This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Section 4.6
20. Step 1:
r radius of spherical balloon
S surface area of spherical balloon
V volume of spherical balloon (b) 16 cos l 1 d
dt Step 2:
At the instant in question, dV
dt 100 ft3/min and r 1 d
dt dr
dS
We want to find the values of and .
dt
dt 6 dl
l 2 dt 62
l 5 ft. Step 3: 1
0.6 1 6
( 2)
102 2 The rate of change of angle 3
radian/sec
20
3
radian/sec.
20 is Steps 4, 5, and 6:
(a) V
dV
dt dr
dt 4 (5)2 100
dr
dt 22. Step 1:
x distance from origin to bicycle
y height of balloon (distance from origin to balloon)
s distance from balloon to bicycle 43
r
3 4 r2 dr
dt Step 2: 1 ft/min dS
dt
dS
dt
dS
dt 4 r2
8r dy
dx
is a constant 1 ft/sec and
is a constant
dt
dt We know that The radius is increasing at the rate of 1 ft/min.
(b) S 17 ft/sec. Three seconds before the instant in question, the
values of x and y are x 0 ft and y 65 ft. Therefore, at
the instant in question x 51 ft and y 68 ft.
Step 3: dr
dt 8 (5)(1) We want to find the value of 40 ft2/min ds
at the instant in question.
dt Step 4:
x2 The surface area is increasing at the rate of s 40 ft2/min. y2 Step 5:
ds
dt 21. Step 1:
l length of rope
x horizontal distance from boat to dock
angle between the rope and a vertical line dl
At the instant in question,
dt 2 ft/sec and l ds
dt 10 ft. 2 dx
dt
dx
dt 2 51 dc
dt l2
102 dx
dt x2 y dy
dt y2 11 ft/sec 682 6x 2
12x [3(2)2 15x)
15) 12(2) dx
dt 15](0.1) 0.3
dr
dt
dp
dt dl
36 dt 10 dy
2y
dt (68)(1) d3
(x
dt (3x 2 dx
d
and .
dt
dt 36
l x2 (51)(17) Step 4, 5, and 6:
l2 dx
2x
2
dt
y The distance between the balloon and the bicycle is
increasing at the rate of 11 ft/sec.
23. (a) Step 3:
We want to find the values of x
1 Step 6: Step 2: (a) x 181 ( 2) 2.5 ft/sec 36 The boat is approaching the dock at the rate of (b) dx
d
(9x) 9
9(0.1) 0.9
dt
dt
dr
dc
0.9 0.3 0.6
dt
dt dc
dt d3
x
dt 3x 2 2.5 ft/sec. 45
x
45 dx
x 2 dt 6x 2
12x 3(1.5)2 12(1.5) 45
(0.05)
1.52 1.5625
dr
dt
dp
dt d
(70x)
dt
dr
dc
dt
dt 70 dx
dt 3.5 70(0.05)
( 1.5625) 3.5
5.0625 182 Section 4.6 24. (a) Note that the level of the coffee in the cone is not
needed until part (b).
Step 1:
V1 volume of coffee in pot
y depth of coffee in pot
Step 2:
dV1
dt Step 2: 10 in3/min At the instant in question, Q Step 3:
We want to find the value of dy
dt dD
dt . dt y dy
dt dy
dt Q
D Step 5: Step 6:
9 0 mL/min . Step 4: 9 10 41 mL/L, 2 dy
.
dt We want to find the value of 9y Step 5:
dV1 233 mL/min, D dQ
2 (mL/L)/min, and
dt Step 3: Step 4:
V1 25. Step 1:
Q
rate of CO2 exhalation (mL/min)
D
difference between CO2 concentration in blood
pumped to the lungs and CO2 concentration in blood
returning from the lungs (mL/L)
y
cardiac output dy
dt 10
9 dy
dt 0.354 in./min D dQ
dt Q dD
dt D2 The level in the pot is increasing at the rate of Step 6: approximately 0.354 in./min. dy
dt (b) Step 1:
V2 volume of coffee in filter
r radius of surface of coffee in filter
h depth of coffee in filter (41)(0) 466
1681 0.277 L/min2 The cardiac output is increasing at the rate of
approximately 0.277 L/min2.
