Solução_Calculo_Stewart_6e

F.
204 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
(b) If W/L is large, the bird would flytoapointC that is closer to B than to D to minimize the energy used flying over water.
If W/L is small, the bird would flytoapointC that is closer to D than to B to minimize the distance of the flight.
E = W √ 25 + x 2 + L(13 − x) ⇒ dE
dx =
Wx
√ − L =0when W √
25 + x
25 + x
2 L = 2
. By the same sort of
x
argument as in part (a), this ratio will give the minimal expenditure of energy if the bird heads for the point x km from B.
(c) For flight direct to D, x =13, so from part (b), W/L =
√
25 + 13 2
13
≈ 1.07. There is no value of W/L for which the bird
should fly directly to B. But note that lim =∞, so if the point at which E is a minimum is close to B,then
x→0 +(W/L)
W/L is large.
(d) Assuming that the birds instinctively choose the path that minimizes the energy expenditure, we can use the equation for
dE/dx =0from part (a) with 1.4k = c, x =4,andk =1: c(4) = 1 · (25 + 4 2 ) 1/2 ⇒ c = √ 41/4 ≈ 1.6.
4.8 Newton's Method
1. (a) The tangent line at x =1intersects the x-axis at x ≈ 2.3,so
x 2 ≈ 2.3. The tangent line at x =2.3 intersects the x-axis at
x ≈ 3,sox 3 ≈ 3.0.
(b) x 1 =5would not be a better first approximation than x 1 =1since the tangent line is nearly horizontal. In fact, the second
approximation for x 1 =5appears to be to the left of x =1.
3. Since x 1 =3and y =5x − 4 is tangent to y = f(x) at x =3, we simply need to find where the tangent line intersects the
x-axis. y =0 ⇒ 5x 2 − 4=0 ⇒ x 2 = 4 . 5
5. f(x) =x 3 +2x − 4 ⇒ f 0 (x) =3x 2 +2,sox n+1 = x n − x3 n +2x n − 4
.Nowx
3x 2 1 =1
n +2
⇒
x 2 =1− 1+2− 4 −1
=1−
3 · 1 2 +2 5 =1.2 ⇒ x3 =1.2 − (1.2)3 +2(1.2) − 4
≈ 1.1797.
3(1.2) 2 +2
7. f(x) =x 5 − x − 1 ⇒ f 0 (x) =5x 4 − 1, sox n+1 = x n − x5 n − x n − 1
.Nowx
5x 4 1 =1
n − 1
⇒
x 2 =1− 1 − 1 − 1
5 − 1
=1− − 1 4
=1.25 ⇒ x3 =1.25 − (1.25)5 − 1.25 − 1
5(1.25) 4 − 1
9. f(x) =x 3 + x +3 ⇒ f 0 (x) =3x 2 +1,sox n+1 = x n − x3 n + x n +3
.
3x 2 n +1
Now x 1 = −1
⇒
x 2 = −1 − (−1)3 +(−1) + 3
3(−1) 2 +1
= −1 −
−1 − 1+3
3+1
= −1 − 1 4 = −1.25.
Newton’s method follows the tangent line at (−1, 1) down to its intersection with
the x-axis at (−1.25, 0), giving the second approximation x 2 = −1.25.
≈ 1.1785.