2.8 Related Rates
Formulas You May Need To Know
V  a3
V r h
2
4 r 

V
3
3
 r h

V
2
3
V  lwh
bh 

V
3
Related rate problems are differentiated with
respect to time. So, every variable, except t is
differentiated implicitly.
Ex. Two rates that are related.
Given y = x2 + 3, find dy/dt when x = 1, given
that dx/dt = 2.
y = x2 + 3
dy
dx Now, when x = 1 and dx/dt = 2, we
= 2x
have
dt
dt
dy
= 2(1)(2) = 4
dt
Procedure For Solving
Related Rate Problems
1. Assign symbols to all given quantities and
quantities to be determined. Make a sketch
and label the quantities if feasible.
2. Write an equation involving the variables
whose rates of change either are given or are
to be determined.
3. Using the Chain Rule, implicitly differentiate
both sides of the equation with respect to t.
4. Substitute into the resulting equation all known
values for the variables and their rates of change.
Solve for the required rate of change.
First, a review problem:
Consider a sphere of radius 10cm.
If the radius changes 0.1cm (a very small amount) how
much does the volume change?
4
V   r3
3
dV  4 r 2 dr
dV  4 10cm   0.1cm
2
dV  40 cm3
The volume would change by approximately 40 cm 3 .

Now, suppose that the radius is changing at an
instantaneous rate of 0.1 cm/sec.
(Possible if the sphere is a soap bubble or a balloon.)
4
V   r3
3
dV
dr
 4 r 2
dt
dt
dV
cm 
2 
 4 10cm    0.1

dt
sec


dV
cm3
 40
dt
sec
The sphere is growing at a rate of 40 cm3 / sec .
Note: This is an exact answer, not an approximation like
we got with the differential problems.

Water is draining from a cylindrical
tank at 3 liters/second. How fast is
the surface dropping?
3
cm
dV
L
 3000
 3
sec
dt
sec
dh
Find
dt
(We need a formula to
relate V and h. )
V   r 2h
dV
2 dh
r
dt
dt
cm3
2 dh
3000
r
sec
dt
(r is a constant.)
cm3
3000
dh
sec

dt
 r2

Steps for Related Rates Problems:
1.
Draw a picture (sketch).
2. Write down known information.
3. Write down what you are looking for.
4. Write an equation to relate the variables.
5. Differentiate both sides with respect to t.
6. Evaluate.

Truck Problem:
Truck A travels east at 40 mi/hr.
Truck B travels north at 30 mi/hr.
How fast is the distance between the
trucks changing 6 minutes later?
r t  d
1
40   4
10
1
30   3
10
32  42  z 2
B
z 5
y3
A
x4
9  16  z 2
25  z 2
5 z

Truck Problem:
Truck A travels east at 40 mi/hr.
Truck B travels north at 30 mi/hr.
How fast is the distance between the
trucks changing 6 minutes later?
r  t x d y  z
2
2
2
1
1 dz
dx
dy
40
2 x 10  42 y 30 10
2z  3
dt
dt
dt
32  42  z 2
dz
4  40  3  30
2 5
9  16  z
dt
2 dz
25

z
250  5
dz
dt
50 
5 z
dt
B
z 5
y3
dy
 30
dt
A
x  4 dx  40
dt
miles
50
hour

Ex. A pebble is dropped into a calm pond, causing
ripples in the form of concentric circles. The radius
r of the outer ripple is increasing at a constant rate
of 1 foot per second. When this radius is 4 ft., what
rate is the total area A of the disturbed water
increasing.
Givens:
dr
= 1 when r = 4
dt
Given equation:
Differentiate:
A = pr
dA
=?
dt
2
dA
dr
= 2pr
dt
dt
dA
= 2p (1)(4)
dt
= 8p
An inflating balloon
Air is being pumped into a spherical balloon at the
rate of 4.5 in3 per second. Find the rate of change
of the radius when the radius is 2 inches.
Given: dV
dt
Equation:
Diff.
& Solve:
= 4.5in / sec
3
r = 2 in.
4 3
V = pr
3
dV
2 dr
= 4pr
dt
dt
dr
Find :
=?
dt
dr
4.5 = 4p 2
dt
dr
.09in /sec =
dt
2
The velocity of an airplane tracked by radar
An airplane is flying at an elevation of 6 miles on a flight
path that will take it directly over a radar tracking station.
Let s represent the distance (in miles)between the radar
station and the plane. If s is decreasing at a rate of 400
miles per hour when s is 10 miles, what is the velocity of
the plane.
s
6
x
Given:
Find:
Equation:
ds
= -400
dt
s = 10
dx
=?
dt
x2 + 62 = s2
Solve:
dx
ds
2 x = 2s
dt
dt
To find dx/dt, we
must first find x
when s = 10
x = s 2 - 36 = 100 - 36 = 8
dx
2(8) = 2(10 )(- 400 )
dt
Day 1
dx
= -500mph
dt
A fish is reeled in at a rate of 1 foot per second
from a bridge 15 ft. above the water. At what
rate is the angle between the line and the water
changing when there is 25 ft. of line out?
x
15 ft.
q
dx
Given:
x = 25 ft. h = 15 ft.
= -1
dt
dq
Find:
=?
dt
Equation:
15
-1
sin
q
=
15
x
sin q =
x
Solve:
dq
- 2 dx
(cosq ) = -15 x
dt
dt
dq
- 15 dx
= 2
dt x cos q dt
dq
=
dt
- 15
(- 1)
2 æ 20 ö
25 ç ÷
è 25 ø
dq
3
=
rad / sec
dt 100
Ex. A pebble is dropped into a calm pond, causing
ripples in the form of concentric circles. The radius
r of the outer ripple in increasing at a constant rate
of 1 foot per second. When this radius is 4 ft., what
rate is the total area A of the disturbed water
increasing.
An inflating balloon
Air is being pumped into a spherical balloon at the
rate of 4.5 in3 per minute. Find the rate of change
of the radius when the radius is 2 inches.
Example
x  y  25
2
• Given
dy
• Find dt
2
dx
4
dt
when x = 3
Note: we must differentiate implicitly
with respect to t
dx
dy
2x  2 y
0
dt
dt
18
Example
• Now substitute in the things we know
dx
4
dt
–
x=3
• Find other values we need dx
– when x = 3,
32 + y2 = 25
y=4
and
dy
2x  2 y
0
dt
dt
19
Example
dx
dy
2x  2 y
0
dt
dt
• Result
dy
244  24
0
dt
dy 32

