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And then r2 might look something like this. This is when t has gotten a little bit larger. We're further down the path. And so one way that you can approximate the slope of the tangent line or the slope between these two points for now is essentially the difference between these two vectors. The difference between thes... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
And so one way that you can approximate the slope of the tangent line or the slope between these two points for now is essentially the difference between these two vectors. The difference between these two vectors is you could view that as delta r. This vector plus that vector is equal to that vector. Or r2 minus r1 is... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
And as that increment between r1 and r2 gets smaller and smaller and smaller, as we have a smaller and smaller t increment, the slope of that delta r is going to more and more approximate the slope of the tangent line, all the way to the point that if you have an infinitely small change in t, so you have a dt, so you g... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
And once again, all of this is a little bit of review. But dr we can write as dr is equal to dx times i plus the infinitesimally small change in x times the i unit vector plus the infinitesimally small change in y times the j unit vector. And you see that if I were to draw a curve, let me just draw another one. Actuall... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
Actually, I don't even have to draw the axes. If our dr looks like that, if that is our dr, then we can break that down into its vertical and horizontal components. This right over here is dy, and that right over there is dx. And so we see that dx times i, actually this is dx times i, and this is dy times j. dy is the ... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
And so we see that dx times i, actually this is dx times i, and this is dy times j. dy is the magnitude, j gives us the direction. dx is the magnitude, i tells us that we're moving in the horizontal direction. Over here, this actually would be a negative. This must be a negative value right over here, and this must be ... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
This must be a negative value right over here, and this must be a positive value based on the way that I drew it. So that gives us a tangent vector. And now we want to, from that tangent vector, figure out a normal vector, a vector that is essentially perpendicular to this vector right over here. And there's actually g... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
And there's actually going to be two vectors like that. There's going to be the vector that kind of is perpendicular in the right direction, because we care about direction, or the vector that's perpendicular in the left direction. We can pick either one, but for this video, I'm going to focus on the one that goes in t... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
We're going to see that that's going to be useful in the next video when we start doing a little bit of vector calculus. And so let's think about what that might be. And what I'll do to make it a little bit clearer, let me draw a dr again. I'll draw a dr like this. This is our dr. And then this right over here, this ri... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
I'll draw a dr like this. This is our dr. And then this right over here, this right over there, we already said this is dy times i. And then this, sorry, that's dy times j. We're going in the vertical direction. dy times j. And then in a different color, this right, I already used that color. Let's see, I haven't used,... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
We're going in the vertical direction. dy times j. And then in a different color, this right, I already used that color. Let's see, I haven't used, well, I haven't used that blue yet. So this right over here is dx times i. So we know from our algebra courses, you take the negative reciprocal. So there's going to be som... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
Let's see, I haven't used, well, I haven't used that blue yet. So this right over here is dx times i. So we know from our algebra courses, you take the negative reciprocal. So there's going to be something about swapping these two things around and then taking the negative one. But we have to figure out, well, we want ... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
So there's going to be something about swapping these two things around and then taking the negative one. But we have to figure out, well, we want the one that goes to the right. So which one should we use? So let's think about it a little bit. If we take dy times i, so we take this length, but in the i direction, we'r... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
So let's think about it a little bit. If we take dy times i, so we take this length, but in the i direction, we're going to get this. We're going to get that. So this is dy times i. And then if we were to just take dx times j, that would take us down. Because dx, it must be negative here, since it's pointed to the left... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
