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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 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 going to give you this delta r right over here. 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 go from r, then you change t a very small amount, that delta r, and we can kind of conceptualize that as dr, then does approximate the a tangent vector.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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 going to give you this delta r right over here. 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 go from r, then you change t a very small amount, that delta r, and we can kind of conceptualize that as dr, then does approximate the a tangent vector. So if you have a very small change in t, then your very small dr, I'll call it, because now we're talking about a differential, your very small differential right over here, that is a tangent vector. So dr is a tangent vector at any given point. And once again, all of this is a little bit of review.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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 go from r, then you change t a very small amount, that delta r, and we can kind of conceptualize that as dr, then does approximate the a tangent vector. So if you have a very small change in t, then your very small dr, I'll call it, because now we're talking about a differential, your very small differential right over here, that is a tangent vector. So dr is a tangent vector at any given point. 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. Actually, I don't even have to draw the axes.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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. 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 magnitude, j gives us the direction.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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 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 a positive value based on the way that I drew it.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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 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 going to be two vectors like that.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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 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 the right direction. 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.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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 the right direction. 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.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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 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.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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, well, I haven't used that blue yet.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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 something about swapping these two things around and then taking the negative one.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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 the one that goes to the right. So which one should we use? So let's think about it a little bit.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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're going to get this. We're going to get that. So this is dy times i.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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. So we have to swap the sine of dx to go upwards.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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 visually inspecting it, to what looks like the perpendicular line. It's consistent with what you learned in algebra class as well, that we're taking the negative reciprocal.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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 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.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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 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.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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 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.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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, which is just going to be dx squared. Same thing as positive dx squared. It's going to be dx squared plus dy squared.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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, 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.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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 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, that's exactly what this thing right over here is.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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, 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.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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 concerned so much with the direction. Another way to view it is it's the magnitude right over here of dr.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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 I'll put a hat on top of it to say that this is a unit normal vector. We'll have magnitude 1.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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 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.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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, divided by the magnitude of this. So divided by ds. And obviously I could distribute it on each of these terms.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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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 to visualize functions that have a single input, a single parameter, like t, and then a two-dimensional vector output, so some kind of expression of t and another expression of t. And this is sort of the three-dimensional analog of that. 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 function, f, at t equals zero.
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Parametric surfaces Multivariable calculus Khan Academy.mp3
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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 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.
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Parametric surfaces Multivariable calculus Khan Academy.mp3
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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 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 one, this one here is also gonna be negative one, and then sine of pi, just like sine of zero, is zero.
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Parametric surfaces Multivariable calculus Khan Academy.mp3
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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 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 negative one, and we get two, then we have three times zero plus zero, so the y component is just zero, and then the z component is also zero. So what that would mean is that this output is gonna be the point that's two along the x-axis, and there's nothing else to it, it's just two along the x-axis. So we'll go ahead and, whoop, remove the graph about, add that point there.
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Parametric surfaces Multivariable calculus Khan Academy.mp3
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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 negative one, and we get two, then we have three times zero plus zero, so the y component is just zero, and then the z component is also zero. So what that would mean is that this output is gonna be the point that's two along the x-axis, and there's nothing else to it, it's just two along the x-axis. 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 get a feel of the function as a whole. 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.
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Parametric surfaces Multivariable calculus Khan Academy.mp3
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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 get a feel of the function as a whole. 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 just be some kind of variable while the output is pi. 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 longer zero.
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Parametric surfaces Multivariable calculus Khan Academy.mp3
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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 just be some kind of variable while the output is pi. 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 longer zero. I should probably erase those guys, actually, so. We're no longer evaluating when t was zero. Okay, so three times sine of t, that's just still the function that we're dealing with.
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Parametric surfaces Multivariable calculus Khan Academy.mp3
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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 longer zero. I should probably erase those guys, actually, so. We're no longer evaluating when t was zero. 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.
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Parametric surfaces Multivariable calculus Khan Academy.mp3
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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, 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.
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Parametric surfaces Multivariable calculus Khan Academy.mp3
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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, 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 that first point that we evaluated.
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Parametric surfaces Multivariable calculus Khan Academy.mp3
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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 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 yourself.
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Parametric surfaces Multivariable calculus Khan Academy.mp3
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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 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.
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Parametric surfaces Multivariable calculus Khan Academy.mp3
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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 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 represents s running freely, sweeps throughout space as you start to let t run freely.
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Parametric surfaces Multivariable calculus Khan Academy.mp3
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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 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 have a fancy word for this in mathematics.
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Parametric surfaces Multivariable calculus Khan Academy.mp3
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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 have a fancy word for this in mathematics. We call it a torus. But it turns out the function here is a fancy way of drawing the torus. And in another video, I'm gonna go through in more detail, if you were just given the torus, how you could find this function, how you can kind of get the intuitive feel for that.
