Calculus
Name: __________________________
Related Rates Packet (SOLUTIONS) – PART II
Date: 12/6
Solutions to Related Rates Exercises (#5, 6, 8--10)
5. A 6 ft tall man walks away from a lamppost that is 15ft tall at a rate of 5 ft sec . How fast is his
shadow lengthening? How fast is his shadow tip moving?
15
dm
 5 ft sec
dt
6
s
m
x
a) How fast is his shadow lengthening?
Similar Triangles Relationship Needed:
Also we know
s  m  x.
i.e. find:
ds
 ?? ft sec
dt
s
6
therefore 15s  6x

x 15
Let’s make a substitution since we know nothing about
15s  6(s  m)

9s  6m
dx
, nor care about it.
dt
Take the derivative and plug in the “moment”:
d
ds
ds
ds
 6 5  9
 30 
 3.333 ft sec
9s  6m  9

dt
dt
dt
dt
dx
b) How fast is his shadow tip moving? ? i.e. find:
 ?? ft sec
dt
s
6
Similar Triangles Relationship Needed:
therefore 15s  6x

x 15
Also we know s  m  x .
ds
Let’s make a substitution since we know nothing about
, nor care about it.
dt
15(x  m)  6x  15x  15m  6x  9x  15m
Take the derivative and plug in the “moment”:
d
dx
dm
dx
dx
 9
 15
 15  5 
 8.333 ft sec
9x  15m  9
dt
dt
dt
dt
dt
Notice… s  m  x …If I took the derivative of this, I’d get
with the above answers. Hmm….
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ds dm dx
. Compare this


dt
dt
dt
Calculus
Name: __________________________
Related Rates Packet (SOLUTIONS) – PART II
Date: 12/6
6. A plane is at an altitude of 4000ft. It is flying west at 700 ft sec . A searchlight, under its path
tracks the plane overhead. How fast is the light pivoting when the plane is 1000ft east of the
searchlight? How fast is the light pivoting when the plane is overhead?
dx
We know:
 700 m
s
dt
We want:
d
 ?? rad when x = 1000 and x = 0
s
dt
Trig Relationship Needed:
x
 x 
or   arc tan 
tan  

4000
 4000 
The “moment(s)”:
1000
 1000 
When x = 1000, tan  
   arc tan 
  .24498rad
4000
 4000 
When x = 0, tan  
0
 0 
 0    arc tan 
  0 rad
4000
 4000 
If we used OPTION 1:
x
4000
d 
x 
tan  
dt 
4000 
1
d
1
dx



2
cos  dt 4000 dt
tan  
Plug in the “moment” when x = 1000
1
d
1


 700
2
cos (.24498) dt
4000
d
 .165 rad
sec
dt
Plug in the “moment” when x = 0
1
d
1


 700
2
cos (0) dt
4000
d
 .175 rad
sec
dt
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If we used OPTION 2:
 x 
  arc tan 

 4000 
d 
 x 
  arc tan 


dt 
 4000  
d

dt
1
2

1
dx

4000 dt
 x 
1

 4000 
Plug in the “moment” when x = 1000
d
1
1


 700
2
dt
 1000  4000
1

 4000 
d
 .165 rad
sec
dt
Plug in the “moment” when x = 0
d
1
1


 700
2
dt
 0  4000
1

 4000 
d
 .175 rad
sec
dt
Calculus
Name: __________________________
Related Rates Packet (SOLUTIONS) – PART II
Date: 12/6
8. A spherical balloon is inflated so that its volume is increasing at the rate of 3 ft 3 / min . How
fast is the diameter of the balloon increasing when the radius is 1 ft?
3
dV
dd
. Find:
when r = 1 ft (i.e. when d = 2 ft)
 3 ft
 ?? ft
min
min
dt
dt
3
V 
4 d
4 3
4
r  V      V 
d 3
3 2
3
24
d
dt
4
dV
4
4

3
2 dd
2 dd
V  24 d   dt  24   3d dt  3  24   3  2 dt


dd
3 ft

dt
2 min
9. A 17-ft ladder is leaning against a wall. If the bottom of the ladder is pulled along the ground
away from the wall at a constant rate of 5 ft/s, how fast will the top of the ladder be moving
down the wall when it is 8 ft above the ground?
dx
dy
Find
 5ft
 ?? when y = 8
s
dt
dt
Moment: x 2  y 2  172  x 2  82  172  x = 15
Important Relationship: x 2  y 2  172
d
dx
dy
dy
 x 2  y 2  172   2x
 2y
 0  2(15)5  2(8)
0 
dt
dt
dt
dt
dy
 9.375 ft
s
dt
10. A softball diamond is a square whose sides are 60 ft long. Suppose that a player running
from first to second base has a speed of 25 ft/s at the instant when she is 10 ft from second
base. At what rate is the player’s distance from home plate changing at that instant?
dx
dy
2nd
Find:
 25 ft
 ?? ft when x = 50 ft.
s
s
dt
dt
d
dx
dy
602  x 2  y 2   2x
 2y
dt
dt
dt
x
2
2
2
“Moment”: When x = 50 ft, 60  50  y so y = 6100
dx
dy
dy
 2  50  25  2 6100

 2y
1st 2x
dt
dt
dt
dy
60 ft
 16.005 ft
s
dt
602  x2  y 2 
y
home
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