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And then we have our final du. And now let me close the brackets. And all of that is times 1 half. So this thing right over here, cosine squared u, just a trig identity, takes us to that. Now this is pretty easy to evaluate. The antiderivative of this is just going to be u evaluated at 2 pi and 0. So it essentially just, let me just write it out.
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Surface integral ex3 part 2 Evaluating the outside surface Multivariable Calculus Khan Academy.mp3
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So this thing right over here, cosine squared u, just a trig identity, takes us to that. Now this is pretty easy to evaluate. The antiderivative of this is just going to be u evaluated at 2 pi and 0. So it essentially just, let me just write it out. This part right over here is just u evaluated from 0 to 2 pi. It's 2 pi minus 0. It just gets evaluated and we get 2 pi.
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Surface integral ex3 part 2 Evaluating the outside surface Multivariable Calculus Khan Academy.mp3
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So it essentially just, let me just write it out. This part right over here is just u evaluated from 0 to 2 pi. It's 2 pi minus 0. It just gets evaluated and we get 2 pi. So out front, you have your 1 half and then times 2 pi. And then this right over here, the antiderivative, this is going to be equal to minus 2 times the antiderivative of cosine of u. Well, that's just sine of u. Evaluated from 0 to 2 pi.
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Surface integral ex3 part 2 Evaluating the outside surface Multivariable Calculus Khan Academy.mp3
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It just gets evaluated and we get 2 pi. So out front, you have your 1 half and then times 2 pi. And then this right over here, the antiderivative, this is going to be equal to minus 2 times the antiderivative of cosine of u. Well, that's just sine of u. Evaluated from 0 to 2 pi. Well, sine of 2 pi is 0. Sine of 0 is 0. So this whole thing is going to evaluate to 0.
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Surface integral ex3 part 2 Evaluating the outside surface Multivariable Calculus Khan Academy.mp3
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Well, that's just sine of u. Evaluated from 0 to 2 pi. Well, sine of 2 pi is 0. Sine of 0 is 0. So this whole thing is going to evaluate to 0. So we could say minus 0. And then we take the antiderivative of this right over here. The antiderivative of this is going to be 1 half u.
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Surface integral ex3 part 2 Evaluating the outside surface Multivariable Calculus Khan Academy.mp3
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So this whole thing is going to evaluate to 0. So we could say minus 0. And then we take the antiderivative of this right over here. The antiderivative of this is going to be 1 half u. And the antiderivative of 1 half cosine of 2u. Well, if we had a 2 out front here, then that would be the derivative of sine of 2u. But we don't have a 2 out here.
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Surface integral ex3 part 2 Evaluating the outside surface Multivariable Calculus Khan Academy.mp3
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The antiderivative of this is going to be 1 half u. And the antiderivative of 1 half cosine of 2u. Well, if we had a 2 out front here, then that would be the derivative of sine of 2u. But we don't have a 2 out here. But we can add a 2. Let me actually write it this way. We can add a 2 right over here.
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Surface integral ex3 part 2 Evaluating the outside surface Multivariable Calculus Khan Academy.mp3
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But we don't have a 2 out here. But we can add a 2. Let me actually write it this way. We can add a 2 right over here. We can add a 2 and then divide by a 2. We can add a 2, and that's a little bit too confusing. Let me make it very clear.
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Surface integral ex3 part 2 Evaluating the outside surface Multivariable Calculus Khan Academy.mp3
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We can add a 2 right over here. We can add a 2 and then divide by a 2. We can add a 2, and that's a little bit too confusing. Let me make it very clear. 1 half cosine of 2u is equal to 1 half times 1 half. Let me write it this way. Is equal to 1 fourth times 2 cosine 2u.
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Surface integral ex3 part 2 Evaluating the outside surface Multivariable Calculus Khan Academy.mp3
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Let me make it very clear. 1 half cosine of 2u is equal to 1 half times 1 half. Let me write it this way. Is equal to 1 fourth times 2 cosine 2u. These are the exact same quantities. And the reason why I wrote it this way is this is clearly the derivative of sine of 2u. So when you take the antiderivative of this, it's the same thing as plus 1 fourth sine of 2u.
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Surface integral ex3 part 2 Evaluating the outside surface Multivariable Calculus Khan Academy.mp3
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Is equal to 1 fourth times 2 cosine 2u. These are the exact same quantities. And the reason why I wrote it this way is this is clearly the derivative of sine of 2u. So when you take the antiderivative of this, it's the same thing as plus 1 fourth sine of 2u. And we're going to evaluate that from 0 to 2 pi. And you confirm for yourself. You take the derivative of this, you do the chain rule, you get the 2 out front.
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Surface integral ex3 part 2 Evaluating the outside surface Multivariable Calculus Khan Academy.mp3
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So when you take the antiderivative of this, it's the same thing as plus 1 fourth sine of 2u. And we're going to evaluate that from 0 to 2 pi. And you confirm for yourself. You take the derivative of this, you do the chain rule, you get the 2 out front. 2 times 1 fourth gives you 1 half. And the derivative of sine of 2u with respect to 2u is cosine of 2u. Now we need to evaluate this at 2 pi and at 0.
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Surface integral ex3 part 2 Evaluating the outside surface Multivariable Calculus Khan Academy.mp3
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You take the derivative of this, you do the chain rule, you get the 2 out front. 2 times 1 fourth gives you 1 half. And the derivative of sine of 2u with respect to 2u is cosine of 2u. Now we need to evaluate this at 2 pi and at 0. When you evaluate it at 2 pi, you get to have 1 half times 2 pi, which is pi. So this is plus pi. Plus 1 fourth times sine of 2 times 2 pi.
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Surface integral ex3 part 2 Evaluating the outside surface Multivariable Calculus Khan Academy.mp3
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Now we need to evaluate this at 2 pi and at 0. When you evaluate it at 2 pi, you get to have 1 half times 2 pi, which is pi. So this is plus pi. Plus 1 fourth times sine of 2 times 2 pi. Sine of 4 pi, that's just going to be 0. So this is going to evaluate to 0. And then minus 1 half times 0.
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Surface integral ex3 part 2 Evaluating the outside surface Multivariable Calculus Khan Academy.mp3
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Plus 1 fourth times sine of 2 times 2 pi. Sine of 4 pi, that's just going to be 0. So this is going to evaluate to 0. And then minus 1 half times 0. And then 1 fourth sine of 2 times 0. This is all going to be 0 when you put 0 there. So this whole thing evaluated just 2 pi.
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Surface integral ex3 part 2 Evaluating the outside surface Multivariable Calculus Khan Academy.mp3
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And then minus 1 half times 0. And then 1 fourth sine of 2 times 0. This is all going to be 0 when you put 0 there. So this whole thing evaluated just 2 pi. And we're in the home stretch, at least for surface 2. And I'll switch back to surface 2's color now, now that we're near the end. So the surface integral for surface 2 is going to be 1 half times 2 pi minus 0 plus pi.
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Surface integral ex3 part 2 Evaluating the outside surface Multivariable Calculus Khan Academy.mp3
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So this whole thing evaluated just 2 pi. And we're in the home stretch, at least for surface 2. And I'll switch back to surface 2's color now, now that we're near the end. So the surface integral for surface 2 is going to be 1 half times 2 pi minus 0 plus pi. So it's 1 half times 3 pi, which is equal to, we can have our drum roll, our mini drum roll, since we're not really done the entire problem right now. But it's equal to 3 pi over 2. So we're making pretty good headway.
