Related rates problems involve finding a
rate at which a quantity changes by relating
that quantity to other quantities whose
rates of change are known.
Because these problems involve rates, they
must be differentiated with respect to
time.
Example : y  y  5 y  x  4
3
2
2
dy
dy
dy
3y
 2 y  5  2x  0
dx
dx
dx
dy
(3 y 2  2 y  5)  2 x
dt
2
dy
2x

dx (3 y 2  2 y  5)
We will determine how to solve
problems involving:
I. Circles
II.Spheres
III.Triangles
IV.Cones
V. Cylinders
The most common way to approach related rates problems is the
following:
1. Identify the known variables, including rates of change and the rate of
change that is to be found. **Drawing a picture or representation of the
problem can help to keep everything in order**
2. Construct an equation relating the quantities whose rates of change
are known to the quantity whose rate of change is to be found.
3. Differentiate both sides of the equation with respect to time. (Often,
the chain rule is employed at this step.)
4. Substitute the known rates of change and the known quantities into the
equation. Solve for the wanted rate of change.
**Errors in this procedure are often caused by plugging in the known values
for the variables before (rather than after) finding the derivative with
respect to time. Doing so will yield an incorrect result.**
Olympic swimmer Mr. Spitz leaps
into a pool, causing ripples in the
form of concentric circles. The radius
is increasing at a constant of 4 feet
per second. When the radius is 8 feet,
at what rate is the total area of
disturbed water changing (assuming
the splash is negligible)?
The variables r and A are related by
A=πr2. The rate of change of the
radius r is dr/dt=4
Equation: A=πr2
Given rate:
dr/dt=4 Find: dA/dt when r=8
d
d
2
[ A]  [r ]
dt
dt
dA
dr
 2r
dt
dt
dA
 2 (8)( 4)
dt
2
dA
 64 ft
sec
dt
Mr. Spitz, the circus clown, is traveling with the
Barnum and Bailey circus, is told to inflate a
balloon. If the volume increases at a constant rate of
50 cm3/sec, at what rate is the radius increasing
when the volume is 972π cm3?
Equation: V= 4/3πr3 Given: dV/dt=50 V = 972π
Find dr/dt
4 3
V  r
3
4 3
972  r
3
729  r 3
r  9cm
dV
2 dr
 4r
dt
dt
dr
50  4 (9)
dt
2
dr
25

 .0491 cm
sec
dt 162
Major Spitz is flying a rescue mission to
extract General Earl during a battle in WWI.
He flies his plane at 200 mph at an altitude of
6 miles. What is the rate of change of the
shortest distance between them when the
horizontal distance is 8 miles?
dx/dt = -200
dz/dt = ??
6 mi
8 mi
z 2  x2  y 2
z  (8) 2  (6) 2
z  100  10mi
z x y
2
2
2
dz
dx
2z
 2x
dt
dt
dz
2(10)  2(8)( 200)
dt
dz 2(8)( 200)

 160mph
dt
2(10)
General Earl is watching for
his rescue plane with
binoculars. Major Spitz
slows down to 100 mph and
descends to 3 miles to spot
General Earl. What is the
rate of change of the angle
the General is watching from
when the horizontal distance
is 4 miles?
y 3
tan   
x x
d
 2 dx
(sec  )
 3 x
dt
dt
2
d
3(cos 2  ) dx

dt
x2
dt
4 )2

3
(
d
5 )( 100)
(
dt
42
d 300

 12 rad
hr
dt
25
Mr. Spitz was best friends with
Augustus Caesar. While studying
by candlelight, Spitz was testing
his speed with implicit
differentiation and timed himself
with an hourglass, the most
advanced technology at the time.
He knows that the sand initially
falls forming a cone, whose radius
is twice its height. The hourglass
fills at a constant rate of 3
cm3/min. When the volume is 36π
cm3, what is the rate of change of
the height?
V  1 r 2 h
3
V  1  ( 2h) 2 h
3
V  1  (4h3 )
3
Note: When V = 36π, h = 3
dV 1
2 dh

 (12)( h )
3
dt
dt
2 dh
3  4 (3)
dt
dh
3
1 cm


sec
dt 36 12
Mr. Spitz is filling his
cylindrical water jug with
some “good H2O” before he
runs the iron man triathlon.
The radius is 4 inches; as
Mr. Spitz fills the jug, the
height is changing at a rate
of .5 in/sec. What rate is the
volume changing at?
V  r h
2
V  (4) h
2
dV
dh
 16
dt
dt
3
dV
in
 16 (.5)  8
sec
dt