2.5 Topics: Related Rates Solutions
1. y  x
A. Given
dx
3
dt
Find
dy
dt
dx
dy
When x = 25
 2 Find
dt
dt
dy
1 dx
dx
dy


2 x
dt 2 x dt
dt
dt
dx
 2 25  2  20
dt
When x = 4
B. Given
y x
1
dy 1
1 dx
 dx
  x 2

dt 2
dt 2 x dt
dy
1
3

3 
dt 2 4
4
2. A point is moving along the graph of y  2 x 2  1 such that
A. x  1
dx
cm
2
. Find
dt
s
B. x  0
dy
for each of the following values of x.
dt
C. x  1
y  2x2  1
dy
dx
 4x 
dt
dt
Answers:
A.
dy
cm
 cm 
 4  1   2
  8
dt
s
 s 
B.
dy
cm
 cm 
 4  0   2
 0
dt
s
s


C.
dy
cm
 cm 
 4 1   2
 8
dt
s
 s 
3. The included angle of the two sides of constant equal length s of an isosceles triangle is  .
A. Show that the area of the triangle is given by A 
1 2
s sin 
2
Consider the following diagram.

s
s
If we draw a perpendicular line from the angle  to the base of the triangle it will bisect the angle 
and it will also bisect the base.
h
b
From my second diagram we can see that
1
b
 2

sin 
 b  2 s sin
2
s
2
 h

cos 
 h  s cos
2 s
2
A
1
1
 
 1 

 1
bh   2 s sin   s cos   s 2  2sin cos   s 2 sin 
2
2
2 
2 2 
2
2 2
s

2
1
b
2
B. If  is increasing at the rate of
Given:
d 1 rad

dt 2 sec
Find:
1

radian per minute, find the rates of change of the area when  
6
2
dA
dt
When:  

dA 1 2
d
 s  cos   
dt 2
dt
1
multiple (just like we do the )
2
A
1 2
s sin 
2
Solution: For  
For  

6

3
and  
6


Equation: A 
3

3
1 2
s sin 
2
NOTE: since s is constant we treat it as a constant
dA 1 2 
  1 1  3 1
3 2
 s  cos      s 2 
s
  
dt 2 
6   2  2  2   2 
8
dA 1 2 
  1 1 1 1 1
 s  cos      s 2       s 2
dt 2 
3  2 2 2 2 8
C. Explain why the rate of change of the area of the triangle is NOT constant even though
Because
and  
d
is constant.
dt
dA 1 2
d
d
dA
 s  cos   
even though s is constant and
is constant we can see that
still depends on cos  .
dt 2
dt
dt
dt
cos  is NOT constant because  is changing (at the given rate of
1
radian per minute).
2
4. A hemispherical water tank with radius 6 meters is filled to a depth of h meters. The volume of water in the tank is given by
1
V   h 108  h 2  0  h  6 . If water is being pumped into the tank at the rate of 3 cubic meters per minute, find the rate of
3
change of the depth of water when h  2 meters.
Given:
dV
m3
3
dt
min
Find:
dh
When: h  2 meters
dt
1
V   h 108  h 2 
3
dV 1 
dh
dh 
   h  2h   108  h 2  1 
dt 3 
dt
dt 
3 dV
dh
  2h 2  108  h 2 
 dt
dt

 m3  
dh dV
3
3


3


dt
dt  108  3h 2   min    108  3  2m 2


dh  m3   3
 3

dt  min   96 m 2
3 m
m

 .0298

min
 32 min

1
Equation: V   h 108  h 2 
3


3
3
  3 m 
  min    108m 2  3 2m 2
 








5. At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 10 cubic feet per minute. The diameter
of the base of the cone is approximately three times the altitude. At what rate is the height of the pile changing when the pile is 15 feet
high?
1
dh
dV
ft 3 Find:
Given:
When: h=15 feet
Eqn: V   r 2 h
 10
3
dt
dt
min
2
1
1  3h 
V   r 2h     h
3
3  2 
3 3
V  h
4
dV 3
dh
dh dV
4
   3h 2



dt 4
dt
dt
dt 9 h 2
h
3h

dh
ft 3
4
 10

dt
min 9 15 ft 2
dh
ft 3
4
8
ft
ft
 10


 .0063
2
dt
min 9  225 ft
405 min
min
6. A ladder 25 feet long is leaning against the wall of a house (see figure). The base of the ladder is pulled away from the wall at a
rate of 2 feet per second.
A. How fast is the top of the ladder moving down the wall when its base is 7 feet, 15 feet and 24 feet from the wall?
Given:
dx
ft
2
dt
sec
x 2  y 2  625
dx
dy
2x   2 y 
0
dt
dt
dy
x dx

dt
y dt
Find:
dy
dt
When: x = 7ft, 15 ft and 24 ft.
Equation:
x 2  y 2  252
Note: When x = 7ft, y = 24 ft and when x = 15 ft, y = 20 ft and when x = 24 ft, y = 7ft
these values were obtained using the Pythagorean theorem (i.e the “equation”)
So, when x = 7ft
dy
x dx
7 ft  ft 
7 ft
ft