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Surface integral ex3 part 2 Evaluating the outside surface Multivariable Calculus Khan Academy.mp3
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So the surface integral for surface 2 is going to be 1 half times 2 pi minus 0 plus pi. So it's 1 half times 3 pi, which is equal to, we can have our drum roll, our mini drum roll, since we're not really done the entire problem right now. But it's equal to 3 pi over 2. So we're making pretty good headway. This was 0. And now this part right over here is 3 pi over 2. And in the next video, we will try to tackle surface 3.
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Surface integral ex3 part 2 Evaluating the outside surface Multivariable Calculus Khan Academy.mp3
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And I brought up in the last video this pretty complicated formula, and to remind you of the circumstances, I was saying that curvature, denoted with this little kappa, is typically calculated as the derivative of the unit tangent vector function. So we'll think of some kind of function that gives unit tangent vectors at every single point, so you know, just a vector with a length of 1 lying tangent to the curve, and that's going to be some kind of function of the same parameter, so big T for unit tangent vector, little t for the parameter, hopefully that's not too confusing. Then we're taking the derivative of this, not with respect to T, the parameter, but with respect to arc length, S. And by arc length, I mean if you take a tiny step along the curve, and consider this to be of size dS, so S typically denotes length along a curve, this will be a tiny change in length, you're wondering how much that unit tangent vector changes. And specifically what I mean by that, if you kind of imagine tangent vectors off sitting in their own space somewhere, where each one of them has unit length, they're all just kind of of the same length, and let's say they're all stemming from the same point, so rather than drawing them stemming from the curve to show that they're tangent, I just want to show them in their own space, the derivative, the change in that tangent vector, would be some other vector that tells you how you move from one to the other, kind of tells you how much it's turning. So the curvature itself is given, not by that vector, because this would be a vector quantity, but by the absolute value of it. And I said in the last video, that it turns out that this quantity, when you work it out for a vector valued function, with components, with components x of t, x of t and y of t, that it happens to equal this complicated formula. And what I'm gonna do here, I'm gonna break it down, and show why this formula isn't actually all that random, it's actually kind of sensible as a description for how much the curve really curves at a point.
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Curvature formula, part 4.mp3
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And specifically what I mean by that, if you kind of imagine tangent vectors off sitting in their own space somewhere, where each one of them has unit length, they're all just kind of of the same length, and let's say they're all stemming from the same point, so rather than drawing them stemming from the curve to show that they're tangent, I just want to show them in their own space, the derivative, the change in that tangent vector, would be some other vector that tells you how you move from one to the other, kind of tells you how much it's turning. So the curvature itself is given, not by that vector, because this would be a vector quantity, but by the absolute value of it. And I said in the last video, that it turns out that this quantity, when you work it out for a vector valued function, with components, with components x of t, x of t and y of t, that it happens to equal this complicated formula. And what I'm gonna do here, I'm gonna break it down, and show why this formula isn't actually all that random, it's actually kind of sensible as a description for how much the curve really curves at a point. So let's start by looking at the numerator here. This x prime of t, first derivative of the first component, times the second derivative of the second component, minus, and then kind of symmetrically, y prime times x double prime. You might be able to recognize this as a certain cross product.
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Curvature formula, part 4.mp3
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And what I'm gonna do here, I'm gonna break it down, and show why this formula isn't actually all that random, it's actually kind of sensible as a description for how much the curve really curves at a point. So let's start by looking at the numerator here. This x prime of t, first derivative of the first component, times the second derivative of the second component, minus, and then kind of symmetrically, y prime times x double prime. You might be able to recognize this as a certain cross product. So the cross product, if we take x prime, y prime, and these are still functions of t, as a vector crossed, with the vector containing the double primes. And if you're unfamiliar with cross products, or you're feeling a little shaky, and you kind of want to review, now would be a good time to pause the video, take a look at the cross product videos, and remind yourself both of how to compute it, and what the underlying intuition is. Because the way that we compute a cross product like this, is you take the components in the diagonal here, you know that rightward diagonal, and multiply them together, that's where you get your x prime, y double prime, and then you subtract off the components in the other diagonal.
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Curvature formula, part 4.mp3
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You might be able to recognize this as a certain cross product. So the cross product, if we take x prime, y prime, and these are still functions of t, as a vector crossed, with the vector containing the double primes. And if you're unfamiliar with cross products, or you're feeling a little shaky, and you kind of want to review, now would be a good time to pause the video, take a look at the cross product videos, and remind yourself both of how to compute it, and what the underlying intuition is. Because the way that we compute a cross product like this, is you take the components in the diagonal here, you know that rightward diagonal, and multiply them together, that's where you get your x prime, y double prime, and then you subtract off the components in the other diagonal. You know, it kind of feels like a determinant. X double prime times y prime. But the way that you interpret this vector, well I'll get to the interpretation in just a moment.
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Curvature formula, part 4.mp3
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Because the way that we compute a cross product like this, is you take the components in the diagonal here, you know that rightward diagonal, and multiply them together, that's where you get your x prime, y double prime, and then you subtract off the components in the other diagonal. You know, it kind of feels like a determinant. X double prime times y prime. But the way that you interpret this vector, well I'll get to the interpretation in just a moment. First let me kind of write out what this is, in terms of our function s. This first vector is just the first derivative of s. So that's s prime of t, the first derivative of s. And we're crossing that with this one, which is the second derivative, vector value derivative, but the second derivative of s. So before we do anything else, let's just start to think about what do both of these vectors mean? How do you interpret the s prime and the s double prime? I'm gonna go ahead and give ourselves a little bit of room here.
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Curvature formula, part 4.mp3
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But the way that you interpret this vector, well I'll get to the interpretation in just a moment. First let me kind of write out what this is, in terms of our function s. This first vector is just the first derivative of s. So that's s prime of t, the first derivative of s. And we're crossing that with this one, which is the second derivative, vector value derivative, but the second derivative of s. So before we do anything else, let's just start to think about what do both of these vectors mean? How do you interpret the s prime and the s double prime? I'm gonna go ahead and give ourselves a little bit of room here. So I'll draw the curve again. We're on our xy plane, and you have some kind of curve. The function s itself is giving vectors whose tips trace out this curve.
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Curvature formula, part 4.mp3
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I'm gonna go ahead and give ourselves a little bit of room here. So I'll draw the curve again. We're on our xy plane, and you have some kind of curve. The function s itself is giving vectors whose tips trace out this curve. As t changes, the tips of these vectors trace out the curve. And now the first derivative, that first derivative vector, s prime of t, is telling you how that tip should move to go along the curve. As you go from one s vector to another s vector, what direction should that tip move?
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Curvature formula, part 4.mp3
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The function s itself is giving vectors whose tips trace out this curve. As t changes, the tips of these vectors trace out the curve. And now the first derivative, that first derivative vector, s prime of t, is telling you how that tip should move to go along the curve. As you go from one s vector to another s vector, what direction should that tip move? And what this means is that at every given point, when you kind of do this in a limiting fashion, and you only look at infinitesimal changes in the original vector, you always get some kind of tangent vector. So all of these are tangent vectors, not necessarily unit. You might have a very long tangent vector to indicate that you're traveling very quickly across that space.
