RELATED RATES
Goal: To use the Chain Rule/implicit differentiation, together with some
known rate of change, to determine an unknown rate of change with respect
to time. (Often the unknown rate is otherwise difficult to measure directly.)
Ex. A spherical balloon is being inflated at a rate of 100 cm3/sec. How fast
is the radius of the balloon increasing when the diameter is 50 cm?
Given: The rate of change, with respect to time, of the volume, dV/dt.
Wanted: The rate of change, w.r.t. time, of the radius, dr/dt, when the
diameter ( = 2r) is 50 cm.
Note: Both of the quantities in the problem, volume V and radius r,
are functions of time t. They are both dependent variables, while t is
the lone independent variable.
Relating equation: To link the two quantities, V and r, together, we need an
equation relating them. The obvious choice here is the usual volume
formula of a sphere.
4
V = π r3
3
Now differentiate, implicitly, both sides of the equation with respect to t.
dV 4
dr
dr
= π (3r 2 )
= 4π r 2
dt
3
dt
dt
Lastly, plug in what we know to solve for the unknown rate.
Known: dV/dt = 100; and r = 25, hence
100 = 4π (25) 2
Therefore,
1
dr
=
cm/sec
dt 25 π
dr
dt
Ex. A 13-ft ladder is leaning against a wall when its base starts to slide away
from the wall. By the time its base is 12 ft from the wall, the base is moving
at the rate of 5 ft/sec.
Let x be the distance between the base of the
wall and the base of the ladder. Let y be the height
the top of the ladder is off the ground. The given is
that dx/dt = +5 at the moment x = 12. The length of
the ladder, 13 feet, is a constant, since it would
never change through out.
(a) How fast is the top of the ladder sliding down the wall then?
We are looking for dy/dt. A simple relation between x and y is via the
Pythagorean Theorem: x2 + y2 = 132. Implicitly differentiate both sides of
the equation with respect to t
dx
dy
2x + 2 y
= 0.
dt
dt
Solve for dy/dt in terms of x, y, and dx/dt
dy − x dx
=
dt
y dt
We know that at the moment x = 12, dx/dt = 5. The only other thing we
need is y, which can be found by setting x = 12 and use the equation
x2 + y2 = 132. We have y = 5. Therefore,
dy − x dx − 12
=
=
⋅ 5 = −12
dt
y dt
5
The minus sign denotes that the ladder is sliding down, i.e., the height y is
decreasing. It is sliding down at a speed of 12 ft/sec.
Answer: −12 ft/sec
(b) At what rate is the area of the triangle (formed by the wall, the
ladder, and the ground) changing at the same time?
Now the unknown is the rate of change of the area, dA/dt. We know
the rates of changes of the 2 sides, therefore, the choice for a relation
between what are known and the unknown is simple: A = xy/2.
Implicitly differentiate both sides
dx 
dA 1  dy
= x + y  .
dt 
dt 2  dt
We are given that when x = 12, dx/dt = 5. We’ve also found in (a)
that, at the same moment, y = 5 and dy/dt = −12. Substitute those values into
the equation above and we have
dA 1
− 144 + 25 − 119
= (12 ⋅ (−12) + 5 ⋅ 5) =
=
= −59.5
dt 2
2
2
Answer: −59.5 ft2/sec
Comment: The calculation above is short and simple (2 steps) only
because we’ve already found y and dy/dt in (a). Otherwise, to solve (b) from
scratch, we would either have to solve (a) first, or rewrite A solely as a
function of x by substituting y = (169 − x2)1/2. Then differentiate the
equation A = x(169 − x2)1/2, which requires a more complex series of
calculation than above.
(c) At what rate is the angle θ, between the ladder and the ground
changing then?
Can you think of a formula that relates the angleθ with the sides x
and y of a right triangle? There are in fact six familiar ones.
Answer: −1 rad/sec
Ex. A water tank in the shape of an inverted circular cone with base radius
2 meters and height 4 meters. If water is being pumped into the tank at a
rate of 2 m3/min, find the rate at which the water level is rising when the
water inside is 3 meters deep.
Answer:
8
m/min
9π
Ex. A plane flying horizontally at an altitude of 1 mile and a speed of 500
miles/h passes directly above a radar station. Find the rate at which the
distance from the plane to the station is increasing when it is 2 miles away
from the station.
Answer: 250 3 miles/hour