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Parametric surfaces Multivariable calculus Khan Academy.mp3
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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 exact same result as we have right up here. 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.
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Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3
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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 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.
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Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3
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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 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 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.
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Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3
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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 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 understandable, I'm going to give you a little bit of a review of something. Imagine that we have some vector field g, and g, at minimum, is defined on the xy-plane.
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Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3
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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 understandable, I'm going to give you a little bit of a review of something. 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 over the path c1, not c, but this path that sits in the xy-plane? 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.
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Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3
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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 over the path c1, not c, but this path that sits in the xy-plane? 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, if you take the dot product of these two things right over here, you're going to get, you're going to get the line integral over our path c1, remember, c1 is this path down here. Let me do it in that same color, so you don't think I'm changing colors on you. 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.
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Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3
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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, if you take the dot product of these two things right over here, you're going to get, you're going to get the line integral over our path c1, remember, c1 is this path down here. Let me do it in that same color, so you don't think I'm changing colors on you. 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 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.
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Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3
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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 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. One way to think about it, these dt's cancel out and you are just left with dx. And this is an important thing to think about because then this allows us to take this line integral into the domain of our parameter. 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.
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Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3
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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. One way to think about it, these dt's cancel out and you are just left with dx. And this is an important thing to think about because then this allows us to take this line integral into the domain of our parameter. 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, dt.
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Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3
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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, 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. These are all equivalent statements. 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.
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Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3
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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. These are all equivalent statements. 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 gonna do c that sits up here that kind of rises above the xy plane. Well, for c, x can still be, the parameterizations for x and y can still be the exact same thing because the x and y values are going to be the exact same thing. The x value, the x and y value there is the exact same thing as the x and y value there.
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Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3
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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 gonna do c that sits up here that kind of rises above the xy plane. Well, for c, x can still be, the parameterizations for x and y can still be the exact same thing because the x and y values are going to be the exact same thing. 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.
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Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3
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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 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.
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Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3
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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 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.
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Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3
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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, but I'll just use r's, because that tends to be the convention. So in order to parameterize c, it's going to be the position vector r, which is going to be a function of t, and x is still just going to be x of t. X of t, i, plus y of t, j, and now we're going to have a z component, and z is going to be a function of x and y, which are in turn functions of t. So z is a function of x, which is a function of t, and a function of y, which is a function of t, k, that tells us how high above to essentially get each of those points, and then once again we know that t is between a and b. T is greater than or equal to a, and less than or equal to b. So we have that parameterization right there, and now we can start to think about, we can start to think about the line integral of f dot dr along this path.
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Stokes' theorem proof part 4 Multivariable Calculus Khan Academy.mp3
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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, y as another variable, plus c times y squared. And I'm thinking of a, b, and c as being constants and x and y as being variables. 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?
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Expressing a quadratic form with a matrix.mp3
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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 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 like an x term sitting on its own, or like a constant out here, like two, and you're adding all of those together.
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Expressing a quadratic form with a matrix.mp3
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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 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 a fancy name to things so that it seems more intimidating than it needs to be.
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Expressing a quadratic form with a matrix.mp3
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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 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'll throw another variable in there, another constant times another variable z. 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 and a vector that contains all of the variable components, x, y, and z.
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Expressing a quadratic form with a matrix.mp3
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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'll throw another variable in there, another constant times another variable z. 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 and a vector that contains all of the variable components, x, y, and z. And the convenience here is then you can have just a symbol, like a v let's say, which represents this whole constant vector. And then you can write down, take the dot product between that, and then have another symbol, maybe a bold-faced x, which represents a vector that contains all of the variables. 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.
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Expressing a quadratic form with a matrix.mp3
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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 and a vector that contains all of the variable components, x, y, and z. And the convenience here is then you can have just a symbol, like a v let's say, which represents this whole constant vector. And then you can write down, take the dot product between that, and then have another symbol, maybe a bold-faced x, which represents a vector that contains all of the variables. 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 vector that contains not just three numbers, but like 100 numbers, and then x would have 100 corresponding variables, and the notation doesn't become any more complicated. It's generalizable to higher dimensions. So the question is, can we do something similar like that with our quadratic form?
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Expressing a quadratic form with a matrix.mp3
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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 vector that contains not just three numbers, but like 100 numbers, and then x would have 100 corresponding variables, and the notation doesn't become any more complicated. It's generalizable to higher dimensions. 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 term, and another one for the yz quadratic term. And it would get out of hand, and as soon as you start introducing things like 100 variables, it would get seriously out of hand, because there's a lot of different quadratic terms. So we want a nice way to express this.