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Curvature formula, part 4.mp3
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As you go from one s vector to another s vector, what direction should that tip move? And what this means is that at every given point, when you kind of do this in a limiting fashion, and you only look at infinitesimal changes in the original vector, you always get some kind of tangent vector. So all of these are tangent vectors, not necessarily unit. You might have a very long tangent vector to indicate that you're traveling very quickly across that space. And now, how would you think about the second derivative vector here, s double prime of t? Well, the way to do that, I like to think of all of the tangent vectors then just kind of living in their own space. If this here is representing s of t, I just want to look in isolation at what s prime of t looks like.
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Curvature formula, part 4.mp3
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You might have a very long tangent vector to indicate that you're traveling very quickly across that space. And now, how would you think about the second derivative vector here, s double prime of t? Well, the way to do that, I like to think of all of the tangent vectors then just kind of living in their own space. If this here is representing s of t, I just want to look in isolation at what s prime of t looks like. So each one of those vectors, you know maybe this first one is just this very long gargantuan indicating you're going very quickly. And then after that, you've got something else here. Maybe it's pointing a little bit down.
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Curvature formula, part 4.mp3
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If this here is representing s of t, I just want to look in isolation at what s prime of t looks like. So each one of those vectors, you know maybe this first one is just this very long gargantuan indicating you're going very quickly. And then after that, you've got something else here. Maybe it's pointing a little bit down. And you're kind of thinking about how all of these derivative vectors change, but I want them all rooted at the same point. I'll just see what happens when they're all rooted at the origin, because that gives us a way to think about what changing the vector should look like. So in particular, as you're moving from this vector to this one, the tip should be kind of moving in this direction.
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Curvature formula, part 4.mp3
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Maybe it's pointing a little bit down. And you're kind of thinking about how all of these derivative vectors change, but I want them all rooted at the same point. I'll just see what happens when they're all rooted at the origin, because that gives us a way to think about what changing the vector should look like. So in particular, as you're moving from this vector to this one, the tip should be kind of moving in this direction. So this second derivative value is going to tell us how the tip of the first derivative should move. And then similarly over here, it tells us how the tip of that should move. And just as an example and kind of a hint as to why this has something to do with curvature, if you have a curve that turns very sharply, turns very sharply, you would have a tangent vector that starts by pointing up and to the right and then very quickly is pointing down, down and to the right.
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Curvature formula, part 4.mp3
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So in particular, as you're moving from this vector to this one, the tip should be kind of moving in this direction. So this second derivative value is going to tell us how the tip of the first derivative should move. And then similarly over here, it tells us how the tip of that should move. And just as an example and kind of a hint as to why this has something to do with curvature, if you have a curve that turns very sharply, turns very sharply, you would have a tangent vector that starts by pointing up and to the right and then very quickly is pointing down, down and to the right. So if they were off, we'd kind of draw them on their own, rooted in their own location. You can maybe see how the second derivative vector is telling it to turn somehow in that direction. And if you capture not just those two, but infinitesimally what's going on as you move from one to the next, you're going to get this kind of turning motion for all of these guys.
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Curvature formula, part 4.mp3
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And just as an example and kind of a hint as to why this has something to do with curvature, if you have a curve that turns very sharply, turns very sharply, you would have a tangent vector that starts by pointing up and to the right and then very quickly is pointing down, down and to the right. So if they were off, we'd kind of draw them on their own, rooted in their own location. You can maybe see how the second derivative vector is telling it to turn somehow in that direction. And if you capture not just those two, but infinitesimally what's going on as you move from one to the next, you're going to get this kind of turning motion for all of these guys. So the second derivative vector would be pulling perpendicular, perpendicular to that first derivative vector as a way of telling it to turn. So just to draw that on its own, if you have a first derivative vector and then the second derivative vector is perpendicular to it, it's telling it how it should turn in some way. But if it was, let's say it was not purely perpendicular, but it was also pointing against, it would be telling that vector, that first derivative vector, to shrink in some way.
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Curvature formula, part 4.mp3
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And if you capture not just those two, but infinitesimally what's going on as you move from one to the next, you're going to get this kind of turning motion for all of these guys. So the second derivative vector would be pulling perpendicular, perpendicular to that first derivative vector as a way of telling it to turn. So just to draw that on its own, if you have a first derivative vector and then the second derivative vector is perpendicular to it, it's telling it how it should turn in some way. But if it was, let's say it was not purely perpendicular, but it was also pointing against, it would be telling that vector, that first derivative vector, to shrink in some way. So that would be an indication that not only is it turning, but it's getting smaller, meaning the trajectory based on S is probably slowing down. And if it was kind of turning it, but also pointing away, that would mean it's telling the first derivative vector to grow. So not only should you be turning on your curve, but you should also be speeding up.
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Curvature formula, part 4.mp3
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But if it was, let's say it was not purely perpendicular, but it was also pointing against, it would be telling that vector, that first derivative vector, to shrink in some way. So that would be an indication that not only is it turning, but it's getting smaller, meaning the trajectory based on S is probably slowing down. And if it was kind of turning it, but also pointing away, that would mean it's telling the first derivative vector to grow. So not only should you be turning on your curve, but you should also be speeding up. But the case we care about most is we're just trying to measure how perpendicular it is, right? And this is where the cross product comes in. Because if you think about how you interpret that cross product, how you interpret the cross product, it's basically the area, if you kind of take these two vectors, tip to tail as they are, and think about the parallelogram that they trace out, let's see, so this blue dotted line should be parallel to the blue vector, the area of this, of this traced out parallelogram for each of them, that's what tells you, that's how you interpret the cross product, the cross product between S prime and S double prime.
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Curvature formula, part 4.mp3
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So not only should you be turning on your curve, but you should also be speeding up. But the case we care about most is we're just trying to measure how perpendicular it is, right? And this is where the cross product comes in. Because if you think about how you interpret that cross product, how you interpret the cross product, it's basically the area, if you kind of take these two vectors, tip to tail as they are, and think about the parallelogram that they trace out, let's see, so this blue dotted line should be parallel to the blue vector, the area of this, of this traced out parallelogram for each of them, that's what tells you, that's how you interpret the cross product, the cross product between S prime and S double prime. So this cross product, by giving you that area, is kind of a measure of just how perpendicular these vectors are, right? Because if one of them, if they point very much in the same direction, right, and they're only slightly perpendicular, that means that the parallelogram they trace out is gonna have a very small area, it's gonna be a smaller area in comparison. So with that, I don't want this video to run too long, so I'm gonna call it in here, but then I'll just continue on in the next one through the same line of reasoning to build to our original formula.
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Curvature formula, part 4.mp3
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And what this means is you're looking for the input points, the values of x and y and all of its other inputs, such that the output, f, is as great as it possibly can be. Now this actually comes up all the time in practice, because usually when you're dealing with a multivariable function, it's not just for fun and for dealing with abstract symbols, it's because it actually represents something. So maybe it represents like profits of a company. Maybe this is a function where you're considering all the choices you can make, like the wages you give your employees, or the prices of your goods, or the amount of debt that you raise for capital, all sorts of choices that you might make. And you want to know what values should you give to those choices, such that you maximize profits, you maximize a thing. And if you have a function that models these relationships, there are techniques, which I'm about to teach you, that you can use to maximize this. Another very common setting, more and more important these days, is that of machine learning and artificial intelligence, where often what you do is you assign something called a cost function to a task.