So this is dy times i. And then if we were to just take dx times j, that would take us down. Because dx, it must be negative here, since it's pointed to the left. So we have to swap the sine of dx to go upwards. So we swap the sine of dx to go upwards. Because obviously here, it was a negative sine. It went leftwards. | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
So we have to swap the sine of dx to go upwards. So we swap the sine of dx to go upwards. Because obviously here, it was a negative sine. It went leftwards. But we wanted to go upwards. So this is going to be negative dx times j. We're now moving in the vertical direction. | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
It went leftwards. But we wanted to go upwards. So this is going to be negative dx times j. We're now moving in the vertical direction. And that, at least visually, this isn't kind of a rigorous proof that I'm giving you. But this is hopefully a good visual representation that that does get you pretty close, just visua... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
We're now moving in the vertical direction. And that, at least visually, this isn't kind of a rigorous proof that I'm giving you. But this is hopefully a good visual representation that that does get you pretty close, just visually inspecting it, to what looks like the perpendicular line. It's consistent with what you ... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
It's consistent with what you learned in algebra class as well, that we're taking the negative reciprocal. We're swapping the x's and the y's, or the change in x and the change in y. And we're taking the negative of one of them. And so we have our normal line just like that, our normal vector. So a normal vector is goi... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
And so we have our normal line just like that, our normal vector. So a normal vector is going to be dyi minus dxj. But then if we want to normalize it, we want to divide by that magnitude. So a normal vector is going to be dy times i minus dx times j. Now, if we want this to be a unit normal vector, we have to divide i... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
So a normal vector is going to be dy times i minus dx times j. Now, if we want this to be a unit normal vector, we have to divide it by the magnitude of a. But what is the magnitude of a? The magnitude of a is going to be equal to the square root of, I'll start with the dx squared. So it's the negative dx squared, whic... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
The magnitude of a is going to be equal to the square root of, I'll start with the dx squared. So it's the negative dx squared, which is just going to be dx squared. Same thing as positive dx squared. It's going to be dx squared plus dy squared. I could have put the negative right in here. But then when you square it, ... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
It's going to be dx squared plus dy squared. I could have put the negative right in here. But then when you square it, that negative would disappear. But this thing right over here, and we saw this when we first started exploring arc length, this thing right over here is the exact same thing as ds. And I know there's n... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
But this thing right over here, and we saw this when we first started exploring arc length, this thing right over here is the exact same thing as ds. And I know there's no ds that we've shown right over here. But we've seen it multiple times. That when you're thinking about, if you think about the length of dr as ds, t... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
That when you're thinking about, if you think about the length of dr as ds, that's exactly what this thing right over here is. So this can also be written as ds. The infinitesimally change in the arc length, but it's a scalar quantity. You're not concerned. You're just concerned with the absolute distance. You're not c... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
You're not concerned. You're just concerned with the absolute distance. You're not concerned so much with the direction. Another way to view it is it's the magnitude right over here of dr. So now we have everything we need to construct our unit normal vector, a unit normal vector at any point. And I'll now write n. And... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
Another way to view it is it's the magnitude right over here of dr. So now we have everything we need to construct our unit normal vector, a unit normal vector at any point. And I'll now write n. And I'll put a hat on top of it to say that this is a unit normal vector. We'll have magnitude 1. It is going to be equal to... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
We'll have magnitude 1. It is going to be equal to a divided by this, or we could even write it this way. So we could write it as, there are multiple ways we can write it. We can write it as, I'll write it in this color, as dy times i minus dx times j. And then all of that times, or let me, not times, divided by ds, di... | Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3 |
It's got a two-dimensional input, two different coordinates to its input, and then a three-dimensional output. Specifically, it's a three-dimensional vector, and each one of these is some expression, it's a bunch of cosines and sines, that depends on the two input coordinates. And in the last video, we talked about how... | Parametric surfaces Multivariable calculus Khan Academy.mp3 |
