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I just say ice so that it's not a lot of friction. Maybe it's standing on a bigger sheet of lake or ice or something. And maybe you're pulling on that ice cube at an angle. Let's say you're pulling at an angle like that. That is my force right there. Let's say my force is equal to, well, that's my force vector, let's say the magnitude of my force vector. So I'll say the magnitude, so I'm going to put two brackets around there, the magnitude of my force vector we can say is let's say it's 10 newtons.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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Let's say you're pulling at an angle like that. That is my force right there. Let's say my force is equal to, well, that's my force vector, let's say the magnitude of my force vector. So I'll say the magnitude, so I'm going to put two brackets around there, the magnitude of my force vector we can say is let's say it's 10 newtons. And let's say the direction of my force vector, any vector has to have a magnitude and a direction. And the direction, let's say it has a 30 degree angle, let's say it's 60 degree angle above horizontal. So that's the direction I'm pulling in.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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So I'll say the magnitude, so I'm going to put two brackets around there, the magnitude of my force vector we can say is let's say it's 10 newtons. And let's say the direction of my force vector, any vector has to have a magnitude and a direction. And the direction, let's say it has a 30 degree angle, let's say it's 60 degree angle above horizontal. So that's the direction I'm pulling in. And then let's say I displace it. This is all a bit of review, hopefully. If you're displacing it, let's say you displace it 5 newtons.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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So that's the direction I'm pulling in. And then let's say I displace it. This is all a bit of review, hopefully. If you're displacing it, let's say you displace it 5 newtons. So let's say the displacement, that's the displacement vector right there, and the magnitude of it is equal to 5 meters. So you've learned from the definition of work, you can't just say, oh, I'm pulling with 10 newtons of force and I'm moving at 5 meters. You can't just multiply the 10 newtons times the 5 meters.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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If you're displacing it, let's say you displace it 5 newtons. So let's say the displacement, that's the displacement vector right there, and the magnitude of it is equal to 5 meters. So you've learned from the definition of work, you can't just say, oh, I'm pulling with 10 newtons of force and I'm moving at 5 meters. You can't just multiply the 10 newtons times the 5 meters. You have to find the magnitude of the component going in the same direction as my displacement. So what I essentially need to do is, if you imagine the length of this vector being 10, that's the total force, but you need to figure out the length of the vector, the component of the force going in the same direction as my displacement. And a little simple trigonometry, you know that this is 10 times the cosine of 60 degrees.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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You can't just multiply the 10 newtons times the 5 meters. You have to find the magnitude of the component going in the same direction as my displacement. So what I essentially need to do is, if you imagine the length of this vector being 10, that's the total force, but you need to figure out the length of the vector, the component of the force going in the same direction as my displacement. And a little simple trigonometry, you know that this is 10 times the cosine of 60 degrees. Or that's equal to, cosine of 60 degrees is 1 half, so that's just equal to 5. So this magnitude, the magnitude of the force going in the same direction of the displacement in this case, is 5 newtons. And then you can figure out the work.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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And a little simple trigonometry, you know that this is 10 times the cosine of 60 degrees. Or that's equal to, cosine of 60 degrees is 1 half, so that's just equal to 5. So this magnitude, the magnitude of the force going in the same direction of the displacement in this case, is 5 newtons. And then you can figure out the work. You could say that the work is equal to 5 newtons times, I'll just write a dot for times, I don't want you to think it's cross product, times 5 meters, which is 25 newton meters, or you could even say 25 joules of work have been done. And this is all really a review of somewhat basic physics, but just think about what happened here. What was the work?