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Multivariable maxima and minima.mp3
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Maybe this is a function where you're considering all the choices you can make, like the wages you give your employees, or the prices of your goods, or the amount of debt that you raise for capital, all sorts of choices that you might make. And you want to know what values should you give to those choices, such that you maximize profits, you maximize a thing. And if you have a function that models these relationships, there are techniques, which I'm about to teach you, that you can use to maximize this. Another very common setting, more and more important these days, is that of machine learning and artificial intelligence, where often what you do is you assign something called a cost function to a task. So maybe you're trying to teach a computer how to understand audio, or how to read handwritten text. What you do is you find a function that basically tells it how wrong it is when it makes a guess. And if you do a good job designing that function, you just need to tell the computer to minimize, so that's kind of the flip side, right?
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Multivariable maxima and minima.mp3
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Another very common setting, more and more important these days, is that of machine learning and artificial intelligence, where often what you do is you assign something called a cost function to a task. So maybe you're trying to teach a computer how to understand audio, or how to read handwritten text. What you do is you find a function that basically tells it how wrong it is when it makes a guess. And if you do a good job designing that function, you just need to tell the computer to minimize, so that's kind of the flip side, right? Instead of finding the maximum, you need to minimize a certain function. And if it minimizes this cost function, that means that it's doing a really good job at whatever task you've assigned it. So a lot of the art and science of machine learning and artificial intelligence comes down to, well, one, finding this cost function and actually describing difficult tasks in terms of a function, but then applying the techniques that I'm about to teach you to have the computer minimize that.
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Multivariable maxima and minima.mp3
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And if you do a good job designing that function, you just need to tell the computer to minimize, so that's kind of the flip side, right? Instead of finding the maximum, you need to minimize a certain function. And if it minimizes this cost function, that means that it's doing a really good job at whatever task you've assigned it. So a lot of the art and science of machine learning and artificial intelligence comes down to, well, one, finding this cost function and actually describing difficult tasks in terms of a function, but then applying the techniques that I'm about to teach you to have the computer minimize that. And a lot of time and research has gone into figuring out ways to basically apply these techniques, really quickly and efficiently. So first of all, on a conceptual level, let's just think about what it means to be finding the maximum of a multivariable function. So I have here the graph of a two-variable function.
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Multivariable maxima and minima.mp3
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So a lot of the art and science of machine learning and artificial intelligence comes down to, well, one, finding this cost function and actually describing difficult tasks in terms of a function, but then applying the techniques that I'm about to teach you to have the computer minimize that. And a lot of time and research has gone into figuring out ways to basically apply these techniques, really quickly and efficiently. So first of all, on a conceptual level, let's just think about what it means to be finding the maximum of a multivariable function. So I have here the graph of a two-variable function. That's something that has a two-variable input that we're thinking of as the xy-plane, and then its output is the height of this graph. And if you're looking to maximize it, basically what you're finding is this peak, kind of the tallest mountain in the entire area. And you're looking for the input value, the point on the xy-plane, directly below that peak, because that tells you the values of the inputs that you should put in to maximize your function.
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Multivariable maxima and minima.mp3
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So I have here the graph of a two-variable function. That's something that has a two-variable input that we're thinking of as the xy-plane, and then its output is the height of this graph. And if you're looking to maximize it, basically what you're finding is this peak, kind of the tallest mountain in the entire area. And you're looking for the input value, the point on the xy-plane, directly below that peak, because that tells you the values of the inputs that you should put in to maximize your function. So how do you go about finding that? Well, this is perhaps the core observation in, well, calculus, not just multivariable calculus. This is similar in the single-variable world, and there's similarities in other settings, but the core observation is that if you take a tangent plane at that peak, so let's just draw in a tangent plane at that peak, it's gonna be completely flat.
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Multivariable maxima and minima.mp3
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And you're looking for the input value, the point on the xy-plane, directly below that peak, because that tells you the values of the inputs that you should put in to maximize your function. So how do you go about finding that? Well, this is perhaps the core observation in, well, calculus, not just multivariable calculus. This is similar in the single-variable world, and there's similarities in other settings, but the core observation is that if you take a tangent plane at that peak, so let's just draw in a tangent plane at that peak, it's gonna be completely flat. But let's say you did this at a different point, right? Because if you tried to find the tangent plane not at that point, but you kind of moved it about a bit to somewhere that's not quite a maximum, if the tangent plane has any kind of slope to it, what that's telling you is that if you take very small directions, kind of in the direction of that upward slope, you can increase the value of your function. So if there's any slope to the tangent plane, you know that you can walk in some direction to increase it.
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Multivariable maxima and minima.mp3
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This is similar in the single-variable world, and there's similarities in other settings, but the core observation is that if you take a tangent plane at that peak, so let's just draw in a tangent plane at that peak, it's gonna be completely flat. But let's say you did this at a different point, right? Because if you tried to find the tangent plane not at that point, but you kind of moved it about a bit to somewhere that's not quite a maximum, if the tangent plane has any kind of slope to it, what that's telling you is that if you take very small directions, kind of in the direction of that upward slope, you can increase the value of your function. So if there's any slope to the tangent plane, you know that you can walk in some direction to increase it. But if there's no slope to it, if it's flat, then that's a sign that no matter which direction you walk, you're not gonna be significantly increasing the value of your function. So what does this mean in terms of formulas? Well, if you kind of think back to how we compute tangent planes, and if you're not very comfortable with that, now would be a good time to take another look at those videos about tangent planes.
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Multivariable maxima and minima.mp3
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So if there's any slope to the tangent plane, you know that you can walk in some direction to increase it. But if there's no slope to it, if it's flat, then that's a sign that no matter which direction you walk, you're not gonna be significantly increasing the value of your function. So what does this mean in terms of formulas? Well, if you kind of think back to how we compute tangent planes, and if you're not very comfortable with that, now would be a good time to take another look at those videos about tangent planes. The slope of the plane in each direction, so you know this would be the slope in the x direction, and then if you look at it from another perspective, this would be the slope in the y direction, each one of those has to be zero. And that, in terms of partial derivatives, means the partial derivative of your function at whatever point you're dealing with, right, so I'll call it x-naught, y-naught, as the point where you're inputting this, has to be zero. And then similarly, the partial derivative with respect to the other variable, with respect to y, at that same point, has to be zero.
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Multivariable maxima and minima.mp3
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Well, if you kind of think back to how we compute tangent planes, and if you're not very comfortable with that, now would be a good time to take another look at those videos about tangent planes. The slope of the plane in each direction, so you know this would be the slope in the x direction, and then if you look at it from another perspective, this would be the slope in the y direction, each one of those has to be zero. And that, in terms of partial derivatives, means the partial derivative of your function at whatever point you're dealing with, right, so I'll call it x-naught, y-naught, as the point where you're inputting this, has to be zero. And then similarly, the partial derivative with respect to the other variable, with respect to y, at that same point, has to be zero. And both of these have to be true, because let's just take a look, I don't know, let's slide it over a little bit here. This tangent plane, if you look at the slope, if you imagine walking in the y direction, you're not increasing your value at all, the slope in the y direction would actually be zero, so that would mean the partial derivative with respect to y would be zero. But with respect to x, when you're moving in the x direction here, the slope is clearly negative, because as you take positive steps in the x direction, the height of your tangent plane is decreasing, which corresponds to if you take tiny steps on your graph, then the height will decrease in a manner proportional to the size of those tiny steps.
