This document discusses how to solve a related rates problem involving the shrinking radius of a circle. It provides the steps to determine the rate at which the area of the circle is changing when the radius is shrinking at 1 meter per second and the radius is currently 2.5 meters. The rate of change of the area is calculated to be 5π square meters per second using derivatives and the formulas for the area of a circle.

1. Below is an animation of a shrinking circle. The radius shrinks from a radius
of 5 to 1. Use the buttons below to play the animation.
A typical related rates question would be along the lines of, “If the radius of
a circle is shrinking at 1 meter per second, how fast is the area of the circle
shrinking at the moment when the radius is 2.5?”. In order to answer the
question, we must first understand the question. Below we have a diagram of
what is occurring:
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2. Circle just before the moment of interest
Circle just after the moment of interest
How fast is the circle’s area changing from blue to green?
Or, how much area was lost when the circle changed its radius in the next moment?
So we have the following:
A(r) = πr2
A(2.5) = π(2.5)2
A(2.5) = 6.25π
(1)
This gives us the area of the circle when the radius is 2.5 (represented by the
blue graph above). We want to know how fast the area is changing from blue
to green, so we would need to compute the area of the green circle as well,
but we do not know the radius of this new circle. We have to use calculus and
that means using derivatives, since we are talking about how fast things change.
Hence we have:
A(r) = πr2
dA
dt
= π
d
dt
(r2
)
dA
dt
= π2r
dr
dt
dA
dt
= 2π(2.5)(1)
dA
dt
= 5π
(2)
You see, this is entirely a different question. The first one answers the question,
“What is the area when the radius is 2.5 meters?” It makes no use of the
2

3. information about how fast the radius is changing. The second one answers,
“How fast is the area changing...?” It does make use of the information about
the rate of the change of the radius. These are very different questions and very
different methodologies are needed to answer the questions. It is important
that we understand what the question is asking and what the symbols used in
calculus identify the quantities. Here is a listing of what the symbols mean for
this particular example:
• A(r) Area as a function of radius r.
•
dA
dt
the rate of the change of the area with respect to time, t.
•
dr
dt
the rate of the change of the radius with respect to time, t.
Translating words into the mathematical shorthand can be very helpful in using
the information as well as knowing what equation you would use to answer the
question. If we consider our question we may translate into the symbols and see
what we need. We have the following:
If the radius of a circle is shrinking at 1 meter per second, how fast is the area
of the circle shrinking at the moment when the radius is 2.5 ?
The part in red gives information about the rate of change for the radius, so we
have
dr
dt
= 1. The part in blue is the question, and it is asking about the rate
of change for the area, so we have
dA
dt
=?. The green part gives us an exact
moment of interest, so r = 2.5 in the formula for
dA
dt
. Since r has a rate of
change with respect to t, we must assume in general that it is a function of t,
hence r = r(t). So our translation comes to the following
A(r) = πr2
(t)
dA
dt
= π
d
dt
r2
(t)
dA
dt
= π 2r
dr
dt
from the general power rule
dA
dt
= 2πr
dr
dt
(3)
Now that we have established a general relationship, we may determine specific
values when we have specific information. In other words, we have a model.
Now using the specific information for this particular problem we have
dA
dt
= 2π(2.5)(1)
dA
dt
= 5π
(4)
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4. Another way to think about this question is how much area is in black between
the circles from one instant to the next, since this is the amount of area we lose.
If we know the radius at the next moment we would have an idea of how much
area was lost. If we use the information in the problem and measure the area
at the next second we have the radius to be 1.5 meters since the radius shrinks
at 1 meter per second. We could do the following:
A(r) = πr2
A(2.5) = π(2.5)2
A(2.5) = 6.25π
A(1.5) = π(1.5)2
A(1.5) = 2.25π
A(2.5) − A(1.5)
1
= 6.25π − 2.25π = 4π
(5)
This gives the amount of area lost traveling from a radius of 2.5 meters to 1.5
meters. As we can see the actual area lost traveling in one whole second is
different than the rate of change at the instant the radius is 2.5 meters. But
using this methodology, using values of r that are closer to 2.5 meters can give
us a better approximation. The underlying assumption here would be that the
rate of change for the radius is a constant 1 meter per second. We would have
the following table
r 2.49 2.499 2.4999 2.49999
Change in Area 4.9π 4.99π 4.999π 4.9999π
As we can see, as the radius draws closer to the moment of interest, namely 2.5
meters, the change in the areas between the circles approaches the 5π value. In
essence, we have used the definition of derivative directly to build this table.
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