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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And then you can figure out the work. You could say that the work is equal to 5 newtons times, I'll just write a dot for times, I don't want you to think it's cross product, times 5 meters, which is 25 newton meters, or you could even say 25 joules of work have been done. And this is all really a review of somewhat basic physics, but just think about what happened here. What was the work? If I write it in the abstract. The work is equal to the 5 newtons, that was the magnitude of my force vector, so it's the magnitude of my force vector, times the cosine of this angle. Times the cosine of the, so let's call that theta, let's say it a little generally.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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What was the work? If I write it in the abstract. The work is equal to the 5 newtons, that was the magnitude of my force vector, so it's the magnitude of my force vector, times the cosine of this angle. Times the cosine of the, so let's call that theta, let's say it a little generally. So times the cosine of the angle. This is the amount of my force in the direction of the displacement, the cosine of the angle between them, times the magnitude of the displacement. So times the magnitude of the displacement.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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Times the cosine of the, so let's call that theta, let's say it a little generally. So times the cosine of the angle. This is the amount of my force in the direction of the displacement, the cosine of the angle between them, times the magnitude of the displacement. So times the magnitude of the displacement. Or if I wanted to rewrite that, I could just write that as the magnitude of the displacement times the magnitude of the force times the cosine of theta. And I've done multiple videos of this in the linear algebra playlist and the physics playlist where I talk about the dot product and the cross product and all of that. But this is the dot product of d and f, of the vectors d and f. So in general, if you're trying to find the work for a constant displacement and you have a constant force, you just take the dot product of those two vectors.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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So times the magnitude of the displacement. Or if I wanted to rewrite that, I could just write that as the magnitude of the displacement times the magnitude of the force times the cosine of theta. And I've done multiple videos of this in the linear algebra playlist and the physics playlist where I talk about the dot product and the cross product and all of that. But this is the dot product of d and f, of the vectors d and f. So in general, if you're trying to find the work for a constant displacement and you have a constant force, you just take the dot product of those two vectors. And if the dot product is a completely foreign concept to you, you might want to watch, I think I've made multiple, four or five videos on the dot product and its intuition and how it compares. But just to give you a little bit of that intuition right here, the dot product, when I take f dot d or d dot f, what it's giving me is I'm multiplying the magnitude. Well, I could just read this out.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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But this is the dot product of d and f, of the vectors d and f. So in general, if you're trying to find the work for a constant displacement and you have a constant force, you just take the dot product of those two vectors. And if the dot product is a completely foreign concept to you, you might want to watch, I think I've made multiple, four or five videos on the dot product and its intuition and how it compares. But just to give you a little bit of that intuition right here, the dot product, when I take f dot d or d dot f, what it's giving me is I'm multiplying the magnitude. Well, I could just read this out. But the idea of the dot product is take how much of this vector is going in the same direction as this vector. And in this case, it's this much. And then multiply the two magnitudes.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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Well, I could just read this out. But the idea of the dot product is take how much of this vector is going in the same direction as this vector. And in this case, it's this much. And then multiply the two magnitudes. And that's what we did right here. So the work is going to be the force vector dot, taking the dot product of the force vector with the displacement vector, and this, of course, is a scalar value. And we'll work out some examples in the future where you'll see that that's true.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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And then multiply the two magnitudes. And that's what we did right here. So the work is going to be the force vector dot, taking the dot product of the force vector with the displacement vector, and this, of course, is a scalar value. And we'll work out some examples in the future where you'll see that that's true. So this is all a review of fairly elementary physics. Now let's take a more complex example, but it's really the same idea. Let's define a vector field.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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And we'll work out some examples in the future where you'll see that that's true. So this is all a review of fairly elementary physics. Now let's take a more complex example, but it's really the same idea. Let's define a vector field. So let's say that I have a vector field f, and we're going to think about what this means in a second. It's a function of x and y, and it's equal to some scalar function of x and y times the i unit vector, or the horizontal unit vector, plus some other function, scalar function of x and y, times the vertical unit vector. So what would something like this be?
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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Let's define a vector field. So let's say that I have a vector field f, and we're going to think about what this means in a second. It's a function of x and y, and it's equal to some scalar function of x and y times the i unit vector, or the horizontal unit vector, plus some other function, scalar function of x and y, times the vertical unit vector. So what would something like this be? This is a vector field. This is a vector field in two-dimensional space, or on the xy-plane. This is a vector field on xy-plane.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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So what would something like this be? This is a vector field. This is a vector field in two-dimensional space, or on the xy-plane. This is a vector field on xy-plane. Or you can even say on or on r2. Either way, I don't want to get too much into the mathiness of it, but what does this do? Well, if I were to draw my xy-plane, so that is my, I'm having trouble drawing a straight line.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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This is a vector field on xy-plane. Or you can even say on or on r2. Either way, I don't want to get too much into the mathiness of it, but what does this do? Well, if I were to draw my xy-plane, so that is my, I'm having trouble drawing a straight line. All right, there we go. That's my y-axis, and that's my x-axis. I'm just drawing the first quadrant, but you could go negative in either direction if you like.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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Well, if I were to draw my xy-plane, so that is my, I'm having trouble drawing a straight line. All right, there we go. That's my y-axis, and that's my x-axis. I'm just drawing the first quadrant, but you could go negative in either direction if you like. What does this thing do? Well, it's essentially saying, look, you give me any x, any y. You give any xy in the xy-plane, and these are going to end up with some numbers, right?
