Calculus
Name: __________________________
Related Rates Packet (SOLUTIONS) – PART III
Date: 12/6
Solutions to Related Rates Exercises (#11, 12, 13, 14)
11. A point, P, is moving along a curve whose equation is: y  x 3  17 When P is at (2,5), y is
increasing at a rate of 2 units/sec. How fast is x changing?
dy
 2 units
s
dt
dx
Find:
 ?? units when x = 2 and y = 5
s
dt
y



3
y  x  17
1
dy 1 3
dx

x  17 2  3x 2 
dt
2
dt


Plug in the “moment”:












1
1 3
dx
2
2  17 2  3  22 
2
dt
5 dx

3
dt
Conclusion: x is increasing at a rate of 5 3 units/sec at the moment when point P is at (2,5).





x



12. A trough is 15 ft long and 4 ft across the top. Its ends are isosceles triangles with height 3 ft.
Water runs into the trough at a rate of 2.5 ft3/min. How fast is the water level rising when it is 2
ft deep?
3
dV
.
 2.5 ft
min
dt
3
dh
Find:
when h = 2ft.
 ?? ft
min
dt
Relationship(s) from Geo:
4
b
Since
3
h
V 
1
h b
bh  length and

2
3 4
4h
 b and the length is 15ft, the volume will be
3
1 4h
V  
 h  15
2 3
Simplified, this is: V  10h2 .
d
dV
dh
V  10h2  
 20h
dt
dt
dt
dh
dh
Plug in your “moment”: 2.5  20  2 

 .0625 ft
min
dt
dt
Take the derivative:
Conclusion: the water level is rising at a rate of 0.0625 ft/min when the water is 2 feet deep.
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

Calculus
Name: __________________________
Related Rates Packet (SOLUTIONS) – PART III
Date: 12/6
13. If a spherical snowball melts so that its surface area decreases at a rate of 1 cm2/min, find the
rate at which the diameter decreases when the diameter is 10 cm. [Surface area = 4r 2 ]
3
dSA
 1 cm
min
dt
Find:
dd
when d = 10cm
 ?? cm
min
dt
d
2
2
d 
Using substitution, we combine these to get: SA  4    SA  d 2
2
d
dSA
dd
SA  d 2  
Take the derivative:
 2d
dt
dt
dt
dd
dd
1 cm
Plug in the “moment: 1  2  10 


 0.016 cm
min
min
dt
dt
20
Important Relationships: SA  4r 2 and r 
Conclusion: the diameter is decreasing at a rate of 0.016 cm/min when the diameter is 10cm.
14. Two cars start moving from the same point. One travels south at 60 mi/h and the other travels
west at 25 mi/h. At what rate is the distance between the cars increasing two hours later?
dy
dx
and
 60 mi
 25 mi
hr
hr
dt
dt
dn
Find:
when t = 2 hrs
 ?? mi
hr
dt
2
2
2
Relationships: d  r  t and x  y  n
x
n
Take derivative:
d
dx
dy
dn
 x2  y 2  n2   2x
 2y
 2n
dt
dt
dt
dt
Plug in the moment:
When t = 2 hrs, x = -50 and y = -120, n 
16900
dn
2  50  25  2  120  60  2 16900
dt
dn
 65 mi hr
dt
Conclusion: the distance between the cars is increasing at a rate of 65mi/hr two hours after they start
moving from the same point.
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