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Related Rates
1
The Basic Question
If one quantity is changing in an equation or situation
at a particular moment, how much is another quantity
changing
The Problem
We are given one rate, a relationship relating variables,
and asked to find another rate
2
Remarks
We usually have three or more variables
For example, we might have two dependent variables
(for example x and y), and one independent variable
(for example t)
We are given the relation between two variables (for
example x and y)
We are given one rate (for example dx/dt) at specific
values (for example x, y, and z values)
We are asked to find the other rate (ex: dy/dt)
3
1
With Related Rates
We use Liebniz’s notation of dy/dt
We take derivatives implicitly
We know standard formulas for perimeter, surface
area, area, and volume for common geometric
figures
We know the Pythagorean Theorem
We draw sketches to understand and solve
We must use the correct units
4
Related Rates Process
1. Draw a picture (essential) include all variables
2. Do not label variables as constants
3. Translate to a math problem – search for equations
that relate quantities
4. Implicitly differentiate the equation
5
Remarks
The key to solving related rates problems is to find
an equation which relates the quantities whose rates
of change are known to the quantity whose rates of
change we would like to find.
Then we differentiate this equation implicitly with
respect to the appropriate variable (typically t, for
time).
Finally, we plug in all of our known values so that
we can solve for the rate of change we want.
6
2
Example
Suppose that the radius r and the
circumference of a circle are both
functions of time
Write an equation that relates dC/dt to
dr/dt
C  2 r
dC
dr
 2
dt
dt
7
Example
Suppose that the radius r and the volume
of a sphere are both functions of time
Write an equation that relates dV/dt to
dr/dt
V
4 r 3
3
dV
dr
 4 r 2
dt
dt
8
Example
Each side of a square is increasing at
a rate of 2 cm/sec
How fast is the perimeter changing?
P  4x
dP
dx
4
dt
dt
x
dP
cm
 42  8
dt
s
9
3
Example
A snowball melts at 2 ft3/h.
Assume it is spherical. Find
the rate the radius is changing
when the radius is 20 inches.
10
Snowball melting
Example
r
r
4
V   r3
3
20 inches
1 ft
5
 ft
1
12 inches 3
Snowball melting 
dV
 2
dt
11
Snowball melting
Example
4
V   r3
3
Implicit differentiate with respect to time
dV
dr
 4 r 2
dt
dt
2
 5  dr
2  4  
 3  dt
dr
9 ft

dt
50 hour
See http://www.youtube.com/watch?v=zXrm-Smy0qU
for another example
12
4
Example
A 5 ft tall Calculus student walks away from a lamp
post at 7 ft/s. The lamp post is 20 ft tall. When she
is 8 ft from the lamp post, find the rate at which the
tip of her shadow is moving.
13
Example
Calculus student
Shadow
Tip
14
Example
Calculus student
Sketch
zx
20
5
x
z
15
5
Example
Calculus student
Compare similar triangles
20
5
zx
z
and develop equation
z zx

20
5
 3z  4 x
16
Example
Calculus student
Differentiate  3 y  4 x
3
dy
dx
4
dt
dt
Solve for
dy
dt
dz 4
28 ft
 7 
dt 3
3 sec
17
Example
A 5 ft tall Calculus student walks away from a lamp
post at 7 ft/s. The lamp post is 20 ft tall. When she
is 8 ft from the lamp post, find the rate at which the
length of her shadow is changing.
18
6
Example
Calculus student
Shadow
Length
19
Example
Calculus student
Sketch
The length of the shadow
is represented by y
20
y
5
x
x y
20
Example
Calculus student
Compare similar triangles
20
5
x y
y
and develop equation
x y y
  3y  x
20
5
21
7
Example
Calculus student
Differentiate  3 y  x
3
dy dx

dt dt
dy
dt
Solve for
dz 1
7 ft
 7 
dt 3
3 sec
22
Example
Calculus student
x y  z
Compare
Problems
20
dx dy dz


