Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 11.11 APPLICATIONS OF TAYLOR POLYNOMIALS ¤ 501
35. (a) If the water is deep, then 2πd/L is large, and we know that tanh x → 1 as x →∞. So we can approximate
tanh(2πd/L) ≈ 1,andsov 2 ≈ gL/(2π) ⇔ v ≈ gL/(2π).
(b) From the table, the first term in the Maclaurin series of
tanh x is x,soifthewaterisshallow,wecanapproximate
tanh 2πd
L
≈ 2πd
L ,andsov2 ≈ gL
2π · 2πd
L
⇔
v ≈ √ gd.
n f (n) (x) f (n) (0)
0 tanh x 0
1 sech 2 x 1
2 −2sech 2 x tanh x 0
3 2sech 2 x (3 tanh 2 x − 1) −2
(c) Since tanh x is an odd function, its Maclaurin series is alternating, so the error in the approximation
tanh 2πd
L
≈ 2πd
L is less than the first neglected term, which is |f 000 3
(0)| 2πd
= 1 3 2πd
.
3! L 3 L
If L>10d, then 1 3 2πd
< 1 3
1
2π · = π3
3 L 3 10 375 , so the error in the approximation v2 = gd is less
than gL
2π · π 3
375 ≈ 0.0132gL.
37. (a) L is the length of the arc subtended by the angle θ,soL = Rθ ⇒
θ = L/R. Nowsec θ =(R + C)/R ⇒ R sec θ = R + C ⇒
C = R sec θ − R = R sec(L/R) − R.
(b) Extending the result in Exercise 17, we have f (4) (x) =secx (18 sec 2 x tan 2 x +6sec 4 x − sec 2 x − tan 2 x),
so f (4) (0) = 5, andsec x ≈ T 4(x) =1+ 1 2 x2 + 5 24 x4 . By part (a),
C ≈ R 1+ 1 2 L
+ 5 4 L
− R = R + 1 2 R 24 R
2 R · L2
R + 5 2 24 R · L4
R − R = L2
4 2R + 5L4
24R . 3
(c) Taking L =100km and R = 6370 km, the formula in part (a) says that
C = R sec(L/R) − R = 6370 sec(100/6370) − 6370 ≈ 0.785 009 965 44 km.
The formula in part (b) says that C ≈ L2
2R + 5L4
24R = 1002
3 2 · 6370 + 5 · 1004 ≈ 0.785 009 957 36 km.
24 · 63703 The difference between these two results is only 0.000 000 008 08 km, or 0.000 008 08 m!
39. Using f(x) =T n (x)+R n (x) with n =1and x = r,wehavef(r) =T 1 (r)+R 1 (r),whereT 1 is the first-degree Taylor
polynomial of f at a. Because a = x n , f(r) =f(x n )+f 0 (x n )(r − x n )+R 1 (r). Butr is a root of f,sof(r) =0
and we have 0=f(x n )+f 0 (x n )(r − x n )+R 1 (r). Takingthefirst two terms to the left side gives us
f 0 (x n )(x n − r) − f(x n )=R 1 (r). Dividing by f 0 (x n ),wegetx n − r − f(xn)
f 0 (x = R1(r) . By the formula for Newton’s
n) f 0 (x n) method, the left side of the preceding equation is x n+1 − r, so|x n+1 − r| =
R 1(r)
f 0 (x n ) . Taylor’s Inequality gives us
|R 1 (r)| ≤ |f 00 (r)|
2!
|x n+1 − r| ≤ M 2K |x n − r| 2 .
|r − x n | 2 . Combining this inequality with the facts |f 00 (x)| ≤ M and |f 0 (x)| ≥ K gives us