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DAVID JORDAN: Hello, and welcome back to recitation. Today the problem I'd like to work with you is about computing partial derivatives and the total differential. So we have a function z which is x squared plus y squared. So it depends on the two variables x and y. Now the variables x and y themselves depend on two auxiliary variables, u and v. So that's the setup that we have. So in part a, we just want to compute the total differential dz in terms of dx and dy. So u and v aren't going to enter into the picture. And then in part b, we're going to compute the partial derivative partial z partial u in two different ways. First, we're going to compute it using the chain rule. And then we're going to compute it using total differentials. And so we'll substitute in some of the work that we had in a to solve that part. So why don't you pause the video now and work on the problem. We'll check back and we'll do it together. Hi, and welcom...

DAVID JORDAN: Hello, and welcome back to recitation. So today, the problem I'd like to work with you is about taking partial derivatives in the presence of constraints. So this is a pretty subtle business. So take your time when you work these problems. So what we have is we have this function w, and it's a function of four variables: x, y, z, and t. OK? But it's not really a function of these four variables because we have a constraint. So we want to study how w changes as we vary the parameters, except that we have imposed this constraint here. So that really we kind of only have three variables, because we have four variables and one constraint. So that's what partial derivatives with constraints help us do. So let's explain first the notation. OK? So it says partial w partial z, and then we have the subscripts x and y. So what's important about this notation is not what you see as much as what you don't see. What you don't see is the va...

DAVID JORDAN: Hello, and welcome back to recitation. The problem I'd like to work with you today is the intersection of two parametrized lines. So we have two lines here. L_1, given with the parametrization in terms of the variable t. And L_2, also given with the parametrization in terms of the variable t. So the first question that I want us to answer is do these lines intersect? And if so, then we want to find out where do they intersect. So why don't you pause the video and work on this. And we can check back in a moment and we'll see how I solved it. OK, welcome back. Let's get started. So we have these two lines in space. Before we start doing any computations, I find it useful to draw a picture. So let's see what's going on. OK. So we have these two lines. We can just find some common points on the lines. So, well, if we put in t equals 0 here, then it looks like we get the point 2 comma 1. OK. And now if we plug in, let's say, t is minus...

DAVID JORDAN: Hello, and welcome back to recitation. The problem I'd like to work with you right now is we have a line which goes through two points that are given to us explicitly, and we have a plane which is given to us by an equation. And what we want to know is where does this line intersect this plane? And so one thing I would suggest to get started is we need to give a parametrization of our line to get started. OK, so why don't you work on that, pause the tape, and we'll come back in a moment and work it out together. OK, welcome back. Let's get started. So let's start off by drawing a cartoon of what's going on here. So we have this plane sitting in space. And we have some line kind of going through space. So maybe it's like this. And there's this single point of intersection. So even from the cartoon, we can kind of, sort of see two things which are going on. Which is that we would expect a point of intersection, or we would expec...

DAVID JORDAN: Hello, and welcome back to recitation. In this question, we're going to be considering a contour plot, which is given to us as followed. The values are not indicated. So the first thing that we want to do is we want to identify-- on this contour plot, there is a unique saddle point, and we want to label that as point A, and there are two points which are either a maximum or a minimum. We can't actually tell because the labels aren't on this contour plot, but we want to go ahead and label those anyways: B and C. So they're either maxima or minima, but we can still find them and we can still identify them. So that's the first part of the problem. The second part is, since this doesn't have the values entered onto the graph, we want to consider what possible configurations could we have? So the second question is: can B and C both be maximal? And can we have B maximal but C minimal? OK. And then in each of these two cases, we want to ske...

DAVID JORDAN: Hello, and welcome back to recitation. Today what we want to work on is drawing level curves. This is for all the artists out there in the audience. We have three functions here: z is 2x plus y, z is x squared plus y squared, and z is x squared minus y squared, and we want to get some practice drawing their level curves. Now, just to remind you, the level curves are not drawn in three dimensions. They're drawn in the xy-plane and they're constructed by setting z to be a constant and then graphing the curve that we get, so we can think about a relief map that we might use if we were hiking. So why don't you get started on that. Pause the video, and we'll check back, and I'll show you how I solve this. Welcome back. So over here, we've got the equation for part a already set up. So z is 2x plus y. So now, what we need to do to get started is just draw the xy-axis. And, you know, there's really not a precise science for drawing these...