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Multivariable maxima and minima.mp3
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And then similarly, the partial derivative with respect to the other variable, with respect to y, at that same point, has to be zero. And both of these have to be true, because let's just take a look, I don't know, let's slide it over a little bit here. This tangent plane, if you look at the slope, if you imagine walking in the y direction, you're not increasing your value at all, the slope in the y direction would actually be zero, so that would mean the partial derivative with respect to y would be zero. But with respect to x, when you're moving in the x direction here, the slope is clearly negative, because as you take positive steps in the x direction, the height of your tangent plane is decreasing, which corresponds to if you take tiny steps on your graph, then the height will decrease in a manner proportional to the size of those tiny steps. So what this gives you here is gonna be a system of equations where you're solving for the value of x-naught and y-naught that satisfies both of these equations. And in future videos, I'll go through specific examples of this, for now I just wanna give a good conceptual understanding. But one very important thing to notice is that just because this condition is satisfied, meaning your tangent plane is flat, just because that's satisfied doesn't necessarily mean that you've found the maximum, that's just one requirement that it has to satisfy.
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Multivariable maxima and minima.mp3
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But with respect to x, when you're moving in the x direction here, the slope is clearly negative, because as you take positive steps in the x direction, the height of your tangent plane is decreasing, which corresponds to if you take tiny steps on your graph, then the height will decrease in a manner proportional to the size of those tiny steps. So what this gives you here is gonna be a system of equations where you're solving for the value of x-naught and y-naught that satisfies both of these equations. And in future videos, I'll go through specific examples of this, for now I just wanna give a good conceptual understanding. But one very important thing to notice is that just because this condition is satisfied, meaning your tangent plane is flat, just because that's satisfied doesn't necessarily mean that you've found the maximum, that's just one requirement that it has to satisfy. But for one thing, if you found the tangent plane at other little peaks, like this guy here, or this guy here, or all of the little bumps that go up, those tangent planes would also be flat. And those little bumps actually have a name, because this comes up a lot, they're called local minima, or local maxima, sorry, so those guys are called local maxima, maxima is just the plural of maximum, and local means that it's relative to a single point. So it's basically if you walk in any direction when you're on that little peak, you'll go downhill.
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Multivariable maxima and minima.mp3
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But one very important thing to notice is that just because this condition is satisfied, meaning your tangent plane is flat, just because that's satisfied doesn't necessarily mean that you've found the maximum, that's just one requirement that it has to satisfy. But for one thing, if you found the tangent plane at other little peaks, like this guy here, or this guy here, or all of the little bumps that go up, those tangent planes would also be flat. And those little bumps actually have a name, because this comes up a lot, they're called local minima, or local maxima, sorry, so those guys are called local maxima, maxima is just the plural of maximum, and local means that it's relative to a single point. So it's basically if you walk in any direction when you're on that little peak, you'll go downhill. So relative to the neighbors of that little point, it is a maximum, but relative to the entire function, you know, these guys are the shorter mountains next to Mount Everest. But there's also another circumstance where you might find a flat tangent plane, and that's at the minima points, right, if you have the global minimum, the absolute smallest, or also just the local minima, these inverted peaks, you'll also find flat tangent planes. So what that means, first of all, is that when you're minimizing a function, you also have to look for this requirement where all the partial derivatives are zero, but it mainly just means that your job isn't done once you've done this, you have to do more tests to check whether or not what you found is a local maximum, or a local minimum, or a global maximum.
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Multivariable maxima and minima.mp3
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So it's basically if you walk in any direction when you're on that little peak, you'll go downhill. So relative to the neighbors of that little point, it is a maximum, but relative to the entire function, you know, these guys are the shorter mountains next to Mount Everest. But there's also another circumstance where you might find a flat tangent plane, and that's at the minima points, right, if you have the global minimum, the absolute smallest, or also just the local minima, these inverted peaks, you'll also find flat tangent planes. So what that means, first of all, is that when you're minimizing a function, you also have to look for this requirement where all the partial derivatives are zero, but it mainly just means that your job isn't done once you've done this, you have to do more tests to check whether or not what you found is a local maximum, or a local minimum, or a global maximum. And these requirements, by the way, often you'll see them written in a more succinct form, where instead of saying all the partial derivatives have to be zero, which is what you need to find, they'll write it in a different form where you say that the gradient of your function f, which of course is just the vector that contains all those partial derivatives, its first component is the partial derivative with respect to the first variable, its second component is the partial derivative with respect to the second variable, and if there was more variables, you would keep going. You'd say that this whole thing has to equal the zero vector, the vector that has nothing but zeros as its components. And it's kind of a common abuse of notation, people will just call that zero vector zero, and maybe they'll emphasize it by making it bold, because the number zero is not a vector, and often making things bold emphasizes that you want to be referring to a vector, but this gives a very succinct way of describing the requirement.
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Multivariable maxima and minima.mp3
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So what that means, first of all, is that when you're minimizing a function, you also have to look for this requirement where all the partial derivatives are zero, but it mainly just means that your job isn't done once you've done this, you have to do more tests to check whether or not what you found is a local maximum, or a local minimum, or a global maximum. And these requirements, by the way, often you'll see them written in a more succinct form, where instead of saying all the partial derivatives have to be zero, which is what you need to find, they'll write it in a different form where you say that the gradient of your function f, which of course is just the vector that contains all those partial derivatives, its first component is the partial derivative with respect to the first variable, its second component is the partial derivative with respect to the second variable, and if there was more variables, you would keep going. You'd say that this whole thing has to equal the zero vector, the vector that has nothing but zeros as its components. And it's kind of a common abuse of notation, people will just call that zero vector zero, and maybe they'll emphasize it by making it bold, because the number zero is not a vector, and often making things bold emphasizes that you want to be referring to a vector, but this gives a very succinct way of describing the requirement. You're just looking for where the gradient of your function is equal to the zero vector, and that way you can just write it on one line. But in practice, every time that you're expanding that out, what that means is you find all of the different partial derivatives. So this is really just a matter of notational convenience and using less space on a blackboard.
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Multivariable maxima and minima.mp3
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And it's kind of a common abuse of notation, people will just call that zero vector zero, and maybe they'll emphasize it by making it bold, because the number zero is not a vector, and often making things bold emphasizes that you want to be referring to a vector, but this gives a very succinct way of describing the requirement. You're just looking for where the gradient of your function is equal to the zero vector, and that way you can just write it on one line. But in practice, every time that you're expanding that out, what that means is you find all of the different partial derivatives. So this is really just a matter of notational convenience and using less space on a blackboard. But whenever you see this, that the gradient equals zero, what you should be thinking of is the idea that the tangent plane, the tangent plane is completely flat. And as I just said, that's not enough, because you might also be finding local maxima or minima points, but in multivariable calculus, there's also another possibility, a place where the tangent plane is flat, but what you're looking at is neither a local maximum nor a local minimum. And this is the idea of a saddle point, which is new to multivariable calculus, and that's what I'll be talking about in the next video.
