Solução_Calculo_Stewart_6e

F.
124 ¤ CHAPTER 3 DIFFERENTIATION RULES
TX.10
17. (a) y =tanx ⇒ dy =sec 2 xdx
(b) When x = π/4 and dx = −0.1, dy =[sec(π/4)] 2 (−0.1) = √ 2 2
(−0.1) = −0.2.
19. y = f(x) =2x − x 2 , x =2, ∆x = −0.4 ⇒
∆y = f(1.6) − f(2) = 0.64 − 0=0.64
dy =(2− 2x) dx =(2− 4)(−0.4) = 0.8
21. y = f(x) =2/x, x =4, ∆x =1 ⇒
∆y = f(5) − f(4) = 2 5 − 2 4 = −0.1
dy = − 2 x dx = − 2 (1) = −0.125
2 42 23. To estimate (2.001) 5 ,we’llfind the linearization of f(x) =x 5 at a =2.Sincef 0 (x) =5x 4 , f(2) = 32,andf 0 (2) = 80,
we have L(x) = 32 + 80(x − 2) = 80x − 128. Thus, x 5 ≈ 80x − 128 when x is near 2 ,so
(2.001) 5 ≈ 80(2.001) − 128 = 160.08 − 128 = 32.08.
25. To estimate (8.06) 2/3 ,we’llfind the linearization of f(x) =x 2/3 at a =8.Sincef 0 (x) = 2 3 x−1/3 =2/ 3 3√
x ,
f(8) = 4, andf 0 (8) = 1 ,wehaveL(x) 3 =4+1 (x − 8) = 1 x + 4 . Thus, 3 3 3 x2/3 ≈ 1 x + 4 when x is near 8, so
3 3
(8.06) 2/3 ≈ 1 3 (8.06) + 4 3 = 12.06
3
=4.02.
27. y = f(x) =tanx ⇒ dy =sec 2 xdx.Whenx =45 ◦ and dx = −1 ◦ ,
dy =sec 2 45 ◦ (−π/180) = √ 2 2
(−π/180) = −π/90,sotan 44 ◦ = f(44 ◦ ) ≈ f(45 ◦ )+dy =1− π/90 ≈ 0.965.
29. y = f(x) =secx ⇒ f 0 (x) =secx tan x, sof(0) = 1 and f 0 (0) = 1 · 0=0. The linear approximation of f at 0 is
f(0) + f 0 (0)(x − 0) = 1 + 0(x) =1.Since0.08 is close to 0, approximating sec 0.08 with 1 is reasonable.
31. y = f(x) =lnx ⇒ f 0 (x) =1/x, sof(1) = 0 and f 0 (1) = 1. The linear approximation of f at 1 is
f(1) + f 0 (1)(x − 1) = 0 + 1(x − 1) = x − 1. Nowf(1.05) = ln 1.05 ≈ 1.05 − 1=0.05, so the approximation
is reasonable.
33. (a) If x is the edge length, then V = x 3 ⇒ dV =3x 2 dx. Whenx =30and dx =0.1, dV =3(30) 2 (0.1) = 270,sothe
maximum possible error in computing the volume of the cube is about 270 cm 3 . The relative error is calculated by dividing
the change in V , ∆V ,byV . We approximate ∆V with dV .
Relative error = ∆V
V ≈ dV V = 3x2 dx
=3 dx 0.1
x 3 x =3 =0.01.
30
Percentage error = relative error × 100% = 0.01 × 100% = 1%.