 4
dt
8
20
Guidelines for Related-Rate Problems
1. Identify given quantities, quantities to be
determined
•
Make a sketch, label quantities
2. Write equation involving variables
3. Using Chain Rule, implicitly differentiate
both sides of equation with respect to t
4. After step 3, substitute known values, solve
for required rate of change
21
R1
Electricity
R2
• The combined electrical
1 1
1
 
resistance R of R1 and R2
R R1 R2
connected in parallel is
given by
• R1 and R2 are increasing at rates of 1
and 1.5 ohms per second respectively.
• At what rate is R changing when R1 =
50 and R2 = 75?
22
Draining Water Tank
• Radius = 20, Height = 40
•
1 2
Volume   r h
3
• The flow rate = 80 gallons/min
• What is the rate of change of
the radius when the height = 12?
dV
 80
dt
dr
 ??
dt
23
Draining Water Tank
• At this point in time
the height is fixed
1 2
Volume   r 12
3
• Differentiate implicitly dV 1
dr 

   2  r   12
with respect to t,
dt 3 
dt 
• Substitute in known
values
• Solve for dr/dt
24
Assignment
• Lesson 3.7
• Page 187
• Exercises 1 – 7 odd, 13 – 27 odd
25
Example #1
• A ladder 10 feet long is resting against a wall. If
the bottom of the ladder is sliding away from the
wall at a rate of 1 foot per second, how fast is the
top of the ladder moving down when the bottom
of the ladder is 8 feet from the wall?
• First, draw the picture:
• We have dx/dt is one foot per second. We
want to find dy/dt.
• X and y are related by the Pythagorean
Thereom
• Differentiate both sides of this equation
with respect to t to get
• When x = 8 ft, we have
• Therefore
• The top of the ladder is sliding down
(because of the negative sign in the result)
at a rate of 4/3 feet per second.
Example #2
• A man 6 ft tall walks with a speed of 8 ft per
second away from a street light atop an 8 foot
pole. How fast is the tip of his shadow moving
along the ground when he is 100 feet from the
light pole.
6 ft
z-x
x
z
18
ft
• Let x be the man’s distance from the pole and z be
the distance of the tip of his shadow from the base
of the pole.
• Even though x and z are functions of t, we do not
attempt to obtain implicit formulas for either.
• We are given that dx/dt = 8 (ft/sec), and we want
to find dz/dt when x = 100 (ft).
• We equate ratios of corresponding sides of the two
similar triangles and find that z/18 = (z-x)/6
• Thus 2z = 3x
• Implicit differentiation now gives 2 dz/dt = 3
dx/dt
• We substitute dx/dt = 8 and find that
(dz/dt = 3/2) * (dx/dt = 3/2) * (8) = 12
So the tip of the man’s shadow is moving at 12 ft
per second.
Try Me!
• A ladder 25 ft long is leaning against a vertical
wall. If the bottom of the ladder is pulled
horizontally away from the wall at 3 ft/sec,
how fast is the top of the ladder sliding down
the wall, when the bottom is 15 ft from the
wall?
Solution
• t = the number of seconds in time that has
elapsed since the ladder started to slide
down the wall.
• y = the number of feet in distance from the
ground to the top of the ladder at t seconds.
• x = the number of feet in the distance from
the bottom of the ladder to the wall at t
seconds.
• Because the bottom of the ladder is pulled
horizontally away from the wall at 3 ft/sec,
dx/dt = 3. We wish to find dy/dt when x = 15.
• From the Pythagorean Thereom, we have y^2
= 625 – x^2
• Because x and y are functions of t, we
differentiate both sides of equation one with
respect to t and obtain 2y dy/dt = -2x dx/dt
giving us dy/dt = -x/y dx/dt
• When x = 15, it follows from equation one that
y = 20.
• Because dx/dt = 3, we get from equation two:
dy/dt = (-15/20) * 3 = -9/4
• Therefore, the top of the ladder is sliding down
the wall at the rate of 2 ¼ ft/sec when the
bottom is 15 ft from the wall.
• The significance of the minus sign is that y is
decreasing as t is increasing.
Was Your Answer Correct?