the region. circular disk. If the rays are parallel to the y-axis, its shadow is
39. The region between the yz-plane and the vertical plane x − 5 a square. If the rays are parallel to the x-axis, its shadow is an
isosceles triangle.
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798 Chapter 12 Vectors and the Geometry of Space
The term vector is used by scientists to indicate a quantity (such as displacement or
velocity or force) that has both magnitude and direction. A vector is often represented by
an arrow or a directed line segment. The length of the arrow represents the magnitude of
D the vector and the arrow points in the direction of the vector. We denote a vector by print-
ing a letter in boldface svd or by putting an arrow above the letter svld.
B
u For instance, suppose a particle moves along a line segment from point A to point B. The
v corresponding displacement vector v, shown in Figure 1, has initial point A (the tail)
− AlB. Notice that the
AC and terminal point B (the tip) and we indicate this by writing v even though it is in a
vector u − ClD has the same length and the same direction as v
FIGURE 1
Equivalent vectors different position. We say that u and v are equivalent (or equal) and we write u − v.
The zero vector, denoted by 0, has length 0. It is the only vector with no specific direction.
Combining Vectors BB,tosoCit,swdiitshpldaicsepmlaecnetmveencttovreicstoAlrBB.lCThaesninthFe ipgaurrteic2le.
C Suppose a particle moves from A to
changes direction and moves from
A
B The rceosmulbtiinngeddeisfpfelacct eomf tehnetsveedcitsoprlaAlcCemisecnatslleisdtthhaet sthuempoafrtAilcBleahnads BmlCovaendd from A to C.
FIGURE 2 The we write
AlC − AlB 1 BlC
In general, if we start with vectors u and v, we first move v so that its tail coincides
with the tip of u and define the sum of u and v as follows.
Definition of Vector Addition If u and v are vectors positioned so the initial
point of v is at the terminal point of u, then the sum u 1 v is the vector from the
initial point of u to the terminal point of v.
The definition of vector addition is illustrated in Figure 3. You can see why this defi
nition is sometimes called the Triangle Law.
uu
u+v u+vv v v+u v uv v
uu v u+ v
v v+ u+
uu
FIGURE 3 FIGURE 4
The Triangle Law The Parallelogram Law
In Figure 4 we start with the same vectors u and v as in Figure 3 and draw another
copy of v with the same initial point as u. Completing the parallelogram, we see that
u 1 v − v 1 u. This also gives another way to construct the sum: if we place u and v so
they start at the same point, then u 1 v lies along the diagonal of the parallelogram with
u and v as sides. (This is called the Parallelogram Law.)
a b Example 1 Draw the sum of the vectors a and b shown in Figure 5.
FIGURE 5 SOLUTION First we move b and place its tail at the tip of a, being careful to draw a
copy of b that has the same length and direction. Then we draw the vector a 1 b [see
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Section 12.2 Vectors 799
Figure 6(a)] starting at the initial point of a and ending at the terminal point of the copy
of b.
Alternatively, we could place b so it starts where a starts and construct a 1 b by the
Parallelogram Law as in Figure 6(b).
TEC Visual 12.2 shows how the Tri- a a
angle and Parallelogram Laws work b
for various vectors a and b. a+b b
a+b
FIGURE 6 (a) (b) ■
It is possible to multiply a vector by a real number c. (In this context we call the real num-
ber c a scalar to distinguish it from a vector.) For instance, we want 2v to be the same
vector as v 1 v, which has the same direction as v but is twice as long. In general, we mul-
tiply a vector by a scalar as follows.
| |Definition of Scalar Multiplication If c is a scalar and v is a vector, then the
scalar multiple cv is the vector whose length is c times the length of v and
whose direction is the same as v if c . 0 and is opposite to v if c , 0. If c − 0 or
v − 0, then cv − 0.
This definition is illustrated in Figure 7. We see that real numbers work like scaling fac-
tors here; that’s why we call them scalars. Notice that two nonzero vectors are parallel
if they are scalar multiples of one another. In particular, the vector 2v − s21dv has the
same length as v but points in the opposite direction. We call it the negative of v.
FIGURE 7 v 2v 1 v _v _1.5v
Scalar multiples of v 2
By the difference u 2 v of two vectors we mean
u 2 v − u 1 s2vd
So we can construct u 2 v by first drawing the negative of v, 2v, and then adding it to
u by the Parallelogram Law as in Figure 8(a). Alternatively, since v 1 su 2 vd − u,
the vector u 2 v, when added to v, gives u. So we could construct u 2 v as in Figu re
8(b) by means of the Triangle Law. Notice that if u and v both start from the same initial
point, then u 2 v connects the tip of v to the tip of u.
FIGURE 8 vu u-v
Drawing u 2 v u-v v
_v u
(a) (b)
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800 Chapter 12 Vectors and the Geometry of Space
a Exam_p2lbe 2 If a a b are the vectors shown in Figure 9, draw a 2 2b.
FIGURE 9 and
_2b b SOLUTION Wea-fir2sbt draw the vector 22b pointing in the direction opposite to b and
a
twice as long. We place it with its tail at the tip of a and then use the Triangle Law to
draw a 1 s22bd as in Figure 10. ■
Components
a-2b For some purposes it’s best to introduce a coordinate system and treat vectors algebra-
FIGURE 10 ically. If we place the initial point of a vector a at the origin of a rectangular coordinate
system, then the terminal point of a has coordinates of the form sa1, a2d or sa1, a2, a3d,
depending on whether our coordinate system is two- or three-dimensional (see Figure 11).
These coordinates are called the components of a and we write
a − ka1, a2 l or a − ka1, a2, a3 l
We use the notation ka1, a2l for the ordered pair that refers to a vector so as not to confuse
it with the ordered pair sa1, a2d that refers to a point in the plane.
y (a¡, a™) z
a
(a¡, a™, a£)
a
O
O x x y
a=ka¡, a™l a=ka¡, a™, a£l
FIGURE 11
y (4, 5) For instance, the vectors shown in Figure 12 are all equivalent to the vector OlP − k3, 2l
(1, 3) P(3, 2) whose terminal point is Ps3, 2d. What they have in common is that the terminal point
0x is reached from the initial point by a displacement of three units to the right and two
FIGURE 12 upward. We can think of all these greepormeseetnritcatvioecntoOlrPs as representations of the algebraic
Representations of a − k 3, 2l vector a − k3, 2l. The particular from the origin to the point Ps3, 2d
ipsociInantlletPhdsraeth1e,eadp2im,oasei3ntdi.soi(noSnevsee, ctFthoiegruvorefectth1oe3r .p)aoL−inettO’PlsP.co−nskiad1e, raa2,nay3lotihs etrherepporessiteinotnatvioenctoAlrBoof fthae,
where the initial point is Asx1, y1, z1d and the terminal point is Bsx2, y2, z2 d. Then we must
have x1 1 a1 − x2, y1 1 a2 − y2, and z1 1 a3 − z2 and so a1 − x2 2 x1, a2 − y2 2 y1,
and a3 − z2 2 z1. Thus we have the following result.
