Lesson 13: Related Rates Problems

Section 2.7
Related Rates

V63.0121.002.2010Su, Calculus I

New York University

May 27, 2010

Announcements

No class Monday, May 31
Assignment 2 due Tuesday, June 1

. . . . . .

Announcements

No class Monday, May 31
Assignment 2 due
Tuesday, June 1

. . . . . .

V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 2 / 18

Objectives

Use derivatives to
understand rates of
change.
Model word problems

. . . . . .


What are related rates problems?
Today we’ll look at a direct application of the chain rule to real-world
problems. Examples of these can be found whenever you have some
system or object changing, and you want to measure the rate of
change of something related to it.

. . . . . .


Problem

Example
An oil slick in the shape of a disk is growing. At a certain time, the
radius is 1 km and the volume is growing at the rate of 10,000 liters per
second. If the slick is always 20 cm deep, how fast is the radius of the
disk growing at the same time?
. . . . . .


A solution

Solution
The volume of the disk is

V = πr2 h.
. r
.
dV
We are given , a certain h
.
dt
value of r, and the object is to
dr
find at that instant.
dt

. . . . . .


Solution

Differentiating V = πr2 h with respect to time we have

0
dV dr dh¡
!
= 2πrh + πr2 ¡
dt dt ¡dt

. . . . . .


Solution


0
dV dr dh¡
! dr 1 dV
= 2πrh + πr2 ¡ =⇒ = · .
dt dt ¡dt dt 2πrh dt

. . . . . .


Solution


0
dV dr dh¡
! dr 1 dV
= 2πrh + πr2 ¡ =⇒ = · .
Now we evaluate:
dr 1 10, 000 L
= ·
dt r=1 km 2π(1 km)(20 cm) s

. . . . . .


Solution


0
dV dr dh¡
! dr 1 dV
= 2πrh + πr2 ¡ =⇒ = · .
Now we evaluate:
dr 1 10, 000 L
= ·
dt r=1 km 2π(1 km)(20 cm) s

Converting every length to meters we have

dr 1 10 m3 1 m
= · =
dt r=1 km 2π(1000 m)(0.2 m) s 40π s

. . . . . .


Outline

Strategy

Examples

. . . . . .


Strategies for Problem Solving

1. Understand the problem
2. Devise a plan
3. Carry out the plan
4. Review and extend

György Pólya
(Hungarian, 1887–1985)
. . . . . .


Strategies for Related Rates Problems

. . . . . .



1. Read the problem.

. . . . . .



2. Draw a diagram.

. . . . . .



2. Draw a diagram.
3. Introduce notation. Give symbols to all quantities that are
functions of time (and maybe some constants)

. . . . . .



2. Draw a diagram.
4. Express the given information and the required rate in terms of
derivatives

. . . . . .



2. Draw a diagram.
derivatives
5. Write an equation that relates the various quantities of the
problem. If necessary, use the geometry of the situation to
eliminate all but one of the variables.

. . . . . .



2. Draw a diagram.
derivatives
6. Use the Chain Rule to differentiate both sides with respect to t.

. . . . . .



2. Draw a diagram.
derivatives
6. Use the Chain Rule to differentiate both sides with respect to t.
7. Substitute the given information into the resulting equation and
solve for the unknown rate.

. . . . . .


Outline

Strategy

Examples

. . . . . .


Another one

Example
A man starts walking north at 4ft/sec from a point P. Five minutes later a
woman starts walking south at 4ft/sec from a point 500 ft due east of P.
At what rate are the people walking apart 15 min after the woman
starts walking?

. . . . . .


Diagram

4
. ft/sec

.
m
.

.
P
.

. . . . . .


Diagram

4
. ft/sec

.
m
.

. 5
. 00
P
.

w
.

4
. ft/sec
. . . . . .


Diagram

4
. ft/sec

.

.
s
m
.

. 5
. 00
P
.

w
.

4
. ft/sec
. . . . . .


Diagram

4
. ft/sec

.

.
s
m
.

. 5
. 00
P
.

w
. w
.

5
. 00
4
. ft/sec
. . . . . .


Diagram

4
. ft/sec

√ .

.
s
m
.
s
. = (m + w)2 + 5002

. 5
. 00
P
.

w
. w
.

5
. 00
4
. ft/sec
. . . . . .


Expressing what is known and unknown

15 minutes after the woman starts walking, the woman has traveled
( )( )
4ft 60sec
(15min) = 3600ft
sec min

while the man has traveled
( )( )
4ft 60sec
(20min) = 4800ft
sec min

ds dm dw
We want to know when m = 4800, w = 3600, = 4, and = 4.
dt dt dt

. . . . . .


Differentiation

We have
( )
ds 1( 2 2
)−1/2 dm dw
= (m + w) + 500 (2)(m + w) +
dt 2 dt dt
( )
m + w dm dw
= +
s dt dt

At our particular point in time

ds 4800 + 3600 672
=√ (4 + 4) = √ ≈ 7.98587ft/s
dt 2 + 5002 7081
(4800 + 3600)

. . . . . .


An example from electricity

Example
If two resistors with resistances
R1 and R2 are connected in
parallel, as in the figure, then . .
the total resistance R,
measured in Ohms (Ω), is given R
. 1 R
. 2
by . .
1 1 1
= +
R R1 R2
(a) Suppose R1 = 80 Ω and R2 = 100 Ω. What is R?
(b) If at some point R′ = 0.3 Ω/s and R′ = 0.2 Ω/s, what is R′ at the
1 2
same time?

. . . . . .


Solution

Solution
R1 R2 80 · 100 4
(a) R = = = 44 Ω.
R1 + R2 80 + 100 9
(b) Differentiating the relation between R1 , R2 , and R we get

1 1 1
− 2
R′ = − R′ −
1 R′
2
R R2
1 R2
2

So when R′ = 0.3 Ω/s and R′ = 0.2 Ω/s,
1 2
( ) ( )
′ 2 R′
1 R′2 R2 R2
1 2 R′
1 R′2
R =R + = +
R2 R2
1 2
(R1 + R2 )2 R2 R2
1 2
( )2 ( )
400 3/10 2/10 107
= 2
+ 2
= ≈ 0.132098 Ω/s
9 80 100 810

. . . . . .


Summary

Related Rates problems are an application of the chain rule to
modeling
Similar triangles, the Pythagorean Theorem, trigonometric
functions are often clues to finding the right relation.
Problem solving techniques: understand, strategize, solve, review.

. . . . . .


Lesson 13: Related Rates Problems

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