These notes closely follow the presentation of the material given in James Stewart’s
textbook Calculus, Concepts and Contexts (2nd edition). These notes are intended
primarily for in-class presentation and should not be regarded as a substitute for
thoroughly reading the textbook itself and working through the exercises therein.
Related Rates Problems
In solving a related rates problem, one attempts to find the rate of change of some
quantity based on the rate of change of some related quantity. The basic strategy for
solving related rates problems is outlined on page 267 of our textbook.
Example Suppose that one leg of a right triangle remains of fixed length while the other
leg grows at a constant rate k  0 and that all the while the triangle remains a
right triangle. Find a formula that gives the rate at which the hypotenuse is
growing.
Solution Let a be the length of the fixed leg of the triangle. Let x be the length of the
other leg and let y be the length of the hypotenuse. Note that a is a constant
quantity but x and y are functions of time ( t ). We are given that dx/dt  k and we
want to determine dy/dt.
Since the triangle remains a right triangle at all times, then by the Pythagorean
Theorem,
a2  x2  y2
for all t  0. (Recall that x and y are functions of t. ) Differentiating both sides of
this equation with respect to t, we obtain
d a 2  x 2   d y 2 
dt
dt
or
dy
2x dx  2y .
dt
dt
Since dx/dt  k, this gives
dy
 kx
y .
dt
For example, suppose that we are given that a  6 inches and k  2 inches per
second and we wish to know at what rate the hypotenuse is growing at the instant
when the growing leg has length 12 inches. The rate of growth of the hypotenuse at
this instant is
dy
.
dt x12
When x  12, the Pythagorean Theorem gives us
6 2  12 2  y 2
which gives
y  6 5.
Thus,
1
4 5
 2  12  4 
5
x12
6 5
5
We conclude that at the instant when x  12 inches, the hypotenuse is growing at a
rate of 4 5 /5  1. 79 inches per second.
dy
dt
2
Example A plane flying horizontally at an altitude of one mile and a speed of 500 miles
per hour passes directly over a radar station. Find the rate at which the distance
from the plane to the station is increasing at the instant when the plane is two miles
away from the station.
Solution Let O be the location of the radar station and let P be the position of the plane
when it is directly over the radar station. Let Q be the position of the plane at any
given time t after flying over the radar station. Then OPQ is a right triangle with
leg OP of length one mile, leg PQ growing at a constant rate of 500 miles per hour
and hypotenuse OQ also growing with time. Let x be the length of PQ and let y be
the length of OQ.
From the work done in the previous example, we have
dy
 500x
y .
dt
We wish to know the value of dy/dt at the instant when y  2 miles.
By the Pythagorean Theorem, when y  2 we have
12  x2  22
which gives that x  3 . Thus,
500 3
dy

 250 3 .
2
dt y2
Our conclusion is that the distance of the plane from the radar station is increasing
at a rate of 250 3  433 miles per hour at the instant when the plane is two miles
from the radar station.
3
Example A water trough is 10 m long and a cross section has the shape of an isosceles
trapezoid that is 30 cm wide at the bottom, 80 cm wide at the top, and has height
50 cm. If the trough is being filled with water at the rate of 0.2 cubic meters per
minute, how fast is the water level rising when the water is 30 cm deep?
Solution Let x be the depth and V be the volume of the water in the trough at any given
time t. Referring to the figure, which shows a cross section of the trough, we
reason that the volume of water in the trough at any time t is given by
V  10 0. 3x  2  1 rx .
2
By properties of similar triangles,
0. 25
r
x  0. 5
which implies that
r  1 x.
2
Thus, we obtain
V  10 0. 3x  1 x x .
2
which is equivalent to
V  3x  5x 2 .
Differentiation with respect to t yields
dV  3  10x dx .
dt
dt
Since we are given that dV/dt  0. 2, then
0. 2  3  10x dx .
dt
Solving for dx/dt gives
dx  0. 2
3  10x
dt
Substitution of x  0. 3 into this equation yields
dx
 1 .
30
dt x0.3
Thus, the water level is rising at a rate of 1/30 of a meter per minute at the instant
when the water level is 0.3 meters.
4
Example Water is leaking out of an inverted conical tank at a rate of 10,000 cubic
centimeters per minute at the same time that water is being pumped into the tank at
a constant rate. The tank has height 6 meters and the diameter at the top is 4
meters. If the water level is rising at a rate of 20 centimeters per minute at the
instant when the water level is 2 meters, find the rate at which water is being
pumped into the tank.
Solution Let k be the constant rate at which water is being pumped into the tank. This is
the unknown. Let x be the water level in the tank at any given time t and let r and V
be, respectively, the radius of the water surface and the volume of water at any
given time t. Since the water in the tank also takes the shape of a cone, then by
referring to the formula for the volume of a cone on the inside front cover of
Stewart, we obtain
V  1 r 2 x.
3
Using properties of similar triangles, we note the relation
2
r
x  6
which implies that
r  1 x.
3
Thus,
2
V  1 1x x
3
3
 1 x 3 .
27
Differentiation with respect to t yields
dV  1   3x 2  dx
27
dt
dt
 1 x 2  dx .
9
dt
Since water is being pumped into the tank at the constant rate k and simultaneously
being pumped out of the tank at the constant rate of 10,000 cm 3 /min, then the net
rate of change of the volume of the water in the tank is dV/dt  k  10, 000
cm 3 /min. (Note that dV/dt is constant.) We thus have
k  10, 000  1 x 2  dx .
(1)
9
dt
Finally, we are given that the water level is rising at a rate of 20 cm/min at the
instant when the water level is 200 centimeters. This is expressed in Leibniz
notation as
dx
 20.
dt x200
Setting x  200 in Equation (1) gives
k  10, 000  1 200 2  20.
9
Solving for k then gives
5
k  10, 000  1 200 2  20
9
 10, 000  800000 
9
 289, 253
We conclude that water is being pumped into the tank at the rate of about
289, 253 cm 3 /min.
6