z 1 Given the points Asx1, y1, z1d and Bsx2, y2, z2 d, the vector a with represen-
position tation AlB is
vector of P
P(a¡, a™, a£) a − k x2 2 x1, y2 2 y1, z2 2 z1l
O Example 3 Find the vector represented by the directed line segment with initial ■
y point As2, 23, 4) and terminal point Bs22, 1, 1d.
x A(x, y, z) B(x+a¡, y+a™, z+a£) SOLUTION By (1), the vector corresponding to AlB is
a − k22 2 2, 1 2 s23d, 1 2 4l − k24, 4, 23l
FIGURE 13
Representations of a − ka1, a2, a3l
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Section 12.2 Vectors 801
The magnitude or length of the vector v is the length of any of its representations and
| |is denoted by the symbol v or i v i. By using the distance formula to compute the length
of a segment OP, we obtain the following formulas.
y The length of the two-dimensional vector a − k a1, a2l is
(a¡+b¡, a™+b™)
| |a − sa12 1 a22
a+b b b™
The length of the three-dimensional vector a − ka1, a2, a3l is
a b¡
a™ a™ | |a − sa12 1 a22 1 a32
0 a¡ b¡ x How do we add vectors algebraically? Figure 14 shows that if a − ka1, a2l and
b − kb1, b2l, then the sum is a 1 b − ka1 1 b1, a2 1 b2l, at least for the case where the
FIGURE 14 components are positive. In other words, to add algebraic vectors we add corresponding
components. Similarly, to subtract vectors we subtract corresponding components. From
the similar triangles in Figure 15 we see that the components of ca are ca1 and ca2. So to
multiply a vector by a scalar we multiply each component by that scalar.
ca ca™ If a − ka1, a2l and b − kb1, b2l, then
a a™ a 1 b − ka1 1 b1, a2 1 b2l a 2 b − ka1 2 b1, a2 2 b2l
ca − kca1, ca2l
a¡ ca¡
FIGURE 15 Similarly, for three-dimensional vectors,
k a1, a2, a3l 1 kb1, b2, b3l − ka1 1 b1, a2 1 b2, a3 1 b3l
k a1, a2, a3l 2 kb1, b2, b3l − ka1 2 b1, a2 2 b2, a3 2 b3l
ck a1, a2, a3l − kca1, ca2, ca3l
| |Example 4 If a − k 4, 0, 3 l and b − k 22, 1, 5 l, find a and the vectors a 1 b,
a 2 b, 3b, and 2a 1 5b.
SOLUTION | |a − s42 1 02 1 32 − s25 − 5
a 1 b − k4, 0, 3l 1 k22, 1, 5l
− k4 1 s22d, 0 1 1, 3 1 5l − k2, 1, 8l
a 2 b − k4, 0, 3l 2 k22, 1, 5l
− k4 2 s22d, 0 2 1, 3 2 5l − k6, 21, 22l
3b − 3k22, 1, 5l − k3s22d, 3s1d, 3s5dl − k26, 3, 15l
2a 1 5b − 2k4, 0, 3l 1 5k22, 1, 5l
− k8, 0, 6l 1 k210, 5, 25l − k22, 5, 31l ■
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802 Chapter 12 Vectors and the Geometry of Space
Vectors in n dimensions are used to list We denote by V2 the set of all two-dimensional vectors and by V3 the set of all three-
various quantities in an organized way. dimensional vectors. More generally, we will later need to consider the set Vn of all
For instance, the components of a six- n-dimensional vectors. An n-dimensional vector is an ordered n-tuple:
dimensional vector
a − k a1, a2, . . . , anl
p − k p1, p2, p3, p4, p5, p6l
where a1, a2, . . . , an are real numbers that are called the components of a. Addition and
might represent the prices of six dif scalar multiplication are defined in terms of components just as for the cases n − 2 and
ferent ingredients required to make a n − 3.
particular product. Four-dimensional
vectors k x, y, z, t l are used in relativity Properties of Vectors If a, b, and c are vectors in Vn and c and d are scalars, then
theory, where the first three compo 1. a 1 b − b 1 a 2. a 1 sb 1 cd − sa 1 bd 1 c
nents specify a position in space and the 3. a 1 0 − a 4. a 1 s2ad − 0
fourth represents time. 5. csa 1 bd − ca 1 cb 6. sc 1 dda − ca 1 da
7. scdda − csdad 8. 1a − a
These eight properties of vectors can be readily verified either geometrically or alge-
braically. For instance, Property 1 can be seen from Figure 4 (it’s equivalent to the Paral
lelogram Law) or as follows for the case n − 2:
a 1 b − k a1, a2 l 1 k b1, b2 l − k a1 1 b1, a2 1 b2 l
Qc − k b1 1 a1, b2 1 a2 l − k b1, b2 l 1 k a1, a2 l
−b1a
(a+b)+c a+b b We can see why Property 2 (the associative law) iPslQtruies by looking at Figure 16 and
=a+(b+c) applying the Triangle Law several times: the vector obtained either by first con-
b+c structing a 1 b and then adding c or by adding a to the vector b 1 c.
Three vectors in V3 play a special role. Let
P a
FIGURE 16 i − k 1, 0, 0l j − k 0, 1, 0l k − k 0, 0, 1l
These vectors i, j, and k are called the standard basis vectors. They have length 1 and
point in the directions of the positive x-, y-, and z-axes. Similarly, in two dimensions we
define i − k1, 0l and j − k 0, 1l. (See Figure 17.)
FIGURE 17 y z y
Standard basis vectors in V2 and V3
(0, 1) kj
j xi
0i
(1, 0) x
(a) (b)
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Section 12.2 Vectors 803
y (a¡, a™) If a − k a1, a2, a3l, then we can write
a a™ j a − k a1, a2, a3l − k a1, 0, 0 l 1 k 0, a2, 0 l 1 k 0, 0, a3l
0 a¡i x − a1k1, 0, 0 l 1 a2k 0, 1, 0 l 1 a3k 0, 0, 1 l
2 a − a1 i 1 a2 j 1 a3 k
(a) a=a¡i+a™ j Thus any vector in V3 can be expressed in terms of i, j, and k. For instance,
z k1, 22, 6l − i 2 2j 1 6k
Similarly, in two dimensions, we can write
(a¡, a™, a£) 3 a − k a1, a2 l − a1 i 1 a2 j
a
y See Figure 18 for the geometric interpretation of Equations 3 and 2 and compare with
a¡i a£k Figure 17.
x a™ j Example 5 If a − i 1 2j 2 3k and b − 4i 1 7 k, express the vector 2a 1 3b in
(b) a=a¡i+a™ j+a£k
terms of i, j, and k.
