Solution
Section 2.8 – Related Rates
Exercise
dy
If y  x 2 and dx  3 , then what is
dt
dt
when x  1
Solution
dy dy dx

dt dx dt
 2 x  3
 6x
dy
 6  1  6
dt x 1
Exercise
dy
 5 , then what is dx
If x  y3  y and
dt
dt
when y  2
Solution
dx  dx dy
dt dy dt


 3 y 2  1  5


 5 3 y2  1


2
dx
 5 3  2   1  55
dt y  2
Exercise
A cube’s surface area increases at the rate of 72 in2 / sec . At what rate is the cube’s volume changing
when the edge length is x = 3 in?
Solution
Cube’s surface: S  6 x 2
dS  12 x dx
dt
dt
72  12 x  3 
V  x3 
x  72  2
26
dV  3x 2 dx
dt
dt
2
dV
 3  3  2   54 in2 / sec
dt x 3
69
Exercise
The radius r and height h of a right circular cone are related to the cone’s volume V by the equation
V  1  r 2h .
3
a) How is dV related to dh if r is constant?
dt
dt
b) How is dV related to dr if h is constant?
dt
dt
c) How is dV related to dr and dh if neither r nor h is constant?
dt
dt
dt
Solution
a) dV  1  r 2 dh
dt 3
dt
b) dV  2  rh dr
dt 3
dt
c) dV  2  rh dr  1  r 2 dh
dt 3
dt 3
dt
Exercise
The voltage V (volts), current I (amperes), and resistance R (ohms) of an electric circuit like the one
shown here are related by the equation V  IR . Suppose that V is increasing at the rate of 1 volt/sec while
I is decreasing at the rate of 1 amp / sec . Let t denote time in seconds.
3
a) What is the value of dV ?
dt
b) What is the value of dI ?
dt
c) What equation relates dR to dV and dI ?
dt
dt
dt
d) Find the rate at which R is changing when V = 12 volts and I = 2 amp. Is R increasing or decreasing?
Solution
a) dV  1 volt / sec
dt
b) dI  1 amp / sec
dt 3
c) dV  R dI  I dR
dt
dt
dt
I dR  dV  R dI
dt
dt
dt
dR  1 dV  V dI
dt I dt I dt
d)
V  IR  R  V


dR  1 1  12  1  1  3  3 ohms / sec
2  3  2
dt 2 
2
70
I
R is increasing
Exercise
Let x and y be differentiable functions of t and let s  x 2  y 2 be the distance between the points (x, 0)
and (0, y) in the xyplane.
a) How is ds related to dx if y is constant?
dt
dt
dy
b) How is ds related to dx and
if neither x nor y is constant?
dt
dt
dt
dy
c) How is dx related to
if s is constant?
dt
dt
Solution


1/2
s  x2  y 2  x2  y 2
 

1/2
2 x dx
a) ds  1 x 2  y 2
dt 2
dt


x
dx
dt
x2  y 2


1/2 
dy 
dx
b) ds  1 x 2  y 2
 2 x dt  2 y dt 
dt 2



dy 
 dx
 x dt  y dt 

x2  y 2 

y
dy
dx 
dt
dt
x2  y 2
x2  y 2
1
x
c) s  x2  y 2
 s2  x2  y 2
dy
0  2 x dx  2 y
dt
dt
dy
2 x dx  2 y
dt
dt
dx   y dy
dt
x dt
71
Exercise
A 13-ft ladder is leaning against a house when its base starts to slide away. By the time the base is 12 ft
from the house, the base is moving at the rate of 5 ft/sec.
a) How fast is the top of the ladder sliding down the wall
then?
b) At what rate is the area of the triangle formed by the
ladder, wall, and the ground changing then?
c) At what rate is the angle  between the ladder and the
ground changing then?
Solution
Given:
L  13 ft
x  12 dx  5 ft / sec
dt
y  132  122  5
a) x2  y 2  132
dy
2 x dx  2 y
0
dt
dt
dy
y
  x dx
dt
dt
dy
  x dx
dt
y dt
  12  5
5
 12 ft / sec
The ladder is sliding down the wall
b) Area of the triangle formed by the ladder and the walls is:
A  1 xy
dA  1  y dx  x dy 
dt 2  dt
dt 
 1   5 5  12  12  
2
 19.5 ft 2 / sec
c) cos   x   sin  d  1 dx
13
dt 13 dt
d   1 dx
13sin  dt
dt

