Below is an animation of a shrinking circle. The radius shrinks from a radiusof 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 ofa circle is shrinking at 1 meter per second, how fast is the area of the circleshrinking at the moment when the radius is 2.5?”. In order to answer thequestion, we must first understand the question. Below we have a diagram ofwhat is occurring:

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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 theblue graph above). We want to know how fast the area is changing from blueto 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 andthat 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

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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 aboutthe rate of the change of the radius. These are very different questions and verydifferent methodologies are needed to answer the questions. It is importantthat we understand what the question is asking and what the symbols used incalculus identify the quantities. Here is a listing of what the symbols mean forthis particular example:

• A(r) Area as a function of radius r.

•

dA

dtthe rate of the change of the area with respect to time, t.

•

dr

dtthe rate of the change of the radius with respect to time, t.

Translating words into the mathematical shorthand can be very helpful in usingthe information as well as knowing what equation you would use to answer thequestion. If we consider our question we may translate into the symbols and seewhat we need. We have the following:

If the radius of a circle is shrinking at 1 meter per second, how fast is the areaof 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

havedr

dt= 1. The part in blue is the question, and it is asking about the rate

of change for the area, so we havedA

dt=?. The green part gives us an exact

moment of interest, so r = 2.5 in the formula fordA

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

(

2rdr

dt

)

from the general power rule

dA

dt= 2πr

dr

dt

(3)

Now that we have established a general relationship, we may determine specificvalues 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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Another way to think about this question is how much area is in black betweenthe 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 mucharea was lost. If we use the information in the problem and measure the areaat the next second we have the radius to be 1.5 meters since the radius shrinksat 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 1.5meters. As we can see the actual area lost traveling in one whole second isdifferent than the rate of change at the instant the radius is 2.5 meters. Butusing this methodology, using values of r that are closer to 2.5 meters can giveus a better approximation. The underlying assumption here would be that therate of change for the radius is a constant 1 meter per second. We would havethe following table

r 2.49 2.499 2.4999 2.49999Change 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 2.5meters, the change in the areas between the circles approaches the 5π value. Inessence, we have used the definition of derivative directly to build this table.

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