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Unformatted text preview: want to ﬁnd
given that
dt
1
ds
= 500. From the ﬁgure, h = s sin 30◦ = s so
dt
2
1 ds
1
dh
=
= (500) = 250 mi/h.
dt
2 dt
2 s
30°
Ground h Exercise Set 4.6 26. 27. 28. dy
= −20. From x2 + 102 = y 2 we get
dt
y =125
dx
y dy
dy
dx
= 2y
so
=
. Use x2 + 100 = y 2 to ﬁnd that
2x
dt
dt
dt
x dt
√
√
x = 15, 525 = 15 69 when y = 125 so
125
500
dx
= √ (−20) = − √ . The boat is approaching the
dt y=125
15 69
3 69
500
dock at the rate of √ ft/min.
3 69
Find dx
dt 126 given that dx
dy
given that
= −12. From x2 + 102 = y 2 we get
dt
dt y=125
dy
x dx
dy
dx
= 2y
so
=
. Use x2 + 100 = y 2 to ﬁnd that
2x
dt
dt
dt
√ y dt
x = 15√
, 525 = 15 69 when y = 125 so
√
15 69
36 69
dy
=
(−12) = −
. The rope must be pulled at the rate
dt √ 125
25
36 69
ft/min.
of
25
Find (a) Let x and y be as shown in the ﬁgure. It is required to ﬁnd
dy
x
x+y
dx
, given that
= −3. By similar triangles, =
,
dt
dt
6
18
1
18x = 6x + 6y , 12x = 6y , x = y , so
2
1 dy
1
3
dx
=
= (−3) = − ft/s.
dt
2 dt
2
2 Pulley
y
10
Boat x Pulley
y
10
Boat x Light 18
Man
Shadow 6
x (b) 29. Find y The tip of the shadow is z = x + y feet from the street light, thus the rate at which it is
dx
dy
dx
3
dy
dz
=
+
. In part (a) we found that
= − when
= −3 so
moving is given by
dt
dt
dt
dt
2
dt
dz
= (−3/2) + (−3) = −9/2 ft/s; the tip of the shadow is moving at the rate of 9/2 ft/s toward
dt
the street light. dx
dt given that
θ=π/4 dθ
2π
π
=
= rad/s. Then
dt
10
5 dθ
dx
= 4 sec2 θ ,
x = 4 tan θ (see ﬁgure) so
dt
dt
π
dx
2π
= 4 sec
= 8π/5 km/s.
dt θ=π/4
4
5 x 4 Ship θ 127 30. Chapter 4 If x, y , and z are as shown in the ﬁgure, then we want
that dx
dy
= −600 and
dt
dt x=2,
y =4 dz
dt given P x=2,
y =4 = −1200. But z 2 = x2 + y 2 so Aircraft x y
z 1
dx
dz
dx
dy dz
dy
= 2x
+ 2y ,
=
x
+y
. When x = 2 and
dt
dt
dt dt
z
√ dt √ dt
y = 4, z 2 = 22 + 42 = 20, z = 20 = 2 5 so
√
dz
1
3000
= √ [2(−600) + 4(−1200)] = − √ = −600 5 mi/h; the
dt x=2,
25
5
y =4
distance between missile and aircraft is decreasing at the rate of
√
600 5 mi/h.
2z 31. 32. Missile dy
dz
dx
= −600 and
given
= −1200 (see
dt x=2,
dt
dt x=2,
y =4
y =4
ﬁgure). From the law of cosines,
z 2 = x2 + y 2 − 2xy cos 120◦ = x2 + y 2 − 2xy (−1/2) = x2 + y 2 + xy , so
dx
dy
dy
dx
dz
= 2x
+ 2y
+x
+y ,
2z
dt
dt
dt
dt
dt
1
dx
dy
dz
=
(2x + y )
+ (2y + x)
. When x = 2 and y = 4,
dt
2z
dt
dt
√
√
z 2 = 22 + 42 + (2)(4) = 28, so z = 28 = 2 7, thus
1
dz
4200
√ [(2(2) + 4)(−600) + (2(4) + 2)(−1200)] = − √ =
=
dt x=2,
2(2 7)
7
y =4
√
−600 7 mi/h; the distance between missile and aircraft is
√
decreasing at the rate of 600 7 mi/h.