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Expressing a quadratic form with a matrix.mp3
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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 term, and another one for the yz quadratic term. And it would get out of hand, and as soon as you start introducing things like 100 variables, it would get seriously out of hand, because there's a lot of different quadratic terms. 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 change what b represents, but you'll see why it's more convenient to write it this way in just a moment.
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Expressing a quadratic form with a matrix.mp3
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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 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 diagonal. And we always think of these as being symmetric matrices. So if you imagine kind of reflecting the whole matrix about this line, you'll get the same numbers.
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Expressing a quadratic form with a matrix.mp3
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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 diagonal. And we always think of these as being symmetric matrices. 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, you turn it on its side. So instead of being a vertical vector, you transpose it to being a horizontal vector on the other side.
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Expressing a quadratic form with a matrix.mp3
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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, you turn it on its side. 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 multiplication, to maybe pause the video, go find the videos about matrix multiplication and kind of refresh or learn about that.
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Expressing a quadratic form with a matrix.mp3
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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 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 this, first let's tackle this right multiplication here. We have a matrix multiplied by a vector.
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Expressing a quadratic form with a matrix.mp3
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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 this, first let's tackle this right multiplication here. 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 corresponding term.
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Expressing a quadratic form with a matrix.mp3
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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 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.
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Expressing a quadratic form with a matrix.mp3
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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, 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.
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Expressing a quadratic form with a matrix.mp3
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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'll have x multiplied by that entire top expression. So x multiplied by ax plus by.
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Expressing a quadratic form with a matrix.mp3
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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 second term of this guy, which is bx plus cy. So y multiplied by bx plus cy.
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Expressing a quadratic form with a matrix.mp3
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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.
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Expressing a quadratic form with a matrix.mp3
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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.
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Expressing a quadratic form with a matrix.mp3
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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 times c times y.
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Expressing a quadratic form with a matrix.mp3
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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 were shooting for. You know, ax squared plus 2bxy plus cy squared.
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Expressing a quadratic form with a matrix.mp3
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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 convenience of this quadratic form being written with a matrix like this is that we can write this more abstractly.
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Expressing a quadratic form with a matrix.mp3
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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 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 you would multiply it on the right. And then you transpose it and multiply it on the left.
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Expressing a quadratic form with a matrix.mp3
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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 you would multiply it on the right. 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.
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Expressing a quadratic form with a matrix.mp3
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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 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 that same expression even if the matrix M is super huge. Let's just see what this would look like in a three-dimensional circumstance.
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Expressing a quadratic form with a matrix.mp3
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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 that same expression even if the matrix M is super huge. 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.
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Expressing a quadratic form with a matrix.mp3
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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. 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 the same as over here, kind of when you reflected about that diagonal. Similarly, C, that's gonna be the same term here, and E would be over here.
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Expressing a quadratic form with a matrix.mp3
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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 the same as over here, kind of when you reflected about that diagonal. 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 by this vector, and then multiplying the corresponding vector that you get by this transposed vector.
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Expressing a quadratic form with a matrix.mp3
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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 by this vector, and then multiplying the corresponding vector that you get by this transposed vector. And you'll get some kind of quadratic form with three variables. And the point is, you'll get a very complicated one, but it's very simple to express things like this. So with that tool in hand, in the next video, I will talk about how we can use this notation to express the quadratic approximations for multivariable functions.