FIGURE 18 SOLUTION Using Properties 1, 2, 5, 6, and 7 of vectors, we have
2a 1 3b − 2si 1 2j 2 3kd 1 3s4i 1 7kd
Gibbs − 2i 1 4j 2 6k 1 12i 1 21k − 14i 1 4j 1 15k ■
Josiah Willard Gibbs (1839–1903), a A unit vector is a vector whose length is 1. For instance, i, j, and k are all unit vec-
professor of mathematical physics tors. In general, if a ± 0, then the unit vector that has the same direction as a is
at Yale College, published the first
book on vectors, Vector Analysis, in 4 u− 1 a− a
1881. More complicated objects,
called quaternions, had earlier been |a| |a|
invented by Hamilton as mathemati-
cal tools for describing space, but | |In order to verify this, we let c − 1y a . Then u − ca and c is a positive scalar, so u has
they weren’t easy for scientists to use.
Quaternions have a scalar part and the same direction as a. Also
a vector part. Gibb’s idea was to use
the vector part separately. Maxwell |u| − |ca| − |c||a| − 1 |a| − 1
and Heaviside had similar ideas, but
Gibb’s approach has proved to be the |a|
most convenient way to study space.
Example 6 Find the unit vector in the direction of the vector 2i 2 j 2 2k.
SOLUTION The given vector has length
| |2i 2 j 2 2k − s22 1 s21d2 1 s22d2 − s9 − 3
so, by Equation 4, the unit vector with the same direction is
1 s2 i 2 j 2 2 kd − 2 i 2 1 j 2 2 k ■
3 3 3 3
Applications
Vectors are useful in many aspects of physics and engineering. In Chapter 13 we will see
how they describe the velocity and acceleration of objects moving in space. Here we look
at forces.
A force is represented by a vector because it has both a magnitude (measured in
pounds or newtons) and a direction. If several forces are acting on an object, the resul
tant force experienced by the object is the vector sum of these forces.
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804 Chapter 12 Vectors and the Geometry of Space
50° 32° Example 7 A 100-lb weight hangs from two wires as shown in Figure 19. Find the
T¡ T™
tensions (forces) T1 and T2 in both wires and the magnitudes of the tensions.
100
50° 32°
T¡ T™
FIGURE 19 100
50° 32° SOLUTION We first express T1 and T2 in terms of their horizontal and vertical compo-
T¡ T™ nents. From Figure 20 we see that
50° 32° | | | |50° 32°
w
5 T1 − 2 TT1¡ cos 50T°™i 1 T1 sin 50° j
FIGURE 20
50° 32°
T2 − T2 cos 32° i 1 T2 sin 32° j
| | | |6
w
The resultant T1 1 T2 of the tensions counterbalances the weight w − 2100 j and so
we must have
T1 1 T2 − 2w − 100 j
Thus
(2| T1 | cos 50° 1 | T2 | cos 32°) i 1 (| T1 | sin 50° 1 | T2 | sin 32°) j − 100 j
Equating components, we get
2| T1 | cos 50° 1 | T2 | cos 32° − 0
| | | | T1 sin 50° 1 T2 sin 32° − 100
| |Solving the first of these equations for T2 and substituting into the second, we get
S D| | | |T1 sin 50° 1 T1 cos 50° sin 32° − 100
cos 32°
| |T1 sin 50° 1 cos 50° sin 32° − 100
cos 32°
So the magnitudes of the tensions are
| |T1 − sin 50° 1 100 < 85.64 lb
tan 32° cos 50°
| | | |and T1 cos 50° <
T2 − cos 32° 64.91 lb
Substituting these values in (5) and (6), we obtain the tension vectors
T1 < 255.05 i 1 65.60 j
T2 < 55.05 i 1 34.40 j ■
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Section 12.2 Vectors 805
1. Are the following quantities vectors or scalars? Explain. | | | | | | 8. I f the vectors in the figure satisfy u − v − 1 and
(a) The cost of a theater ticket u 1 v 1 w − 0, what is w ?
(b) The current in a river
(c) The initial flight path from Houston to Dallas u
(d) The population of the world w
2. W hat is the relationship between the point (4, 7) and the v
vector k 4, 7 l? Illustrate with a sketch.
s9e–g1m4 e nFtinAdlBa. vector AalBwaitnhdrtehpereesqeunitvaatlieonntgrievpernesbeynttahteiodnirsetcatretidnlginaet
3. N ame all the equal vectors in the parallelogram shown. Draw
AB the origin.
E 9. As22, 1d, Bs1, 2d 10. As25, 21d, Bs23, 3d
11. As3, 21d, Bs2, 3d 12. As3, 2d, Bs1, 0d
13. As0, 3, 1d, Bs2, 3, 21d 14. As0, 6, 21d, Bs3, 4, 4d
DC 15–18 Find the sum of the given vectors and illustrate geometrically.
4. Write each combination of vectors as a single vector. 15. k21, 4l, k6, 22l 16. k3, 21l, k21, 5l
(a) AlB 1 BlC (b) ClD 1 DlB 17. k3, 0, 1l, k0, 8, 0l 18. k1, 3, 22l, k0, 0, 6l
(c) DlB 2 AlB (d) DlC 1 ClA 1 AlB
19–22 Find a 1 b, 4a 1 2b, | a |, and | a 2 b |.
AB
19. a − k23, 4l, b − k9, 21l
D 20. a − 5 i 1 3 j, b − 2i 2 2 j
C 21. a − 4 i 2 3 j 1 2 k, b − 2 i 2 4 k
22. a − k8, 1, 24l, b − k5, 22, 1l
5. C opy the vectors in the figure and use them to draw the
following vectors. 23–25 Find a unit vector that has the same direction as the given
vector.
(a) u 1 v (b) u 1 w 23. k6, 22l 24. 25 i 1 3 j 2 k
(c) v 1 w (d) u 2 v 25. 8 i 2 j 1 4 k
(e) v 1 u 1 w (f) u 2 w 2 v
26. Find the vector that has the same direction as k6, 2, 23l but has
uvw length 4.
6. C opy the vectors in the figure and use them to draw the 27–28 What is the angle between the given vector and the positive
following vectors. direction of the x-axis?
27. i 1 s3 j 28. 8 i 1 6 j
(a) a 1 b (b) a 2 b
(c) 21 a (d) 23b | | 29. I f v lies in the first quadrant and makes an angle y3 with the
(e) a 1 2b (f) 2b 2 a positive x-axis and v − 4, find v in component form.
30. I f a child pulls a sled through the snow on a level path with a
ba
force of 50 N exerted at an angle of 388 above the horizontal,
7. I n the figure, the tip of c and the tail of d are both the midpoint find the horizontal and vertical components of the force.
of QR. Express c and d in terms of a and b. 31. A quarterback throws a football with angle of elevation 40° and
speed 60 ftys. Find the horizontal and vertical components of
P the velocity vector.
b
ac R
d
Q
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806 Chapter 12 Vectors and the Geometry of Space
32–33 Find the magnitude of the resultant force and the angle it 38. The tension T at each end of a chain has magnitude 25 N (see
makes with the positive x-axis. the figure). What is the weight of the chain?