1  5
13sin 
sin   5
13
 5
13 5
13
 1 rad / sec

1
 
72
2
Exercise
Water is flowing at the rate of 6 m3 / min from a reservoir shaped like a hemispherical bowl of radius 13
m. Answer the following questions, given that the volume of water in a hemispherical bowl of radius R is
V   y 2  3R  y  when the water is y meters deep.
3
a) At what rate the water level changing when the water is 8 m deep?
b) What is the radius r of the water’s surface when the water is y m deep?
c) At what rate is the radius r changing when the water is 8 m deep?
Solution
Given: dV  6 m3 / min R  13 m
a) V   y
3
dt
2

 3R  y    Ry 2  3 y3

dV  2 Ry   y 2 dy
dt
dt
Factor  y
dV   y  2R  y  dy
dt
dt
dy
1

 6    241 m / min
dt  8  2 13  8 

r 2  169  169  26 y  y 2
13  y
2
b) The hemispherical is on the circle: r 2  13  y   132

 169  169  26y  y 2
 26y  y 2
r  26 y  y 2

c) r  26 y  y 2
dr
1
dt y 8 2

1/2



dr  1 26 y  y 2 1/2  26  2 y  dy
dt 2
dt
26  2 y dy
1
2
dt
26 y  y 2
26  2  8
 1    5  0.005526

24
288
2
26 8  8
   
73
or
5.526  103
Exercise
A spherical balloon is inflated with helium at the rate of 100 ft 3 / min . How fast is the balloon’s radius
increasing at the instant the radius is 5 ft? How fast the surface area increasing?
Solution
Given:
dV  100 ft 3 / min
dt
r  5 ft
If V  4  r 3  dV  4 r 2 dr
3
dt
dt
dr  1 dV
dt 4 r 2 dt
1

100 
2
4  5
 1 ft / min
S  4 r 2 
dS  8 r dr  8  51  40 ft 2 / min
dt
dt
The rate of the surface area is increasing.
Exercise
A balloon rising vertically above a level, straight road at a constant rate of 1 ft/sec. Just when the balloon
is 65 ft above the ground, a bicycle moving at a constant rate of 17 ft/sec passes under it. How fast is the
distance s  t  between the bicycle and the balloon increasing 3 sec later?
Solution
Given:
dy
 1 ft / sec
dt
y  65 ft
dx  17 ft / sec
dt
Bicycle increasing 3 sec: x  vt  17  3  51 ft
s 2  x2  y 2 
2s ds  2 x dx  2 y
dt
dt
s  512  652  83 ft
dy
dt
ds  1  x dx  y dy 
dt s  dt
dt 
 1  5117   65 1 
83
 11 ft / sec
74
Exercise
A dinghy is pulled toward a dock by a rope from the bow through a ring on the dock 6 ft above the bow.
The rope is hauled in at rate of 2 ft/sec.
a) How fast is the boat approaching the dock when 10 ft of rope are out?
b) At what rate is the angle  changing at this instant?
Solution
ds  2 ft / sec
dt
h  6 ft
Given:
a) s  10 ft
s 2  x2  62  x  s 2  36
2s ds  2 x dx
dt
dx 
dt
dt
s
ds
dt
s 2  36
dx

dt s 10
10
102  36
 2   2.5
ft / sec
b) cos   6   sin  d   6 ds
s
dt
s 2 dt
d 
6
ds
dt sin  s 2 dt
102  36
sin   x 
 8
10
s
10
d 
6  2   0.15 rad / sec
dt .8 102
Exercise
The coordinates of a particle in the metric xyplane are differentiable functions of time t with
dx  1 m / sec and dy  5 m / sec . How fast is the particle’s distance from the origin changing as it
dt
dt
passes through the point (5, 12)?
Solution
Given:
dx  1 m / sec
dt
s 2  x2  y 2
dy
 5 m / sec
dt

s  x2  y 2  52  122  13
dy
dt
ds  1  x dx  y dy 
dt s  dt
dt 
2s ds  2 x dx  2 y
dt
dt
ds
 1  5  1  12  5   5 m / sec
dt  5,12 13
75
Exercise
Coffee is draining from a conical filter into a cylindrical coffeepot at the rate of 10 in3 / min .
a) How fast is the level in the pot rising when the coffee in the cone is 5 in. deep?
b) How fast is the level in the cone falling then?
Solution
r pot  3 dV  10 in3 / min
dt
a) Let h be the height of the coffee in the pot.
Volume of the coffee: V   r 2h  9 h
dV  9 dh
dt
dt
dh  1 dV  1 10   10 in / min
dt 9 dt 9
9
b) Radius of the filter: r  h
2

2
3
Volume of the filter: V  1  r 2 h  1  h h   h
3
3 2
12
dV   h2 dh
dt
4
dt
dh  4 dV  4  10    8 in / min
dt  h2 dt  5 2
5
 