dt dt dt
y
5
7
7 28

3 3
x
z
See http://www.youtube.com/watch?v=JKXgOIJ_N9E
for another example
23
Example
A car traveling west on I-10 passes
a highway patrolman parked 90 feet
north of the interstate
When a car is 150 feet from the
patrolman’s car, radar indicates the
car is approaching at 72 ft/sec
How fast is the car traveling?
24
8
Example
police car
Sketch
x
90
y
25
Example
police car
Find the equation and
differentiate
x
90
y 2  902  x2
y
2y
y
dy
dx
 2x
dt
dt
dy
dx
x
dt
dt
26
Example x
police car
90
y
y 2  902  1502  y  120
dy
dx
x
dt
dt
dy
120  150  72 
dt
dy
feet
 90
dt
sec
(about 61 mph)
y
27
9
Example
police car
For more information try
https://www.youtube.com/playlist?list=PL5KkMZvBpo
5Ahgc5EnzYPYjBqz3utcPNv
This link is to a video series of related rates
examples, including the police car problem
28
Example
A 26 foot ladder leans against a vertical
wall
The ladder is sliding down
When the foot of the ladder is 10 ft from
the base of the wall it is moving at 4 ft/sec
How fast is the top of the ladder moving
down the wall at that instant?
29
Example
slipping ladder
x 2  y 2  262
Sketch
26
y
x
dx
dy
 2y
0
dt
dt
dx
dy
2 x  2 y
dt
dt
2x
x  10  y  262  102  24
dx
4
dt
30
10
Example
x
dx
dy
 y
dt
dt
slipping ladder
x  10, y  24
dx
4
dt
4 10   24
dy
dt
dy
5 ft

dt
3 sec
See http://www.youtube.com/watch?v=m7qZDl_82Wo
for another example
31
Example
The cost C (in dollars) of manufacturing
x number of high-quality computer laser
printers is
C(x) = 15x4/3 + 54x2/3 + 600,000
Currently, the level of production is 1728 printers
and that level is increasing at the rate of 350
printers each month.
Find the rate at which the cost is increasing each
month
32
Example
laser printers
C  x   15 x 4 3  54 x 2 3  600000
dx
dC
=350 when x  1728 and we are to find
dt
dt
dC
dx
  20 x1 3  36 x 1 3 
dt
dt


dC
13
1 3
 20 1728  36 1728
350
dt
 $85,050 per month
33
11
Example
The monthly revenue R (in dollars) of a telephone
polling service is related to the number x of completed
responses by the function
R  x   25 3.5x 2  25x  12,000 on 0,1500
If the number of completed
responses is increasing at the rate
of 10 forms per month, find the rate
at which the monthly revenue is
changing when x = 750
Stop polling
me!
34
Example
telephone polling
R  x   25 3.5 x  25 x  12, 000 on 0,1500
2
dx
forms
 10
when x  750
dt
month
R  25  3.5 x 2  25 x   12,000
12
1 2
dR 25
dx
  3.5 x 2  25 x   7 x  25
dt
2
dt
1 2
25
2

3.5  750   25  750 
 7  750  2510
2


 $469.73 per month
Example
35
A pile of sand is shaped like a
right circular cone.
A pile of sand is shaped like a right circular cone. It
is shaped such that the base radius of the pile is one
third of its height.
Sand is being added to the pile at the steady rate of
10 ft3/min.
How fast is the height of
the pile rising when the
pile is 4 feet tall?
36
12
Example
A pile of sand is shaped like a
right circular cone.
We are given
1
V   r 2h
3
1
r h
3
dV
ft 3
 10
dt
min
dh
when h  4 ft
dt
We are asked to find
1 1 
 3
V    h h 
h
3 3 
27
2
Writing V in terms of h
Taking the derivative
implicitly
37
A pile of sand is shaped like a
right circular cone.
Example
Solving for
dV  2 dh
 h
dt 9
dt
dV  2 dh
 h
dt 9
dt
dh
dt
10 