32. y 33. y 37° 37°
20 lb 200 N
45° 300 N 60°
0 30° 0
x x
16 lb
39. A boatman wants to cross a canal that is 3 km wide and wants
34. The magnitude of a velocity vector is called speed. Suppose
that a wind is blowing from the direction N45°W at a speed to land at a point 2 km upstream from his starting point. The
of 50 kmyh. (This means that the direction from which the current in the canal flows at 3.5 kmyh and the speed of his boat
wind blows is 45° west of the northerly direction.) A pilot is is 13 kmyh.
steering a plane in the direction N60°E at an airspeed (speed in (a) In what direction should he steer?
still air) of 250 kmyh. The true course, or track, of the plane is
the direction of the resultant of the velocity vectors of the plane (b) How long will the trip take?
and the wind. The ground speed of the plane is the magnitude
of the resultant. Find the true course and the ground speed of 40. Three forces act on an object. Two of the forces are at an angle
the plane. of 100° to each other and have magnitudes 25 N and 12 N. The
third is perpendicular to the plane of these two forces and has
35. A woman walks due west on the deck of a ship at 3 miyh. The magnitude 4 N. Calculate the magnitude of the force that would
ship is moving north at a speed of 22 miyh. Find the speed and exactly counterbalance these three forces.
direction of the woman relative to the surface of the water.
41. F ind the unit vectors that are parallel to the tangent line to the
36. A crane suspends a 500-lb steel beam horizontally by support parabola y − x 2 at the point s2, 4d.
cables (with negligible weight) attached from a hook to each
end of the beam. The support cables each make an angle of 60° 42. (a) Find the unit vectors that are parallel to the tangent line to
with the beam. Find the tension vector in each support cable the curve y − 2 sin x at the point sy6, 1d.
and the magnitude of each tension.
(b) Find the unit vectors that are perpendicular to the tangent
line.
(c) Sketch the curve y − 2 sin x and the vectors in parts (a)
and (b), all starting at sy6, 1d.
43. If A, B, and C are the vertices of a triangle, find
AlB 1 BlC 1 ClA
44. L et C be the point on the line sOelgAm, ben−t AOBlBt,haant disctw−icOelCas, sfahrow
from B as it is from A. If a−
2 1
that c − 3 a 1 3 b.
60° 60° 45. (a) Draw the vectors a − k3, 2l, b − k2, 21l, and c − k7, 1l.
(b) Show, by means of a sketch, that there are scalars s and t
such that c − sa 1 t b.
(c) Use the sketch to estimate the values of s and t.
(d) Find the exact values of s and t.
37. A block-and-tackle pulley hoist is suspended in a warehouse 46. Suppose that a and b are nonzero vectors that are not parallel
by ropes of lengths 2 m and 3 m. The hoist weighs 350 N. The and c is any vector in the plane determined by a and b. Give
ropes, fastened at different heights, make angles of 50° and
38° with the horizontal. Find the tension in each rope and the a geometric argument to show that c can be written as
magnitude of each tension.
c − sa 1 t b for suitable scalars s and t. Then give an argu-
ment using components.
50° 38° | | 47. I f r − kx, y, zl and r0 − kx0, y0, z0l, describe the set of all
2m 3m points sx, y, zd such that r 2 r0 − 1.
| | | | | | 48. If r − kx, yl, r1 − kx1, y1l, and r2 − kx2, y2l, describe the
set of all points sx, yd such that r 2 r1 1 r 2 r2 − k,
where k . r1 2 r2 .
49. F igure 16 gives a geometric demonstration of Property 2 of
vectors. Use components to give an algebraic proof of this
fact for the case n − 2.
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Section 12.3 The Dot Product 807
50. Prove Property 5 of vectors algebraically for the case n − 3. and an array of corner mirrors on the moon, to calculate very
Then use similar triangles to give a geometric proof. precisely the distance from the earth to the moon.)
51. U se vectors to prove that the line joining the midpoints of z
two sides of a triangle is parallel to the third side and half
its length. b y
a
52. Suppose the three coordinate planes are all mirrored and a x
light ray given by the vector a − ka1, a2, a3l first strikes the
xz-plane, as shown in the figure. Use the fact that the angle of
incidence equals the angle of reflection to show that the direc-
tion of the reflected ray is given by b − ka1, 2a2, a3l. Deduce
that, after being reflected by all three mutually perpendicular
mirrors, the resulting ray is parallel to the initial ray. (American
space scientists used this principle, together with laser beams
So far we have added two vectors and multiplied a vector by a scalar. The question arises:
is it possible to multiply two vectors so that their product is a useful quantity? One such
product is the dot product, whose definition follows. Another is the cross product, which
is discussed in the next section.
1 Definition If a − ka1, a2, a3l and b − kb1, b2, b3l, then the dot product of a
and b is the number a ? b given by
a ? b − a1b1 1 a2b2 1 a3b3
Thus, to find the dot product of a and b, we multiply corresponding components and
add. The result is not a vector. It is a real number, that is, a scalar. For this reason, the dot
product is sometimes called the scalar product (or inner product). Although Defini-
tion 1 is given for three-dimensional vectors, the dot product of two-dimensional vectors
is defined in a similar fashion:
ka1, a2l ? kb1, b2l − a1b1 1 a2b2
Example 1 k2, 4l ? k3, 21l − 2s3d 1 4s21d − 2
k21, 7, 4l ? k6, 2, 212 l − s21ds6d 1 7s2d 1 4(212 ) − 6
si 1 2j 2 3kd ? s2j 2 kd − 1s0d 1 2s2d 1 s23ds21d − 7 ■
The dot product obeys many of the laws that hold for ordinary products of real num-
bers. These are stated in the following theorem.
2 Properties of the Dot Product If a, b, and c are vectors in V3 and c is a
scalar, then
| |1. a ? a − a 2 2. a ? b − b ? a
3. a ? sb 1 cd − a ? b 1 a ? c 4. scad ? b − csa ? bd − a ? scbd
5. 0 ? a − 0
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808 Chapter 12 Vectors and the Geometry of Space
These properties are easily proved using Definition 1. For instance, here are the proofs
of Properties 1 and 3:
| |1. a ? a − a12 1 a22 1 a32 − a 2
3. a ? sb 1 cd − ka1, a2, a3l ? kb1 1 c1, b2 1 c2, b3 1 c3l
− a1sb1 1 c1d 1 a2sb2 1 c2d 1 a3sb3 1 c3d
− a1b1 1 a1c1 1 a2b2 1 a2c2 1 a3b3 1 a3c3
− sa1b1 1 a2b2 1 a3b3d 1 sa1c1 1 a2c2 1 a3c3 d
−a?b1a?c
The proofs of the remaining properties are left as exercises. ■
z The dot product a ? b can be given a geometric interpretation in terms of the angle
B between a and b, which is defined to be the angle between the representations of a and
b
¨ a-b b that start at tOhleAoarnigdinO,lBwhinereFi0gu<re1<. No.teInthoattheifr words, is the angle between the
O a line segments a and b are parallel vectors, then
x
− 0 or − .