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Expressing a quadratic form with a matrix.mp3
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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, and since it's two-dimensional, it'll have some kind of two-dimensional input, and the output will be a vector each of whose components is some kind of function of x and y, right? So I'll just write p of x, y for that x component, and q of x, y for that y component, and each of these are just scalar-valued functions. It's actually quite common to use p and q for these values. It's one of those things where sometimes you'll even see a theorem about vector calculus in terms of just p and q, kind of leaving it understood to the reader that, yeah, p and q always refer to the x and y components of the output of a vector field, and in this specific case, the function that I chose, it's actually the one that I used in the last video, p is equal to x times y, and q is equal to y squared minus x squared, and in the last video, I was talking about interpreting the partial derivative of v, the vector-valued function, with respect to one of the variables, which has its merits, and I think it's a good way to understand vector-valued functions in general, but here, that's not what I'm gonna do. It's actually, another useful skill is to just think in terms of each specific component, so if we just think of p and q, we have four possible partial derivatives at our disposal here, two of them with respect to p, so you can think about the partial derivative of p with respect to x, or the partial derivative of p with respect to y, and then similarly, q, you could think about partial derivative of q with respect to x, x, this should be a partial, or the partial derivative of q with respect to y, so four different values that you could be looking at and considering and understanding how they influence the change of the vector field as a whole, and in this specific example, let's actually compute these, so derivative of p with respect to x, p is this first component, we're taking the partial of this with respect to x, y looks like a constant, constant times x, derivative is just that constant. If we took the derivative with respect to y, the roles are reversed, and its partial derivative is x, because x looks like that constant, but q, its partial derivative with respect to x, y looks like a constant, negative x squared goes to negative two x, but then when you're taking it with respect to y, y squared now looks like a function whose derivative is two y, and negative x squared looks like the constant, so these are the four possible partial derivatives, but let's actually see if we can understand how they influence the function as a whole, what it means in terms of the picture that we're looking at up here, and in particular, let's focus on a point, a specific point, and let's do this one here, so it's something that's sitting on the x-axis, so this is where y equals zero, and x is something positive, so this is probably when x is around two-ish, let's say, so the value we want to look at is x, y, when x is two, and y is zero, so if we start plugging that in here, what that would mean, this guy goes to zero, this guy goes to two, this guy, negative two times x, is gonna be negative two, and then negative two times y is gonna be zero, and let's start by just looking at the partial derivative of p with respect to x, so what that means is that we're looking for how the x component of these vectors change as you move in the x direction, so for example, around this point, we're kind of thinking of moving in the x direction vaguely, so we want to look at the two neighboring vectors and consider what's going on with the x direction, but these vectors, this one points purely down, this one also points purely down, and so does this one, so no change is happening when it comes to the x component of these vectors, which makes sense, because the value at that point is zero, the partial derivative of p with respect to x is zero, so we wouldn't expect a change, but on the other hand, if we're looking at partial derivative of p with respect to y, this should be positive, so this should suggest that the change in the x component as you move in the y direction is positive, so we go up here, and now we're not looking at change in the x direction, not looking at change in the x direction, but instead we're wondering what happens as we move generally upwards, so we're gonna kind of compare it to these two guys, and in that case, the x component of this one is a little bit to the left, the one below it, it's a little bit to the left, then we get to our main guy here, and it's zero, the x component is zero, because it's pointing purely down, and up here it's pointing a little bit to the right, so as y increases, the x component of these vectors also increases, and again, that makes sense, because this partial derivative is positive, this too suggests that as you're changing y, the value of p, the x component of our function, I should probably keep that on screen, the x component of our vector-valued function is increasing, because that's positive.
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Partial derivatives of vector fields, component by component.mp3
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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 integral. 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 of r with respect to u crossed with the partial of r with respect to v over the magnitude of the same thing. Partial of r with respect to u crossed with the partial of r with respect to v. And now we've played with ds a lot. 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.
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Vector representation of a surface integral Multivariable Calculus Khan Academy.mp3
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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 of r with respect to u crossed with the partial of r with respect to v over the magnitude of the same thing. Partial of r with respect to u crossed with the partial of r with respect to v. And now we've played with ds a lot. 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 obviously the du dv can be written as dv du. You could write it as da, a little chunk of area in the uv plane or in the uv domain. 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.
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Vector representation of a surface integral Multivariable Calculus Khan Academy.mp3
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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 obviously the du dv can be written as dv du. You could write it as da, a little chunk of area in the uv plane or in the uv domain. 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 should be, or I'm guessing there's a huge simplification that's popping out at you right now.
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Vector representation of a surface integral Multivariable Calculus Khan Academy.mp3
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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 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 quantities. You divide by something and multiply by something.
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Vector representation of a surface integral Multivariable Calculus Khan Academy.mp3
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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 quantities. 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 product. This is going to give us a vector right over here.
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Vector representation of a surface integral Multivariable Calculus Khan Academy.mp3
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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 product. 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.
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Vector representation of a surface integral Multivariable Calculus Khan Academy.mp3
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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 with respect to v, du dv. Let me scroll over to the right a little bit. du dv.
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Vector representation of a surface integral Multivariable Calculus Khan Academy.mp3
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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 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 integral like this written.
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Vector representation of a surface integral Multivariable Calculus Khan Academy.mp3
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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 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. I'm just going to rewrite that chunk.
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Vector representation of a surface integral Multivariable Calculus Khan Academy.mp3
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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. 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.
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Vector representation of a surface integral Multivariable Calculus Khan Academy.mp3
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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 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. Very small changes in the vector given a small change in u. And then we're multiplying that times du dv.
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Vector representation of a surface integral Multivariable Calculus Khan Academy.mp3
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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. Very small changes in the vector given a small change in u. And then we're multiplying that times du dv. Now du and dv are just scalar quantities. They're infinitesimally small, but for the sake of this argument, you can just view that they're not vectors. They're just scalar quantities.
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Vector representation of a surface integral Multivariable Calculus Khan Academy.mp3
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