Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

SECTION 11.11 APPLICATIONS OF TAYLOR POLYNOMIALS D 95 ·
1 3 1 s Bth I · s ·
27. sinx = x- 3!x +Six - · · ·. y e A ternatmg enes
Estimation Theorem, the error in the approximation
sin x = x - ~x 3 is less than I ;,x 5 1 < 0.01 ¢:}
lx 5 l < 120(0.01) ¢:} lxl < (1.2) 115 ~ 1.037. The curves
y = x - ix 3 andy= sinx- 0.01 intersect at x ~ 1.043, so
the graph confirms our estimate. Since both the sine function
0.9
r-------------~~~~--~
y=sinx+ o.oV /
/ / y=x-kx 3
/ ~ 'fooox-O.QJ
0.9 "-L,__L,.:...__ _.__________./ 1.2
0.8
and the given approximation are odd functions, we need to check the esti~ate only for x > 0. Thus, the desired range of
. values for xis - 1.037 < x < 1.037.
x 3 x 5 x 7
29. arctan x = x - 3
+
5 - 7
+ · · · . By the Alternating Series
Estimation Theorem, the error is less than l -~ x 7 1 < 0.05 ¢:}
y =arctan x + 0.05
jx 7 j < 0.35 ¢:} lxl < (0.35? 17 ~ 0.8607. The curves
y = x- tx 3 + kx 5 andy= arctanx + 0.05 intersect at
x ~ 0.9245, so the graph confirms our estimate. Since both the
arctangent function and the given approximation are odd functions,
we need to check the estimate only for x > 0. Thus, the desired
range of values for xis - 0.86 < x < 0.86.
- 1
y = x - .!. x 3 + .!. xs
3 5
31. Let s(t) be the position function of the car, and for convenience set s(O) = 0. The velocity of the car is v(t) = s' (t) and th~
acceleration is a(t) = s" (t), so the second degree Taylor Polynomial is T2(t) = s(O) + v(O)t + a~O) t 2 = 20t + t 2 . We
estimate the distance traveled during the next second to be s(1) ~ T2(1) = 20 + 1 =21m. The function T 2 (t) would not be
accurate over a full minute, since the car could not possibly maintain an acceleration of 2 m/s 2 for that long (if it did, its final
speed would be 140 m/ s ~ 313 mi/ hl).
33. E = ~2 - (D!d)2 = ~2- D2(1:d/ D)2 = ; 2 [1 - (1+ ~r2l
We use the Binomial Series to expand (1 + d/ D ) - 2 :.
when D is much larger than d; that is, when P is far away from the dipole.
35. (a) If the water is deep, then 27rd/ L is large, and we know that tanh x -+ 1 as x -+ oo. So we can approximate
tanh(27rd/ L) ~ 1, andsov 2 ~ f!L/ (27r) ¢:} v ~ ..jgLj (21r).
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