We wish to ﬁnd (a) Let x, y , and z be the distances shown in the ﬁrst ﬁgure. Find P x Aircraft 120°
y
z Missile dz
dt given that
x=2,
y =0 dx
= −75 and
dt dy
= −100. In order to ﬁnd an equation relating x, y , and z , ﬁrst draw the line segment that
dt
joins the point P to the car, as shown in the second ﬁgure. Because triangle OP C is a right
triangle, it follows that P C has length x2 + (1/2)2 ; but triangle HP C is also a right triangle
2
dx
dy
dz
= 2x
+ 2y
+ 0,
x2 + (1/2)2 + y 2 = x2 + y 2 + 1/4 and 2z
so z 2 =
dt
dt
dt
√
1
dx
dy
dz
=
x
+y
. Now, when x = 2 and y = 0, z 2 = (2)2 + (0)2 + 1/4 = 17/4, z = 17/2
dt
z
dt
dt
√
1
dz
=√
so
[2(−75) + 0(−100)] = −300/ 17 mi/h
dt x=2,
( 17/2)
y =0
North 1
2 mi
y West 1
mi
2 P
x East z
Helicopter (b) decreasing, because P
y Car x
O
z H dz
< 0.
dt C Exercise Set 4.6 33. (a) 128 We want dy
dt given that
x=1,
y =2 dx
dt = 6. For convenience, ﬁrst rewrite the equation as
x=1,
y =2 8 82
16 dy dy
y3
dy
dx
dx
+ y then 3xy 2
+ y3
=
y,
=
so
16
55
dt
dt
5 dt dt
dt
y − 3xy 2
5
23
dy
=
(6) = −60/7 units/s.
16
dt x=1,
(2) − 3(1)22
y =2
5
dy
<0
falling, because
dt
xy 3 = (b)
34. Find dx
dt so 3x2
35. given that
(2,5) dy
dt = 2. Square and rearrange to get x3 = y 2 − 17
(2,5) 2y dy dx
dx
dy dx
= 2y ,
=2,
dt
dt dt
3x dt dt (2,5) (2) = 5
units/s.
3 The coordinates of P are (x, 2x), so the distance between P and the point (3, 0) is
√
dD
dx
given that
= −2.
D = (x − 3)2 + (2x − 0)2 = 5x2 − 6x + 9. Find
dt x=3
dt x=3
5x − 3
dD
dD
dx
=√
, so
2 − 6x + 9 dt
dt
dt
5x 36. 5
6 = (a) x=3 12
= √ (−2) = −4 units/s.
36 Let D be the distance between P and (2, 0). Find
D=
dD
dt (x − 2)2 + y 2 = x=3 (x − 2)2 + x = √ dD
dt given that
x=3 x2 − 3x + 4 so dx
dt = 4.
x=3 dD
2x − 3
=√
;
dt
2 x2 − 3x + 4 3
3
= √ = units/s.
4
24 (b) Let θ be the angle of inclination. Find dθ
dt given that
x=3 dx
dt = 4.
x=3 √
x+2
dθ
x+2
y
x
dx dθ
dx
=
so sec2 θ
=− √
,
= − cos2 θ √
.
tan θ =
x−2
x−2
dt
2 x(x − 2)2 dt dt
2 x(x − 2)2 dt
When x = 3, D = 2 so cos θ = 37. Solve 1
dθ
and
2
dt =−
x=3 15
5
√ (4) = − √ rad/s.
42 3
23 dx
dy
dy dx
dx
dy
=3
given y = x ln x. Then
=
= (1 + ln x) , so 1 + ln x = 3, ln x = 2, x = e2 .
dt
dt
dt
dx dt
dt dy
dy
dx
dx
16
dx
+ 18y
= 0; if
=
= 0, then (32x +...
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 Spring '14
 The Land

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