In this section we look at problems where changing quantities are related. For instance, a growing oil slick is changing in diameter and volume at the same time. How are the rates of change of these quantities related? The chain rule for derivatives is the key.

1. Section 2.7
Related Rates
V63.0121.027, Calculus I
October 20, 2009
Announcements
Midterm average 57.69/75 (77%), median 59/75 (79%),
standard deviation 11%
Solutions soon.
. . . . . .

2. “Is there a curve?”
Midterm
Mean was 77% and
standard deviation
was 11%
So scores average are
good
Scores above 66/75
(88%) are great
For final letter grades,
refer to syllabus
. . . . . .

3. 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.
. . . . . .

4. 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?
. . . . . .

5. A solution
The volume of the disk is
V = π r2 h .
. r
.
dV
We are given , a certain h
.
dt
value of r, and the object is
dr
to find at that instant.
dt
. . . . . .

6. Solution
Solution
Differentiating V = π r2 h with respect to time we have
0
dV dr dh¡
!
= 2π rh + π r2 ¡
dt dt ¡dt
. . . . . .

7. Solution
Solution
Differentiating V = π r2 h with respect to time we have
0
dV dr dh¡
! dr 1 dV
= 2π rh + π r2 ¡ =⇒ = · .
dt dt ¡dt dt 2π rh dt
. . . . . .

8. Solution
Solution
Differentiating V = π r2 h with respect to time we have
0
dV dr dh¡
! dr 1 dV
= 2π rh + π r2 ¡ =⇒ = · .
dt dt ¡dt dt 2π rh dt
Now we evaluate:
dr 1 10, 000 L
= ·
dt r=1 km 2π(1 km)(20 cm) s
. . . . . .

9. Solution
Solution
Differentiating V = π r2 h with respect to time we have
0
dV dr dh¡
! dr 1 dV
= 2π rh + π r2 ¡ =⇒ = · .
dt dt ¡dt dt 2π rh dt
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
. . . . . .

10. Outline
Strategy
Examples
. . . . . .

11. 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)
. . . . . .

12. Strategies for Related Rates Problems
. . . . . .

13. Strategies for Related Rates Problems
1. Read the problem.
. . . . . .

14. Strategies for Related Rates Problems
1. Read the problem.
2. Draw a diagram.
. . . . . .

15. 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)
. . . . . .

16. 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
. . . . . .

17. 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 the
problem. If necessary, use the geometry of the situation to
eliminate all but one of the variables.
. . . . . .

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 the
problem. If necessary, use the geometry of the situation to
eliminate all but one of the variables.
6. Use the Chain Rule to differentiate both sides with respect to
t.
. . . . . .

19. 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 the
problem. If necessary, use the geometry of the situation to
eliminate 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.
. . . . . .

20. Outline
Strategy
Examples
. . . . . .

21. 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?
. . . . . .

22. Diagram
4
. ft/sec
m
.
.
. . . . . .

23. Diagram
4
. ft/sec
m
.
. 5
. 00
w
.
4
. ft/sec
. . . . . .

24. Diagram
4
. ft/sec
.
s
m
.
. 5
. 00
w
.
4
. ft/sec
. . . . . .

25. Diagram
4
. ft/sec
.
s
m
.
. 5
. 00
w
. w
.
5
. 00
4
. ft/sec
. . . . . .

26. Diagram
4
. ft/sec
√
s
.= (m + w)2 + 5002
.
s
m
.
. 5
. 00
w
. w
.
5
. 00
4
. ft/sec
. . . . . .

27. 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
We want to know when m = 4800, w = 3600, = 4, and
dt dt
dw
= 4.
dt
. . . . . .

28. Differentiation
We have
( )
ds 1( )−1/2 dm dw
= (m + w)2 + 5002 (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)
. . . . . .