26. Step 1:
y Step 2:
At the instant in question,
h (233)( 2)
(41)2 dV2
dt 10 in3/min and
(x , y ) 5 in. Step 3:
We want to find dh
.
dt x Step 4:
Note that 3
, so r
6 r
h 12
rh
3 Then V2 h
.
2 h3
.
12 x
y Step 5:
dV2
dt h 2 dh
4 dt xcoordinate of particle’s location
ycoordinate of particle’s location
angle of inclination of line joining the particle to the
origin. Step 2: Step 6: At the instant in question, (5)2 dh
10
4 dt
dh
8
in./min
dt
5
dh
Note that
0, so the rate at which the level is
dt dx
dt Step 3: falling is positive. The level in the cone is falling at the Step 4: 8
rate of
5 Since y 0.509 in./min. 10 m/sec and x We want to find x 0, 3 m. d
.
dt x 2, we have tan
tan 1 x. y
x x2
x x and so, for Section 4.6
Step 5:
d
dt 1
1 28. Step 1:
x xcoordinate of particle
y ycoordinate of particle
D distance from origin to particle dx
x 2 dt Step 6:
d
dt 1
32 1 183 (10) Step 2: 1 radian/sec At the instant in question, x The angle of inclination is increasing at the rate of dx
dt 1 radian/sec. 1 m/sec, and dy
dt 5 m, y 12 m, 5 m/sec. Step 3: 27. Step 1: We want to find y dD
.
dt Step 4:
x2 D
(x , y ) y2 Step 5:
θ dD
dt x x
1 2 dx
2x
dt
y2 x2 dy
2y
dt dx
dt x2 y dy
dt y2 Step 6:
x
y (5)( 1) dD
dt xcoordinate of particle’s location
ycoordinate of particle’s location
angle of inclination of line joining the particle to the
origin (12)( 5) 52 5 m/sec 122 The particle’s distance from the origin is changing at the
rate of 5 m/sec. 29. Step 1: Step 2:
At the instant in question, dx
dt 8 m/sec and x Street
light 4 m. Step 3:
We want to find d
.
dt 16 ft Step 4: 6 ft Since y x, we have tan and so, for x y
x x 0,
1 tan [ ( x) 1/2 x tan 1 ( x) ( x) 1/2 . Shadow 1/2 , x x
s s distance from streetlight base to man
length of shadow Step 2:
Step 5:
d
dt 1 1
[( x)
1
1
x 1 1/2 2 x(x 1
dx
( x) 3/2( 1)
2
dt dx
1
2( x)3/2 dt 1
2 10 ft. Step 3:
We want to find ds
.
dt By similar triangles, dx
1) dt 16s 6s 6x, or s s s
6 x
16 . This is equivalent to 3
x.
5 Step 5:
1 2 5 ft/sec and x Step 4: Step 6:
d
dt dx
dt At the instant in question, 4( 4 ( 8)
1) 2
radian/sec
5 ds
dt 3 dx
5 dt The angle of inclination is increasing at the rate of Step 6: 2
radian/sec.
5 ds
dt 3
( 5)
5 3ft/sec The shadow length is changing at the rate of 3 ft/sec. 184 Section 4.6
Step 2: 30. Step 1:
s distance ball has fallen
x distance from bottom of pole to shadow At the instant in question,
dV
dt Step 2:
At the instant in question, s
ds
dt 1
32
2 16 12
2 4 ft and Step 4: dx
.
dt 43
r and S
3 We have V Step 4:
x 30
By similar triangles,
50 s x 1500 50x
1 1500s x
. This is equivalent to
50 sx, or sx 1500. We will use 2 ds dS
dt 1500(4) 2(16) 1500 ft/sec. 31. Step 1:
x
position of car (x 0 when car is right in front of
you)
camera angle. (We assume is negative until the car
passes in front of you, and then positive.) At the first instant in question, x
1
(264)
2 A half second later, x 0 ft and dx
dt 264 ft/sec. 2 1/3 dV 3V
4 dt 4
4000
(10)3
.
