RELATED RATES

By: Kyler HallAutrey: AP CalculusMarch 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. Identify all given quantities and the quantities that need to be determined.

2. Make a sketch of the problem and label the quantities.

3. 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.

Example of a Related Rates Problem: Rate of Change of the Volume of a Cube

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

dx= 2

Rate of Change of the Volume of a Cube continued…

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 = X 3

Rate of Change of the Volume of a Cube continued…

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:

23dV dx

Xdt dt

2(3)(2) (6)dV

dt

3V X

72dV

dt cm3/sec

Example of a Related Rates problem: Rate of Change of a plane

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?

Rate of Change of a plane continued…

First draw the shape and label the quantities indicated by the problem:

Then make an equation that would include any rates of change indicated by the problem:

x

yz

?dx

dt

0dy

dt

?x

5y

10z

240dz

dt

miles

miles hour

miles

hour

miles

2 2 2x y z

Rate of Change of a plane continued…

Now take the derivative, using the Chain rule in terms of time t, of the formula found in the last step:

Now plug in the values given by the problem:

2 2 2x y z

2 2 2dx dy dz

x y zdt dt dt

Which can be rewritten as:

dx dy dzx y z

dt dt dt

5(0) 10(240)dx

xdt

Note: The rate of change of y is equal to 0 since y is a constant.

Rate of Change of a plane continued…

With this problem another variable, x, has to b found because there are two variables that do not have numerical values:

Now solve for the rate of change of x:

2 2 2x y z 2 2 2(5) (10)x

2 75x 5 3x

5 3 10(240)dx

dt

2400

5 3

dx

dt

480 3160 3

3

dx

dt mile

shour

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.

L

W

4dL

dt

?dW

dt

200A LW

meters/second

m2

0dA

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 dLL W

dt dt dt

Since A equals 200 and A = LW then LW = 200. Therefore:

200L

W

200dA dW dLW

dt W dt dt

FRQ example continued…3. Now insert any known values into the equation and solve

for W.

200 10 4

2W

W

2

1004

4 100

WW

W

2 25

5

W

W

meters

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.

L

W

4dL

dt

?dW

dt

200A LW

meters/second

m2

0dA

dt meters/second

Since A is a constant

D

W = 10 meters

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.

2 2 2

2 2 2

W L D

dW dL dDW L D

dt dt dtdW dL dD

W L Ddt dt dt

FRQ example continued…

3. Insert any known values and find values that are missing that aren’t needed to answer the problem.

2 2 2

2 2 2

2

200 10

20

(20) (10)

500

500 10 5

dW dL dDW L D

dt dt dtLW W

W

W L D

D

D

D

10( .05) 20(4) 10 5

5 80 10 5

75

10 5

75 5

10 5 5

75 5 3 5

50 2

dD

dt

dD

dt

dD

dt

dD

dt

dD

dt

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.

ANSWERS

a)

b)

3

8

dOY

dt feet/second

33

16

dA

dt feet2/second

HERE’S HOW TO DO THE PROBLEM!!

Part A (Find the rate of change of OY)

Y

O X

0dXY

dt

1

2

dOX

dt

feet/second

feet/second

15XY feet

2

2 2 2

2 2

2

9

9 15

144

12

OX

OX OY XY

OY

OY

OY

2 2 2

19 12 15(0)

2

OX OY XY

dOX dOY dXYOX OY XY

dt dt dtdOY

dt

PART B (FINDING THE RATE OF CHANGE OF THE AREA)

feet/second

912

29 3

24 8

dOY

dtdOY

dt

1

21 1

2 21 3 1 1

9 122 8 2 2

A OX OY

dA dOY dOXOX OY

dt dt dtdA

dt

27 15 27 60

16 4 16 16

33

16

dA

dt

dA

dt

feet2/second

TRY IT YOURSELF!! 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:

1

20

dD

dt cm/min

Here’s how to do the problem!!

R

1dSA

dt cm2/min

24

2

2

SA R

D R

DR

224

2

2

1 20

1

20

DSA SA D

dSA dDD

dt dtdD

dt

dD

dt

TRY IT YOUSELF!!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) At the instant when the radius of the cone is 3 units, what is the rate of change of the area of its base?

b) At the instant when the radius of the cone is 3 units, what is the rate of change of its height H?

Answers

a) 3π in2/sec

b) 8 in/sec

Here’s how to do the problem!!

H

R

28

3

12

1

2

dV

dt

r

V

dR

dt

When:

cubic inches per second

inches

cubic inches

Inches per second

Part A (Finding the rate of change of the area of the base.)

2baseA R

Part B (Finding the rate of change of the height H)

2

12 3 3

2

dA dRR

dt dt

dA

dt

inches2/second

2

2

1

31 2

3 3

1 2 128 3 4 3

3 3 2

V R H

dV dH dRR HR

dt dt dt

dH

dt

2

2

3

3 12

3

4

VH

R

H

H

inches

28 3 4

24 3

8

dH

dt

dH

dtdH

dt

inches/second

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.