So what we're gonna do, we're just gonna visualize things in the output space, and we're gonna try to think of all the possible points that could be outputs. So, for example, let's just start off simple. Let's get a feel for this function by evaluating it at a simple pair of points. So let's say we evaluate this functi... | Parametric surfaces Multivariable calculus Khan Academy.mp3 |
So let's say we evaluate this function, f, at t equals zero. I think it would probably be pretty simple. And then s is equal to pi. So let's think about what this would be. We go up and we say, okay, t of zero, cosine of zero is one, so this whole thing is gonna be one, same with this one. And sine of zero is zero, so ... | Parametric surfaces Multivariable calculus Khan Academy.mp3 |
So let's think about what this would be. We go up and we say, okay, t of zero, cosine of zero is one, so this whole thing is gonna be one, same with this one. And sine of zero is zero, so this over here is gonna be zero, and this is also gonna be zero. Now cosine of pi is negative one, so this here is gonna be negative... | Parametric surfaces Multivariable calculus Khan Academy.mp3 |
Now cosine of pi is negative one, so this here is gonna be negative one, this one here is also gonna be negative one, and then sine of pi, just like sine of zero, is zero. So this whole thing actually ends up simplifying quite a bit so that the top is three times one plus negative one, one times negative one is negativ... | Parametric surfaces Multivariable calculus Khan Academy.mp3 |
So we'll go ahead and, whoop, remove the graph about, add that point there. So that's what would correspond to this one particular input, zero and pi. And you know, you could do this with a whole bunch, and you might add a couple other points based on other inputs that you find, but this will take forever to start to g... | Parametric surfaces Multivariable calculus Khan Academy.mp3 |
And another thing you can do is say, okay, maybe rather than thinking of evaluating at a particular point, imagine one of the inputs was constant. So let's imagine that s stayed constant at pi, okay? But then we let t range freely, so that means we're gonna have some kind of different output here, and we're gonna let t... | Parametric surfaces Multivariable calculus Khan Academy.mp3 |
So what that means is we keep all of these, this negative one, negative one, and zero for what sine of pi is, but the output now, it's gonna be three cosine of t, cosine of t, plus negative one times the cosine of t, so it's gonna be minus cosine of t. The next part, it's gonna still be three sine of t. This is no long... | Parametric surfaces Multivariable calculus Khan Academy.mp3 |
Okay, so three times sine of t, that's just still the function that we're dealing with. Three sine of t, and then minus one times sine of t, so minus sine of t. Keep drawing it in green just to be consistent. And then the bottom stays at zero. And this whole thing actually simplifies. Three cosine t minus cosine t, tha... | Parametric surfaces Multivariable calculus Khan Academy.mp3 |
And this whole thing actually simplifies. Three cosine t minus cosine t, that's just two cosine t. And then same deal for the other one. It's gonna be two sine of t. So this whole thing actually simplifies down to this. So this is, again, when we're letting s stay constant and t ranges freely. And when you do that, wha... | Parametric surfaces Multivariable calculus Khan Academy.mp3 |
So this is, again, when we're letting s stay constant and t ranges freely. And when you do that, what you're gonna end up getting is a circle that you draw. And you can maybe see why it's a circle, because you have this cosine sine pattern. It's a circle with radius two, and it should make sense that it runs through th... | Parametric surfaces Multivariable calculus Khan Academy.mp3 |
It's a circle with radius two, and it should make sense that it runs through that first point that we evaluated. So that's what happens if you let just one of the variables run. But now let's do the same thing, but think instead of what happens is s varies and t stays constant. I encourage you to work it out for yourse... | Parametric surfaces Multivariable calculus Khan Academy.mp3 |
I encourage you to work it out for yourself. I'll go ahead and just kind of draw it, because I kind of want to give the intuition here. So in that case, you're gonna get a circle that looks like this. So again, I encourage you to try to think through for the same reasons. Imagine that you let s run freely, keep t const... | Parametric surfaces Multivariable calculus Khan Academy.mp3 |
So again, I encourage you to try to think through for the same reasons. Imagine that you let s run freely, keep t constant at zero. Why is it that you would get a circle that looks like this? And in fact, if you let both t and s run freely, a very nice way to visualize that is to imagine that this circle, which represe... | Parametric surfaces Multivariable calculus Khan Academy.mp3 |
And in fact, if you let both t and s run freely, a very nice way to visualize that is to imagine that this circle, which represents s running freely, sweeps throughout space as you start to let t run freely. And what you're gonna end up getting when you do that is a shape that goes like this. And this is a donut. We ha... | Parametric surfaces Multivariable calculus Khan Academy.mp3 |
