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CHRISTINE BREINER: Welcome back to recitation. In this video, I'd like us to work on the following problem that has to do with tangent planes and approximations. So we know that a rectangle-- we'll say a rectangle has sides x and y, and we know that to find the area of the rectangle then, we just take x times y, and I would like us to approximate the area for x equal to 2.1 and y equal to 2.8. And obviously, with this type of equation, it's not hard to just compute this, but I'd like us to use the tangent plane approximation to determine the value, and then we'll compare it to the actual value, just to give us an idea of how we can use the tangent plane to approximate things. And obviously, I'd like to do this near x equal 2 and y equal 3. So when you're doing your tangent plane approximation, do the approximation at x equal 2 and y equal 3. And then when you're done with that, I would like you to answer this question. So near x equal 2 and...

CHRISTINE BREINER: Welcome back to recitation. In this video, what I want to work on is using what we know about matrix multiplication and finding inverses of matrices to solve a system of equations. So we've set up the system already as if it's already in matrix form. And what I'd like us to do is, for this particular A-- this 3-by-3 matrix A-- find a vector x, so that A*x equals b. Where b is equal to these two things. So you're going to do two problems. You're going to do when b equals 1, 2, negative 3. And you're going to do when b is equal to [0, 0, 0]. So you want to find vector x so that A*x equals this value here. And what I'd like you to do is I'd like you to use the strategy that you saw in the lecture, which is find A inverse, and then take A inverse b. So we really want to practice understanding how to find the inverse of a matrix. So why don't you work on this, pause the video. When you feel comfortable, confident, that you...

CHRISTINE BREINER: Welcome back to recitation. Today we're going to talk about something you've been seeing in the lectures. Specifically, we're going to talk about continuity and differentiability. And we're going to use an example to see a little bit what the difference is. That how you can be continuous and not necessarily differentiable. And how it's a little stronger to have differentiability. So we're going to deal with a piecewise function. I'm going to ask a question, I'll give you a little bit of time to work on it, and then we'll come back. So the question is the following: for what values of a and b is the following function either first, continuous, or second, differentiable? So the function is defined in the following way: f of x is going to be equal to the function x squared plus 1 when x is bigger than 1. And it's going to be equal to a linear function where you get to pick the a, which is the slope, and you get to pi...

CHRISTINE BREINER: Welcome back to recitation. In this video, what I'd like us to do is work on understanding simply connected regions in three dimensions. Well, there's one two-dimensional one, but the rest are three dimensions. So what I want you to do is for each of the following-- there are six different regions-- determine whether or not each of them is simply connected. So the first one is R^3. The second one is if I take R^3 and I remove the entire z-axis. The third one is if I take R3 and I remove 0. The fourth one is if I take R^3 and remove a circle. The fifth one is R^2 minus a line segment. And the sixth one is a solid torus. So a solid torus looks like a doughnut, and it includes the inside of the doughnut. This looks like a doughnut, hopefully, to you. And it's not hollow. It includes the inside. So what I'd like you to do, again, is determine whether or not each of these regions is simply connected. And why don't you pause the video whil...

CHRISTINE BREINER: Welcome back to recitation. In this segment, we're going to talk about the product rule for three functions and then we're going to do an example. And what I want to do first is remind you the product rule for two functions, because we're going to use that to figure out the product rule for three functions. So throughout this segment, we are going to assume that u and v and w are all functions of x. So I'm going to drop the of x just so it's a little easier to write. This notation should be familiar with things you saw in the lecture. So, for two functions, let me remind you. If u*v, the product, and you take its derivative, so prime will denote d/dx. Then we can take the derivative of the first times the second function left alone, plus the derivative of the second function times the first left alone. So this should again be familiar from class. And now what we want to do is expand that to the product of three functions, u times v t...

CHRISTINE BREINER: Welcome back to recitation. In this video, I'd like us to do the following two problems, both related to the same position vector. So we're starting off with a position vector defined as r of t is equal to 1 minus 2 t squared i plus t squared j plus negative 2 plus 2 t squared k. So that's our position vector, and I'd like us to do the following two things. And you'll notice this problem is pretty much just a computational problem. We're going to make sure that we know what these things I'm about to talk about are, how you define them, and how you get from the position vector to each of these things. So we want to compute the velocity, the speed, the acceleration, and find the unit tangent vector for r of t. And then, the second part, we want to compute the arc length of the trajectory from t equals 0 to t equals 2. So I'll give you a moment to do that problem. Why don't you pause the video, work on the problem. When ...