 .583
2
 
dt
y dt
24 ft  sec 
12 sec
sec
when x = 15 ft
dy
x dx
15 ft  ft 
3 ft
ft


 1.5
2
 
dt
y dt
20 ft  sec 
2 sec
sec
when x = 24 ft
dy
x dx
24 ft  ft 
48 ft
ft


 6.857
2
 
dt
y dt
7 ft  sec 
7 sec
sec
B. Consider the triangle formed by the side of the house, the ladder and the ground. Find the rate at which the area of the triangle is
changing when the base of the ladder is 7 feet from the wall.
1
Given:
Equation:
dx
ft Find: dA When: x = 7 ft
A  xy
2
2
dt
sec
dt
A
1
xy 
2
dA 1  dy
dx  1 
ft  1  49 ft 2
ft 2  527 ft 2
 7 ft 
  x   y     7 ft   
 48

  
  24 ft  2
dt 2  dt
dt  2 
sec  2  12 sec
sec 
24 sec
 12 sec 
C. Find the rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 7 feet from
the wall.
d
x
dx
ft
Given:
Find:
When: x = 7 ft.
Equation: Tan 
2
dt
y
dt
sec
x
Tan 
y
x
   arc tan   
 y
d

dt
1
x
1  
 y
2

y
dx
dy
 x
1
dt
dt 
2
2
y
y  x2
dy 
 dx
 y  x 
1

dt
dt 
 7 


24   2   7     
2
2 
2
 12  
y
 24    7  
y2
d
1 
49 
1  576  49 
1  625  1 radians
radians
deg

 .083
 4.77
 48   




dt 625 
12  625  12  625  12  12 sec
sec
sec

y
25 ft
2 ft/sec
x
7. An air traffic controller spots two planes at the same altitude converging on a point as they fly at right angles to each other (see
diagram below). One plane is 225 miles from the point moving at 450 miles per hour. The other plane is 300 miles from the point
moving at 600 miles per hour.
A. At what rate is the distance between the planes decreasing?
Given:
dx
mi
 450
dt
hr
dy
mi
 600
dt
hr
Find:
ds
When: x = 225 miles, y = 300 miles Equation: x 2  y 2  s 2
dt
x2  y2  s2
dx
dy
ds
2x  2 y
 2s
dt
dt
dt
ds

dt
x
dx
dy 225mi  450 mi   300mi  600 mi 
y




mi
hr 
hr 


dt
dt 
 750
s
375mi
hr
Note: s = 375 was obtained by plugging in the values of x and y into the “equation”.
B) How much time does the air traffic controller have to get one of the planes on a different path?
time 
distance 375miles 1

 hr  30 minutes
miles 2
rate
750
hr
300
s
y
x
225
8. A man 6 feet tall walks at a rate of 5 feet per second away from a light that is 15 feet above the ground (see the diagram below).
When he is 10 feet from the base of the light…
A) at what rate is the tip of his shadow moving?
Note: I have labeled the given diagram in such a way that the man is “x” feet away from the light pole and the tip of the man’s
shadow is “y” feet away from the light pole and the length of his shadow is “y – x” feet.
dx
ft
5
dt
s
Given:
15
6

y yx
5
y x
3
dy
dt
Find:
When x = 10 feet
Equation:
15
6

(using similar triangles)
y yx
 15  y  x   6 y  15 y  15 x  6 y  9 y  15 x 
y
5
x
3
dy 5 dx 5  ft  25 ft
ft

 5  
 8.3
dt 3 dt 3  s 
3 s
s
B) at what rate is the length of his shadow changing?
dx
ft
5
dt
s
Given:
NOTE:
d  y  x
dt
Find:

d  y  x
dt
When x = 10 feet.
dy dx 25 ft
ft 25 ft 15 ft 10 ft
ft


5 


 3.3
dt dt
3 s
s
3 s 3 s
3 s
s
OR this could be solved utilizing the equation from part A.
d  y  x d  2 y 
15
6
6y
2y

  y  x 
  y  x 

   
y yx
15
5
dt
dt  5 
15
6
x
y
d  y  x
dt

2 dy 2  25 ft  10 ft
 

5 dt 5  3 s  3 s