Video: Finding the Rate of Change in the Area of a Shrinking Circular Disc Using Related Rates

The circular disc preserves its shape as it shrinks. What is the rate of change of its area with respect to radius when the radius is 59 cm?

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Video Transcript

The circular disc preserves its shape as it shrinks. What is the rate of change of its area with respect to radius when the radius is 59 centimetres?

We begin by recalling the formula that allows us to calculate the rate of change of a function at a given point when π‘₯ is equal to π‘Ž. It’s 𝑓 prime of π‘Ž equals the limit as β„Ž approaches zero of 𝑓 of π‘Ž plus β„Ž minus 𝑓 of π‘Ž over β„Ž, where 𝑓 prime is the derivative of the function. But we don’t seem to have a function here. So let’s consider what we do know about the area of a circle. It’s given by the formula 𝐴 equals πœ‹π‘Ÿ squared. We could write this as 𝐴 of π‘Ÿ. 𝐴 is a function of π‘Ÿ. That means that the rate of change of 𝐴 with respect to π‘Ÿ is the derivative of 𝐴 with respect to π‘Ÿ.

Now, we’re trying to find the rate of change when the radius is equal to 59. So we’re going to let 𝐴 be equal to 59. We want to find 𝐴 prime of 59. And by definition, that must be equal to the limit as β„Ž approaches zero of 𝐴 of 59 plus β„Ž minus 𝐴 of 59 all over β„Ž. Let’s work out what 𝐴 of 59 plus β„Ž and 𝐴 of 59 actually are. 𝐴 of π‘Ÿ is πœ‹π‘Ÿ squared. So 𝐴 of 59 plus β„Ž is πœ‹ times 59 plus β„Ž squared. We distribute our parentheses. And we see that this is equal to πœ‹ times 3481 plus 118β„Ž plus β„Ž squared. Similarly, 𝐴 of 59 is πœ‹ times 59 squared, which is 3481πœ‹. We can replace 𝐴 of 59 plus β„Ž and 𝐴 of 59 with these two expressions in our definition for the derivative. And when we factor by πœ‹, we see the numerator is πœ‹ times 3481 plus 118β„Ž plus β„Ž squared minus 3481. Now, of course, these give us zero.

So we’re looking for the limit as β„Ž approaches zero of πœ‹ times 118β„Ž plus β„Ž squared all over β„Ž. And you might now spot we can actually divide through by β„Ž. And our derivative is now the limit as β„Ž approaches zero of πœ‹ times 118 plus β„Ž. We’re now ready to perform direct substitution. We let β„Ž be equal to zero. And when we do, we find that 𝐴 prime of 59 equals 118πœ‹. The rate of change of the circular disc’s area with respect to its radius is 118πœ‹ centimetres squared per centimetre. Now, you might be inclined to think that the answer should be negative. We’re told that the circular disc is shrinking. However, that’s a bit of a trick. The area changes in the same positive or negative direction as the radius. So, in fact, it is indeed a positive rate of change.

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