FIGURE 1 A The formula in the following theorem is used by physicists as the definition of the dot
product.
y
3 Theorem If is the angle between the vectors a and b, then
a ? b − | a | | b | cos
Proof If we apply the Law of Cosines to triangle OAB in Figure 1, we get
4 | AB |2 − | OA |2 1 | OB |2 2 2 | OA | | OB | cos
| | | | | | | | | | | |(Observe that the Law of Cosines still applies in the limiting cases when − 0 or , or
a − 0 or b − 0.) But OA − a , OB − b , and AB − a 2 b , so Equation 4
becomes
5 | a 2 b |2 − | a |2 1 | b |2 2 2 | a | | b | cos
Using Properties 1, 2, and 3 of the dot product, we can rewrite the left side of this
equation as follows:
| |a 2 b 2 − sa 2 bd ? sa 2 bd
−a?a2a?b2b?a1b?b
− | a |2 2 2a ? b 1 | b |2
Therefore Equation 5 gives
| a |2 2 2a ? b 1 | b |2 − | a |2 1 | b |2 2 2 | a | | b | cos
Thus 22a ? b − 22 | a | | b | cos
or a ? b − | a | | b | cos ■
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Section 12.3 The Dot Product 809
Example 2 If the vectors a and b have lengths 4 and 6, and the angle between them is
y3, find a ? b.
SOLUTION Using Theorem 3, we have
| || | b cossy3d 1
a?b− a − 4 ? 6 ? 2 − 12 ■
The formula in Theorem 3 also enables us to find the angle between two vectors.
6 Corollary If is the angle between the nonzero vectors a and b, then
cos − a?b
|a||b|
Example 3 Find the angle between the vectors a − k2, 2, 21l and b − k5, 23, 2 l.
SOLUTION Since
| | | |a − s22 1 22 1 s21d2 − 3 and b − s52 1 s23d2 1 22 − s38
and since
a ? b − 2s5d 1 2s23d 1 s21ds2d − 2
we have, from Corollary 6,
cos − a?b − 2
3 s38
|a||b|
So the angle between a and b is
S D
− cos21 2 < 1.46 sor 84°d ■
3 s38
Two nonzero vectors a and b are called perpendicular or orthogonal if the angle
between them is − y2. Then Theorem 3 gives
a ? b − | a | | b | cossy2d − 0
and conversely if a ? b − 0, then cos − 0, so − y2. The zero vector 0 is considered
to be perpendicular to all vectors. Therefore we have the following method for determin-
ing whether two vectors are orthogonal.
7 Two vectors a and b are orthogonal if and only if a ? b − 0.
Example 4 Show that 2i 1 2j 2 k is perpendicular to 5i 2 4j 1 2k. ■
SOLUTION Since
s2i 1 2j 2 kd ? s5i 2 4j 1 2kd − 2s5d 1 2s24d 1 s21ds2d − 0
these vectors are perpendicular by (7).
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810 Chapter 12 Vectors and the Geometry of Space
a¨ b a · b>0 Because cos . 0 if 0 < , y2 and cos , 0 if y2 , < , we see that
¨ acute a ? b is positive for , y2 and negative for . y2. We can think of a ? b as mea-
ab a · b=0 suring the extent to which a and b point in the same direction. The dot product a ? b is
¨=π/ 2 positive if a and b point in the same general direction, 0 if they are perpendicular, and
¨b a · b<0 negative if they point in generally opposite directions (see Figure 2). In the extreme case
a ¨ obtuse where a and b point in exactly the same direction, we have − 0, so cos − 1 and
FIGURE 2 a ? b − |a||b|
| | | |If a and b point in exactly opposite directions, then we have − and so cos − 21
and a ? b − 2 a b .
TEC Visual 12.3A shows an anima- Direction Angles and Direction Cosines
tion of Figure 2.
The direction angles of a nonzero vector a are the angles ␣, , and ␥ (in the interval
z f0, gd that a makes with the positive x-, y-, and z-axes, respectively. (See Figure 3.)
The cosines of these direction angles, cos ␣, cos , and cos ␥, are called the direction
cosines of the vector a. Using Corollary 6 with b replaced by i, we obtain
8 cos ␣ − a?i − a1
ç a |a||i| |a|
a¡ ∫
(This can also be seen directly from Figure 3.)
å y Similarly, we also have
x | | | |9 a2 a3
cos  − a cos ␥ − a
FIGURE 3
By squaring the expressions in Equations 8 and 9 and adding, we see that
10 cos2␣ 1 cos2 1 cos2␥ − 1
We can also use Equations 8 and 9 to write
a − k a1, a2, a3 l − k| a| cos ␣, |a| cos , |a| cos ␥l
| |− a kcos ␣, cos , cos ␥l
Therefore
| |11 1 a − k cos ␣, cos , cos ␥ l
a
which says that the direction cosines of a are the components of the unit vector in the
direction of a.
Example 5 Find the direction angles of the vector a − k 1, 2, 3 l.
| |SOLUTION Since a − s1 2 1 2 2 1 3 2 − s14 , Equations 8 and 9 give
cos ␣ − 1 cos  − 2 cos ␥ − 3
s14 s14 s14
S D S D S Dand so1 2 3
␣ − cos21 s14 < 74°  − cos21 s14 < 58° ␥ − cos21 s14 < 37°
■
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 12.3 The Dot Product 811
TEC Visual 12.3B shows how Fig- Projections
ure 4 changes when we vary a and b. FpveoigcinutotrrePw4. iIstfhhSorweispsrterhesepenrfeotasoettinootnaftPtilhoSenispsePclrQaplelaennddditchPulelRavroefcfrttoowmropRvreotcojteotchrtseioalninaoenfdcbobnowtnatiiotnhiantghaenPdlsQaims, tdeheeinnnoitttiehadel
by proja b. (You can think of it as a shadow of b).
R
| |The scalar projection of b onto a (also called the component of b along a) is defined
b
to be the signed magnitude of the vector projection, which is the number b cos ,
a where is the angle between a and b. (See Figure 5.) This is denoted by compa b.
Observe that it is negative if y2 , < . The equation
PS Q
proja b
R a ? b − | a || b | cos − | a |(| b | cos )
b shows that the dot product of a and b can be interpreted as the length of a times the sca-
lar projection of b onto a. Since
a
a?b a
SP Q |b| cos − − ? b
proja b |a| |a|
FIGURE 4 the component of b along a can be computed by taking the dot product of b with the unit
Vector projections vector in the direction of a. We summarize these ideas as follows.
R Scalar projection of b onto a: | |compa b− a?b
Vector projection of b onto a: a
b S Dproja b −
a a?b a − a?b a
¨ SQ
P ͉ b ͉ cos ¨ =compa b |a| |a| | a |2
FIGURE 5 Notice that the vector projection is the scalar projection times the unit vector in the direc-
Scalar projection tion of a.
