lesson 13: related rates problems
TRANSCRIPT
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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 3 / 18
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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.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 4 / 18
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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?
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 5 / 18
. . . . . .
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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 6 / 18
. . . . . .
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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 7 / 18
. . . . . .
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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 7 / 18
. . . . . .
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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 7 / 18
. . . . . .
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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 7 / 18
. . . . . .
Outline
Strategy
Examples
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 8 / 18
. . . . . .
Strategies for Problem Solving
1. Understand the problem2. Devise a plan3. Carry out the plan4. Review and extend
György Pólya(Hungarian, 1887–1985)
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 9 / 18
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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.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 10 / 18
. . . . . .
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 10 / 18
. . . . . .
Strategies for Related Rates Problems
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.
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 10 / 18
. . . . . .
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 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.
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 10 / 18
. . . . . .
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
derivatives
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.
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 10 / 18
. . . . . .
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 10 / 18
. . . . . .
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 andsolve for the unknown rate.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 10 / 18
. . . . . .
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 10 / 18
. . . . . .
Outline
Strategy
Examples
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 11 / 18
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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?
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 12 / 18
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Diagram
.
.
..P
.m
.500
.w.w
.500.s
.4 ft/sec
.4 ft/sec
.s =
√(m+ w)2 + 5002
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 13 / 18
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Diagram
.
.
..P
.m
.500
.w
.w
.500.s
.4 ft/sec
.4 ft/sec
.s =
√(m+ w)2 + 5002
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 13 / 18
. . . . . .
Diagram
.
.
..P
.m
.500
.w
.w
.500
.s
.4 ft/sec
.4 ft/sec
.s =
√(m+ w)2 + 5002
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 13 / 18
. . . . . .
Diagram
.
.
..P
.m
.500
.w.w
.500.s
.4 ft/sec
.4 ft/sec
.s =
√(m+ w)2 + 5002
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 13 / 18
. . . . . .
Diagram
.
.
..P
.m
.500
.w.w
.500.s
.4 ft/sec
.4 ft/sec
.s =
√(m+ w)2 + 5002
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 13 / 18
. . . . . .
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 14 / 18
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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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 15 / 18
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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?
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 16 / 18
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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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 17 / 18
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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.
V63.0121.002.2010Su, Calculus I (NYU) Section 2.7 Related Rates May 27, 2010 18 / 18