Let's continue with our proof of Stokes' Theorem. And this time we're gonna focus on the other side of Stokes' Theorem. We're gonna try to figure out what is the line integral over the boundary C, where this is C right over here, the boundary of our surface, of F dot dr. What we're gonna see is that we're gonna get the... | Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3 |
But before we do that, I'm gonna take a little bit of a detour to kind of build up to this. So let's just take this and put it to the side for now. Actually, let me actually just delete it right now. And what I'm actually gonna do is I'm gonna focus on this region down here, this region in the xy-plane. And in particul... | Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3 |
And what I'm actually gonna do is I'm gonna focus on this region down here, this region in the xy-plane. And in particular, so this is path C, which is the boundary of our surface. I'm gonna focus on the path that is the boundary of this region, this path that sits in the xy-plane. And I will call that path C1. And so ... | Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3 |
And I will call that path C1. And so one, we can think about a parameterization of just that path in the xy-plane. We could say that C1 could be parameterized as x is equal to x of t, and y is also a function of t. And t is obviously our parameter, and it can go between a and b. So maybe when t is equal to a, it sits r... | Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3 |
So maybe when t is equal to a, it sits right here, and then as t gets larger and larger and larger, it goes all the way around, and eventually, when t is equal to b, it gets to that exact same point. So that's our parameterization right there. And now, just to make the rest of this proof a little bit more understandabl... | Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3 |
Imagine that we have some vector field g, and g, at minimum, is defined on the xy-plane. So, and it could be defined other places, but let's say that g is equal to m of xy i plus n of xy, n of xy j. Now, what would the line integral, this is all a review, we've seen this a long time ago, what would be the line integral... | Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3 |
What would be the line integral over the path c1, and I'll even, I like to write that sometimes, of, and I'm using g, so I don't get confused with f, our original vector field, of g, of g, our vector field here, along that path, g dot dr. G dot dr. Well, dr, dr is just going to be equal to dx, dxi, dxi plus dy, dyj. So... | Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3 |
The line integral over our path c1, but when you take this dot product, you have the, you multiply the x components and then add that to the product of the y components. So you have m times dx, you have m times dx, m times dx, plus, plus n times dy, n times dy. I just took the dot product of g and dr, n times dy. And w... | Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3 |
And when you evaluate these things, the one way to think about it is that dx is the same thing, dx is the same thing as, let me write it up here in a different color, dx is the same thing as dx, the derivative of x with respect to t, dt, and same logic for y. dy is equal to the derivative of y with respect to t, dt. On... | Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3 |
So then this will be equal to, this will be equal to, this will be equal to the integral in the domain of our parameter. So now we are in the t domain, and t is going to vary between a and b. We are in the t domain between a and b. This is going to be equal to m, m times, instead of writing dx, I'm gonna write dx, dt, ... | Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3 |
This is going to be equal to m, m times, instead of writing dx, I'm gonna write dx, dt, dt. So it's going to be dx, let me write it this way, dx, the derivative of x with respect to t, dt, that's the first expression, plus, plus n, and then the exact same thing, times dy, dt, n times dy, dt, n times dy, dt, dt, dt. The... | Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3 |
Now, with all of that out of the way, and this is really, all of this is really just a reminder so that the rest of this proof becomes a little bit intuitive. With that out of the way, let's come up with a parameterization for this path up here, for c. For c. Remember, we just did c1 down in the xy plane, now we're gon... | Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3 |
The x value, the x and y value there is the exact same thing as the x and y value there. The only difference is we now have a z component and we know, we defined it way up here. Our z component is going to be a function, is going to be a function of x and y. It tells us how high to go. So we can parameterize, we can pa... | Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3 |
It tells us how high to go. So we can parameterize, we can parameterize c as, we can parameterize c as, maybe I'll write it as a vector. So let me write it, let me parameterize, so I'll write c as a vector. C, actually no, I'll write it, let me write it this way. Let me write c, c can be written as, let me do that purp... | Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3 |