CHRISTINE BREINER: Welcome back to recitation. In this video, I really just want to practice matrix multiplication, which is potentially something new for some of you, and maybe some of you have been doing it for a while and are very good at it. But I want to make sure that everyone is feeling confident in their ability to multiply matrices. So we have three matrices here. We have A, B, and C. And what I want you to do is I want you to compute what makes sense below. I have four products of matrices below. a is A times B, b is B times A, c is B times C, and d is A times C. So I want you to multiply the matrices that make sense to multiply, and then the ones that don't, make sure you understand why. Give yourself a brief explanation of why you can't multiply them. So why don't you work on that, pause the video, and when you feel confident in your answers, bring the video back up, and you can check them against my work. OK, welcome back. Well, we wanted to make ...

CHRISTINE BREINER: Welcome back to recitation. We're going to practice using some of the tools you developed recently on taking derivatives of exponential functions and taking derivatives of logarithmic functions. So I have three particular examples that I want us to look at. And I'd like us to find derivatives of the following functions. The first one is f of x is equal to x to the pi plus pi to the x. The second function is g of x is equal to natural log of cosine of x. And the third one is-- that's an h not a natural log-- h of x is equal to natural log of e to the x squared. So you have three functions you want to take the derivative of with respect to x. I'm going to give you a moment to to work on those and figure those out using the the tools you now have. And then we'll come back and I will work them out for you as well. OK, so let's start off with the derivative of the first one. OK, now, the reason in particular that I did this one-- it m...

CHRISTINE BREINER: Welcome back to recitation. In this video, what I'd like you to do is use least squares to fit a line to the following data, which includes three points: the point (0, 1), the point (2, 1), and the point (3, 4). And I've drawn a rough picture where these points are on a graph, and I'll be talking a little bit about that after you try this problem. So why don't you try to solve the problem based on the technique you learned in class, and then bring the video back up. I'll show you again what you're actually doing, and then I will also solve the problem and you can see, you can compare your answer to mine. OK, welcome back. So the first thing I want to do when we're talking about least squares to fit a line to these data, what I'd like to do is I'm going to draw a line on here, and we're going to talk about what the least squares actually is trying to do. And so I'm going to draw this line, say, something like t...

CHRISTINE BREINER: Welcome back to recitation. I want us to work a little more on finding anti-derivatives. In particular, in this video we want to find an anti-derivative of a trigonometric function, a power of the tangent function. So I would like you to find an anti-derivative of tangent theta quantity to the fourth. And the hint I will give you is that you're going to need some fairly familiar, hopefully, by now, trigonometric identities to get this to work. And then you will need some other strategies that you've also been developing. So I'll give you a while to work on it and then I'll be back and I'll show you how I did it. OK. Welcome back. We want to, again, we want to find an anti-derivative for tangent theta quantity to the fourth. And I mentioned that what we're going to need is a particular trigonometric-- well, I didn't say particular, sorry-- but we will need some trigonometric identities to make this work. And the one in particu...

CHRISTINE BREINER: Welcome back to recitation. In this video I want to show how we can use change of variables, to polar coordinates in particular, to evaluate an integral that, without the change in variables, we don't have the techniques to do. So I'm going to show us how to evaluate the integral from minus infinity to infinity of e to the minus x squared dx. And I'll just point out that if you do anything in probability you will see this integral a great deal. So this is a distribution and you'll see it a great deal if you ever do anything in probability theory. But how are we going to use polar coordinates to evaluate this? Well, the object is going to be to introduce a little bit more into this integral, so that when I actually introduce that in I'm going to have an r squared term in the exponent and I'm going to have-- of course by the change of variables in the Jacobian-- multiplied by r. And that's what's going to save us. So let me...

CHRISTINE BREINER: Welcome back to recitation. In this video I'd like us to do two things. The first thing we're going to do is we're going to graph the curve r equals 1 plus cosine theta over 2, for theta between 0 and 4 pi, and we're going to graph it in the xy-plane. And then after we've done that, we're going to take a look at some components of that curve and we're going to calculate the area of some components that close up. So what I'd like you to do first is get a good picture of this curve, in the xy-plane. I'll give you a little while to do that. So why don't you pause the video, get a good picture of that curve, then come back when you're ready and I'll show you how I graph it, and then we'll get into these area problems. OK. Welcome back. So the goal, again, was to graph a certain curve described by r and theta, but in the xy-plane. For theta between 0 and pi over 4. And when I do these problems, we want to m...