Chapter 2
Differentiation:
Related Rates – Day 1
1
Related Rates


When we talk about a quantity change with
respect to time (whether something is increasing,
decreasing, growing, shrinking, etc.), we are
basically talking about the derivative of that
quantity with respect to time.
For example, let’s say little Johnny is growing at a
rate of 2 inches per year. We can write this rate
mathematically:
dh
 2 in. / year
dt

Basically, this tells me that the change in height
with respect to time is 2 in. per year.
Related Rates

Determine the mathematical expressions for each
of the following:
1.
The radius of a circle is increasing at 4 ft/min
dr
 4 ft / min
dt
2.
3.
The volume of a cone is decreasing at 2 in3/second
dV
 2 in 3 / s
dt
The account is shrinking by $50,000/day
da
 50,000 dollars / day
dt
Related Rates


Often times, we want to know how one variable
is related to another variable with respect to
time. In other words, we want to know how
rates are related; thus, we study related rates!!!
In order to determine the relationship among
rates, we need to differentiate with respect to
time; in other words, we need to be able to
determine:
d
dt
Related Rates

For example, let’s say that a rectangle is 10in x
6in whose sides are changing. The formulas for
both the perimeter and area in terms of l and w
are:
P  2l  2w
1.
A  lw
The rate of change for the perimeter is:
d
dP
dl
dw
P  2l  2w   2  2
dt
dt
dt
dt
2.
The rate of change for the area is:
d
dA
dw
dl
A  lw   l  w
dt
dt
dt
dt
Related Rates

For example, let’s say that a rectangle is 10in x
6in whose sides are changing.
1.
The formulas for the rates of change if the length and
width are increasing at a rate of 2 in/s are:
dl
 2 in / s
dt
2.
dw
 2 in / s
dt
Given this information, we can determine the rate of
change of perimeter:
dP
dl
dw
Since
2 2
dt
dt
dt
dP
 22 in / s   2(2 in / s )  8 in / s
dt
Related Rates

For example, let’s say that a rectangle is 10in x
6in whose sides are changing.
1.
The formulas for the rates of change if the length and
width are increasing at a rate of 2 in/s are:
dl
 2 in / s
dt
2.
dw
 2 in / s
dt
Given this information, we can also determine the rate
of change of area:
dA
dw
dl
Since
l
w
dt
dt
dt
dA
 (10 in )( 2 in / s )  (6in )( 2 in / s )  32 in 2 / s
dt
Related Rates

Related rates can be difficult for some to master,
so I’ve come up with 5 steps to help you solve
these problems:
1.
2.
3.
4.
5.
Make a sketch of the situation, introducing all known
and unknown variables and information
Identify the given rates of change and determine the
rate of change to be found (all rates of change should
be written as derivatives)
Write the equation that relates the variables. You
may have to write several equations.
Differentiate with respect to time, t
Substitute known information to solve for the desired
rate of change.
Related Rates

Note: It is a common MISTAKE to try to
substitute known information before
differentiating implicitly. Do not try to substitute
the known information until you have completed
the implicit differentiation!!!
Example 1:


An oil tanker hits an iceberg and starts to spill oil creating
a circular region. The oil is rapidly spilling out of the
tanker and the radius of the circle increases at rate of 2
ft/s. How fast is the area of the oil increasing when the
radius is 60 ft?
Step 1: Make a sketch

At this exact
point in time,
the radius is 60
ft (r = 60 ft).
Example 1:


An oil tanker hits an iceberg and starts to spill oil creating
a circular region. The oil is rapidly spilling out of the
tanker and the radius of the circle increases at rate of 2
ft/s. How fast is the area of the oil increasing when the
radius is 60 ft?
Step 2: Identify given rate(s) and rate(s) to be found
dr
 change in radius  2 ft / s
dt
dA
 change in area
dt
Example 1:


An oil tanker hits an iceberg and starts to spill oil creating
a circular region. The oil is rapidly spilling out of the
tanker and the radius of the circle increases at rate of 2
ft/s. How fast is the area of the oil increasing when the
radius is 60 ft?
Step 3: Write the equation(s) that relate the variables
 The area of a circle formula:
A   r2
Example 1:


An oil tanker hits an iceberg and starts to spill oil creating
a circular region. The oil is rapidly spilling out of the
tanker and the radius of the circle increases at rate of 2
ft/s. How fast is the area of the oil increasing when the
radius is 60 ft?
Step 4: Differentiate with respect to time

d
A   r2
dt
dA
dr
 2 r
dt
dt

Example 1:


An oil tanker hits an iceberg and starts to spill oil creating
a circular region. The oil is rapidly spilling out of the
tanker and the radius of the circle increases at rate of 2
ft/s. How fast is the area of the oil increasing when the
radius is 60 ft?
Step 5: Substitute and solve for the desired rate
dr
Recall that
 2 ft / s and r  60 ft
dt
dA
dr
dA
 2 r 
 2 (60 ft )( 2 ft / s )
dt
dt
dt
dA
ft 2
 240
dt
s
Example 2:


A ladder 20 ft. long leans against a vertical wall. If the
bottom of the ladder slides away from the wall at a rate of
2 ft. per second, how fast is the ladder sliding down the
wall when the top of the ladder is 12 feet above the
ground?
Step 1: Make a sketch

At this exact point in time,
the length of the ladder is 20
ft and the height of the ladder
is 12 ft (y = 12 ft). So, by the
Pythagorean Theorem, the
base must be 16 ft away from
the wall (x = 16 ft).
Example 2:


A ladder 20 ft. long leans against a vertical wall. If the
bottom of the ladder slides away from the wall at a rate of
2 ft. per second, how fast is the ladder sliding down the
wall when the top of the ladder is 12 feet above the
ground?
Step 2: Identify given rate(s) and rate(s) to be found
dx
 change in x  2 ft / s
dt
dy
 change in y
dt
Example 2:


A ladder 20 ft. long leans against a vertical wall. If the
bottom of the ladder slides away from the wall at a rate of
2 ft. per second, how fast is the ladder sliding down the
wall when the top of the ladder is 12 feet above the
ground?
Step 3: Write the equation(s) that relate the variables
 The Pythagorean Theorem will related the variables:
x  y  20
2
2
2
Example 2:


A ladder 20 ft. long leans against a vertical wall. If the
bottom of the ladder slides away from the wall at a rate of
2 ft. per second, how fast is the ladder sliding down the
wall when the top of the ladder is 12 feet above the
ground?
Step 4: Differentiate with respect to time


d 2
x  y 2  20 2
dt
dx
dy
2x  2 y
0
dt
dt
Example 2:


A ladder 20 ft. long leans against a vertical wall. If the
bottom of the ladder slides away from the wall at a rate of
2 ft. per second, how fast is the ladder sliding down the
wall when the top of the ladder is 12 feet above the
ground?
Step 5: Substitute and solve for the desired rate
dx
dy
dy
2x  2 y
 0  2 x(2 ft / s )  2 y
0
dt
dt
dt
dy
2(16 ft )( 2 ft / s )  2(12 ft )  0
dt
dy
24 ft   64 ft 2 / s
dt
2
dy
64 ft / s
8

  ft / s
dt
24 ft
3
Homework

Related Rates





Read Sections 2.6
P. 154: 1 – 35 (odd) and 36
WS: Related Rates Homework
WS: Related Rates Turvey
WS: Related Rates Problem Set
Chapter 2
Differentiation:
Related Rates – Day 2
21
Example 3:


A space shuttle is launched and is rising at a rate of 880
ft/s when it is 4000 ft up. If a camera is mounted on the
ground 3000 feet away from where the shuttle is
launched, how fast must a camera angle change at that
instant in order to keep the shuttle in sight?
Step 1: Make a sketch
 At this exact point in
time, the height of the
rocket is 4000 ft (y =
4000 ft) and the height
of the horizontal
distance from the rocket
to the camera is 3000 ft
θ
(this is constant, so we
don’t give it a variable!)
Example 3:


A space shuttle is launched and is rising at a rate of 880
ft/s when it is 4000 ft up. If a camera is mounted on the
ground 3000 feet away from where the shuttle is
launched, how fast must a camera angle change at that
instant in order to keep the shuttle in sight?
Step 2: Identify given rate(s) and rate(s) to be found
dy
 change in height  880 ft / s
dt
d
 change in the angle
dt
Example 3:


A space shuttle is launched and is rising at a rate of 880
ft/s when it is 4000 ft up. If a camera is mounted on the
ground 3000 feet away from where the shuttle is
launched, how fast must a camera angle change at that
instant in order to keep the shuttle in sight?
Step 3: Write the equation(s) that relate the variables
 The equation that will relate y with θ is tan θ :
y
tan  
3000
Example 3:


A space shuttle is launched and is rising at a rate of 880
ft/s when it is 4000 ft up. If a camera is mounted on the
ground 3000 feet away from where the shuttle is
launched, how fast must a camera angle change at that
instant in order to keep the shuttle in sight?
Step 4: Differentiate with respect to time
d 
y 
tan  


dt 
3000 ft 
d
1
dy
sec 


dt 3000 ft dt
2
Example 3:


A space shuttle is launched and is rising at a rate of 880
ft/s when it is 4000 ft up. If a camera is mounted on the
ground 3000 feet away from where the shuttle is
launched, how fast must a camera angle change at that
instant in order to keep the shuttle in sight?
Step 5: Substitute and solve for the desired rate
d
1 dy
sec 

dt 3000 dt
2
Before we can substitute and solve, we need to know dy/dt
and θ. We were given dy/dt, but we still need to solve for
θ at this point in time.
4
4000
When y  4000, tan  
3000
  tan  
3
  0.927 radians
1
Example 3:


A space shuttle is launched and is rising at a rate of 880
ft/s when it is 4000 ft up. If a camera is mounted on the
ground 3000 feet away from where the shuttle is
launched, how fast must a camera angle change at that
instant in order to keep the shuttle in sight?
Step 5: Substitute and solve for the desired rate
d
1 dy
d
1  880 ft 
2
sec 

 sec (0.927)



dt 3000 dt
dt 3000  s 
d
1  880 ft 
2


cos (0.927)   0.1057 ft / s
dt 3000  s 
2
Example 4:

You are trying to fill a tank in the shape of a cone with an
altitude of 12 ft and a radius of 6 ft with water. The water is
getting pumped into the tank at a rate of 10 gallons per minute.
Find the rate at which the water rises when the water is 3 feet
deep. Note 1: 1 gallon = .1337 cu ft.; Note 2: The water fills
rapidly at first, but slows down as the cone gets wider.

Step 1: Make a sketch

h
At this exact point in time,
the height of the water is 3 ft
(h = 3 ft); however, the
height of the entire cone is 12
ft. and the radius of the
entire cone is 6 ft.
Example 4:

You are trying to fill a tank in the shape of a cone with an
altitude of 12 ft and a radius of 6 ft with water. The water is
getting pumped into the tank at a rate of 10 gallons per minute.
Find the rate at which the water rises when the water is 3 feet
deep.

Step 2: Identify given rate(s) and rate(s) to be found
dr
NOTE: There are too
 change in the radius many unknown
dt
variables!!! So, we will
eventually need to
dh
 change in height
eliminate one!!!
dt
dV
 change in the volume  10 gal / min
dt
Example 4:

You are trying to fill a tank in the shape of a cone with an
altitude of 12 ft and a radius of 6 ft with water. The water is
getting pumped into the tank at a rate of 10 gallons per minute.
Find the rate at which the water rises when the water is 3 feet
deep.

Step 3: Write the equation(s) that relate the variables
 The equation that will relate r and h with V is:
 2
V r h
3

At this point, we need to eliminate one of the variables;
the rate of volume is given and we are looking for the
rate that water rises (height), so we need to try to
eliminate the radius!
Example 4:

In order to eliminate the radius from the equation (i.e.
write volume using only height), we must go back to
geometry and consider similar triangles:
h

h
Set up a ratio of similar triangles and solve for r.
2
 3

 h
 h 
r 6
h

 r  so V    h   h  h
12
3 2
3 4 
h 12
2
2
Example 4:

You are trying to fill a tank in the shape of a cone with an
altitude of 12 ft and a radius of 6 ft with water. The water is
getting pumped into the tank at a rate of 10 gallons per minute.
Find the rate at which the water rises when the water is 3 feet
deep.

Step 3: Write the equation(s) that relate the variables
 The equation that will relate h with V is:
 3
V h
12
Note: When we complete step 3, the equation should
ONLY include the variables for which the rates are
GIVEN and the variable whose rate we are finding!!!
Example 4:

You are trying to fill a tank in the shape of a cone with an
altitude of 12 ft and a radius of 6 ft with water. The water is
getting pumped into the tank at a rate of 10 gallons per minute.
Find the rate at which the water rises when the water is 3 feet
deep.

Step 4: Differentiate with respect to time
d 
 3
V h 

dt 
12 
dV  2 dh
 h 
dt 4
dt
Example 4:

You are trying to fill a tank in the shape of a cone with an
altitude of 12 ft and a radius of 6 ft with water. The water is
getting pumped into the tank at a rate of 10 gallons per minute.
Find the rate at which the water rises when the water is 3 feet
deep.

