Solução_Calculo_Stewart_6e

F.
230 ¤ CHAPTER 4 PROBLEMS PLUS
TX.10
(c) Using part (a) with D =1and T 1 =0.26,wehaveT 1 = D √
⇒ c 1 = 1
c ≈ 3.85 km/s. T 4h2 + D
0.26 3 =
2
1
c 1
⇒
4h 2 + D 2 = T 2 3 c 2 1 ⇒ h = 1 2
T
2
3 c 2 1 − D2 = 1 2
from part (b) and T 2 =
sin θ = c 1
c 2
⇒ sec θ =
2hc 2
T 2 =
c 1 c
2
2 − c 2 1
2h sec θ
c 1
+
D − 2h tan θ
c 2
c 2
and tan θ =
c
2
2 − c 2 1
(0.34)2 (1/0.26) 2 − 1 2 ≈ 0.42 km. To find c 2 ,weusesin θ = c1
c 2
from part (a). From the figure,
c 1
,so
c
2
2 − c 2 1
+ D c 2 2 − c2 1
− 2hc1 . Using the values for T 2 [given as 0.32],
c 2 c
2
2 − c 2 1
2hc 2
h, c 1,andD, we can graph Y 1 = T 2 and Y 2 =
c 1 c
2
2 − c 2 1
+ D c 2 2 − c2 1 − 2hc 1
c 2
c
2
2 − c 2 1
and find their intersection points.
Doingsogivesusc 2 ≈ 4.10 and 7.66,butifc 2 =4.10,thenθ =arcsin(c 1 /c 2 ) ≈ 69.6 ◦ , which implies that point S is to
the left of point R in the diagram. So c 2 =7.66 km/s.
21. Let a = |EF| and b = |BF| asshowninthefigure. Since = |BF| + |FD|,
|FD| = − b. Now
|ED| = |EF| + |FD| = a + − b = √ r 2 − x 2 + − (d − x) 2 + a 2
= √
r 2 − x 2 + − (d − x) 2 + √ r 2 − x 2 2
= √ r 2 − x 2 + − √ d 2 − 2dx + x 2 + r 2 − x 2
Let f(x) = √ r 2 − x 2 + − √ d 2 + r 2 − 2dx.
f 0 (x) = 1 2 (r2 − x 2 ) −1/2 (−2x) − 1 2 (d2 + r 2 − 2dx) −1/2 (−2d) =
−x
√
r2 − x + d
√ 2 d2 + r 2 − 2dx .
f 0 (x) =0
⇒
x
√
r2 − x = d
√ 2 d2 + r 2 − 2dx ⇒ x 2
r 2 − x = d 2
2 d 2 + r 2 − 2dx
⇒
d 2 x 2 + r 2 x 2 − 2dx 3 = d 2 r 2 − d 2 x 2 ⇒ 0=2dx 3 − 2d 2 x 2 − r 2 x 2 + d 2 r 2 ⇒
0=2dx 2 (x − d) − r 2 (x 2 − d 2 ) ⇒ 0=2dx 2 (x − d) − r 2 (x + d)(x − d) ⇒ 0=(x − d)[2dx 2 − r 2 (x + d)]
But d>r>x,sox 6=d. Thus, we solve 2dx 2 − r 2 x − dr 2 =0for x:
x = −(−r2 ) ± (−r 2 ) 2 − 4(2d)(−dr 2 )
2(2d)
= r2 ± √ r 4 +8d 2 r 2
. Because √ r
4d
4 +8d 2 r 2 >r 2 , the “negative” can be
discarded. Thus, x = r2 + √ r 2 √ r 2 +8d 2
4d
of |ED| occurs at this value of x.
= r2 + r √ r 2 +8d 2
4d
[r >0] = r √
r + r2 +8d 4d
2 . The maximum value