RELATED RATES
Section 2.6
When you are done with your
homework, you should be able
to…
• Find a related rate
• Use related rates to solve real-life
problems
Find the
derivative of
with respect to y
x
A. x '  
y
x
B. x '  
2y
2
x
C. x '  
2 xy
D. None of the above
x y2
2
Find the volume of a cone with a
radius of 24 inches and a height of
10 inches. Round to the nearest
hundredth.
6031.86
0.03
FINDING RELATED RATES
•
We use the chain rule to implicitly find
the rates of change of two or more
related variables that are changing with
respect to time.
Some common formulas used in
this section
• Volume of a…
4 3
V

r
– Sphere:
3
– Right Circular Cylinder: V   r 2 h
1 2
– Right Circular Cone: V   r h
3
1
– Rectangular Pyramid: V  lwh
3
– Pythagorean Theorem: a 2  b2  c 2
GUIDELINES FOR SOLVING
RELATED-RATE PROBLEMS
1.
Identify all given quantities and quantities to be
determined. Make a sketch and label the
quantities.
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 time t.
4. After completing step 3, substitute into the
resulting equation all known values for the
variables and their rates of change. Then solve
for the required rate of change.
Find the rate of change of the
distance between the origin and a
moving point on the graph of
y = sin x if dx/dt = 2 cm/sec.
Find the rate of change of the
volume of a cone if dr/dt is 2
inches per minute and h = 3r
when r = 6 inches. Round to the
nearest hundredth. How is this
problem different?
678.68
0.03
Vertical Motion. A ball is dropped
from a height of 100 feet. One
second later, another ball is
dropped from a height of 75 feet.
Which ball hits the ground first?
Do we need to use calculus to
solve this problem?
HINT: The equation for the position function is
s  t   16t 2  v0t  s0 , when using earth's gravity and feet.
First ball:
0  16t  100
2
16t  100
2
t2 
100
16
100
16
10
t
4
t  2.5 seconds to hit the ground
t
Second ball:
0  16t 2  75
16t 2  75
t2 
75
16
t
75
16
75
4
t  2.165 seconds to hit the ground
t
t  1  3.165 seconds to hit the ground
Conclusion: The first ball will hit the ground first, and
the second ball will hit the ground 3.165  2.5  0.665
second later.
Angle of Elevation. A fish is reeled in
at a rate of 1 foot per second from a
point 10 feet above the water. At what
rate is the angle between the line and
the water changing when there is a
total of 25 feet of line out?
10'
x

10
sin  
 we found 
x
dx
  1 ft / sec  given 
dt
d
d 10 
sin     
dt
dt  x 
d
10 dx
cos 
 2
dt
x dt
d
10 dx
 1

  2  sec   
 sec  
dt
x dt
 cos 

d
10
25
  2  1
dt
25
252  102
dx


plugged
in


1

 ,  trig identities using triangle 
dt


d 10 1
 
dt 25 5 21
d 2 21

dt
525
d
 0.017 rad/ sec
dt
Consider the following situation:
A container, in the shape of an
inverted right circular cone, has a
radius of 5 inches at the top and
a height of 7 inches. At the
instant when the water in the
container is 6 inches deep, the
surface level is falling at the rate
of -1.3 in/s. Find the rate at which
the water is being drained.