2.5 Related Rates
1A) Equation: x  4 y  3
Given that
Given)
dx
dy
 1 , find
when x  2
dt
dt
dx
 1;
dt
Find)
dy
;
dt
When)
x2
Equation) x  4 y  3
Work: We will need to take the derivative with respect to time and solve for the “find”.
x  4y  3
d
d
 x  4y  3
dt
dt
dx
dy
 4  0
dt
dt
4dy
dx

dt
dt
dy
1 dx
dy

*Note that x did not appear, so
did not depend on When.
dt
4 dt
dt
dy
1
1
  1  
dt
4
4
2B)
Given that
Given)
Work:
 2 2 2 
dy
dx
 3, Find
when  x, y   
 2 , 2  .
dt
dt


dy
 3;
dt
Find)
x2  y 2  2x
d 2
d
x  y2    2x

dt
dt
2x 
2
dx
dy
dx
 2y   2
dt
dt
dt
y
dy dx xdx


dt dt dt
dx
;
dt
 2 2 2 
 2 , 2 


When)  x, y   
Equation)
x2  y 2  2x
y
dy
dx
 1  x 
dt
dt
y dy dx
 
1  x dt dt
2
dx
2
3 
dt
 2 2 2
 
2
2
2
2  3  dx
dt
2
2
dx
1  3 
dt
3)
3
dx
dt
Oil from a ruptured tanker spreads in a circular pattern whose radius increases at a constant rate of 2 ft/sec. How fast is the
area of the spill increasing when the radius of the spill is 60 ft?
r
Given)
dr
ft
2
;
dt
sec
Work:
dA
dr
   2r 
dt
dt
   2  60 ft   2
dA
ft 2
 240
dt
sec
ft
sec
Find)
dA
;
dt
When) r  60 ft
Equation) A    r
2
Example:
4)
A baseball diamond is a square whose sides are 90ft long. Suppose that a player running from second base to third base has a
speed of 30 ft/sec at the instant when he is 20 ft from third base. At what rate is the players distance from home plate
changing at that instant?
W
X
Y
90ft
Work:
Given:
dx
ft
 30
dt
sec
Find:
dy
dt
When: x  20 ft
Equation:
x 2  902  y 2
d 2
d
x  902   y 2

dt
dt
2x
dx
dy
 0  2y
dt
dt
x dx dy

y dt dt
x 2  90 2  y 2
20 2  90 2  y 2
20 ft 
ft  dy
  30

sec  dt
10 85 ft 
400  8100  y 2
*Note you have to solve the original equation for Y.
8500  y 2
 8500  y
10 85  y
y  10 85
Thus:
60 ft dy

85 sec dt
dy
ft
 6.51
dt
sec

***Note: Do not substitute a number in for a variable if the variable is changing. Wait until you have isolated your “find”
after you have taken the derivative.
5)
A camera 3000 ft away from the launching pad of a rocket is photographing the takeoff. If the rocket is rising vertically at
880 ft/sec when it is 4000 ft above the launch pad, how fast must the camera angle change at that instant to keep the camera
aimed at the rocket?
h
θ
3000 ft
Work:
dh
ft
d
 880
Find:
dt
sec
dt
d
d  h 
 tan    
dt
dt  3000 
Given:
sec2  
When: h  4000 ft
d
1
dh

1 
dt 3000
dt
d
1
1
dh

 2 
dt 3000 sec  dt
d
1
1


  880 
dt 3000  5 2
 
3
d
1
9

  880
dt 3000 25
d
66 rad d
rad 180
6.05

 .1056


dt 625 sec dt
sec  rad
sec
Equation:
tan  
h
3000
5)
Alternate method for solving #5
Work: Given:
dh
ft
 880
dt
sec
tan  
d

dt
d

dt
Find:
h
3000
  arctan
1
 h 
1 

 3000 
2

1 dh
3000 dt
2

1
 880
3000
1
 4000 
1 

 3000 
d
dt
When: h  4000 ft
h
3000
1
1

 88
16 300
1
9
1 1
3 1
66 rad

  22    22 
25 75
25 25
625 sec
9

Equation:
tan  
h
3000
Example:
6) As shown in the figure above, water is draining from a conical tank with height 12 feet and diameter 8 feet into a cylindrical tank
that is changing at the rate of (h-12) feet per minute. The volume V of a cone with radius r and height h is
1
V   r 2h
3
Work:
Given:
A)
dh
ft
 h  12
dt
min
Find:
dv
dt
Equation:
1
V   r 2h
3
Write an expression for the volume of water in the conical tank as a function of h.
4
* using similar triangles we find what r is equal to.
1
V   r 2h
3
12
2
1 1 
V    h h
3 3 
1
V   h3
27
B)
When: h  3 ft
r 4

h 12
1
r h
3
At what rate is the volume of water in the conical tank changing when h=3? Indicate units of measure.
d
1
dh
V    3h 2 
dt
27
dt
dv 1
ft 

   3  3 ft 2    9

dt 27
 min 
dv
ft 3
 9
dt
min
C)
Let y be the depth, in feet, of the water in the cylindrical tank. At what rate is y changing when h=3 ? Indicate units of
measure.
Given:
dv
ft 3
 9
dt
min
Find:
dy
dt
When: h  3 ft
Equation: V  Area of Base  Height
V  400 ft 2  y
dv
dy
1
dv dy
 400 ft 2 

 
2
dt
dt 400 ft dt dt

9 ft
dy
1
9 ft 3 dy




2
400 min dt
400 ft min
dt
Angular Rate of Change: In a free-fall experiment, an object is dropped from a height of 256 feet. A camera on the ground
500 feet from the point of impact records the fall of the object.
S
256ft
θ
500ft
A)
Find the position function giving the height of the object at time t, assuming the object is released at time t = 0. At what
time will the object reach ground level?
s  t   16t 2  V0t  S0
s  t   16t 2  0  t  256
s  t   16t 2  256
(* Position function)
When will the object reach the ground?
0  16t 2  256
16t 2  256
t 2  16
t  4
t  4 sec
B)
Find the rates of change of the angle of elevation of camera when t = 1 and t = 2.
Find
d
dt
When t  1sec; t  2sec
Equation: tan 

s
500
s
500
16t 2  256
  arctan 
500
d
1
1


  32t 
2
dt
 16t 2  256  500
1 

500


  arctan
When t = 1:
d

dt
1
 16 12  256 
1 



500


2

1
  32 1   0.0520
500
When t = 2:
d

dt
1
 16  2   256 
1 



500


2
2

1
  32  2    0.1115511183
500