RELATED RATES
By: Kyler Hall
Autrey: AP Calculus
March 7, 2011
THINGS TO KNOW BEFORE STARTING A RELATED
RATES PROBLEM!

Chain Rule – If g is differentiable at x and f is
differentiable at g(x), then the composite
function F = fog defined by F(x) = f’(g(x)) is
differentiable at x and F’ is given by the
product:
F’(x) = f’(g(x)) • g’(x)
STEPS FOR SOLVING RELATED RATES
PROBLEMS
1.
2.
3.
Identify all given quantities and the quantities
that need to be determined.
Make a sketch of the problem and label the
quantities.
Write an equation that uses all of the variables
that have rates of change that are labeled or
need to be found.
STEPS FOR SOLVING RELATED RATES
PROBLEMS CONTINUED…
4. Using the Chain Rule, implicitly differentiate
both sides of the equation with respect to time
t.
5. After completing the previous step, substitute
the quantities into the equation as well as their
rates of change, if they are given. Then solve for
the rate of change that is required to answer
the problem and include any units given by the
problem.

All edges of a cube are expanding at a rate of
6 centimeters per second. How fast is the
volume changing when each edge is 2
centimeters?
- First, the quantities need to be identified,
once this has been done they need to be
labeled onto a picture of the cube
X=2
X
X
dy
=2
dx

Now, an equation needs to be written
that includes all of the variables and
rates of change either labeled or that
need to be found . In the case of this
problem the formula for the volume of a
cube needs to be written:
V = (l)(w)(h)
But since this since the problems labels the sides as
X the problem can be written:
V=X3

Now using the Chain rule, take the
derivate in terms of time t of the
equation, plug in the values given into
the equation of the derivate, and then
solve for the missing variable and include
its units:
V  X3
dV
2 dx
 3X
dt
dt
dV
 (3)(2) 2 (6)
dt
dV
 72 cm3/sec
dt

An airplane is flying at an altitude of 5 miles
and passes directly over a radar antenna.
When the plane is 10 miles away, the radar
detects that the distances is changing at a rate
of 240 miles per hours. What is the speed of the
plane?

First draw the shape and label the quantities indicated
by the problem:
dx
x?
dt
x
y  5 miles
y
dy
 0 miles
hour
dt

z
?
z  10 miles
dz
 240 miles
hour
dt
Then make an equation that would include any rates of
change indicated by the problem:
x2  y 2  z 2

Now take the derivative, using the Chain rule
in terms of time t, of the formula found in the
last step:
x2  y 2  z 2
2x

dx
dy
dz
 2y
 2z
dt
dt
dt
Which can
be rewritten
as:
x
dx
dy
dz
y
z
dt
dt
dt
Now plug in the values given by the problem:
dx
x  5(0)  10(240)
dt
Note: The rate of change of y
is equal to 0 since y is a
constant.

With this problem another variable, x, has to b
found because there are two variables that do
not have numerical values:
x y z
x 2  (5)2  (10)2
2

2
2
x 2  75
x5 3
Now solve for the rate of change of x:
 dx 
5 3    10(240)
 dt 
dx 2400

dt 5 3
dx 480 3
miles

 160 3 hour
dt
3
Example of an FRQ Related Rates
Problem
1977 AB 6
A rectangle has a constant area of 200 square
meters and its length is increasing at the rate of 4
meters per second.
a)
Find the width W at the instant the width is
decreasing at the rate of 0.5 meters per
second.
b)
At what rate is the diagonal D of the rectangle
changing at the instant when the width is 10
meters.
FRQ example continued…
Part A: (Finding the Width of W)
1. First indentify the quantities and label them on a
picture of the problem.
dL
 4 meters/second
L dt
W
A  LW  200 m2
dW
?
dt
dA
0
dt
meters/second
Since A is a constant
FRQ example continued…
2. Then write an equation that includes all of the rates
of change indicated by the problem, and use the
chain rule to take the derivative.
A  LW
dA
 dW
 L
dt
 dt

