Calculus
Name: __________________________
Related Rates Packet (SOLUTIONS) – PART I
Date: 12/6
Solutions to Related Rates Exercises (#1—4, 7)
d
y  3x  5
dt 
dy
dx
3
0
dt
dt
1. Equation: y  3 x  5
a. Given that dx / dt  2 , find dy / dt when x = 1.
b. Given that dy / dt  1 , find dx / dt when x = 0.
dy
dx
3
0
dt
dt
dx
b) 1  3
dt
dx 1

dt
3
dy
dx
3
0
dt
dt
a)
dy
 32  0  6
dt
Note: The derivatives themselves do not depend on the x or y coordinates… x=1 and x=0 were extraneous pieces of
information! 
2. A 6m ladder is against a wall. If its bottom is pushed toward the wall at a constant rate of
1m
,
2 s
how fast is the top of the ladder sliding when it (the top of the ladder) reaches 5m up the wall?
dx 1
dy

m s Find:
 ?? m s when y = 5.
dt
2
dt
Triangle Relationship Needed: Pythagorean Theorem
x 2  y 2  36
Take the derivative:
d
 x 2  y 2  36 
dt
dx
dy
2x
 2y
0
dt
dt
Plug in the “moment” y = 5….need to figure out x at this time too!
x 2  52  36
2
2
x  y  36 
x  11
2x
dx
dy
 2y
0
dt
dt

2 11
1
dy
 25
0
2
dt
Conclusion: The top of the ladder is sliding up at a rate of
the wall.
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
dy
11

ms
dt
10
11
m s when the ladder is 5 meters up
10
Calculus
Name: __________________________
Related Rates Packet (SOLUTIONS) – PART I
Date: 12/6
3. A winch on a dock is 20ft above the water level. It is attached to a boat at a point at water level
and reels the boat in at 2 ft sec . How fast is the boat moving when the rope the rope is 30 ft long?
20.05ft long?
dx
 2 ft s
dt
dr
Find:
 ?? ft s
dt
when r = 30 and when r = 20.05
Triangle Relationship Needed: Pythagorean Thm.
x 2  202  r 2
Take the derivative:
d
dx
dr
 x2  202  r 2   2x
 0  2r
dt
dt
dt
Plug in the “moment” r = 30….need to figure out x at this time too!
x 2  202  302  x  500
2x
dx
60
dx
dr
dx

ft s  2.683 ft s
 2 500
 0  2r
 0  2  30  2 
dt
dt
dt
dt
500
Conclusion: The boat is moving toward the dock at a rate of 2.683 ft s when rope is 30 ft long.
Plug in the “moment” r = 20.05….need to figure out x at this time too!
x 2  202  20.052  x  2.0025
dx
dr
dx
 2 20.0025
 0  2r
 0  2  20.05  2 
dt
dt
dt
dx
60

ft s  28.337 ft s
dt
20.0025
2x
Conclusion: The boat is moving toward the dock at a rate of 28.337 ft s when rope is 20.05 ft
long.
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Calculus
Name: __________________________
Related Rates Packet (SOLUTIONS) – PART I
Date: 12/6
3
4. Water is flowing into a cone (whose height is 16cm and radius is 4cm) at a rate of 2 cm min .
How fast is the water level rising when it is 10 cm deep?
dV
 2 cm3 min
dt
Find:
dh
 ?? cm min when h = 10
dt
Volume Relationship Needed:
1
V  r 2h Uh-oh! We have 2 variables, but we only want “h”
3
4
r
16
h
Similar Triangles Relationship Needed:
16 4
1
 r  h

h
r
4
Now substitute:
2
V 
Take the derivative:
d 
1

V 
h3  

dt 
48

1 1 

h h
3  4 

V 
1
h3
48
dV
1
dh

3h2
dt
48
dt
Plug in the “moment” when h = 10:
dh .32
1

 .102 cm min
2 dh
 dt
2
3(10)

48
dt
Conclusion: The water level is rising at a rate of .102 cm min when the water is 10 cm deep.
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Calculus
Name: __________________________
Related Rates Packet (SOLUTIONS) – PART I
Date: 12/6
7. A stone dropped into a still pond sends out a circular ripple whose radius increases at a constant
rate of 3 ft/s.
How rapidly is the area enclosed by the ripple increasing at the end of 10 s?
We know:
r
dr
 3 ft
s
dt
We want:
dA
 ?? ft when t = 10
s
dt
Geometry relationship: A  r 2
Take the derivative and use the moment (when t = 10, the radius will be 30
ft):
d
dA
dr
 A  r 2  

 2r
dt
dt
dt
dA
 2  30  3  180 ft
s
dt
Conclusion: The area enclosed by the ripple is increasing at the rate of 180 ft
seconds.
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s
at the end of 10