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Multivariable maxima and minima.mp3
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And what I want to do in this video is just get a little bit of a gut sense of what it means to take a derivative of a vector-valued function. In this case, it will be with respect to our parameter t. So let me draw some new stuff right here. So let's say I have the vector-valued function R of t, and this is no different than what I did in the last video, x of t times unit vector i plus y of t times the unit vector j. If we're dealing in three dimensions, we'd add a z of t times k, but let's keep things relatively simple. And let's say that this describes a curve, and let's say the curve we're dealing with between t is between a and b, and this curve will look something like... let me do my best effort to draw the curve. Let's draw some random curve here. So let's say the curve looks something like that.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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If we're dealing in three dimensions, we'd add a z of t times k, but let's keep things relatively simple. And let's say that this describes a curve, and let's say the curve we're dealing with between t is between a and b, and this curve will look something like... let me do my best effort to draw the curve. Let's draw some random curve here. So let's say the curve looks something like that. This is when t is equal to a, so it's going to go in this direction. This is when t is equal to b right here. This is t is equal to a, so this right here would be x of a.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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So let's say the curve looks something like that. This is when t is equal to a, so it's going to go in this direction. This is when t is equal to b right here. This is t is equal to a, so this right here would be x of a. This right here is y of a. And similarly, this up here, this is x of b, and this over here is y of b. Now, we saw in the last video that the endpoints of these position vectors are what's describing this curve.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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This is t is equal to a, so this right here would be x of a. This right here is y of a. And similarly, this up here, this is x of b, and this over here is y of b. Now, we saw in the last video that the endpoints of these position vectors are what's describing this curve. So r of a we saw in the last video. It describes that point right there. I don't want to review that too much.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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Now, we saw in the last video that the endpoints of these position vectors are what's describing this curve. So r of a we saw in the last video. It describes that point right there. I don't want to review that too much. But what I want to do is think about what is the difference between two points. So let's say that we take some random point here. Let's say some random t here.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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I don't want to review that too much. But what I want to do is think about what is the difference between two points. So let's say that we take some random point here. Let's say some random t here. Let's call that r of t. Actually, I'm going to do a different point just because I want to make it a little bit clearer. So let's say that that right there is r of some t, some particular t right there. That is r of t. It's going to be a plus something.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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Let's say some random t here. Let's call that r of t. Actually, I'm going to do a different point just because I want to make it a little bit clearer. So let's say that that right there is r of some t, some particular t right there. That is r of t. It's going to be a plus something. So that's some particular t. And let's say that we want to figure out, and let's say we increase t by a little bit, by h. So let's say that r of t plus h, well, if we view the parameter t as time, we can kind of view we've moved forward in time by some amount. So our little particle has moved a little bit. And let's say that we're over here.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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That is r of t. It's going to be a plus something. So that's some particular t. And let's say that we want to figure out, and let's say we increase t by a little bit, by h. So let's say that r of t plus h, well, if we view the parameter t as time, we can kind of view we've moved forward in time by some amount. So our little particle has moved a little bit. And let's say that we're over here. So that is that right there in yellow is r of t plus h. Just a slightly larger value for h. Now, one question we might ask ourselves is how quickly is r changing with respect to t? So the first thing we might want to say is what's the difference between these two? If I were to take, and I want to visualize it, if I were to take r, the position vector that we get by evaluating r at t plus h, and from that I were to subtract r of t, what do we get?
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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And let's say that we're over here. So that is that right there in yellow is r of t plus h. Just a slightly larger value for h. Now, one question we might ask ourselves is how quickly is r changing with respect to t? So the first thing we might want to say is what's the difference between these two? If I were to take, and I want to visualize it, if I were to take r, the position vector that we get by evaluating r at t plus h, and from that I were to subtract r of t, what do we get? Well, you might want to review some of your vector algebra, but we're essentially just going to get this vector. We're going to let me do it in a nice vibrant color. We're going to get this vector right there that I'm doing in magenta.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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If I were to take, and I want to visualize it, if I were to take r, the position vector that we get by evaluating r at t plus h, and from that I were to subtract r of t, what do we get? Well, you might want to review some of your vector algebra, but we're essentially just going to get this vector. We're going to let me do it in a nice vibrant color. We're going to get this vector right there that I'm doing in magenta. So that magenta vector right there is the vector r of t plus h minus r of t. And it should make sense because when you add vectors, you go heads to tails. You could alternatively write this as r of t plus this character right here, plus r of t plus h minus r of t. When you add two vectors, you're adding, let me make it very clear, I'm adding this vector to this vector right here. You put the tail of the second vector at the head of the first.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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We're going to get this vector right there that I'm doing in magenta. So that magenta vector right there is the vector r of t plus h minus r of t. And it should make sense because when you add vectors, you go heads to tails. You could alternatively write this as r of t plus this character right here, plus r of t plus h minus r of t. When you add two vectors, you're adding, let me make it very clear, I'm adding this vector to this vector right here. You put the tail of the second vector at the head of the first. So this is the first vector, and then I put the tail of the second there, and then the sum of those two, as we predicted, should be equal to this last one. It should be equal to r of t plus h. And we see that is the case, and even algebraically, you would see that obviously this guy and that guy are going to cancel out. So hopefully that satisfies you.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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You put the tail of the second vector at the head of the first. So this is the first vector, and then I put the tail of the second there, and then the sum of those two, as we predicted, should be equal to this last one. It should be equal to r of t plus h. And we see that is the case, and even algebraically, you would see that obviously this guy and that guy are going to cancel out. So hopefully that satisfies you. And I want to be clear, this all of a sudden, this isn't a position vector. We're not saying that, hey, let's nail this guy's tail at the origin and use this guy to describe a unique position. Now all of a sudden, it's just kind of a pure vector.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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So hopefully that satisfies you. And I want to be clear, this all of a sudden, this isn't a position vector. We're not saying that, hey, let's nail this guy's tail at the origin and use this guy to describe a unique position. Now all of a sudden, it's just kind of a pure vector. It's describing just a change between two other position vectors. So this guy is right out here, but this vector literally describes the change. But let's say we care, and how would this look algebraically if we were to expand it like that?