DAVID JORDAN: Hello, and welcome back to recitation. So today I want to practice doing, computing some integrals with you at polar coordinates. So we have these three integrals set up here. And the way the integrals are given to you, they're given to you in rectangular coordinates. So the first thing you need to do is to re-express these in polar coordinates. And so for the first one, part a, I want you to completely compute the integral. For part b and c, we have this function f, which we haven't specified yet. So the exercise is to set up the integral. We won't actually compute it, we'll just set it up completely. So rewrite it in terms of r and theta. So why don't you pause the video and get started on that. And check back with me and we'll work it out together. Welcome back. Let's get started. So for all of these, when we're transferring from rectangular to polar coordinates, the most difficult part is understanding what region we'r...

DAVID JORDAN: Hello, and welcome back to recitation. So the problem that I want to work with you now is to compute some integrals, but we want to compute them in the presence of a density function. So the region that we're considering is very simple. It's just the unit square. So we have the origin here, we have the line x equals 1, we have the line y equals 1, and we just want to compute in this region. And so we want to use this density function to find various sort of physical characteristics of this region. So first, we want to find its mass, and so we are going to need to recall how you get mass from density. We want to find the center of mass. That is, where is the point on which we could balance this if we cut it out? If we tried to balance it on our fingers, where is the average mass concentrated? We want to find the moment of inertia about the origin, and we want to find the moment of inertia about the x-axis. So we're going to have to remember our fo...

DAVID JORDAN: Hello, and welcome back to recitation. In this problem, what I'd like us to do is I'd like us to sketch the graphs, in three dimensions, of these functions. So z here is a function of x and y. On this second one, z is also a function of x and y. It just happens not to depend on y. When you graph these, I'd suggest to consider slices, so what happens if you consider x equals 0 or if you consider z equals 0. As you graph these, let's see what you can do. So why don't you pause the video, and I'll check back with you in a moment, and I can show you how I solved these. OK, welcome back. So why don't we start by looking at this function: z is the square root of x squared plus y squared. OK, I'll try to always draw my axes in the same way as we do in lecture. So x is pointing towards us, y to the right, and z up. So as I suggested, I think a nice way to get started with these problems is to just try setting the variables x and y var...

DAVID JORDAN: Hello, and welcome back to recitation. So in this problem, we're considering a function f of three variables, f of x, y, z, and it's differentiable. And we're not told a formula for f. We just know that it's differentiable at this point P, which is 1, minus 1, 1, and we're told that the gradient of f at that point is this particular vector 2i plus j minus 3k, at that point P. So all we understand about f is how it looks around the point P. Now, we also have this relation between the variables, so x, y and z aren't unrelated. They're related by this constraint that z is x squared plus y plus 1. So with this information, we want to compute the gradient of a new function g, and the new function g is a function of two variables, and this function g is obtained from f by just plugging in our relation for y. So we can use our constraint to solve for y, and then this function g is just f with that constraint applied. And what we really w...

DAVID JORDAN: Hello, welcome back to recitation. The problem I'd like to work on with you now is a long one. So it's going to be practice computing line integrals. So to begin with, we have this function of two variables. f is x to the fifth plus 3x y cubed. And we have this-- C is the upper semi-circle going from (1, 0) to (-1, 0). So it's this upper semi-circle here that we often consider. And so the first thing that we want to do is to just compute the gradient, capital F, to be the gradient of this function f. And then parts b through d, we're going to compute this line integral of this vector field f along this curve C. We're going to compute it in three different ways. So first of all, we're going to compute it directly, just using the definition. And then in Part c, we're going to compute it using the path independence of line integrals and we're going to replace the path C with a simpler path. And then finally in Part d, we're g...