Ch 4.6 Related Rates
Graphical, Numerical, Algebraic by
Finney Demana, Waits, Kennedy
Differentiation With Respect to
Different Variables
2


r
h
The volume of a cone can be expressed as: V =
3
a) Find dV/dt if V, r, and h are all functions of time.
b) Find dV/dt if r is constant so that V and h are functions
of time.
c) Find dV/dt if V is constant so that only r and h are
functions of time.
Differentiation With Respect to
Different Variables
  r2  h
V=
The volume of a cone can be expressed as:
3
a) Find dV/dt if V, r, and h are all functions of time.
dV
  2 dh
dr 
=  r
+ 2rh 
dt
3  dt
dt 
b) Find dV/dt if r is constant so that V and h are functions of time.
dV
 r 2 dh
=

dt
3 dt
c) Find dV/dt if V is constant so that r and h are functions of time.
  2 dh
dr 
0 =  r
+ 2rh 
3  dt
dt 
Related Rate Example
A hot-air balloon rising straight up from a level field is tracked by
a range finder 500 feet from the lift-off point. At the moment the
range finder’s elevation angle is π/4, the angle is increasing at
the rate of .14 radians/min. How fast is the balloon rising at that
moment?
Related Rate Example
A hot-air balloon rising straight up from a level field is tracked by
a range finder 500 feet from the lift-off point. At the moment the
range finder’s elevation angle is π/4, the angle is increasing at
the rate of .14 radians/min. How fast is the balloon rising at that
moment?
d
dh

rad
At  =
,
= .14
, Find
4
min
dt
dt
h
1. tan  =
 h = 500 tan 
500
dh
d
2
2.
= 500 sec 
dt
dt
h
dh
2 


3.
At

=
,
=
500
sec
.14 

4
4
4
R.F.
dt
500 ft
dh
= 1000  .14  = 140 ft
sec
dt
 
Steps to Solving a Related Rate Problem
1. Identify the variable whose rate of change we want and the
variables whose rate of change we have.
2. Draw a picture and label with values and variables.
3. Write an equation relating the variable wanted with known
variables.
4. Differentiate implicitly with respect to time.
5. Substitute known values
6. Interpret the solution and reread the problem to make sure
you’ve answered all the questions.
z
.6
y
x
.8
Related Rate Example
Water runs into a conical tank at the rate of 9 ft3/min. The tank
stands point down and has a height of 10 ft and a base radius
of 5 ft. How fast is the water level rising when the water is 6 ft
deep?
Related Rate Example
Water runs into a conical tank at the rate of 9 ft3/min. The tank
stands point down and has a height of 10 ft and a base radius
of 5 ft. How fast is the water level rising when the water is 6 ft
dV
dh
deep?
= 9, r = 5, Find
at h = 6
dt
1. V =
5 ft
dt
1
r
1
h
  r 2  h, but
= , so r =
3
h
2
2
1 h
1
h3
 3
2. V =      h =  
=
h
3 2
3
4
12
dV

dh
3.
=  h2 
dt
4
dt

dh
4. 9 =  36 
4
dt
dh
1
=
= .31831 ft
sec
dt

2
r
10 ft
h
Related Rate Example
A 13 ft ladder is leaning against a wall. Suppose that the base of
the ladder slides away from the wall at the constant rate of 3
ft/sec.
1. Show how the motion of the two ends of the ladder can be
represented by parametric equations.
2. What values of t make sense in this problem.
3. Graph using a graphing calculator in simultaneous mode. State
an appropriate viewing window. (Hint: hide the coordinate
axis).
4. Use analytic methods to find the rates at which the top of the
ladder is moving down the wall at t = .5, 1, 1.5 and 2 sec. How
fast is the top of the ladder moving as it hits the ground?
Related Rate Example
A 13 ft ladder is leaning against a wall. Suppose that the base of
the ladder slides away from the wall at the constant rate of 3
ft/sec.
1. Show how the motion of the two ends of the ladder can be
represented by parametric equations.
x1  t  = 3t, y1  t  = 0 x1  t  = distance from base of wall at time = t
x 2  t  = 0, y2  t  = 132 - 3t  = height of ladder at time = t
2
2. What values of t make sense in this problem.
{ 0 < t < 13/3 }
Related Rate Example
3. Graph using a graphing calculator in simultaneous mode. State
an appropriate viewing window. (Hint: hide the coordinate
axis).
t: [0,4] x: [0,13] y: [0, 13]
4. Use analytic methods to find the rates at which the top of the
ladder is moving down the wall at t = .5, 1, 1.5 and 2 sec. How
fast is the top of the ladder moving as it hits the ground?
y 2 + x 2 = 169
dy
x dx
=dt
y dt
dy
dy
dy
dy
.5  -.348, 1  -.712, 1.5   -1.107,  2   -1.561
dt
dt
dt
dt
dy
When ladder hits ground,
(4.3)  - 24.05 ft
sec
dt