RELATED RATES
DERIVATIVES WITH RESPECT
TO TIME
How do you take the derivative with respect to
time when “time” is not a variable in the equation?
• Consider a circle that is growing on the coordinate plane:
• Growing Circle Animation
• Equation of a circle centered at the origin with radius of 2:
– x2 + y2 = 4
In each case find the derivative with
respect to ‘t’. Then find dy/dt.
1. 3x 2  5 y 4  6 y  2
2. 3 tan x  5sin y  16
3. 3x3  5xy 4  6 y 5  12
4. 4 ln x  cos( xy )  8
What is a related rate?
TABLE OF CONTENTS
 AREA AND VOLUME
 PYTHAGOREAN THEOREM AND
SIMILARITY
 TRIGONOMETRY
 MISCELLANEOUS EQUATIONS
AREA AND VOLUME
RELATED RATES
Example 1
Suppose a spherical balloon is
inflated at the rate of 10 cubic
inches per minute. How fast is the
radius of the balloon increasing
when the radius is 5 inches?
Ex 1: Answer
Volume of a Sphere:
Given:
Find:
dV
 10 in 3 /min
dt
dr
?
dt
when r = 5 inches
1
dr
in/min 
10
dt
4 3
V  r
3
dV
2 dr
 4 r
dt
dt
dr
10  4  5 
dt
2
10
dr

100 dt
Example 2
A shrinking spherical
balloon loses air at
the rate of 1 cubic
inch per minute. At
what rate is its radius
changing when the
radius is
(a) 2 inches?
(b) 1 inch?
Ex 2: Answer
Volume of a Sphere:
Given:
Find:
dV
 1 in 3 /min
dt
dr
?
dt
when a) r = 2 inches
b) r = 1 inch
4 3
V  r
3
dV
2 dr
 4 r
dt
dt
1
dr
a) 

16 dt
1
dr
b) 

4
dt
Example 3
The area of a
rectangle, whose
length is twice its
width, is increasing
at the rate of 8cm 2 / s
Find the rate at
which the length is
increasing when the
width is 5 cm.
Ex 3: Answer
Area of a rectangle:
A  lw
Given: l = 2w
dA
 8 cm 2 / s
dt
Find:
dl
?
dt
when w = 5 cm
l = 10 cm
l
Al
2
1 2
A l
2
dA
dl
l
dt
dt
dl
8  10
dt
4
dl
cm/s 
5
dt
Example 4
• Gravel is being dumped from a conveyor belt
at a rate of 30 ft3/min and its coarseness is
such that it forms a pile in the shape of a
cone whose base diameter and height are
always equal. How fast is the height of the
pile increasing when the pile is 10 ft high?
Ex 4: Answer
Volume of a Cone:
1 2
V  r h
3
dV
 30 ft 3 / min
dt
Given:
d = h or 2r = h
Find: dh  ?
dt
when h = 10 ft
Eliminate ‘r’ from the
equation and simplify
2
1 1 
V    h h
3 2 
1
3
V  h
12
Ex 4: Answer (con’t)
Take the derivative
1
3
V  h
12
dV 1 2 dh
 h
dt 4
dt
Substitute in the
specific values and
solve.
1
2 dh
30   10 
4
dt
6
dh
ft/min 
5
dt
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Example 5
An inverted conical
container has a height of 9
cm and a diameter of 6
cm. It is leaking water at a
rate of 1 cubic centimeter
per minute. Find the rate
at which the water level h
is dropping when h equals
3cm.
Ex 5: Answer
Volume of a Cone:
3
1 2
V  r h
3
9
dV
Given:
 1 cm3 / min
dt
Find: dh  ?
dt
when h = 3 cm
Since the base radius is 3
and the height of the cone
is 9, the radius of the water
level will always be 1/3 of
the height of the water.
That is r = 1/3h
Ex 5: Answer (con’t)
Volume of a Cone:
2
1 1 
V    h h
3 3 
dV 1 2 dh
 h
dt 9
dt
1 2
V  r h
3
3
9
1
V   h3
27
1
2 dh
1    3
9
dt
1
dh
cm/min 

dt
Table of contents
PYTHAGOREAN THEOREM
AND SIMILARITY
Example 6
A 13 meter long
ladder leans against a
a vertical wall. The
base of the ladder is
pulled away from the
wall at a rate of 1 m/s.
Find the rate at which
the top of the ladder is
falling when the base
of the ladder is 5m
away from the wall.
Ex 6: Answer
13
y
Given: Length of ladder – 13 m
dx
 1 m/s
dt
x
Find: dy  ?
dt
when x = 5 m
Use Pythagorean Theorem to relate the sides of the triangle!
Ex 6: Answer (con’t)
13
y
By the Pythagorean Thm:
x  y  13
2
x
Find ‘y’ when
x = 5 using
Pythagorean
Thm.
52  y 2  169
y  12
2
2
dx
dy
2x  2 y
0
dt
dt
dy
2  5 1  2 12   0
dt
dy 5