9
 4
2
dh
dt
45 ft
dh
90


8 min
dt 16
38
Example
A bird is flying horizontally 40 feet
above your head at 20 ft/s
How fast is the angle of elevation
changing when your horizontal
distance from the bird is 30 feet?
40 ft
39
13
bird flying
Example
Sketch
x 2  40
40

x
x  30  x2  40  50
40
bird flying
Example
dx
ft
 20
dt
sec
40
From sketch tan  
x
Given x  30 ft,
50
40

Differentiate
30
From sketch sec 
sec2
d
40 dx
 2
dt
x dt
5
25
 sec2 
3
9
41
Example
Substitute know values into sec2
bird flying
d
40 dx
 2
dt
x dt
40
 5  d

20 
 
2 
2
dt
 
 30 
2
d
8 radians

dt
25 second
42
14
Two Ships
A ship is 12 miles due north of port and is traveling
north (away from port) at a rate of 4 miles per hour
(mph). Another ship is 5 miles due east of port and is
traveling west (back towards the port) at a rate of 3
mph. What is the rate of change of the distance
between the ships?
43
Ship A
c  122  52  13
c = 13
We want to find
a=12
Port
b=5
dc
dt
Ship B
We note that Ship A is moving away from port
faster than Ship B is moving towards it, so we
believe the distance between the two ships is
increasing.
44
We are given that Ship A is moving away from
port at a rate of 4 mph. This means that the
rate of change of distance a between Ship A
and the port is increasing, so
da
4
dt
We know that Ship B is moving towards port at a
rate of 3 mph. This means that the rate of change
of distance (b) between Ship B and the port is
decreasing, so
db
 3
dt
45
15
Finally, we need an equation that relates a, b, and c.
The most obvious one that comes to mind is the
Pythagorean Theorem which relates the lengths of
the sides of a right triangle:
c 2  a 2  b2
Using implicit differentiation with respect to time,
t, we get:
2c
dc
da
db
 2a
 2b
dt
dt
dt
46
2c
dc
da
db
 2a
 2b
dt
dt
dt
Since we want to solve for we can divide both
sides by and then plug in all known values.
da
db
b
dt
dt

c
dc 2(12)(4)  2(5)(3)

 2.54
dt
2(13)
dc a da b db


dt c dt c dt
a
This tells us that the distance between the two ships
is increasing at a rate of 2.54 mph
47
What if Ship A was traveling due south (back
towards port) at a rate of 4 mph?
In this case, all of the values are da
 4
exactly the same except
dt
Then
dc 2(12)(4)  2(5)(3)

 4.85
dt
2(13)
This time the distance between the two ships is
decreasing at a rate of 4.85 mph. Since both ships are
heading back towards port, the distance between
them is getting smaller, and at a quicker rate.
48
16
Pythagorean Theorem Problems
There are some special triangles that have sides that are
integers. If you recognize one these in your problem, it
can save work in your solution.
3, 4, 5
5, 12, 13
8, 15, 17
7, 24, 25
9, 40, 41
12, 35, 37
You don't need to remember these because you can
use the Pythagorean Theorem to solve the triangle.
Also multiples of these triangles are often used.
49
Frequently Used Formulas
Area of a triangle:
Area of a circle:
Area of a square:
Volume of a sphere:
Surface area of a sphere:
Surface area of a cylinder
Volume of a cylinder:
Volume of a cone:
Volume of a cube:
Surface Area of a Cube:
(1/2)bh
 r2
s2
(4/3)  r3
4  r2
2 r h + 2 r2 (both ends)
2 r h
(no ends)
2 r h +  r2 (one end)
 r2h
(1/3)  r2h
s3
6 s2
50
Remarks
Related Rates Tutorials and Animations
http://astro.ocis.temple.edu/~dhill001/relate
drates/relatedrates.html
http://www2.sccfl.edu/lvosbury/CalculusI_Folder/RelatedRate
Problems.htm
http://people.hofstra.edu/Stefan_Waner/Real
World/tutorials/frames4_4.html
http://www.karlscalculus.org/calc8_1.html
51
17
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