AP®︎/College Calculus BC
- 2011 Calculus BC free response #1a
- 2011 Calculus BC free response #1 (b & c)
- 2011 Calculus BC free response #1d
- 2011 Calculus BC free response #3a
- 2011 Calculus BC free response #3 (b & c)
- 2011 Calculus BC free response #6a
- 2011 Calculus BC free response #6b
- 2011 Calculus BC free response #6c
- 2011 Calculus BC free response #6d
2011 Calculus BC free response #3 (b & c)
Volume of a solid of rotation and Chain Rule for rates of change. Created by Sal Khan.
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- At6:12he mentions that thinking of (dv/dk)(dk/dt)=(dv/dt) as fractions cancelling out isn't 'rigorous' (a notion he always repeats when doing this calculation. Is there a video in which he explains a more rigorous way to think of this?(1 vote)
- I dont think so. In general, it seems that he prefers to use intuitive, geometric arguments to justify mathematical results. If you want a more rigorous discussion, try out Chris Tisdell on Youtube.(3 votes)
- Could you also solve this problem through radian integration along the angle of the z-axis? As in, the integral from z-axis angle=0 to z-axis angle=2pi of R.
That would seem to have the benefit of applying generally to any solid of revolution, you just plug in R.(1 vote)
We're on part b. The region R is rotated about the x-axis to form a solid. Find the volume, V, of the solid in terms of k. So this is our region R, and it's going to be rotated around the x-axis. It's going to be rotated around the x-axis to form a solid. So that solid is going to look something like this. It's going to look like this. This is going to be this end of the solid when it's rotated around, and it's going to look something like this. Let me do my best attempt to draw our solid. And then that side is going to look something like that. It'll look a little bit like a loudspeaker. And so let me shade it in a little bit. So that's going to be our solid and you can imagine that maybe this right over here is our y-axis. That is our y-axis. And then going straight through the center of the solid, you have your x-axis. If the solid was slightly transparent, you would see the other side of this opening right over here, and the x-axis goes straight through the center. So how do we find the volume of this solid? Well, we could think about each of the individual disks of this solid right over here. So think about it this way. Let's take a disk right over here. Let me take this disk right over here so that you can imagine this little bit of a cross section. It's a little bit of a cross section of this solid. So this disk right over here, it's going to have some area, some surface area, and then it's going to have some depth. I'll draw the edge. You can imagine the depth is kind of the edge of the quarter, while the surface area is kind of the face of the quarter, the area of the face of the quarter. So what's going to be the volume of this white disk right over here? Well, it's going to be the surface area of the disk times the depth. So what is the surface area? Well, surface area of a circle is pi times the radius squared. So it's going to be pi times-- and what's this radius? Well, the radius is the height between the x-axis and the function. So this radius is e to the 2x power. So the area of each of these disks, for this x right over here, for this value of x, is going to be pi times e to the 2x squared. And then if we want the volume of this entire disk, we then multiply the area times the depth. So times dx. So this would be the volume of this little disk over here that has a very small infinitesimally wide depth. What we want to do is sum all of these disks. We want to have a bunch of these disks, that we want to sum up all of them. So we want to take an infinite sum of these infinitely thin disks. So essentially we're just going to take the sum of all of these from x is equal to 0, all the way to x is equal to k. Right, because this right over here, this is x is equal to k, this is x is equal to 0, we take the sum of all the disks. We will get our actual volume. So we can actually evaluate this integral analytically, it's not too bad. Because either-- let me rewrite it. Pi times-- e to the 2x squared is e to the 4x, just 2x times 2. And then, if we want to take the integral, you might be able to do this just by inspection, but if you want to say, hey look, I have a little function here, I want its derivative sitting here, what we can do is we can multiply this expression by 4, and we can also divide it by 4. If you multiply something by 4 and divide by 4, you haven't changed its value. You've multiplied it by 1. But what this does is it gives us-- let me rewrite it a little bit. We can rewrite this as pi over 4 times 4e to the 4x. I just rewrote this. I multiplied and divided by 4, multiplied and divided by 4. And the reason why I did that is so that we have the derivative of 4x sitting right over here, we have this 4. Pi over 4 is a constant, so we can actually take it out of the integral. And so we can say that this expression right over here is the same as this expression right over here, from pi over 4 times the integral from 0 to k of 4e to the 4x dx. And so the whole reason why I multiplied and divided by 4 is I have something here, I have its derivative. So I can really just pretend like this is an e to the x and you can take the integral, or the anti-derivative, with respect to 4x. So the inside is e to the 4x. And you can verify. If you take the derivative of e to the 4x, it is the derivative of 4x, which is 4, times e to the 4x. So this right here is the anti-derivative of that. And we're going to evaluate it. We are going to evaluate it from 0 to k. All of that times pi over 4. And so when you evaluate it at k, it's e to the 4k. So this is going to be pi over 4 times e to the 4k minus e to the 4 times 0, which is e to the 0, which is minus 1. So this is our volume, pi over 4 times e to the 4k minus 1. So that is part b. We did that. Now part c. I think we have time for this. The volume, V, found in part b, changes as k changes. If dk, the derivative of k with respect to t is 1/3, determine the derivative of V with respect to t when k is equal to 1/2. So this is straight up from the chain rule. So the chain rule will tell us-- and if you view differentials as just very, very small numbers, it actually seems like a little bit of common sense-- the rate of change of V, so a very small change in V with respect to a very small change in t is equal to a very small change in V with respect to a very small change in k times a very small change in k divided by a very small change in t. Or the derivative of V with respect to t is equal to the derivative of V with respect to k times the derivative of k with respect to t. And the reason why I said this makes sense if you view a differential as a small number is you would-- and that's not a very rigorous way of doing it, but it's a good kind of common sense way of just thinking about things-- is that these guys would cancel out if these were numbers, and then you would have dV dt on both sides. So that's why I said that this is slightly common sense. You can kind of view it that way. But this essentially gives us everything we need to know. We need to solve for dV dt. They already gave us dk dt at 1/3 when k is equal to 1/2. So they're already telling us the rate of change of k with respect to t. We already know that this thing right over here is going to be 1/3. And we can figure out the derivative of V with respect to k very easily because we have V as a function of k right over here. So let's figure that out. So the derivative of V with respect to k is equal to pi over 4 times the derivative in here. And the derivative of this is just 4e to the 4k. The derivative of negative 1 is just 0. So this is times 4e to the 4k, is the derivative. These guys cancel out. So it's equal to pi e to the 4k, that's the derivative of V with respect to k. We want to find it when k is equal to 1/2, so we could write V prime at 1/2, or when k is equal to 1/2 dV dk. dV dk is going to be equal to pi times e to the 4 times 1/2 which is equal to pi e squared. So this part right over here is pi e squared. And so the derivative of V with respect to t when k is equal to 1/2 and the derivative of k with respect to t is 1/3 is going to be just pi e squared times 1/3. So let me just rewrite it over here. So dV dt is going to be equal to 1/3 times this, or we could write pi e squared over 3. And we're done.