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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Now all of a sudden, it's just kind of a pure vector. It's describing just a change between two other position vectors. So this guy is right out here, but this vector literally describes the change. But let's say we care, and how would this look algebraically if we were to expand it like that? So this is going to be equal to, what's r of t plus h? That's the same thing as x of t plus h times the unit vector i plus y of t plus h times the unit vector j. That's just that piece.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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But let's say we care, and how would this look algebraically if we were to expand it like that? So this is going to be equal to, what's r of t plus h? That's the same thing as x of t plus h times the unit vector i plus y of t plus h times the unit vector j. That's just that piece. That piece right there is that piece. Minus, so this piece, so minus, I'll do it in the second line. I could have done it out here, but I'm running out of space.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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That's just that piece. That piece right there is that piece. Minus, so this piece, so minus, I'll do it in the second line. I could have done it out here, but I'm running out of space. Minus x of t, right? R of t is just x of t times i plus, but I'll just distribute the minus sign. So it's minus y of t times j.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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I could have done it out here, but I'm running out of space. Minus x of t, right? R of t is just x of t times i plus, but I'll just distribute the minus sign. So it's minus y of t times j. Actually, let me write it this way, plus this. So you realize that this is really just this guy right here. I'm just evaluating at t. So you have x of t and y of t, and then later we can distribute.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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So it's minus y of t times j. Actually, let me write it this way, plus this. So you realize that this is really just this guy right here. I'm just evaluating at t. So you have x of t and y of t, and then later we can distribute. If you distribute this minus sign, you get a minus x of t and a minus y of t. And in vector addition, you might need a little review on this if you haven't seen it in a while, you can just add the corresponding components. You can add the x components and you can add the y components. So this is going to be equal to, let me rewrite it over here, because I think I'm going to need some space later on.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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I'm just evaluating at t. So you have x of t and y of t, and then later we can distribute. If you distribute this minus sign, you get a minus x of t and a minus y of t. And in vector addition, you might need a little review on this if you haven't seen it in a while, you can just add the corresponding components. You can add the x components and you can add the y components. So this is going to be equal to, let me rewrite it over here, because I think I'm going to need some space later on. So let me rewrite it over here. So I have R of t plus h minus R of t is equal to, now I'm just going to group the x and the y components. This is equal to the x components added together, but this is a negative, so we're going to subtract this guy from that guy.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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So this is going to be equal to, let me rewrite it over here, because I think I'm going to need some space later on. So let me rewrite it over here. So I have R of t plus h minus R of t is equal to, now I'm just going to group the x and the y components. This is equal to the x components added together, but this is a negative, so we're going to subtract this guy from that guy. So we have x of t plus h minus x of t, and then all of that times our unit vector in the x direction. And then we'll have plus y of t plus h minus y of t times the unit vector in the j direction. I'm just rearranging things right now.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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This is equal to the x components added together, but this is a negative, so we're going to subtract this guy from that guy. So we have x of t plus h minus x of t, and then all of that times our unit vector in the x direction. And then we'll have plus y of t plus h minus y of t times the unit vector in the j direction. I'm just rearranging things right now. So this will tell us what is our change between any two R's for a given change in distance. And our change in distance here is h between any two position vectors. Now, what I set out at the beginning of this video, I said, well, I want to figure out the change, and we're going to think about the instantaneous change with respect to t. So I want to see, well, how much did this change over a period of h?
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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I'm just rearranging things right now. So this will tell us what is our change between any two R's for a given change in distance. And our change in distance here is h between any two position vectors. Now, what I set out at the beginning of this video, I said, well, I want to figure out the change, and we're going to think about the instantaneous change with respect to t. So I want to see, well, how much did this change over a period of h? Instead of writing h, we could have written delta t. It would have been the same thing. So I want to divide this by h. So I want to say, look, my vector has changed this much, but I want to say it's over a period of h. And this is analogous to when we do slope. We say rise over run over delta y or change in y over change in x.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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Now, what I set out at the beginning of this video, I said, well, I want to figure out the change, and we're going to think about the instantaneous change with respect to t. So I want to see, well, how much did this change over a period of h? Instead of writing h, we could have written delta t. It would have been the same thing. So I want to divide this by h. So I want to say, look, my vector has changed this much, but I want to say it's over a period of h. And this is analogous to when we do slope. We say rise over run over delta y or change in y over change in x. This is kind of the change in our function per change in x. Let's just divide everything, or I shouldn't say change in x, per change in t. So here our change in t is h. The difference between t plus h and t is just going to be h. And so we're going to divide everything by h. When you multiply a vector by some scalar or divide it by some scalar, you're just taking each of its components and multiplying or dividing by that scalar. And we get that right there.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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We say rise over run over delta y or change in y over change in x. This is kind of the change in our function per change in x. Let's just divide everything, or I shouldn't say change in x, per change in t. So here our change in t is h. The difference between t plus h and t is just going to be h. And so we're going to divide everything by h. When you multiply a vector by some scalar or divide it by some scalar, you're just taking each of its components and multiplying or dividing by that scalar. And we get that right there. So this, for any finite difference right here, h, this will tell us how much our vector changes per h. But if we want to find the instantaneous change, just like what we did when we first learned differential calculus, we said, okay, this is kind of analogous to a slope. This would be good. This would work out well for us if the path under question looked something like this, if it was a linear path.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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And we get that right there. So this, for any finite difference right here, h, this will tell us how much our vector changes per h. But if we want to find the instantaneous change, just like what we did when we first learned differential calculus, we said, okay, this is kind of analogous to a slope. This would be good. This would work out well for us if the path under question looked something like this, if it was a linear path. If our path looked something like this, we could just calculate this and we'll essentially have the average change in our position vectors. So you can imagine two position vectors. That's one of them.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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This would work out well for us if the path under question looked something like this, if it was a linear path. If our path looked something like this, we could just calculate this and we'll essentially have the average change in our position vectors. So you can imagine two position vectors. That's one of them. Well, actually, they'd all be parallel. Well, they don't always. Well, the position vectors, they don't have to be parallel, but I won't.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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That's one of them. Well, actually, they'd all be parallel. Well, they don't always. Well, the position vectors, they don't have to be parallel, but I won't. They don't have to be parallel. They could be like that. And then this would just describe the change between these two per h, or how quickly are the position vectors changing per our change in our parameter.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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Well, the position vectors, they don't have to be parallel, but I won't. They don't have to be parallel. They could be like that. And then this would just describe the change between these two per h, or how quickly are the position vectors changing per our change in our parameter. The h you could also consider as kind of a delta t. Sometimes people find the h simpler. Sometimes they find the delta t. But anyway, I'm concerned with the instantaneous. We're dealing with curves.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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And then this would just describe the change between these two per h, or how quickly are the position vectors changing per our change in our parameter. The h you could also consider as kind of a delta t. Sometimes people find the h simpler. Sometimes they find the delta t. But anyway, I'm concerned with the instantaneous. We're dealing with curves. We're dealing with calculus. This would have been okay if we were just in an algebraic linear world. So what do we do?
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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We're dealing with curves. We're dealing with calculus. This would have been okay if we were just in an algebraic linear world. So what do we do? Well, maybe we can just take the limit as h approaches zero. So let me scroll this over. So let's just take the limit.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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So what do we do? Well, maybe we can just take the limit as h approaches zero. So let me scroll this over. So let's just take the limit. Let me do this in a nice, vibrant color. Let's take the limit as h approaches zero of both sides of this. So here, too, I'm going to take the limit as h approaches zero.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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So let's just take the limit. Let me do this in a nice, vibrant color. Let's take the limit as h approaches zero of both sides of this. So here, too, I'm going to take the limit as h approaches zero. And here, too, I'm going to take the limit as h approaches zero. So I just want to say, well, what happens? How much do I change per a change in my parameter t?