 dL 
 W  

 dt 
Since A equals 200 and A
= LW then LW = 200.
Therefore:
200
L
W
dA  200  dW


dt  W  dt

 dL 
 W  

 dt 
FRQ example continued…
3.
Now insert any known values into the equation and solve for W.
200  1 
0
     4 W
W  2
100
4W  
W
4W 2  100
W  25
W  5 meters
2
FRQ example continued…
Part B (Find the rate of change of diagonal D.
1.
Identify the quantities and label them onto a picture of the problem.
dL
 4 meters/second
L dt
D
W
dW
?
dt
W = 10 meters
A  LW  200 m2
dA
 0 meters/second
dt
Since A is a constant
FRQ example continued…
2.
Write an equation that include all the variables that have rates of
change indicated by the problem, and use the chain rule to take its
derivate.
W L D
dW
dL
dD
2W
 2L
 2D
dt
dt
dt
dW
dL
dD
W
L
D
dt
dt
dt
2
2
2
FRQ example continued…
3. Insert any known values and find values that are
missing that aren’t needed to answer the
problem.
 dD 
dW
dL
dD
W
L
D
dt
dt
dt
200  LW  10W
W  20
W 2  L2  D 2
(20) 2  (10) 2  D 2
D 2  500
D  500  10 5
10(.05)  20(4)  10 5 

dt


 dD 
5  80  10 5 

 dt 
75
dD

10 5 dt
75
5 dD


10 5
5 dt
dD 75 5 3 5


dt
50
2
meters/second
TRY IT YOURSELF!!
A ladder 15 feet long is leaning against a building
so that the end X is on level ground and end Y is
on the wall. X is moved away from the building at
the constant rate of .5 feet per second.
a)
Find the rate in feet per second at which the
length OY is changing when X is 9 feet from the
building.
b)
Find the rate of change in square feet per
second of the area of the triangle XOY when X is
9 feet from the building.
A NSWERS
a)
b)
dOY
3
  feet/second
dt
8
dA 33

feet2/second
dt 16
H ERE ’ S HOW TO DO THE PROBLEM !!
Y
dXY
0
dt
O
X
feet/second
dOX 1

dt
2
XY  15 feet
feet/second
Part A (Find the rate of change of OY)
OX  9
OX 2  OY 2  XY 2
OX 2  OY 2  XY 2
dOX
dOY
dXY
OX
 OY
 XY
dt
dt
dt
1
 dOY 
9    12 
  15(0)
2
 dt 
9  OY 2  152
2
OY 2  144
OY  12
dOY
9
12

dt
2
dOY
9
3


dt
24
8
PART B (F INDING
AREA )
1
A   OX  OY 
feet/second
THE RATE OF CHANGE OF THE
2
dA 1
dOY 1
dOX
 OX
 OY
dt 2
dt
2
dt
dA 1
 3 1
1
  9      12   
dt 2
 8 2
2
dA  27   15   27   60 
       
dt  16   4   16   16 
dA 33

feet2/second
dt 16


If a 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.
Answer:
dD
1

dt 20
cm/min
dSA
 1 cm2/min
dt
R
2
SA  4 R 2
D  2R
D
R
2
D
SA  4    SA   D 2
2
dSA
dD
 2 D
dt
dt
 dD 
1  20 

dt


dD
1

dt 20
The volume V of a cone is increasing at the rate
of 28π cubic inches per second. At the instant
when the radius R on the cone is 3 units, its
volume is 12π cubic inches, and the radius is
increasing at ½ inches per second.
a)
b)
At the instant when the radius of the cone is 3
units, what is the rate of change of the area of
its base?
At the instant when the radius of the cone is 3
units, what is the rate of change of its height H?
a) 3π in2/sec
b) 8 in/sec
dV
 28
dt
cubic inches
per second
When:
H
r 3
R
inches
V  12
dR 1

dt 2
cubic inches
Inches per
second
Part A (Finding the rate of change of the area
of the base.)
2
Abase   R
dA
 dR 
 2 R 

dt
 dt 
dA
1
 2  3    3
dt
2

inches2/second
Part B (Finding the rate of change of the
height H)
3V
H
 R2
312

H
2
 3 
H 4
inches
1
V  R2 H
3
dV 1
2  dH
 R 
dt 3
 dt
 2
 dR 
   HR 

3
dt



1
 dH  2
1
28    3 
    4  3  
3
 dt  3
2
 dH 
28  3 
  4
 dt 
dH
24  3
dt
dH
 8 inches/second
dt
WORKS CITED




Stewart, James. Calculus: Early Transcendentals.
Sixth Edition. United States: Thomson Brooks/Cole,
2008. 241-246. Print.
Leithold, First. The Calculus with Analytic Geometry.
Third edition. New York, NY: Harper & Row, 1976.
156-157. Print.
Larson, First, First Hostetler, and First Edwards.
Calculus I with Precalculus: A One-Year Course.
Boston: Houghton Mifflin Company, 2002. 300-305.
Print.
Larson, First, and First Edwards. Calculus. Ninth
edition. Belmont, CA: Brooks/Cole, Cengage
Learning, 2010. 150-156. Print.