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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I'm just drawing the first quadrant, but you could go negative in either direction if you like. What does this thing do? Well, it's essentially saying, look, you give me any x, any y. You give any xy in the xy-plane, and these are going to end up with some numbers, right? These are going to, with xy here, you're going to get some value. And when you put xy here, you're going to get some value, so you're going to get some combination of the i and j unit vector. So you're going to get some vector.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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You give any xy in the xy-plane, and these are going to end up with some numbers, right? These are going to, with xy here, you're going to get some value. And when you put xy here, you're going to get some value, so you're going to get some combination of the i and j unit vector. So you're going to get some vector. So what this does is defines a vector that's associated with every point on the xy-plane. So you could say, hey, if I take this point on the xy-plane and I were to pop it into this, I'll get something times i plus something times j, and when you add those two, maybe I get a vector that looks like something like that. And you could do it on every point.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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So you're going to get some vector. So what this does is defines a vector that's associated with every point on the xy-plane. So you could say, hey, if I take this point on the xy-plane and I were to pop it into this, I'll get something times i plus something times j, and when you add those two, maybe I get a vector that looks like something like that. And you could do it on every point. I'm just taking random samples. Maybe when I go here, the vector looks something like that, maybe when I go here, the vector looks like this, maybe when I go here, the vector looks like that, maybe when I go up here, the vector goes like that. I'm just randomly picking points.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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And you could do it on every point. I'm just taking random samples. Maybe when I go here, the vector looks something like that, maybe when I go here, the vector looks like this, maybe when I go here, the vector looks like that, maybe when I go up here, the vector goes like that. I'm just randomly picking points. It defines a vector on all of the xy-coordinates where these functions, these scalar functions, are properly defined, and that's why it's called a vector field. It defines what a potential, maybe force would be, or some other type of, well, we could say force, at any point. If you happen to have something there, maybe that's what the function is.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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I'm just randomly picking points. It defines a vector on all of the xy-coordinates where these functions, these scalar functions, are properly defined, and that's why it's called a vector field. It defines what a potential, maybe force would be, or some other type of, well, we could say force, at any point. If you happen to have something there, maybe that's what the function is. And I could keep doing this forever and filling in all the gaps, but I think you get the idea. It associates a vector with every point on the xy-plane. Now, this is called a vector field, so it probably makes a lot of sense that this could be used to describe any type of field.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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If you happen to have something there, maybe that's what the function is. And I could keep doing this forever and filling in all the gaps, but I think you get the idea. It associates a vector with every point on the xy-plane. Now, this is called a vector field, so it probably makes a lot of sense that this could be used to describe any type of field. It could be a gravitation field, it could be an electric field, it could be a magnetic field, and this could be essentially telling you how much force there would be on some particle in that field. That's exactly what this would describe. Now, let's say that in this field I have some particle traveling on the xy-plane, and let's say it starts here, and by virtue of all of these crazy forces that are acting on it, and maybe it's on some tracks or something, so it won't always move exactly in the direction that the field is trying to move it at.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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Now, this is called a vector field, so it probably makes a lot of sense that this could be used to describe any type of field. It could be a gravitation field, it could be an electric field, it could be a magnetic field, and this could be essentially telling you how much force there would be on some particle in that field. That's exactly what this would describe. Now, let's say that in this field I have some particle traveling on the xy-plane, and let's say it starts here, and by virtue of all of these crazy forces that are acting on it, and maybe it's on some tracks or something, so it won't always move exactly in the direction that the field is trying to move it at. So let's say it moves in a path that moves something like this. And let's say that this path, or this curve, is defined by a position vector function. So let's say that that's defined by r of t, which is just x of t times i plus y of t times our unit vector j.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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Now, let's say that in this field I have some particle traveling on the xy-plane, and let's say it starts here, and by virtue of all of these crazy forces that are acting on it, and maybe it's on some tracks or something, so it won't always move exactly in the direction that the field is trying to move it at. So let's say it moves in a path that moves something like this. And let's say that this path, or this curve, is defined by a position vector function. So let's say that that's defined by r of t, which is just x of t times i plus y of t times our unit vector j. That's r of t right there. Well, in order for this to be a finite path, this is true before t is greater than or equal to a and less than or equal to b. This is the path that the particle just happens to take due to all of these wacky forces.