Example 6 Find the scalar projection and vector projection of b − k 1, 1, 2 l
onto a − k 22, 3, 1 l.
| |SOLUTION Since a − ss22d2 1 3 2 1 12 − s14 , the scalar projection of b onto a
is
a?b s22ds1d 1 3s1d 1s2d 3
| |compa b − a − s14 1 − s14
The vector projection is this scalar projection times the unit vector in the direction of a:
K L| |
proja b − 3 a − 3 a − 2 3 , 9 , 3 ■
s14 a 14 7 14 14
One use of projections occurs in physics in calculating work. In Section 6.4 we
R defined the work done by a constant force F in moving an object through a distance d as
F
W − Fd, but this applies only when the force is directed along the PllinRepoofinmtiontgioinnosfotmhee
object. Suppose, however, that the constant force is a vector F −
¨ S Q other direction, as in DFig−urPelQ6.. If the force moves the object from P to Q, then the dis-
P D placement vector is The work done by this force is defined to be the product
FIGURE 6 of the component of the force along D and the distance moved:
W − s| F | cos d | D |
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
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812 Chapter 12 Vectors and the Geometry of Space
But then, from Theorem 3, we have
12 W − | F | | D | cos − F ? D
Thus the work done by a constant force F is the dot product F ? D, where D is the dis
placement vector.
35° Example 7 A wagon is pulled a distance of 100 m along a horizontal path by a
F constant force of 70 N. The handle of the wagon is held at an angle of 358 above the
35°
horizontal. Find the work done by the force.
D
SOLUTION If F and D are the force and displacement vectors, as pictured in Figure 7,
FIGURE 7 then the work done is
W − F ? D − | F | | D | cos 35°
− s70ds100d cos 35° < 5734 N ∙m − 5734 J ■
Example 8 A force is given by a vector F − 3i 1 4j 1 5k and moves a particle from
the point Ps2, 1, 0d to the point Qs4, 6, 2d. Find the work done.
SOLUTION The displacement vector is D − PlQ − k 2, 5, 2 l, so by Equation 12, the
work done is
W − F ? D − k 3, 4, 5 l ? k 2, 5, 2 l
− 6 1 20 1 10 − 36
If the unit of length is meters and the magnitude of the force is measured in newtons,
then the work done is 36 J. ■
1. W hich of the following expressions are meaningful? Which are 11–12 If u is a unit vector, find u ? v and u ? w.
11. 12. u
meaningless? Explain.
(a) sa ? bd ? c (b) sa ? bdc uv v
w
| | (c) a sb ? cd (d) a ? sb 1 cd
| | (e) a ? b 1 c (f) a ? sb 1 cd
2–10 Find a ? b. w
2. a − k5, 22l, b − k3, 4l
3. a − k1.5, 0.4l, b − k24, 6l
4. a − k 6, 22, 3l, b − k 2, 5, 21 l 13. (a) Show that i ? j − j ? k − k ? i − 0.
(b) Show that i ? i − j ? j − k ? k − 1.
5. a − k 4, 1, 1 l , b − k 6, 23, 28 l
4 14. A street vendor sells a hamburgers, b hot dogs, and c soft
drinks on a given day. He charges $4 for a hamburger, $2.50
6. a − k p, 2p, 2pl, b − k 2q, q, 2q l for a hot dog, and $1 for a soft drink. If A − k a, b, c l and
P − k 4, 2.5, 1 l, what is the meaning of the dot product A ? P ?
7. a − 2 i 1 j, b − i 2 j 1 k
15–20 Find the angle between the vectors. (First find an exact
8. a − 3 i 1 2 j 2 k, b − 4 i 1 5 k expression and then approximate to the nearest degree.)
| | | | 9. a − 7, b − 4, the angle between a and b is 30° 15. a − k 4, 3 l, b − k 2, 21 l
| | | | 10. a − 80, b − 50, the angle between a and b is 3y4 16. a − k 22, 5 l, b − k 5, 12 l
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Section 12.3 The Dot Product 813
17. a − k1, 24, 1l, b − k0, 2, 22l 38. I f a vector has direction angles ␣ − y4 and  − y3, find the
18. a − k21, 3, 4l, b − k5, 2, 1l third direction angle ␥.
19. a − 4i 2 3j 1 k, b − 2i 2 k
20. a − 8 i 2 j 1 4 k, b − 4 j 1 2 k 39–44 Find the scalar and vector projections of b onto a.
39. a − k 25, 12 l, b − k 4, 6 l
21–22 Find, correct to the nearest degree, the three angles of the
triangle with the given vertices. 40. a − k 1, 4 l, b − k 2, 3 l
21. Ps2, 0d, Qs0, 3d, Rs3, 4d
22. As1, 0, 21d, Bs3, 22, 0d, Cs1, 3, 3d 41. a − k4, 7, 24l, b − k3, 21, 1l
23–24 Determine whether the given vectors are orthogonal, 42. a − k21, 4, 8l, b − k12, 1, 2l
parallel, or neither.
23. (a) a − k9, 3l, b − k22, 6l 43. a − 3 i 2 3 j 1 k, b − 2 i 1 4 j 2 k
(b) a − k4, 5, 22l, b − k3, 21, 5l
(c) a − 28 i 1 12 j 1 4 k, b − 6 i 2 9 j 2 3 k 44. a − i 1 2 j 1 3 k, b − 5 i 2 k
(d) a − 3 i 2 j 1 3 k, b − 5 i 1 9 j 2 2 k
24. (a) u − k25, 4, 22l, v − k3, 4, 21l 45. S how that the vector ortha b − b 2 proja b is orthogonal to a.
(b) u − 9 i 2 6 j 1 3 k, v − 26 i 1 4 j 2 2 k (It is called an orthogonal projection of b.)
(c) u − kc, c, cl, v − kc, 0, 2cl
46. F or the vectors in Exercise 40, find ortha b and illustrate by
25. Use vectors to decide whether the triangle with vertices drawing the vectors a, b, proja b, and ortha b.
Ps1, 23, 22d, Qs2, 0, 24d, and Rs6, 22, 25d is right-angled.
47. If a − k 3, 0, 21 l, find a vector b such that compa b − 2.
26. Find the values of x such that the angle between the vectors
k 2, 1, 21 l, and k 1, x, 0 l is 458. 48. Suppose that a and b are nonzero vectors.
(a) Under what circumstances is compa b − compb a?
27. Find a unit vector that is orthogonal to both i 1 j and i 1 k. (b) Under what circumstances is proja b − projb a?
28. Find two unit vectors that make an angle of 608 with
49. Find the work done by a force F − 8 i 2 6 j 1 9 k that moves
v − k 3, 4 l. an object from the point s0, 10, 8d to the point s6, 12, 20d along
a straight line. The distance is measured in meters and the force
29–30 Find the acute angle between the lines. in newtons.
29. 2x 2 y − 3, 3x 1 y − 7
30. x 1 2y − 7, 5x 2 y − 2 50. A tow truck drags a stalled car along a road. The chain makes
an angle of 30° with the road and the tension in the chain is
31–32 Find the acute angles between the curves at their points of 1500 N. How much work is done by the truck in pulling the
intersection. (The angle between two curves is the angle between car 1 km?
their tangent lines at the point of intersection.)