C, actually no, I'll write it, let me write it this way. Let me write c, c can be written as, let me do that purple color. C we can say is, x is x of t, and actually let me write this as a vector. So I will write it, and I'll use a vector r, not to be confused with this r right over here. So these are two different r's... | Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3 |
Hey guys, there's one more thing I need to talk about before I can describe the vectorized form for the quadratic approximation of multivariable functions, which is a mouthful to say. So let's say you have some kind of expression that looks like a times x squared, and I'm thinking of x as a variable, times b times xy, ... | Expressing a quadratic form with a matrix.mp3 |
Now this kind of expression has a fancy name. It's called a quadratic form, quadratic form. And that always threw me off. I always kind of was like, what does form mean? You know, I know what a quadratic expression is, and quadratic typically means something is squared or you have two variables, but why do they call it... | Expressing a quadratic form with a matrix.mp3 |
I always kind of was like, what does form mean? You know, I know what a quadratic expression is, and quadratic typically means something is squared or you have two variables, but why do they call it a form? And basically it just means that the only things in here are quadratic. You know, it's not the case that you have... | Expressing a quadratic form with a matrix.mp3 |
You know, it's not the case that you have like an x term sitting on its own, or like a constant out here, like two, and you're adding all of those together. Instead it's just you have purely quadratic terms. But of course mathematicians don't want to call it just a purely quadratic expression. Instead they have to give... | Expressing a quadratic form with a matrix.mp3 |
Instead they have to give a fancy name to things so that it seems more intimidating than it needs to be. But anyway, so we have a quadratic form, and the question is, how can we express this in a vectorized sense? And for analogy, let's think about linear terms, where let's say you have a times x, plus b times y, and I... | Expressing a quadratic form with a matrix.mp3 |
If you see something like this, where every variable is just being multiplied by a constant, and then you add terms like that to each other, we can express this nicely with vectors, where you pile all of the constants into their own vector, a vector containing a, b, and c, and you imagine the dot product between that a... | Expressing a quadratic form with a matrix.mp3 |
And this way, your notation just kinda looks like a constant times a variable, just like in the single variable world, when you have a constant number times a variable number, it's kinda like taking a constant vector times a variable vector. And the importance of writing things down like this is that v could be a vecto... | Expressing a quadratic form with a matrix.mp3 |
So the question is, can we do something similar like that with our quadratic form? Because you can imagine, let's say we started introducing the variable z, then you would have to have some other term, you know, some other constant times the xz quadratic term, and then some other constant times the z squared quadratic ... | Expressing a quadratic form with a matrix.mp3 |
So we want a nice way to express this. And I'm just gonna kind of show you how we do it, and then we'll work it through to see why it makes sense. So usually, instead of thinking of b times xy, we actually think of this as two times some constant times xy, and this of course doesn't make a difference. You would just ch... | Expressing a quadratic form with a matrix.mp3 |
You would just change what b represents, but you'll see why it's more convenient to write it this way in just a moment. So the vectorized way to describe a quadratic form like this is to take a matrix, a two by two matrix, since this is two dimensions, where a and c are on the diagonal, and then b is on the other diago... | Expressing a quadratic form with a matrix.mp3 |
So if you imagine kind of reflecting the whole matrix about this line, you'll get the same numbers. So it's important that we have that kind of symmetry. And now what you do is you multiply the vector, the variable vector that's got xy on the right side of this matrix, and then you multiply it again, but you kind of, y... | Expressing a quadratic form with a matrix.mp3 |
So instead of being a vertical vector, you transpose it to being a horizontal vector on the other side. And this is a little bit analogous to having two variables multiplied in. You have two vectors multiplied in, but on either side. And this is a good point, by the way, if you are uncomfortable with matrix multiplicat... | Expressing a quadratic form with a matrix.mp3 |
And this is a good point, by the way, if you are uncomfortable with matrix multiplication, to maybe pause the video, go find the videos about matrix multiplication and kind of refresh or learn about that. Because moving forward, I'm just gonna assume that it's something you're familiar with. So going about computing th... | Expressing a quadratic form with a matrix.mp3 |
We have a matrix multiplied by a vector. Well, the first component that we get, we're gonna multiply the top row by each corresponding term in the vector. So it'll be a times x, a times x, plus b times y, plus b times that second term y. And then similarly for the bottom term, we'll take the bottom row and multiply the... | Expressing a quadratic form with a matrix.mp3 |