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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So here, too, I'm going to take the limit as h approaches zero. And here, too, I'm going to take the limit as h approaches zero. So I just want to say, well, what happens? How much do I change per a change in my parameter t? But what's kind of the instantaneous change? As the difference gets smaller and smaller and smaller. This is exactly what we first learned when we learned about instantaneous slope or instantaneous velocity or slope of a tangent line.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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How much do I change per a change in my parameter t? But what's kind of the instantaneous change? As the difference gets smaller and smaller and smaller. This is exactly what we first learned when we learned about instantaneous slope or instantaneous velocity or slope of a tangent line. Well, this thing looks a little bit undefined to me right now. We haven't defined limits for vector-valued functions. We haven't defined derivatives for vector-valued functions.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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This is exactly what we first learned when we learned about instantaneous slope or instantaneous velocity or slope of a tangent line. Well, this thing looks a little bit undefined to me right now. We haven't defined limits for vector-valued functions. We haven't defined derivatives for vector-valued functions. But lucky for us, all of this stuff here looks pretty familiar. This is actually the definition of our derivative. And these are scalar-valued functions right here.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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We haven't defined derivatives for vector-valued functions. But lucky for us, all of this stuff here looks pretty familiar. This is actually the definition of our derivative. And these are scalar-valued functions right here. They're multiplied by vectors in order for us to get vector-valued functions. But this right here, by definition, this is the derivative. This is x prime of t. Or this is dx dt.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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And these are scalar-valued functions right here. They're multiplied by vectors in order for us to get vector-valued functions. But this right here, by definition, this is the derivative. This is x prime of t. Or this is dx dt. This right here is y prime of t. Or we could write that as dy dt. So all of a sudden, we can define, we can say, and I'm being a little hand-wavy here, but I want to give you the intuition more than anything. We can say that the derivative, we can call this expression right here as the derivative of my vector-valued function r with respect to t. Or we could call it dr dt.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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This is x prime of t. Or this is dx dt. This right here is y prime of t. Or we could write that as dy dt. So all of a sudden, we can define, we can say, and I'm being a little hand-wavy here, but I want to give you the intuition more than anything. We can say that the derivative, we can call this expression right here as the derivative of my vector-valued function r with respect to t. Or we could call it dr dt. Notice I keep the vector signs there. This is its derivative, and all it's going to be equal to, all it's going to be equal to, r prime of t, is going to be equal to, well, this is just the derivative of x with respect to t, is equal to x prime of t times the x unit vector, the horizontal unit vector, plus y prime of t times the y unit vector, times j, the unit vector in the horizontal direction. That's a pretty nice and simple outcome.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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We can say that the derivative, we can call this expression right here as the derivative of my vector-valued function r with respect to t. Or we could call it dr dt. Notice I keep the vector signs there. This is its derivative, and all it's going to be equal to, all it's going to be equal to, r prime of t, is going to be equal to, well, this is just the derivative of x with respect to t, is equal to x prime of t times the x unit vector, the horizontal unit vector, plus y prime of t times the y unit vector, times j, the unit vector in the horizontal direction. That's a pretty nice and simple outcome. But the harder thing, maybe, is to kind of visualize what it represents. So if we think about what happens, let me draw a big graph, just to get the visualization going in a healthy way. So let's say my curve looks something like this.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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That's a pretty nice and simple outcome. But the harder thing, maybe, is to kind of visualize what it represents. So if we think about what happens, let me draw a big graph, just to get the visualization going in a healthy way. So let's say my curve looks something like this. That's my curve. Let's say that this is, we want to figure out the instantaneous change at this point right here, so that is r of t. And then if we take r of t plus h, we saw this already, t plus h might be something like right there. So this is r of t plus h. Now, right now, the difference between these two, and this is just the numerator when you take the difference, or how fast we're changing from this vector to that vector, in terms of t, and it's hard to visualize here.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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So let's say my curve looks something like this. That's my curve. Let's say that this is, we want to figure out the instantaneous change at this point right here, so that is r of t. And then if we take r of t plus h, we saw this already, t plus h might be something like right there. So this is r of t plus h. Now, right now, the difference between these two, and this is just the numerator when you take the difference, or how fast we're changing from this vector to that vector, in terms of t, and it's hard to visualize here. And I'm going to do a whole video so we can think about the magnitude here. That might be some vector. Well, the difference between these two is just going to be that.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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So this is r of t plus h. Now, right now, the difference between these two, and this is just the numerator when you take the difference, or how fast we're changing from this vector to that vector, in terms of t, and it's hard to visualize here. And I'm going to do a whole video so we can think about the magnitude here. That might be some vector. Well, the difference between these two is just going to be that. But then when you divide it by h, it's going to be a larger vector. If we assume h is a small number, let's say h is less than 1, we're going to get a larger vector. But this is kind of the average change over this time.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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Well, the difference between these two is just going to be that. But then when you divide it by h, it's going to be a larger vector. If we assume h is a small number, let's say h is less than 1, we're going to get a larger vector. But this is kind of the average change over this time. But as h gets smaller and smaller and smaller, this r prime of t, its direction is going to be tangential to the curve. And I think you can visualize that. As these two guys get closer and closer and closer, the dr's get smaller, so the change, the dr, the difference between the two, the delta r's get smaller and smaller.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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But this is kind of the average change over this time. But as h gets smaller and smaller and smaller, this r prime of t, its direction is going to be tangential to the curve. And I think you can visualize that. As these two guys get closer and closer and closer, the dr's get smaller, so the change, the dr, the difference between the two, the delta r's get smaller and smaller. You can imagine if h was even smaller, if it was right here. All of a sudden the difference between those two vectors is getting smaller. And it's getting more and more tangential to the curve.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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As these two guys get closer and closer and closer, the dr's get smaller, so the change, the dr, the difference between the two, the delta r's get smaller and smaller. You can imagine if h was even smaller, if it was right here. All of a sudden the difference between those two vectors is getting smaller. And it's getting more and more tangential to the curve. But then we're also dividing by a smaller h. So the actual derivative, as the limit as h approaches 0, it might be, maybe it's even a bigger number there. And actually, the magnitude of this vector, it's a little hard to visualize, it's going to be dependent on a parametrization for the curve. It's not dependent on the shape of the curve.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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And it's getting more and more tangential to the curve. But then we're also dividing by a smaller h. So the actual derivative, as the limit as h approaches 0, it might be, maybe it's even a bigger number there. And actually, the magnitude of this vector, it's a little hard to visualize, it's going to be dependent on a parametrization for the curve. It's not dependent on the shape of the curve. The direction of this vector is dependent on the shape of this curve. So the direction, this will be tangent to the curve. Or you could imagine that this vector is on the tangent line to the curve.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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It's not dependent on the shape of the curve. The direction of this vector is dependent on the shape of this curve. So the direction, this will be tangent to the curve. Or you could imagine that this vector is on the tangent line to the curve. The magnitude of it is a little bit harder to understand. I'll try to give you a little bit of intuition on that in the next video. But this is what I want you to understand right now, because we're going to be able to use this in the future when we do the line integral over vector-valued functions.
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Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3
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It's defined over r2 over the xy-plane. So it's a function of x and y. It associates a vector with every point on the plane. And let's say my vector field is y times the unit vector i. Let's say minus x times the unit vector j. And so you can imagine if we were to draw our x and y axes. I'll do it over here.
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Using a line integral to find the work done by a vector field example Khan Academy.mp3
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And let's say my vector field is y times the unit vector i. Let's say minus x times the unit vector j. And so you can imagine if we were to draw our x and y axes. I'll do it over here. If we were to draw our x and y axes, this associates a vector, a force vector. Let's say this is actually a force vector with every point in our xy-plane. So this is x, this is y.
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Using a line integral to find the work done by a vector field example Khan Academy.mp3
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