31. y − x 2, y − x 3 51. A sled is pulled along a level path through snow by a rope.
32. y − sin x, y − cos x, 0 < x < y2 A 30-lb force acting at an angle of 40° above the horizontal
moves the sled 80 ft. Find the work done by the force.
33–37 Find the direction cosines and direction angles of the vector.
(Give the direction angles correct to the nearest degree.) 52. A boat sails south with the help of a wind blowing in the direc
33. k 2, 1, 2 l 34. k 6, 3, 22 l tion S36°E with magnitude 400 lb. Find the work done by the
35. i 2 2 j 2 3k 36. 21 i 1 j 1 k wind as the boat moves 120 ft.
37. k c, c, c l, where c . 0
53. Use a scalar projection to show that the distance from a point
P1sx1, y1d to the line ax 1 by 1 c − 0 is
| |a x1 1 by1 1 c
sa 2 1 b 2
Use this formula to find the distance from the point s22, 3d to
the line 3x 2 4y 1 5 − 0.
54. I f r − kx, y, z l, a − ka1, a2, a3 l, and b − kb1, b2, b3l, show
that the vector equation sr 2 ad ? sr 2 bd − 0 represents a
sphere, and find its center and radius.
55. F ind the angle between a diagonal of a cube and one of its
edges.
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814 Chapter 12 Vectors and the Geometry of Space
56. F ind the angle between a diagonal of a cube and a diagonal of 60. S uppose that all sides of a quadrilateral are equal in length and
one of its faces. opposite sides are parallel. Use vector methods to show that the
diagonals are perpendicular.
57. A molecule of methane, CH4, is structured with the four hydro
gen atoms at the vertices of a regular tetrahedron and the car 61. Use Theorem 3 to prove the Cauchy-Schwarz Inequality:
bon atom at the centroid. The bond angle is the angle formed |a ? b| < |a||b|
by the H—C—H combination; it is the angle between the lines 62. The Triangle Inequality for vectors is
fthat join the carbon atom to two of the hydrogen atoms. Show |a 1 b| < |a| 1 |b|
that the bond angle is about 109.5°. Hint: Take the vertices of (a) Give a geometric interpretation of the Triangle Inequality.
the tetrahedron to be the points s1, 0, 0d, s0, 1, 0d, s0, 0, 1d, and (b) Use the Cauchy-Schwarz Inequality from Exercise 61 to
gs1, 1d, ( 1 1 1 ).
1, as shown in the figure. Then the centroid is 2 , 2 , 2 prove the Triangle Inequality. [Hint: Use the fact that
z | |a 1 b 2 − sa 1 bd ∙ sa 1 bd and use Property 3 of the
H
dot product.]
CH H
y 63. The Parallelogram Law states that
xH | a 1 b |2 1 | a 2 b |2 − 2 | a |2 1 2 | b |2
| | | | 58. I f c − a b 1 b a, where a, b, and c are all nonzero vectors, (a) Give a geometric interpretation of the Parallelogram Law.
show that c bisects the angle between a and b. (b) Prove the Parallelogram Law. (See the hint in Exercise 62.)
59. Prove Properties 2, 4, and 5 of the dot product (Theorem 2).
64. Show that if u 1 v and u 2 v are orthogonal, then the vectors
u and v must have the same length.
65. If is the angle between vectors a and b, show that
proja b ? projb a − sa ? bd cos2
Given two nonzero vectors a − k a1, a2, a3 l and b − k b1, b2, b3 l, it is very useful to be
able to find a nonzero vector c that is perpendicular to both a and b, as we will see in the
next section and in Chapters 13 and 14. If c − kc1, c2, c3 l is such a vector, then a ? c − 0
and b ? c − 0 and so
1 a1c1 1 a2c2 1 a3c3 − 0
2 b1c1 1 b2c2 1 b3c3 − 0
To eliminate c3 we multiply (1) by b3 and (2) by a3 and subtract:
3 sa1b3 2 a3b1dc1 1 sa2b3 2 a3b2dc2 − 0
Equation 3 has the form pc1 1 qc2 − 0, for which an obvious solution is c1 − q and
c2 − 2p. So a solution of (3) is
c1 − a2b3 2 a3b2 c2 − a3b1 2 a1b3
Substituting these values into (1) and (2), we then get
c3 − a1b2 2 a2b1
This means that a vector perpendicular to both a and b is
k c1, c2, c3 l − k a2b3 2 a3b2, a3b1 2 a1b3, a1b2 2 a2b1 l
The resulting vector is called the cross product of a and b and is denoted by a 3 b.
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Section 12.4 The Cross Product 815
Hamilton 4 Definition If a − k a1, a2, a3 l and b − k b1, b2, b3 l, then the cross product
of a and b is the vector
The cross product was invented by
the Irish mathematician Sir William a 3 b − k a2b3 2 a3b2, a3b1 2 a1b3, a1b2 2 a2b1l
Rowan Hamilton (1805–1865), who
had created a precursor of vectors, Notice that the cross product a 3 b of two vectors a and b, unlike the dot product, is
called quaternions. When he was five
years old Hamilton could read Latin, a vector. For this reason it is also called the vector product. Note that a 3 b is defined
Greek, and Hebrew. At age eight he
added French and Italian and when only when a and b are three-dimensional vectors.
ten he could read Arabic and Sanskrit.
At the age of 21, while still an under- In order to make Definition 4 easier to remember, we use the notation of determinants.
graduate at Trinity College in Dublin,
Hamilton was appointed Professor Z ZA determinant of order 2 is defined byab − ad 2 bc
of Astronomy at the university and c d
Royal Astronomer of Ireland!
(Multiply across the diagonals and subtract.) For example,
Z Z2
26 1 − 2s4d 2 1s26d − 14
4
A determinant of order 3 can be defined in terms of second-order determinants as
Z Z Z Z Z Z Z Za1
follows: b1 a2 a3 − a1 b2 b3 2 a2 b1 b3 1 a3 b1 b2
c1 b2 b3 c2 c3 c1 c3 c1 c2
c2 c3
5
Observe that each term on the right side of Equation 5 involves a number ai in the first
row of the determinant, and ai is multiplied by the second-order determinant obtained
from the left side by deleting the row and column in which ai appears. Notice also the
Z Z Z Z Z Z Z Zminus sign in the second term. For example,12210131 3 0
3 0 1 −1 4 2 22 25 2 1 s21d 25 4
25 4 2
− 1s0 2 4d 2 2s6 1 5d 1 s21ds12 2 0d − 238
If we now rewrite Definition 4 using second-order determinants and the standard basis
vectors i, j, and k, we see that the cross product of the vectors a − a1 i 1 a2 j 1 a3 k and
b − b1 i 1 b2 j 1 b3 k is
Z Z Z Z Z Z6 a2 a3 a1 a3 a1 a2
a3b− b2 b3 i2 b1 b3 j1 b1 b2 k
In view of the similarity between Equations 5 and 6, we often write
Z Z i j k
7 a 3 b − a1 a2 a3
b1 b2 b3
Although the first row of the symbolic determinant in Equation 7 consists of vectors, if
we expand it as if it were an ordinary determinant using the rule in Equation 5, we obtain
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
816 Chapter 12 Vectors and the Geometry of Space
Equation 6. The symbolic formula in Equation 7 is probably the easiest way of remem
bering and computing cross products.
