Related Rates:
Consider the following problem:
The mechanics at a Toyota Automotive are reboring a 6-in deep cylinder to
fit a new piston. The machine they are using increases the cylinder’s radius onethousandth of an inch every 3 minutes. How rapidly is the cylinder volume
increasing when the bore’s radius is 3.800 in.?
r
Step 1: Picture and Variables
Let
3
V = volume (in ) of the bore at time t (min)
r = radius (in) of the bore at time t.
Step 2: Numerical Information:
r = 3.800 in and
MATH 1310
Ronald Brent © 2016 All rights reserved.
6 in
dr
1

in/min
d t 3000
Lecture 21
1 of 11
Step 3: What do we want to find? Ans:
dV
dt
Step 4: How are the variables related? The cylinder’s volume is:
V  ( r 2 )6
Step 5: Differentiate with respect to t.
dV
dr
 12  r
dt
dt
Step 6: Evaluate:
dV
1
 12   3.800 
so
dt
3000
dV 3.8 
3

 0.0478 in /min
dt
250
MATH 1310
Ronald Brent © 2016 All rights reserved.
Lecture 21
2 of 11
The method for solving related-rate problems is as follows:
1.
First draw a picture labeling all appropriate variables. Use t for time.
Assume all variable are differentiable functions of time.
2.
Write down the given numerical information (in terms of the symbols
you have chosen.)
3.
Write down what you are being asked to find. (Usually a rate, expressed
as a derivative.)
4.
Write an equation that relates the variables. You may have to combine
two or more equations to get a single equation that relates the variable
whose rate you want to the variables whose rates you know.
5.
Differentiate with respect to t. Then express the rate you want in terms
of the rate and variables whose values you know.
6.
Evaluate. Use known values to find the unknown rate.
MATH 1310
Ronald Brent © 2016 All rights reserved.
Lecture 21
3 of 11
Example:
A spherical balloon is being filled with air at a rate such that its radius is
increasing at a rate of 2 inches per second. How quickly is the volume
increasing when the radius is 6 inches?
Step 1: Picture and Variables
3
V = volume (in ) of the balloon at time t (s)
r = radius (in) of the balloon at time t.
r (t )
Step 2: Numerical Information:
r  6 in
dr
 2 in/s
dt
Step 3: We wish to find
MATH 1310
Ronald Brent © 2016 All rights reserved.
dV
.
dt
Lecture 21
4 of 11
Step 4: How are the variables related?
4 3
V (r)   r
3
Step 5: Differentiate with respect to t.
d
d 4
3
2 dr
V (t )    [ r (t )]   4  [ r (t )]
dt
dt  3
dt

Step 6: Evaluate:
When r  6 , and
dr
2
dt
dV
3
 4 (6 2 )( 2)  288   905 in /s
dt
MATH 1310
Ronald Brent © 2016 All rights reserved.
Lecture 21
5 of 11
Example:
A rocket is being launched. It takes off with a velocity of 550 miles
per hour. 25 miles away, there is a photographer video-taping the launch. At
what rate it the angle of elevation of the camera changing when the rocket
achieves an altitude of 25 miles?
Step 1: Picture and Variables
Let h(t ) be the height (mi) of the
rocket at time t (hr).
h (t )
 (t ) is the angle of elevation at time t.
 (t )
25 mi
Step 2: Numerical Information:
dh
 550 mi/hr
dt
MATH 1310
Ronald Brent © 2016 All rights reserved.
h = 25 mi
Lecture 21
6 of 11
Step 3: We wish to find
d
.
dt
Step 4: How are the variables related?
tan  
h (t )
25
Step 5: Differentiate with respect to t.
d
d h (t ) 1 d h
d 1 d h
or sec 2 

tan  

dt
d t 25 25 d t
d t 25 d t
d
Solving for
gives
dt
d cos 2  d h

dt
25 d t
Step 6: Evaluate:
d
 22 cos 2 
dt
MATH 1310
Ronald Brent © 2016 All rights reserved.
Lecture 21
7 of 11
To determine cos 2  , recall that the problem asks for
d
when the rocket is 25
dt
mile high, so the triangle looks like:
Since   45 

cos 
1
, and
2
4
1
hence cos 2   , and so
2
d
1
 22  11 radians per hour.
dt
2
25
 (t )
25
MATH 1310
Ronald Brent © 2016 All rights reserved.
Lecture 21
8 of 11
Example:
3
A conical tank is being filled at a rate of 9 ft /min. The tank stands point
down and has a height of 10 ft and a base radius of 5 ft. How fast is the water
level rising when the tank is 6 ft deep?
5 ft
Step 1: Picture and Variables
x
10 ft
y
Let V = volume (ft3) of the water in tank at time t (min)
x = radius (ft) of the surface of the water at time t.
y = depth (ft) of water and time t.
MATH 1310
Ronald Brent © 2016 All rights reserved.
Lecture 21
9 of 11
Step 2: Numerical Information:
y = 6 ft and
dV
3
 9 ft /min
dt
Step 3: We wish to find
dy
.
dt
Step 4: How the variable are related: The water forms a cone of volume:
1
V   x2 y
3
dx
Since no information is given about x or
, we need to eliminate x from
dt
the problem. The similar triangles give
x 5
1

or x  y
y 10
2
So
MATH 1310
Ronald Brent © 2016 All rights reserved.
Lecture 21
10 of 11
2

1
y
V      y  y 3
3 2
12
Step 5: Differentiate with respect to t.
dy  2 dy
dV 
 (3 y 2
)  (y
)
dt
4
dt
d t 12
Step 6: Evaluate:
9

4
(6 2
dy
) so
dt
dy 1
  0.32 ft/min.
dt 
MATH 1310
Ronald Brent © 2016 All rights reserved.
Lecture 21
11 of 11