m/s
dt 12
Ex 7: A balloon and a bicycle
• A balloon is rising vertically above a
level straight road at a constant rate of 1
ft/sec. Just when the balloon is 65 ft
above the ground, a bicycle moving at a
constant rate of 17 ft/sec passes under
it. How fast is the distance s(t) between
the bicycle and balloon increasing 3 sec
later?
Ex 7: Balloon and Bicycle solution
• Given:
dy
 1 ft/s
• rate of balloon dt
dx
• rate of cyclist
 17 ft/s
dt
s
• Find: ds
dt
y
 ? ft/s
• when x = ? and y = ?
x
• Distance = rate * time
Ex 7: Balloon and Bicycle solution
x y s
2
s
y
x
2
2
dx
dy
ds
2x  2 y
 2s
dt
dt
dt
ds
2  5117   2  68 1  2 85 
dt
ds
 11 ft s
dt
Ex 8: The airplane problem• A highway patrol plane flies 3 mi above a
level, straight road at a steady pace 120
mi/h. The pilot sees an oncoming car and
with radar determines that at the instant the
line of sight distance from plane to car is 5
mi, the line of sight distance is decreasing at
the rate of 160 mi/h. Find the car’s speed
along the highway.
Ex 8: Airplane - solution
Given:
rate of plane:
dp
 120 mi
hr
dt
when s=5:
Find:
ds
mi


160
dx
hr
dt

?
rate of the car:
dt
Ex 8: Airplane – solution(con’t)
p
s
x
3
3
+
p
s
(x+p)
3
Ex 8: Airplane – solution(con’t)
2
2
x

p

3

s


2
ds
 dx dp 
2  x  p      0  2s
dt
 dt dt 
 dx

2  4    120   2  5  160 
 dt

 dx

8   120   1600
 dt

dx
 120  200
dt
dx
 80mph
dt
s
(x+p)
3
Example 9
A 6 foot-tall man is walking straight away
from a 15 ft-high streetlight. At what rate is
his shadow lengthening when he is 20 ft
away from the streetlight if he is walking
away from the light at a rate of 4 ft/sec.
Ex 9: Answer
15
6
x
s
Set up a proportion
using the sides of
the large triangle
and the sides of the
small triangle.
Given: streetlight – 15 ft
man – 6 ft
dx
 4 ft/s
dt
Find: ds
?
dt
when x = 20 ft
Ex 9: Answer (con’t)
15
6
x
s
15s  6 x  6s
9s  6 x
ds
dx
9 6
dt
dt
15
6

xs s
ds 6
  4
dt 9
ds 8
 ft/s
dt 3
Table of contents
RELATED RATES WITH
TRIGONOMETRY
Example 10
A ferris wheel with a
radius of 25 ft is
revolving at the rate of
10 radians per minute.
How fast is a
passenger rising when
the passenger is 15 ft
higher than the center
of the ferris wheel?
Ex 10: Answer
25

Given: Radius – 25 ft
y
d
 10 rad/min
dt
Find: dy
dt
y
sin  
25
25sin   y
?
when y = 15 ft.
d dy
25cos 

dt dt
Ex 10: Answer
25

y
Find cos  when y = 15 ft
252  152  x 2
x  20
20
cos  
25
4
cos  
5
dy
4
25  10 
dt
5
dy
 200 ft/min
dt
Example 11
A baseball diamond is a square with sides
90 ft long. Suppose a baseball player is
advancing from second to third base at a rate
of 24 ft per second, and an umpire is standing
on home plate. Let  be the angle between
the third base line and the line of sight from
the umpire to the runner. How fast is 
changing when the runner is 30 ft from 3rd
base?
Ex 11: Answer
Given: Side length – 90 ft.
x
dx
 24 ft/s
dt
90

x
tan  
90
d
?
dt
when x = 30 ft.
Find:
d
1 dx
sec 

dt 90 dt
2
Ex 11: Answer (con’t)
Solve equation for d/dt.
x
90
d
1
dx
 cos 2 
dt 90
dt
Find cos  when x = 30:

h  302  902
h  9000
d 1  90 
 
 24

dt 90  9000 
2
90
cos  
9000
d
6

rad/s
dt
25
Table of contents
MISCELLANEOUS
EQUATIONS
Example 12
An environmental study of a certain community
indicates that there will be
Q( p)  p 2  3 p  1200
units of a harmful pollutant in the air when the
population is p thousand. The population is
currently 30,000 and is increasing at a rate of
2,000 per year. At what rate is the level of air
pollution increasing?
Ex 12: Answer
Given: Q( p)  p 2  3 p  1200
dQ
dp
 2 thous/yr.
dt
Find: dQ
dt
?
when p =30thous/yr.
dp
dp
 2p
3
dt
dt
dt
dQ
 2  30  2  3  2
dt
dQ
 126 thous./yr.
dt