Z ZExample 1 If a − k1, 3, 4l and b − k2, 7, 25l, then
ij k
a3b− 1 3 4
2 7 25
Z Z Z Z Z Z34i21 4 j1 1 3 k
25 2 25 2 7
−7
− s215 2 28d i 2 s25 2 8d j 1 s7 2 6d k − 243i 1 13j 1 k ■
Example 2 Show that a 3 a − 0 for any vector a in V3.
Z ZSOLUTION If a − k a1, a2, a3 l, then
ijk
a 3 a − a1 a2 a3
a1 a2 a3
− sa2a3 2 a3a2d i 2 sa1a3 2 a3a1d j 1 sa1a2 2 a2a1d k
− 0i 2 0j 1 0k − 0 ■
We constructed the cross product a 3 b so that it would be perpendicular to both a
and b. This is one of the most important properties of a cross product, so let’s emphasize
and verify it in the following theorem and give a formal proof.
8 Theorem The vector a 3 b is orthogonal to both a and b.
Proof In order to show that a 3 b is orthogonal to a, we compute their dot product as
follows:Zsa 3 bd ? a − Z Za3 Z Za3 Za2
a2 b3 a1 2 a1 b3 a2 1 a1 b2 a3
b2 b1 b1
axb − a1sa2b3 2 a3b2 d 2 a2sa1b3 2 a3b1d 1 a3sa1b2 2 a2b1d
− a1a2b3 2 a1b2a3 2 a1a2b3 1 b1a2a3 1 a1b2a3 2 b1a2a3
n −0
ab A similar computation shows that sa 3 bd ? b − 0. Therefore a 3 b is orthogonal to
¨ both a and b.
■
FIGURE 1 If a and b are represented by directed line segments with the same initial point (as
The right-hand rule gives the direction in Figure 1), then Theorem 8 says that the cross product a 3 b points in a direction
perpendicular to the plane through a and b. It turns out that the direction of a 3 b is
of a 3 b.
given by the right-hand rule: if the fingers of your right hand curl in the direction of
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 12.4 The Cross Product 817
a rotation (through an angle less than 180°) from a to b, then your thumb points in the
direction of a 3 b.
| |Now that we know the direction of the vector a 3 b, the remaining thing we need to
complete its geometric description is its length a 3 b . This is given by the following
theorem.
9 Theorem If is the angle between a and b (so 0 < < ), then
| a 3 b | − | a | | b | sin
TEC Visual 12.4 shows how a 3 b Proof From the definitions of the cross product and length of a vector, we have
changes as b changes.
| | a 3 b 2 − sa2b3 2 a3b2d2 1 sa3b1 2 a1b3d2 1 sa1b2 2 a2b1d2
− a 2 b32 2 2a2a3b2b3 1 a32b 2 1 a32 b12 2 2a1a3b1b3 1 a 2 b32
2 2 1
1 a 12b 2 2 2a1 a2 b1b2 1 a 2 b12
2 2
− sa12 1 a 2 1 a32dsb12 1 b 2 1 b32d 2 sa1b1 1 a2b2 1 a3b3d2
2 2
− | a |2| b |2 2 sa ? bd2
| | | | | | | | − a 2 b 2 2 a 2 b 2 cos2 (by Theorem 12.3.3)
| | | | − a 2 b 2 s1 2 cos2d
− | a |2| b |2 sin2
Taking square roots and observing that ssin 2 − sin because sin > 0 when ■
0 < < , we have
| a 3 b | − | a | | b | sin
Geometric characterization of a 3 b Since a vector is completely determined by its magnitude and direction, we can now
| | | |say that a 3 b is the vector that is perpendicular to both a and b, whose orientation is
determined by the right-hand rule, and whose length is a b sin . In fact, that is
exactly how physicists define a 3 b.
10 Corollary Two nonzero vectors a and b are parallel if and only if
a3b−0
| |Proof Two nonzero vectors a and b are parallel if and only if − 0 or . In either
case sin − 0, so a 3 b − 0 and therefore a 3 b − 0. ■
b ͉ b͉ sin ¨ The geometric interpretation of Theorem 9 can be seen by looking at Figure 2. If a
¨a | | | |and b are represented by directed line segments with the same initial point, then they
FIGURE 2 determine a parallelogram with base a , altitude b sin , and area
A − | a| (| b | sin ) − | a 3 b |
Thus we have the following way of interpreting the magnitude of a cross product.
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818 Chapter 12 Vectors and the Geometry of Space
The length of the cross product a 3 b is equal to the area of the parallelogram
determined by a and b.
Example 3 Find a vector perpendicular to the plane that passes through the points
Ps1, 4, 6d, Qs22, 5, 21d, and Rs1, 21, 1d.
SOLUTION The vector PlQ 3 PlR is perpendicular to both PlQ and PlR and is therefore
perpendicular to the plane through P, Q, and R. We know from (12.2.1) that
PlQ − s22 2 1d i 1 s5 2 4d j 1 s21 2 6d k − 23i 1 j 2 7k
PlR − s1 2 1d i 1 s21 2 4d j 1 s1 2 6d k − 25 j 2 5k
Z ZWe compute the cross product of these vectors: i jk
PlQ 3 PlR − 23 1 27
0 25 25
− s25 2 35d i 2 s15 2 0d j 1 s15 2 0d k − 240 i 2 15 j 1 15k
So the vector k240, 215, 15 l is perpendicular to the given plane. Any nonzero scalar
multiple of this vector, such as k28, 23, 3 l, is also perpendicular to the plane.
■
Example 4 Find the area of the triangle with vertices Ps1, 4, 6d, Qs22, 5, 21d,
and Rs1, 21, 1d.
SOLUTION In Example 3 we computed that PlQ 3 PlR − k240, 215, 15 l. The area of
the parallelogram with adjacent sides PQ and PR is the length of this cross product:
| |PlQ 3 PlR − ss240d2 1 s215d2 1 152 − 5s82
The area A of the triangle PQR is half the area of this parallelogram, that is, 5 s82 . ■
2
If we apply Theorems 8 and 9 to the standard basis vectors i, j, and k using − y2,
we obtain
i 3 j − k j 3 k − i k 3 i − j
j 3 i − 2k k 3 j − 2i i 3 k − 2j
Observe that
i3j±j3i
Thus the cross product is not commutative. Also
whereas i 3 si 3 jd − i 3 k − 2j
si 3 id 3 j − 0 3 j − 0
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