3
3
1/3
4000
( 8)
3 2 33. Step 1:
p xcoordinate of plane’s position
x xcoordinate of car’s position
s distance from plane to car (lineofsight) At the instant in question,
dp
dt 0, 120 mph, s d
at each of the two instants.
dt Step 4:
(x Step 4:
1 x
132 p)2 32 s2 Step 5:
2(x Step 5:
1 1 dx
132 dt x2
132 dx
dt p) dp
dt 2s ds
dt Step 6:
Note that, at the instant in question, Step 6:
1 d
0:
dt 132: 5 mi, and dx
.
dt We want to find We want to find When x 3 dV
4 dt Step 3: Step 3: When x 1/3 3V
4 area is decreasing at the rate of 1.6 cm2/min. p 132 ft and 264 ft/sec. 1 2
3 4 Step 2: Step 2: d
dt 3V 2/3
4 16
3
1.6 cm2/min
3
4
1000
dS
Since
0, the rate of decrease is positive. The surface
dt 1500 ft/sec The shadow is moving at a velocity of tan 4 Step 5: Note that V dt Step 6: dx
dt 3V 1/3
, so S
4 Step 6:
1500s dx
dt 4 r 2. These equations can be combined by noting that r dS
dt . Step 5:
dx
dt dS
.
dt We want to find Step 3: 50x 1
(20) = 10 cm.
2 Step 3: 16 ft/sec. We want to find 8 cm3/min and r 8 mL/min 02
132 1
d
dt 1
(264)
132 1
1 132
132 2 32. Step 1:
r radius of balls plus ice
S surface area of ball plus ice
V volume of ball plus ice 1
(264)
132 52 x
2 radians/sec 1 radian/sec 2(4 32 dx
dt
dx
8
dt
dx
dt 0) 4 mi.
120 2(5)( 160) 120
120
dx
dt The car’s speed is 80 mph. 1600
200
80 mph ds
dt 160 mph. 185 Section 4.6
Step 6: 34. Step 1:
s shadow length
sun’s angle of elevation d
dt (20)( 2) (10)(1)
102 202 0.1 radian/sec
5.73 degrees/sec Step 2: To the nearest degree, the angle is changing at the rate At the instant in question,
s 60 ft and d
dt 0.27 /min of 0.0015 radian/min. 6 degrees per second. 36. Step 1: Step 3: A ds
.
dt We want to find
Step 4: c
a 80
or s
s tan 80 cot 120° Step 5:
ds
dt O 80 csc d
dt 2
a
b
c Step 6:
80
and
60 Note that, at the moment in question, since tan
0 2 4
and so csc
5 , we have sin distance from O to A
distance from O to B
distance from A to B At the instant in question, a
b 5 nautical miles, da
dt 14 knots, and c2
c2 3 nautical miles, a2
a2 db
dt 21 knots. Step 3: 2.25 in./min We want to find 7.1 in./min
ds
Since
dt B Step 2: 5
.
4 52
80
(0.0015 )
4
12 in
ft
0.1875
1 ft
min ds
dt b dc
.
dt Step 4:
Law of Cosines: 0, the rate at which the shadow length is b2
b2 2ab cos 120
ab decreasing is positive. The shadow length is decreasing at
Step 5: the rate of approximately 7.1 in./min. 2c 35. Step 1:
a distance from origin to A
b distance from origin to B
angle shown in problem statement 10 m, and b da
dt 2 m/sec, a2 db
dt dc
dt
dc
14
dt
dc
dt 2(7) 1 m/sec, 20 m. Step 3:
We want to find d
.
dt tan 37.
a
tan 1
b Step 5:
d
dt b 1
1 a2
b da
dt a
b2 2b db
dt a db
dt b da
dt b2 (5)2 ab 2(5)(14) 2(3)(21) (3)2 (5)(3) (5)(21) 49 (3)(14) 413
29.5 knots The ships are moving apart at a rate of 29.5 knots. Step 4:
a
or
b da
dt Note that, at the instant in question,
c At the instant in question, 2a Step 6: Step 2: a dc
dt db
dt b da
dt a db
dt a2 b2 dy
dt dy dx
dx
10(1 x 2) 2(2x)
dx dt
dt
dx
Since
3 cm/sec, we have
dt
dy
60x
cm/sec.
dt
(1 x2)2 (a) dy
dt [1 60( 2)
( 2)2]2 (b) dy
dt (1 60(0)
0 2)2 (c) dy
dt 60(20)
(1 202)2 120
52 (1 24
cm/sec
5 0 cm/sec
0.00746 cm/sec dx
20x
x 2 ) 2 dt 7 ...
View
Full
Document
 Spring '08
 ALL

Click to edit the document details