Related Rates
Kirsten Maund
Dahlia Sweeney
Background
Calculus was invented to predict
phenomena of change: planetary motion,
objects in freefall, varying populations, etc.
In many practical applications, several
related rates vary together. Naturally, the
rates at which they vary are also related to
each other. With calculus, we can describe
and calculate such related rates.
What are related rates?
 A related rates problem involves two or
more quantities that vary with time and an
equation that expresses some relationship
between them.
 Typically, the values of these quantities at
some instant are given together with all
their time rates of change but one. The
problem is usually to find the time rate of
change this is not given, at some instant
specified in the problem.
How to Solve Related Rates
 One common method for solving such a
problem is to begin with implicit differentiation
of the equation that relates the given
quantities.
 For example, suppose that x and y are each
functions of time such that:
x^2 + y^2 = a^2 (a is a constant)
 Differentiate both sides of this equation with
respect to time t. This produces the equation:
2x dx/dt + 2y dy/dt = 0
 If the values of x, y, and dx/dt at a certain
instant t are known, then the last equation
can be solved for the value of dy/dt at time
t.
 Note that it is not necessary to know x and
y as functions of t.
 It is typical for a related rates problem to
contain insufficient information to express
x and y as functions of t.
WARNING!
The most common error to be avoided
is the premature substitution of the
given data, before rather than after
implicit differentiation.
Strategy for Solving
 Step 1: Make a drawing of the
situation if possible.
 Step 2: Use letters to represent the
variables involved in the situation say x, y.
 Step 3: Identify all rates of change
given and those to be determined,
Use the calculus notation (dx/dt,
dy,dt, etc) to represent them.
 Step 4: Determine an equation that
involves both
 The variables from step two
 The derivative of step three
 Step 5: Differentiate (by implicit
differentiation) the equation of step
four
 Step 6: Substitute all know values
into the differentiated equation
 Step 7: Use algebraic manipulation
,if necessary, to solve for the
unknown rate or quantity
Formulas You May Need To Know
V  a3
V r h
2
4 r 

V
3
3
 r h

V
2
3
V  lwh
bh 

V
3
Example #1
 A ladder 10 feet long is resting against a wall. If
the bottom of the ladder is sliding away from the
wall at a rate of 1 foot per second, how fast is
the top of the ladder moving down when the
bottom of the ladder is 8 feet from the wall?
 First, draw the picture:
 We have dx/dt is one foot per second.
We want to find dy/dt.
 X and y are related by the
Pythagorean Thereom
 Differentiate both sides of this equation
with respect to t to get
 When x = 8 ft, we have
 Therefore
 The top of the ladder is sliding down
(because of the negative sign in the
result) at a rate of 4/3 feet per second.
Example #2
 A man 6 ft tall walks with a speed of 8 ft
per second away from a street light atop
an 8 foot pole. How fast is the tip of his
shadow moving along the ground when he
is 100 feet from the light pole.
6 ft
z-x
x
z
18 ft
 Let x be the man’s distance from the pole and




z be the distance of the tip of his shadow from
the base of the pole.
Even though x and z are functions of t, we do
not attempt to obtain implicit formulas for
either.
We are given that dx/dt = 8 (ft/sec), and we
want to find dz/dt when x = 100 (ft).
We equate ratios of corresponding sides of
the two similar triangles and find that z/18 =
(z-x)/6
Thus 2z = 3x
 Implicit differentiation now gives 2 dz/dt =
3 dx/dt
 We substitute dx/dt = 8 and find that
(dz/dt = 3/2) * (dx/dt = 3/2) * (8) = 12
So the tip of the man’s shadow is moving at 12
ft per second.
Try Me!
 A ladder 25 ft long is leaning against a
vertical wall. If the bottom of the ladder is
pulled horizontally away from the wall at 3
ft/sec, how fast is the top of the ladder
sliding down the wall, when the bottom is
15 ft from the wall?
Solution
 t = the number of seconds in time that has
elapsed since the ladder started to slide
down the wall.
 y = the number of feet in distance from the
ground to the top of the ladder at t
seconds.
 x = the number of feet in the distance from
the bottom of the ladder to the wall at t
seconds.
 Because the bottom of the ladder is pulled
horizontally away from the wall at 3 ft/sec,
dx/dt = 3. We wish to find dy/dt when x =
15.
 From the Pythagorean Thereom, we have
y^2 = 625 – x^2
 Because x and y are functions of t, we
differentiate both sides of equation one
with respect to t and obtain 2y dy/dt = -2x
dx/dt giving us dy/dt = -x/y dx/dt
 When x = 15, it follows from equation one
that y = 20.
 Because dx/dt = 3, we get from equation
two: dy/dt = (-15/20) * 3 = -9/4
 Therefore, the top of the ladder is sliding
down the wall at the rate of 2 ¼ ft/sec
when the bottom is 15 ft from the wall.
 The significance of the minus sign is that y
is decreasing as t is increasing.
Was Your Answer Correct?
Bibliography
http://www.math.dartmouth.edu/~klbooks
ite/2.17/217examples/217ladder.htm
 http://www.math.dartmouth.edu/~klbooks
ite/2.17/217examples/217baseball.htm

© Maund and Sweeney 2011