And then similarly for the bottom term, we'll take the bottom row and multiply the corresponding term. So b times x, b times x, plus c times y, c times y. So that's what it looks like when we do that right multiplication. And of course, we've gotta keep our transposed vector over there on the left side. So now we have,... | Expressing a quadratic form with a matrix.mp3 |
And of course, we've gotta keep our transposed vector over there on the left side. So now we have, this is just a two by one vector now. This is a one by two. You could think of it as a horizontal vector or a one by two matrix. But now when we multiply these guys, you just kind of line up the corresponding terms. You'l... | Expressing a quadratic form with a matrix.mp3 |
You could think of it as a horizontal vector or a one by two matrix. But now when we multiply these guys, you just kind of line up the corresponding terms. You'll have x multiplied by that entire top expression. So x multiplied by ax plus by. Ax plus by. And then we add that to the second term y multiplied by the secon... | Expressing a quadratic form with a matrix.mp3 |
So x multiplied by ax plus by. Ax plus by. And then we add that to the second term y multiplied by the second term of this guy, which is bx plus cy. So y multiplied by bx plus cy. And all of these are numbers, so we can simplify it. Once we start distributing, the first term is x times a times x. So that's ax squared. | Expressing a quadratic form with a matrix.mp3 |
So y multiplied by bx plus cy. And all of these are numbers, so we can simplify it. Once we start distributing, the first term is x times a times x. So that's ax squared. And then the next term is x times b times y. So that's b times xy. Over here we have y times b times x. | Expressing a quadratic form with a matrix.mp3 |
So that's ax squared. And then the next term is x times b times y. So that's b times xy. Over here we have y times b times x. So that's the same thing as b times xy. So that's kind of why we have, why it's convenient to write a two there because that naturally comes out of our expansion. And then the last term is y tim... | Expressing a quadratic form with a matrix.mp3 |
Over here we have y times b times x. So that's the same thing as b times xy. So that's kind of why we have, why it's convenient to write a two there because that naturally comes out of our expansion. And then the last term is y times c times y. So that's cy squared. So we get back the original quadratic form that we we... | Expressing a quadratic form with a matrix.mp3 |
And then the last term is y times c times y. So that's cy squared. So we get back the original quadratic form that we were shooting for. You know, ax squared plus 2bxy plus cy squared. That's how this entire term expands. As you kind of work it through, you end up with the same quadratic expression. Now, the convenienc... | Expressing a quadratic form with a matrix.mp3 |
You know, ax squared plus 2bxy plus cy squared. That's how this entire term expands. As you kind of work it through, you end up with the same quadratic expression. Now, the convenience of this quadratic form being written with a matrix like this is that we can write this more abstractly. And instead of writing the whol... | Expressing a quadratic form with a matrix.mp3 |
Now, the convenience of this quadratic form being written with a matrix like this is that we can write this more abstractly. And instead of writing the whole matrix M, you could just let a letter like M represent that whole matrix. And then take the vector that represents the variable, maybe like a bold-faced x, and yo... | Expressing a quadratic form with a matrix.mp3 |
And then you transpose it and multiply it on the left. So typically you denote that by putting a little t as a superscript. So x transposed multiplied by the matrix from the left. And this expression, this is what a quadratic form looks like in vectorized form. And the convenience is the same as it was in the linear ca... | Expressing a quadratic form with a matrix.mp3 |
And this expression, this is what a quadratic form looks like in vectorized form. And the convenience is the same as it was in the linear case. Just like v could represent something that had 100 different numbers in it, and x would have 100 different constants, you could do something similar here, where you can write t... | Expressing a quadratic form with a matrix.mp3 |
Let's just see what this would look like in a three-dimensional circumstance. So, actually I'll need more room. So I'll go down even further. So we have x transpose multiplied by the matrix multiplied by x, bold-faced x. And let's say instead this represented, you know, you have x, then y, then z, our transposed vector... | Expressing a quadratic form with a matrix.mp3 |
So we have x transpose multiplied by the matrix multiplied by x, bold-faced x. And let's say instead this represented, you know, you have x, then y, then z, our transposed vector. And then our matrix, let's say it was A, B, C, D, E, F. And because it needs to be symmetric, whatever term is in this spot here needs to be... | Expressing a quadratic form with a matrix.mp3 |
Similarly, C, that's gonna be the same term here, and E would be over here. So there's only really six free terms that you have, but it fills up this entire matrix. And then on the right side, we would multiply that by x, y, z. Now, I won't work it out in this video, but you can imagine actually multiplying this matrix... | Expressing a quadratic form with a matrix.mp3 |
