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(a) The volume of a growing spherical cell is $ V = \frac {4}{3} \pi r^3, $ where the radius $ r $ is measured in micrometers $ (1 \mu m = 10^{-6} m). $ Find the average rate of change of $ V $ with respect to $ r $ when $ r $ changes from

(i) 5 to 8 $ \mu m $ (ii) 5 to 6 $ \mu m $ (iii) 5 to 5.1 $ \mu m $

(b) Find the instantaneous rate of change of $ V $ with respect to $ r $ when $ r = 5 \mu m. $

(c) Show that the rate of change of the volume of a sphere with respect to its radius is equal to its area. Explain geometrically why this result is true. Argue by analogy with Exercise 13(c).

a) 540, 381,320 micrometer squared

b) 314 micrometer $^{3}$

c) see solution

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this problem involves a sphere and we're going to start out by finding some average rates of change. And an average rate of change is like a standard slope like y tu minus y one over x two minus X one. So for finding the rate of change from 5 to 8 for the radius measurements. But we need to do is take the volume for eight minus the volume for five, divided by eight minus five. And so we have 4/3 pi times eight cubed, minus 4/3 pi times five cubed, divided by three and will spare you all the details from all the fraction work. You end up with 2048 hi over three minus 500 pi over three, divided by three And that gives you 172 pi and the units. So the volume was measured in cubic micrometers and the radius is in micrometer, so we would have cubic micrometers per micrometer. Now we're going to go with another interval. This time it's going to be from 5 to 6 micrometers. So the same process the volume at six minus the volume at five, divided by six minus five So we end up with 864 3rd pi minus 503rd pi divided by one. And that gives us 121 and 1/3 pie as our average rate of change. We're going to go through this process 1/3 time for another interval. This time it is from 5 to 5.1. So you notice that we're inching closer to 55 Seems to be the number of the day here. And so we're going to have the volume at 5.1, minus the volume at five, divided by 5.1 minus five and we end up with now. I went ahead and approximated this one. I think that we're just kind of getting the point. Here. We end up with approximately 1 76.87 pi minus 503rd pi over 0.1. Your appoint one, and that's approximately 102.1 pie. So the thing to notice here is that as our interval gets smaller and smaller and we're getting closer and closer in toward R equals five, we're getting closer to rates of change of looks like getting closer and closer to 100 pie. So that's what Part B is all about. In Part B. We actually find the instantaneous rate of change at five. So let's start by finding the derivative of the function. And remember, the function was volume equals 4/3 pi r cubed So it's derivative would be four pi r squared. And then when we evaluate that at five, we have four times pi times five squared and that is indeed 100 pie. So those other rates of change were inching their way closer to that, and that would be micrometers cubed per micrometer. Okay, For part C, we noticed that the rate of change of volume was four pyre squared, and we know the surface area of a sphere is also four pi r squared. So we noticed that those are equal and we're going to take a geometric investigation into what's going on here. So you have your original sphere that had a radius of our and then the sphere goes grows bigger and now it has a radius of R plus. Delta are, Let's say so. Let's look at the change in volume. The change in volume would be the volume of the old Excuse me, the volume of the new larger sphere minus the volume of the old smaller sphere. And so the volume of the new sphere is 4/3 times pi times its radius R plus delta R cubed on the volume of the old sphere is 4/3 pi times its radius are cute. Now we're going to work this out, Algebraic Lee, give ourselves a little more space here. We can factor the 4/3 pi out of both terms and we have 4/3 pi times Now we need to multiply this binomial out. We need to expand it and we can use the binomial theorem or just multiply it out the long way are plus, delta R times are close. Delta R times are plus Delta are and you're going to end up with our cubed plus three are square Delta are plus three R Delta R squared plus Delta R cubed and we still have minus r cubed at the end. From this last term. Here we factored the 4/3 pi out of it. Notice that we can cancel the r cubed and the minus r cubed. And so this is Delta the This is the change in volume. What if we were interested in the change in volume over the change in X? So let's divide this by Delta X. We should call a Delta Are changin radius notice that every term has a factor of Delta are in it. So if we were to cancel those Delta ours, what we would have left is 4/3 pi times three R squared plus three R Delta R plus Delta R squared. That is the change in volume over the change in radius. Now what if Delta R is really small? It's very close to zero then this quantity is very close to zero and this quantity is very close to zero. And what we have remaining is 4/3 pi times three r squared and that is four pi r squared.