lesson 13: related rates problems

Section 2.7Related Rates

V63.0121.002.2010Su, Calculus I

New York University

May 27, 2010

Announcements

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

. . . . . .

. . . . . .

Announcements

I No class Monday, May 31I Assignment 2 due

Tuesday, June 1

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

. . . . . .

Objectives

I Use derivatives tounderstand rates ofchange.

I Model word problems


. . . . . .

What are related rates problems?

Today we’ll look at a direct application of the chain rule to real-worldproblems. Examples of these can be found whenever you have somesystem or object changing, and you want to measure the rate ofchange of something related to it.


. . . . . .

Problem

Example

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


. . . . . .

A solution

SolutionThe volume of the disk is

V = πr2h.

We are givendVdt

, a certainvalue of r, and the object is to

finddrdt

at that instant.

. .r.h


. . . . . .

Solution

Differentiating V = πr2h with respect to time we have

dVdt

= 2πrhdrdt

+ πr20

dhdt

=⇒ drdt

=1

2πrh· dVdt

.

Now we evaluate:

drdt

∣∣∣∣r=1 km

=1

2π(1 km)(20 cm)· 10,000L

s

Converting every length to meters we have

drdt

∣∣∣∣r=1 km

=1

2π(1000m)(0.2m)· 10m

3

s=

140π

ms


. . . . . .

Outline

Strategy

Examples


. . . . . .

Strategies for Problem Solving

1. Understand the problem2. Devise a plan3. Carry out the plan4. Review and extend

György Pólya(Hungarian, 1887–1985)


. . . . . .

Strategies for Related Rates Problems

1. Read the problem.2. Draw a diagram.3. Introduce notation. Give symbols to all quantities that are

functions of time (and maybe some constants)4. Express the given information and the required rate in terms of

derivatives5. Write an equation that relates the various quantities of the

problem. If necessary, use the geometry of the situation toeliminate all but one of the variables.

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.


. . . . . .


1. Read the problem.

2. Draw a diagram.3. Introduce notation. Give symbols to all quantities that are







. . . . . .


1. Read the problem.2. Draw a diagram.

3. Introduce notation. Give symbols to all quantities that arefunctions of time (and maybe some constants)

4. Express the given information and the required rate in terms ofderivatives

5. Write an equation that relates the various quantities of theproblem. If necessary, use the geometry of the situation toeliminate all but one of the variables.




. . . . . .



functions of time (and maybe some constants)

4. Express the given information and the required rate in terms ofderivatives





. . . . . .




derivatives





. . . . . .









. . . . . .






6. Use the Chain Rule to differentiate both sides with respect to t.

7. Substitute the given information into the resulting equation andsolve 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 awoman 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 womanstarts walking?


. . . . . .

Diagram

.

.

..P

.m

.500

.w.w

.500.s

.4 ft/sec

.4 ft/sec

.s =

√(m+ w)2 + 5002


. . . . . .

Diagram

.

.

..P

.m

.500

.w

.w

.500.s

.4 ft/sec

.4 ft/sec

.s =

√(m+ w)2 + 5002


. . . . . .

Diagram

.

.

..P

.m

.500

.w

.w

.500

.s

.4 ft/sec

.4 ft/sec

.s =

√(m+ w)2 + 5002


. . . . . .

Diagram

.

.

..P

.m

.500

.w.w

.500.s

.4 ft/sec

.4 ft/sec

.s =

√(m+ w)2 + 5002


. . . . . .

Expressing what is known and unknown

15 minutes after the woman starts walking, the woman has traveled(4ftsec

)(60secmin

)(15min) = 3600ft

while the man has traveled(4ftsec

)(60secmin

)(20min) = 4800ft

We want to knowdsdt

whenm = 4800, w = 3600,dmdt

= 4, anddwdt

= 4.


. . . . . .

Differentiation

We have

dsdt

=12

((m+ w)2 + 5002

)−1/2(2)(m+ w)

(dmdt

+dwdt

)=

m+ ws

(dmdt

+dwdt

)At our particular point in time

dsdt

=4800+ 3600√

(4800+ 3600)2 + 5002(4+ 4) =

672√7081

≈ 7.98587ft/s


. . . . . .

An example from electricity

ExampleIf two resistors with resistancesR1 and R2 are connected inparallel, as in the figure, thenthe total resistance R,measured in Ohms (Ω), is givenby

1R

=1R1

+1R2

.. .

. .

.R1 .R2

(a) Suppose R1 = 80Ω and R2 = 100Ω. What is R?(b) If at some point R′

1 = 0.3Ω/s and R′2 = 0.2Ω/s, what is R′ at the

same time?


. . . . . .

Solution

Solution

(a) R =R1R2

R1 + R2=

80 · 10080+ 100

= 4449Ω.

(b) Differentiating the relation between R1, R2, and R we get

− 1R2R

′ = − 1R21R′1 −

1R22R′2

So when R′1 = 0.3Ω/s and R′

2 = 0.2Ω/s,

R′ = R2

(R′1

R21+

R′2

R22

)=

R21R

22

(R1 + R2)2

(R′1

R21+

R′2

R22

)

=

(4009

)2(3/10802

+2/101002

)=

107810

≈ 0.132098Ω/s


. . . . . .

Summary

I Related Rates problems are an application of the chain rule tomodeling

I Similar triangles, the Pythagorean Theorem, trigonometricfunctions are often clues to finding the right relation.

I Problem solving techniques: understand, strategize, solve, review.


lesson 13: related rates problems

Technology

related rates problems1

nyu section

related ratesv63

rates of change

rh dt v63

r2 dt dtdt v63

dt dtdt dt

functions of time