(a) Sodium chlorate crystals are easy to grow in the shape of cubes by allowing a solution of water and sodium chlorate to evaporate slowly. If $ V $ is the volume of such a cube with side length $ x. $ calculate $ dV/dx $ when $ x = 3 $ mm and explain its meaning.
(b) Show that the rate of change of the volume of a cube with respect to its edge length is equal to half the surface area of the cube. Explain geometrically why this result is true by analogy with Exercise 11(b).
The situation here is that the crystal is in the shape of a cube and we're interested in finding the rate of change of the volume with respect to X when X is three. So the volume of a cube is X cubed, and the derivative of that DVD X would be three x squared. So then DVD X for X equals three. We're going to substitute a three in there for X and we get three times three squared and that would be 27 and the units on volume would be millimeters cubed and the units on X would be millimeters. So what this is telling us is that at this particular size of cube, the volume is growing at a rate of 27 cubic millimeters for every millimeter change in X and in part B of this problem, we're looking at surface area. So to get the surface area of a cube, you're going to add the area of every surface. And for this cube, every surface has an area of X squared and there are six of those. So the surface area is six x squared. Notice that the derivative of the volume is half of the surface area. Now let's take a look at that from a geometric perspective. So suppose that the the cube grows bigger and the side length is now Delta X, or the side length is now X plus Delta X. So we added Delta X to each side so the new volume would be X plus Delta X quantity. Cute and the old volume was just execute. So the change in volume is a new volume minus the old volume. And let's work that out. Algebraic Lee. So the change in volume is going to be. We have to cube this by no meal, and so you're gonna have to work that out either using the binomial theorem or just multiply X plus Delta X Times X plus Delta X Times X plus Delta X. Do that very carefully and you're going to get X cubed plus three X squared Delta, X plus three X times Delta X squared plus Delta X cubed and then we still have the minus X cubed at the end, subtracting the volume of the old one. Notice that we can cancel the X cubed in the minus x cubed. So this is our change in volume and suppose we want to know the change in volume over the change in X. So let's go ahead and divide this by changing X, Delta X. And while we're doing that, let's factor out a Delta X from the numerator. So have Delta X Times three X squared plus three x Delta X plus still two X squared. That's over Delta X so we can cancel the Delta X from the top in the bottom. And all we have left is that the change in volume with respect to a change in inside length is three X squared plus three x times Delta X plus Delta X squared. So if Delta X is small, then the second term is going to be very close to zero, and the third term is going to be very close to zero. And the rate of change of volume with respect to X is going to be approximately three X squared