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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.
2 41. 3 5 6 2x y y 2. 3 tan 5sin 16x y
3 4 53. 3 5 6 12x xy y 4. 4 ln cos( ) 8x xy
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:
when r = 5 inches
34
3V r
310 in /mindV
dt
?dr
dt
24dV dr
rdt dt
210 4 5
dr
dt
10
100
dr
dt
1 in/min
10
dr
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:
when a) r = 2 inches
b) r = 1 inch
34
3V r
31 in /mindV
dt
?dr
dt
24dV dr
rdt dt
1a)
16
dr
dt
1b)
4
dr
dt
Example 3
The area of a rectangle, whose length is twice its width, is increasing at the rate of
Find the rate at which the length is increasing when the width is 5 cm.
28 /cm s
Ex 3: AnswerArea of a
rectangle:
Given: l = 2w
Find: when w = 5 cm l = 10 cm
A l w
28 cm /dA
sdt
?dl
dt
2
21
2
lA l
A l
dA dll
dt dt
8 10dl
dt
4 cm/s
5
dl
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:
Given:
d = h or 2r = h
Find:
when h = 10 ft
21
3V r h
330 ft / mindV
dt
?dh
dt
Eliminate ‘r’ from the equation and simplify
21 1
3 2V h h
31
12V h
Ex 4: Answer (con’t)
Take the derivative
21
4
dV dhh
dt dt
2130 10
4
dh
dt
6 ft/min
5
dh
dt
Table of contents
Substitute in the specific values and solve.
31
12V h
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:
Given:
Find:
when h = 3 cm
21
3V r h
3
9
31 cm / mindV
dt
?dh
dt 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:21
3V r h
3
9
21 1
3 3V h h
31
27V h
21
9
dV dhh
dt dt 21
1 39
dh
dt
1 cm/min
dh
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
?dy
dt
13y
x
Given: Length of ladder – 13 m
Find:
when x = 5 m
Use Pythagorean Theorem to relate the sides of the triangle!
1 m/sdx
dt
Ex 6: Answer (con’t)
2 2 0dx dyx ydt dt
2 2 213x y
2 25 169
12
y
y
13y
x
By the Pythagorean Thm:
Find ‘y’ when x = 5 using Pythagorean Thm. 2 5 1 2 12 0
dy
dt
5 m/s
12
dy
dt
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:
• rate of balloon
• rate of cyclist
• Find:
• when x = ? and y = ?
• Distance = rate * time
s
x
y
17 /sdx
ftdt
1 ft/sdy
dt
? ft/sds
dt
Ex 7: Balloon and Bicycle - solution
s
x
y
2 2 2x y s
2 2 2dx dy dsx y sdt dt dt
2 51 17 2 68 1 2 85ds
dt
11ds
ft sdt
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:
when s=5:
120dp mi
hrdt
160ds mi
hrdt
Find:
rate of the car: ?dx
dt
Ex 8: Airplane – solution(con’t)
p
3
p+x
3s
s 3
(x+p)
Ex 8: Airplane – solution(con’t)
s 3
(x+p)
2 2 23x p s
2 0 2dx dp ds
x p sdt dt dt
2 4 120 2 5 160dx
dt
8 120 1600dx
dt
120 200dx
dt
80dx
mphdt
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
Given: streetlight – 15 ft
man – 6 ft
Find:
when x = 20 ft
4 ft/sdx
dt
x s
15
6
?ds
dtSet up a proportion
using the sides of the large triangle and the sides of the small triangle.
Ex 9: Answer (con’t)
9 6ds dx
dt dt
15 6
x s s
15 6 6
9 6
s x s
s x
x s
15
6
64
98
ft/s3
ds
dtds
dt
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
Given: Radius – 25 ft
Find:
when y = 15 ft.
10 rad/mind
dt
25
y
?dy
dt
sin25
y
25sin y 25cosd dy
dt dt
Ex 10: AnswerFind cos when y = 15 ft
2 2 225 15
20
x
x
25
y
20cos
254
cos5
425 10
5
dy
dt
200 ft/mindy
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.
Find:
when x = 30 ft.
24 ft/sdx
dt
90
x
?d
dt
tan90
x 2 1sec
90
d dx
dt dt
Ex 11: Answer (con’t)
Solve equation for d/dt.
Find cos when x = 30:
21cos
90
d dx
dt dt
90
x
2 230 90
9000
h
h
90cos
9000
2
1 9024
90 9000
d
dt
6 rad/s
25
d
dt
Table of contents
MISCELLANEOUS EQUATIONS
Example 12 An environmental study of a certain community
indicates that there will be
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?
2( ) 3 1200Q p p p
Ex 12: Answer
2 3dQ dp dp
pdt dt dt
2 30 2 3 2dQ
dt
Given:
2 thous/yr.dp
dt
Find:
when p =30thous/yr.
?dQ
dt
126 thous./yr.dQ
dt
2( ) 3 1200Q p p p