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(a) Find the average rate of change of the area of a circle with respect to its radius $ r $ as $ r $ changes from

(i) 2 to 3 $ \space \space \space $ (ii) 2(0 2.5 $ \space \space \space $ (iii) 2 to 2.1

(b) Find the instantaneous rate of change when $ r = 2. $

(c) Show that the rate of change of the area of a circle with respect to its radius (at any $ r $ ) is equal to the circumference of the circle. Try to explain geometrically why this is true by drawing a circle whose radius is increased by an amount $ \Delta r. $ How can you approximate the resulting change in area $ \Delta A $ if $ \Delta r $ is small?

a) $5 \pi, 4.5 \pi, 4.1 \pi$

b) 4$\pi$

c) $\Delta A \approx(2 \pi r \Delta r),$ when $\Delta r$ is small

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in this problem. We're looking at the area of a circle and we're finding average rate of change for various increments. Changes in the radius starting with the radius change from 2 to 3. And any time you're looking for an average rate of change, you're just using a basic slope formula. So if the radius is changing from 2 to 3, what we want to do is find the area for a radius of three, minus the area for a radius of two and divide that by three minus two. That would be our slope formula. So we end up with the area where a radius of three is pi times three squared. So that would be nine pie on the area for a radius of two would be pi times two square, so that would be four pi. We're dividing that by one, and so we get five pi. For the second part, we're finding the rate of change. The average rate of change as a radius changes from 2 to 2.5. So same idea the area for 2.5 minus the area for two over 2.5, minus two on the area for 2.5 is 6.25 Pi in the area for two is four pi and we're dividing that by 40.5 and this gives us 4.5 pi. We're going through that process one more time, and the radius is changing from 2 to 2.1. So we have the area for radius of 2.1, minus the area for radius of to over 2.1 minus two. And that gives us 4.41 pie minus four pi over 0.1. And that works out to be 4.1 pie. Okay, so for part B, we're finding the instantaneous rate of change when R equals two. So we're finding a prime of to First, we need to find a prime of X. I remember A of X was the area of the circle. I should call it A of our high r squared is the area of the circle. So it's derivative would be two pi r. So then a prime of to would be two pi times to, and that would be four pi notice that our previous answers were getting closer and closer to four pi. The smaller our interval got we had five pi. We had 4.5 pi and we had 4.1 pie and if our interval had gotten even smaller, we would have been even closer to four pi. All right, for part C noticed that the rate of change of the area is two pi r and we also know that the circumference of a circle is two pi r. So we're noticing that those are the same and we want to look at a geometric argument for that. So we have our circle with Radius R, and it grows bigger. And now the radius is r plus Delta are so our new circle has area. High times are plus Delta R squared and our old circle had area high r squared. So the Delta A The change in area is the new area minus the old area, and that would be pi times are plus Delta R squared minus pi times R squared. Let's simplify that. We can factor a pie out of both and then we're going to multiply out the binomial and we get R squared plus two are Delta are plus Delta R squared minus R squared the minus R squared comes from this last term. Who factored out the pie earlier. Notice that we can cancel the are scored in the minus r squared. So what we have left for the change in area is pi times a quantity to our delta are plus Delta R squared. Now suppose we were interested in Delta. A over Delta are change in area over change in time. We're going to take what we just got for Delta A and divide it by Delta R. And while we're doing that, let's factor out of Delta. Are so we have pie times. Delta are times to R plus Delta are over, Delta are canceled, The Delta are from the top of the bottom and we have pie times to R Plus Delta are now. Suppose we're looking at a very small they'll toe are. If Delta are is extremely small, then that's essentially zero. On what we have left is two pi r the circumference of a circle. So if you were interested in approximating a change in area for a very small delta, are you would just approximate it by finding the circumference