Transcript Section 2.6 - jpiichsabcalculus

Section 2.6
Related Rates
Read Guidelines For Solving Related Rates
Problems on p. 150.
Example 1
A 17 foot ladder is sliding down a wall. The
base of the ladder is moving away from the wall
at a rate of 2 feet per second. How fast is the top
of the ladder moving down the wall when the
base of the ladder is 8 feet away from the wall?
Step 1: Draw a sketch and label known and
unknown quantities. Write down what is
given and what is to be determined.
s = 17 ft.
y
x
dx
Given:
 2 ft/s
dt
dy
Find:
when x  8
dt
Step 2: Write an equation involving the variables
whose rates of changes either are given or
to be determined.
x 2  y 2  17 2
Step 3: Differentiate each side with respect to t.
d 2
d
2
 x  y    289
dt
dt
dx
dy
2x  2 y  0
dt
dt
Step 4: Substitute all known values for the variables
and their rates of change. Then solve for
the required rate of change.
x 2  y 2  17 2
x  8 so y 15
dy
2  8  2   2 15   0
dt
dy
16
   1.067 ft/s
dt
15
The top of the ladder is moving down the wall at a
rate of about −1.067 ft/s when the base of the ladder
is 8 ft. from the wall.
Example 2
An adventurer rides down a zip-line at a speed of
80 mph. If the angle of depression of the zipline is 75°, how fast is the zip-liner’s altitude
changing?
dz
Given:
 80 mph
75°
dt
z
h
dh
Find:
dt
75°
z
h
The adventurer’s altitude
is decreasing by a rate of
about 77.274 mph when
the angle of depression
is 75°.
h
sin 75 
z
z sin 75  h
dh
dz
 sin 75
dt
dt
dh
 sin 75  80 
dt
dh
 77.274 mph
dt
Example 3
A 6 foot tall man walks away from a 22 foot
street light at a speed of 8 feet per second. What
is the rate of change of the length of his shadow
when he is 19 feet away from the light? Also, at
what rate is the tip of his shadow moving?
22
6
x
s
dx
Given:
 8 ft/s
dt
ds
 a  Find: when x  19
dt
6
s
3
s


22 x  s 11 x  s
3 x  3s  11s
3 x  8s
3
s x
8
22
6
x
s
The length of the man’s
shadow is increasing at a
rate of 3 ft/s.
3
s

11 x  s
3 x  3s  11s
3 x  8s
3
s x
8
ds 3 dx

dt 8 dt
ds 3
  8   3 ft/s
dt 8
dx
ds
Given:
 8 ft/s and
 3 ft/s
dt
dt
22
6
dy
 b  Find: when x  19
x
s
dt
y=x+s
dy dx ds


dt dt dt
 8  3  11
The tip of his shadow is moving at rate of 11 ft/s
when he is 19 ft. from the street light.
Example 4
A large spherical balloon is being inflated and its
volume is increasing at a rate of 3.5 cubic feet
per minute. What is the rate of change of the
radius when the radius is 7 feet?
r
dV
Given:
 3.5 cu. ft. per min.
dt
dr
Find:
when r  7 feet
dt
r
4 3
V  r
3
dV 4
2 dr
2 dr
   3 r
 4r
dt 3
dt
dt
dr
3.5  4  7 
dt
dr
3.5

2
dt 4  7 
2
The radius of the spherical
balloon is increasing at a
rate of about 0.006 ft/sec
when the radius is 7 ft.
dr
 0.006 ft/min
dt
Example 5
An upside-down conical tank full of water has a
“base” radius of 3 meters and a height of 5
meters. The is being drained at a rate of 2 cubic
meters per meter. What is the rate of change of
the height of the water when the height is 4
meters?
3
r
h
dV
Given:
 2 cu. meters per min
dt
5
dh
Find:
when h  4
dt
3
r
5
h
3
r
5
h
r h

3 5
3
r h
5
1 2
V  r h
3
2
1 3 
V   h  h
3 5 
3 3
V  h
25
3
3 3
V  h
25
r
h
5
dV
3
2 dh
   3 h
dt 25
dt
9
dh
2   16 
25
dt
dh
 0.111 meters/min
dt
The height of the water is decreasing at a rate of
about 0.111 meter/min when its height is 4 meters.
The Grand Finale!!!!
Example 6
An upside-down conical tank full of water has a
“base” radius of 5 feet and a height of 7 feet. The
water is being drained into a cylindrical tank with
radius of 5 feet and height 6 feet. The radius of the
water in the conical tank is decreasing at a rate of 2
feet per minute. At what rate does the water level
in the cylindrical tank rise when the water level in
the conical tank is 3 feet?
5
dr
Given:
 2 ft/min
dt
r
7
h1
dh2
Find:
when h1  3 ft
dt
5
5
h2
6
5
r
7
h1
5
5
h2
1 2
2
Vcone  r h1 and Vcyl  r h2
3
5
h1 r

r
7
7
5
h1
7
h1  r
1 27 
5
6
Vcone  r  r 
3
5 
7 3
Vcone  r
15
2
Vcyl    5  h2  25h2
5
Vcone
r
7
h1
5
5
h2
6
7 3
 r and Vcyl  25h2
15
dVcone 7
2 dr
   3 r
dt
15
dt
dVcone 7 2
14 2
 r  2    r
dt
5
5
dVcyl
dh2
 25
dt
dt
dVcyl
dVcone

dt
dt
5
dVcyl
r
7
h1
5
5
h2
6
dVcone

dt
dt
7
h1  r
5
7
14 2
3 r
  r
5
5
7
r
15
2

dh2
14 15  
dh2
      
 0.514 ft/min 25
 5 7 
dt
dt


The water level in the cylindrical tank is increasing
at rate of about 0.514 ft./min when the water level of
the conical tank is 3 ft.