Step 5: Substitute and solve for the desired rate
dV
Recall
 10 gal / min and h  3 ft
dt

Before we can solve, we need to convert from gallons
to feet:
3
0.1337 ft
The conversion ratio is
gal
3
3
1.337 ft
gal 0.1337 ft

10

min
min
gal
Example 4:

You are trying to fill a tank in the shape of a cone with an
altitude of 12 ft and a radius of 6 ft with water. The water is
getting pumped into the tank at a rate of 10 gallons per minute.
Find the rate at which the water rises when the water is 3 feet
deep.

Step 5: Substitute and solve for the desired rate
dV  2 dh
 h 
dt 4
dt
1.337 ft 3 
2 dh
 (3 ft )
min
4
dt
1.337 ft 3 9 ft 2 dh


min
4
dt
4
1.337 ft 3 dh


2
9 ft
min
dt
0.189 ft dh

min
dt
Homework

Related Rates





Read Sections 2.6
P. 154: 1 – 35 (odd) and 36
WS: Related Rates Homework
WS: Related Rates Turvey
WS: Related Rates Problem Set
Chapter 2
Differentiation:
Related Rates – Day 3
37
Example 5:

An 11 foot by 20 foot swimming pool is being filled at a
constant rate of 5 ft3 per minute. The pool has a depth of 4 feet
in the shallow end and 9 feet in the deep end. (Assume the
pool’s depth increases constantly between the shallow end and
the deep end.) How fast is the height of the water increasing
when the water has filled 2 feet of the deep end?

Step 1: Make a sketch

At this exact point in time,
the height of the water is 2 ft
(h = 2 ft)
Example 5:

An 11 foot by 20 foot swimming pool is being filled at a
constant rate of 5 ft3 per minute. The pool has a depth of 4 feet
in the shallow end and 9 feet in the deep end. (Assume the
pool’s depth increases constantly between the shallow end and
the deep end.) How fast is the height of the water increasing
when the water has filled 2 feet of the deep end?

Step 2: Identify given rate(s) and rate(s) to be found
dh
 change in the height
dt
3
dV
5 ft
 change in volume 
dt
min
db
 change in the base
dt
NOTE: There are too
many unknown
variables. So, we
will eventually need
to eliminate one.
Example 5:

Step 3: Write the equation(s) that relate the variables
 This is a tough one!
 The shape of the pool is a 3-dimensional trapezoid or a
trapezoidal prism. It will be very difficult to work with
the equation:
b1  b2
V
wh
2



So what do we do?
We need to look at the shape in another way.
Well, if we think about how water rises in the shape, it
starts at the bottom of the deep end.
Example 5:

Step 3: Write the equation(s) that relate the variables
 While the water is rising from 0 to 5 feet (since 9ft – 4ft
= 5 ft), our 3-dimensional shape is a triangular prism –
This part is
a much easier shape to solve:
superfluous
Now finding the volume
of a triangular prism is
much easier:
Width is constant, so:
1
V  bhw
2
11
V  bh
2
Example 5:

There are still too many variables. We know dV/dt and we
are looking for dh/dt, so we should try to eliminate b (in
other words, we need to write the equation for volume in
terms of height alone).

b 20
 b  4h

h 5
11
11
2
so V  bh  (4h)h  22h
2
2
Set up a ratio of similar
triangles and solve for b.
Example 5:

An 11 foot by 20 foot swimming pool is being filled at a
constant rate of 5 ft3 per minute. The pool has a depth of 4 feet
in the shallow end and 9 feet in the deep end. (Assume the
pool’s depth increases constantly between the shallow end and
the deep end.) How fast is the height of the water increasing
when the water has filled 2 feet of the deep end?

Step 4: Differentiate with respect to time

d
2
V  22h
dt

dV
dh
 44h
dt
dt
Example 5:

An 11 foot by 20 foot swimming pool is being filled at a
constant rate of 5 ft3 per minute. The pool has a depth of 4 feet
in the shallow end and 9 feet in the deep end. (Assume the
pool’s depth increases constantly between the shallow end and
the deep end.) How fast is the height of the water increasing
when the water has filled 2 feet of the deep end?

Step 5: Substitute and solve for the desired rate
dV
dh
 44h
dt
dt
5 ft 3
dh
 44 ft (2 ft ) 
min
dt
5 ft
dh

88 min dt
ft
dh
0.057

min dt
Homework

Related Rates





Read Sections 2.6
P. 154: 1 – 35 (odd) and 36
WS: Related Rates Homework
WS: Related Rates Turvey
WS: Related Rates Problem Set