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PROFESSOR: Welcome back to recitation. In this video we're going to do another solid of revolution problem. So what I'd like to do in this problem is to find the volume of the solid generated by rotating the region bounded by the following curves: y equals 0, x equal 4, and y equals square root of x, around the line x equals 6. And you can choose your favorite method to do this. What I would like first, when you're doing this kind of problem, is get a rough sketch of the region. So you have some picture of what's actually going to happen. You don't necessarily need a three-dimensional picture. But at least have the two-dimensional region and understand the where the rotation line is, with respect to that region. So I will give you a little bit of time to work on that problem. And when I come back I'll show you how I do it. OK, welcome back. So again, what we're doing in this video is we're going to be looking, finding the volume of a solid ...

PROFESSOR: Welcome back to recitation. In this video what I'd like us to do is practice Taylor series. So I want us to write the Taylor series for the following function, f of x equals 3 x cubed plus 4 x squared minus 2x plus 1. So why don't you pause the video, take some time to work on that, and then I'll come back and show you what I get. All right, welcome back. Well, we want to find the Taylor series for this polynomial f of x equals 3 x cubed plus 4 x squared minus 2x plus 1. So what I'm going to do is I'm just going to write down Taylor's-- or the expression we have for the sum, for the Taylor series in general and then I'm going to start computing what I need and I'm going to see what I get. So what do I need to remember? Well let's remind ourselves what the formula is. We should get f of x is equal to the sum from n equals 0 to infinity of the nth derivative of f at 0 over n factorial times x to the n. So that's what we wan...

Hi. Welcome back to recitation. We've been talking about Taylor series for a number of functions and rules by which you can compute Taylor series. I have here an example that I don't think we did in lecture. So this is the function f of x equals secant of x. Now, unlike some of the other ones you've seen, there's not a really simple formula for the whole Taylor series of secant x. So what I'd like you to do is not to find, you know, a formula for the general term, but rather, just to use some of the tools that we've learned to compute the first few terms of the Taylor series for f of x equals secant x. Say, up through the x to the fourth term, if you wanted, or even a little further if you were feeling ambitious. So why don't you pause the video, have a go at that, come back, and we can do it together. So welcome back. I asked you to compute the first few terms of a Taylor series. One thing you can always do in this case, is you can go and you ...

Hi. Welcome back to recitation. In class, one of the things you've talked about recently was computing surface areas of solids of rotation. So I have a nice problem relating to that here. So the circle with center (R, 0) and radius little r-- so this is center big R, 0, and radius little r, which is less than big R-- is rotated around the y-axis. And the question is, what's the surface area of the resulting solid? So we have here the circle. Its center is at the point big R, 0, and its radius is little r, so this is the equation of that circle. And we're going to rotate this circle around this axis. So we're going to spin it around. And what you're going to get is a donut, or what mathematicians call a torus. So here's a little schematic of it here, with one dotted little cross section corresponding to this circle. So the question is, what is the surface area of this torus? So why don't you pause the video, take a few minutes to work this probl...

PROFESSOR: Welcome back to recitation. In this video, what I'd like us to do is, do a little bit of practice with sigma notation. So this will be just a few short problems to make sure that you're comfortable with what all the pieces in the sigma notation actually do. We're going to start with two problems here. And the first one is going to be a fill-in-the-blanks type of problem. And the object is, I've given you a sum on the left-hand side, and then I've given you two other sums, but I've left in each place two blanks, and I've filled in the rest. You have enough information to fill in the two blanks. So what I'd like you to do in this problem is fill in the two blanks so that the sums are equal. And the object is obviously is to do this without writing out all the terms and adding up and then going backwards. So you really want to try and understand what each part of the sigma notation does. The second problem I'd like you to do is ...

PROFESSOR: Welcome back to recitation. Today what we're going to do is use what we know about first and second derivatives and what we know about functions from way back in algebra and precalculus, to sketch a curve. So I want you to sketch the curve y equals x over 1 plus x squared. Doesn't have to be perfect, but try and use what you know about these derivatives, first and second derivatives of this function, and what you've talked about in the lecture to get a pretty good sketch of this. I'll give you a little time to work on it and then I'll be back and I'll work on it for you. Welcome back. So hopefully you feel good about the sketch you've drawn. But just to check everything, we can go through it together. And what I'm going to do, just to keep track of things, is I'm going to put an axis in this region and then I'm going to do all my work sort of off to the side and come back slowly. So we'll try and keep track of everyth...

PROFESSOR: Welcome back to recitation. Today what I want to do is something maybe a little bit more theoretical, but the goal is to show that something that you are going to be repeatedly doing when you use quadratic approximations is, in fact, true. So I'm going to explain the situation, give a quick example, and then show you what we're setting out to do. So the situation is as follows: we're going to-- any time you see a Q of f, that's going to represent the quadratic approximation to f at x equals 0. So what I've done is say, Q of f I'm going to define to be the thing on the right, which is exactly the formula you were given in class for the quadratic approximation of a function f at x equals 0. So f is approximately the thing on the right near 0. Our goal is to show that if I want to take the quadratic approximation of a product of two functions, that I can do it in a different way. I can do it in the way written on the right hand side, which ...

Hi. Welcome back to recitation. You've been talking in class about partial fraction decomposition as a tool for integration. So remember that the point of partial fraction decomposition is that whenever you have a rational function, that is, one polynomial divided by another, that in principle, partial fraction decomposition lets you write any such expression as a sum of things, each of which is relatively easy to integrate. So the technique here is purely algebraic. And then you just apply integral rules that we've already learned. So I have here four rational functions for you. And what I'd like you to do in each case, is try to decompose it into the general form that Professor Jerison taught you. So don't, I'm not asking you to-- if you'd like, you're certainly welcome to go ahead and compute the antiderivatives after you do that, but I'm not going to do it for you, or I'm not going to ask you to do it. So what I'd like you to do...

Hi. Welcome back to recitation. We've been talking in class a little bit about parametric equations and arc length. So let's do an example of a problem where you compute an arc length of a curve given by some parametric equations. So in particular, I have here the parametric equations y equals t minus 1 over t, and x equals t plus 1 over t, for 1 less than or equal to t, less than or equal to 2. So those parametric equations trace out some piece of a curve in the plane. And what I'd like you to do is write down an integral whose value is equal to the arc length of that curve. So the integral you're going to get is going to be pretty hard to evaluate. So I wouldn't recommend you spend a whole lot of time trying to evaluate it. But just so we see that we, you know, we can do this, and so we have the integral whose value is the arc length. So take a few minutes, work on that, pause the video, come back, and we can work on it together. All right. So I hope...

PROFESSOR: Welcome back to recitation. In this video we'd like to do another optimization problem. This one's a little bit harder than the distance problem. So the question is the following: consider triangles formed by lines passing through the point x-- (8, 4), sorry, the x-axis and the y-axis. Find the dimensions that minimize area. So what does this fist sentence mean? It really means use this point to draw a line through this point-- I'll give you an example, it's kind of a wiggly line, but hopefully it looks like a line to you-- and it makes a triangle with this line, the x-axis, and the y-axis. We can certainly calculate the area of that triangle. So the problem is asking you to find the dimensions of the triangle that minimize the area with the constraint that the line, the hypotenuse goes through the point (8, 4). I'm going to give you a couple minutes to work on it. Why don't you pause video here and then when you're ready, restart th...