This is one of those things that's pretty good practice for some important concepts coming up in multivariable calc, and it's also just good to sit down and take a complicated thing and kind of break it down piece by piece. So a vector field like the one that I just showed is represented by a vector-valued function, an... | Partial derivatives of vector fields, component by component.mp3 |
In the last video, we figured out how to construct a unit normal vector to a surface. And so now we can use that back in our original surface integral to try to simplify a little bit, or at least give us a clue of how we can calculate these things. And also think about different ways to represent this type of a surface... | Vector representation of a surface integral Multivariable Calculus Khan Academy.mp3 |
So if we just substitute what we came up as our normal vector, our unit normal vector right here, we will get, we will get, so once again, it's the surface integral of F dot, and F dot all of this business right over here, and I'm going to write it all in white just so it doesn't take me too much time. So the partial o... | Vector representation of a surface integral Multivariable Calculus Khan Academy.mp3 |
We know that another way to write ds, and I gave the intuition, hopefully, for that several videos ago when we first explored what a surface integral was all about, we know that ds can be represented as the magnitude of the partial of r with respect to u crossed with the partial of r with respect to v, du dv. And obvio... | Vector representation of a surface integral Multivariable Calculus Khan Academy.mp3 |
And actually since now this integral is in terms of uv, we're no longer taking a surface integral, we're now taking a double integral over the uv domain. So you could say kind of a region in uv. So I'll say r to say that it's a region in the uv plane that we're now thinking about. But there's probably a huge, or there ... | Vector representation of a surface integral Multivariable Calculus Khan Academy.mp3 |
But there's probably a huge, or there should be, or I'm guessing there's a huge simplification that's popping out at you right now. We're dividing by the magnitude of the cross product of these two vectors, and then we're multiplying by the magnitude of the cross product of these two vectors. Those are just scalar quan... | Vector representation of a surface integral Multivariable Calculus Khan Academy.mp3 |
You divide by something and multiply by something. Well, that's just the same thing as multiplying or dividing by 1. So these two characters cancel out, and our integral simplifies to the double integral over that region, the corresponding region in the uv plane, of f, of our vector field f dotted with this cross produ... | Vector representation of a surface integral Multivariable Calculus Khan Academy.mp3 |
This is going to give us a vector right over here. That's going to give us a vector. It gives us actually a normal vector, and then we divide by its magnitude. It gives you a unit normal vector. So this, you're going to take the dot product of f with r, the partial of r with respect to u, crossed with the partial of r ... | Vector representation of a surface integral Multivariable Calculus Khan Academy.mp3 |
It gives you a unit normal vector. So this, you're going to take the dot product of f with r, the partial of r with respect to u, crossed with the partial of r with respect to v, du dv. Let me scroll over to the right a little bit. du dv. And we'll see in a few videos from now that this is essentially how we go about a... | Vector representation of a surface integral Multivariable Calculus Khan Academy.mp3 |
du dv. And we'll see in a few videos from now that this is essentially how we go about actually calculating these things. If you have a parametrization, you can then get everything in terms of a double integral, in terms of uv, this way. Now, the last thing I want to do is explore another way that you'll see a surface ... | Vector representation of a surface integral Multivariable Calculus Khan Academy.mp3 |
Now, the last thing I want to do is explore another way that you'll see a surface integral like this written. It all comes from really writing this part in a different way, but hopefully it gives you a little bit more intuition of what this thing is even saying. So I'm just going to rewrite this chunk right over here. ... | Vector representation of a surface integral Multivariable Calculus Khan Academy.mp3 |
I'm just going to rewrite that chunk. And I'm going to use slightly different notation because it will hopefully help make a little bit more sense. So the partial of r with respect to u I can write as the partial of r with respect to u. And we're taking the cross product. Let me make my u's a little bit more u-like so ... | Vector representation of a surface integral Multivariable Calculus Khan Academy.mp3 |
And we're taking the cross product. Let me make my u's a little bit more u-like so we don't confuse them with v's. And we're taking the cross product of that with the partial of r with respect to v. So very small changes in our vector, in our parametrization right here, our position vector, given a small change in v. V... | Vector representation of a surface integral Multivariable Calculus Khan Academy.mp3 |
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