Exercise
A particle moves along the parabola y  x 2 in the first quadrant in such a way that its x-coordinate
(measure in meters) increases at a steady 10 m/sec. How fast is the angle of inclination  of the line
joining the particle to the origin changing when x = 3 m?
Solution
Given:
y  x2
v  dx  10 m / sec x  3 m
dt
y x2

x
x
x
d tan   d x
dt
dt
sec2  d  dx
dt dt
d  1 dx  cos2  dx
dt sec2  dt
dt
tan  

3

 2
2
 9 3

2

 10 


 1 rad / sec
76
Exercise
A light shines from the top of a pole 50 ft high. A ball is dropped from the same height from a point 30 ft
away from the light. How fast is the shadow of the ball moving along the ground 1 sec later? (Assume
2
the ball falls a distance s  16t 2 ft in t sec.)
Solution
Y
s  16t 2
s  h  50
s
h
XQ OX 30  XQ



h
50
50
50 XQ  30h  h XQ
30
O
50  h  XQ  30h
XQ  30h
50  h
30  50  s 

50   50  s 


30 50  16t 2

50  50  16t 2
 1500  480t
16t 2
2
2
 1500  480t
16t 2 16t 2
 1500  30
16t 2
d XQ  1500 32t
2
dt
16t 2
 
 1500 32t
256t 4
  375
2t 3
d XQ
  375  1500 ft / sec
3
dt
1
t
2 1
2
P
50
Triangles XOY and XQP are similar:
2
77
Q
X
Exercise
A spherical iron ball 8 in. in diameter is coated with a layer of ice of uniform thickness. If the ice melts at
the rate of 10 in3 / min , how fast is the thickness of the ice decreasing when it is 2 in. thick? How fast is
the outer surface area of ice decreasing?
Solution
Given:
D  8 in  r  4 in
1
dV  10 in3 / min
dt
think  2 in
V  4  r 3  dV  4 r 2 dr
dt
dt
3
dr  1 dV
dt 4 r 2 dt
dr
1

 10    725 in / min
dt r  6 4 6 2
 
S  4 r 2
dS  8 r dr
dt
dt
dS
 8  6   5   10 in2 / min
r

6
3
dt
72


The outer surface are of the ice is decreasing at  10 in2 / min
3
Exercise
On a morning of a day when the sun will pass directly overhead, the shadow of an 80ft building on level
ground is 60 ft long. At the moment in question, the angle  the sun makes with the ground is increasing
at the rate of 0.27 / min. At what rate is the shadow decreasing?
Solution
x  60 ft
Given:
h  80 ft
d  0.27 min  0.27  rad 1  3 rad / min
dt
180 min 2000
tan   80  d tan   d 80
x
dt
dt x
sec2  d   80 dx
dt
x 2 dt
dx   x 2 sec2  d
80
dt
dt

 3
2
602 5
80
cos  
60
602 802
3

 2000
 0.589 ft / min
78
 60  3
100
5
Exercise
A baseball diamond is a square 90 ft on a side. A player runs from first base to second at a rate of 16
ft/sec.
a) At what rate is the player’s distance from third base changing when the player is 30 ft from first
base?
b) At what rates are angles  and  changing at that time?
1
2
c) The player slides into second base at the rate of 15 ft/sec. At what rates are angles  and 
1
changing as the player touches base?
Solution
dx  16 ft / sec
dt
Given: d1  90 ft d2  30 ft
x
nd
s
x: Distance between player and 2 base
s: Distance between player and 3rd base
a) x  90  30  60 ft
s 2  x2  902  s  602  902  11700  30 13
2s ds  2 x dx
dt
dt
ds  x dx
dt s dt
 60  16 
30 13
 8.875 ft / sec
d
1   90 ds
b) sin   90  cos 
1
1 dt
s
s 2 dt
d
1
dt
d

90
cos   x
1
s
ds
s 2 cos  dt
1
1
dt
  90 ds
s 2 x dt
s
  90 ds
s  x dt
90

 8.875
30 13  60 
 0.123 rad / sec
d
2   90 ds
cos   90   sin 
2
2 dt
s
s 2 dt
d
2 
dt
90
s 2 sin 
ds  90 ds
dt s  x dt
2
79
sin   x
2
s
2

90
 8.875
30 13  60 
 0.123 rad / sec
d
c)
1
dt

90
ds
ds  x dx
dt s dt
s 2 cos  dt
1
  90 x dx
s 2 x s dt
s
  90 dx
s 2 dt
90
dx

2
x  8100 dt
Player slides into second base  x  0
d
90
1

 15  16 rad / sec
2
dt x  0
0  8100
d
2
dt
90
ds  90 x dx  90 dx
dt s 2 x s dt s 2 dt
2
s
90
dx

x 2  8100 dt

s 2 sin 
Player slides into second base  x  0
d
2
dt
x 0

90
02  8100
 15   16
rad / sec
80