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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So let's say that that's defined by r of t, which is just x of t times i plus y of t times our unit vector j. That's r of t right there. Well, in order for this to be a finite path, this is true before t is greater than or equal to a and less than or equal to b. This is the path that the particle just happens to take due to all of these wacky forces. So when the particle is right here, maybe the vector field acting on it, maybe it's putting a force like that. But since the thing is on some type of tracks, it moves in this direction. And then when it's here, maybe the vector field is like that, but it moves in that direction because it's on some type of tracks.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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This is the path that the particle just happens to take due to all of these wacky forces. So when the particle is right here, maybe the vector field acting on it, maybe it's putting a force like that. But since the thing is on some type of tracks, it moves in this direction. And then when it's here, maybe the vector field is like that, but it moves in that direction because it's on some type of tracks. Now, everything I've done in this video is to build up to a fundamental question. What was the work done on the particle by the field? To answer that question, we could zoom in a little bit.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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And then when it's here, maybe the vector field is like that, but it moves in that direction because it's on some type of tracks. Now, everything I've done in this video is to build up to a fundamental question. What was the work done on the particle by the field? To answer that question, we could zoom in a little bit. Let's say I'm going to zoom in on only a little small snippet of our path. And let's just try to figure out what the work is done in a very small part of our path because it's constantly changing. The field is changing directions.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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To answer that question, we could zoom in a little bit. Let's say I'm going to zoom in on only a little small snippet of our path. And let's just try to figure out what the work is done in a very small part of our path because it's constantly changing. The field is changing directions. My object is changing directions. So let's say when I'm here, and let's say I move a small amount of my path. So let's say I move, this is an infinitesimally small dr.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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The field is changing directions. My object is changing directions. So let's say when I'm here, and let's say I move a small amount of my path. So let's say I move, this is an infinitesimally small dr. I have a differential. It's a differential vector, infinitely small displacement. And let's say over the course of that, the vector field is acting in this local area.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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So let's say I move, this is an infinitesimally small dr. I have a differential. It's a differential vector, infinitely small displacement. And let's say over the course of that, the vector field is acting in this local area. Let's say it looks something like that. It's providing a force that looks something like that. So that's the vector field in that area, or the force directed on that particle right when it's at that point.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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And let's say over the course of that, the vector field is acting in this local area. Let's say it looks something like that. It's providing a force that looks something like that. So that's the vector field in that area, or the force directed on that particle right when it's at that point. There's an infinitesimally small amount of time and space, so you could say, OK, over that little small point, we have this constant force. What was the work done over this small period? You could say, what's the small interval of work?
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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So that's the vector field in that area, or the force directed on that particle right when it's at that point. There's an infinitesimally small amount of time and space, so you could say, OK, over that little small point, we have this constant force. What was the work done over this small period? You could say, what's the small interval of work? You could say, d work, or a differential of work. Well, by the same exact logic that we did with the simple problem, it's the magnitude of the force in the direction of our displacement times the magnitude of our displacement. And we know what that is just from this example up here.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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You could say, what's the small interval of work? You could say, d work, or a differential of work. Well, by the same exact logic that we did with the simple problem, it's the magnitude of the force in the direction of our displacement times the magnitude of our displacement. And we know what that is just from this example up here. That's the dot product. It's the dot product of the force and our super small displacement. So that's equal to the dot product of our force and our super small displacement.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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And we know what that is just from this example up here. That's the dot product. It's the dot product of the force and our super small displacement. So that's equal to the dot product of our force and our super small displacement. Now, just by doing this, we're just figuring out the work over maybe a really small, super small dr. But what we want to do is we want to sum them all up. We want to sum up all of the dr's to figure out the total work done, and that's where the integral comes in.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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So that's equal to the dot product of our force and our super small displacement. Now, just by doing this, we're just figuring out the work over maybe a really small, super small dr. But what we want to do is we want to sum them all up. We want to sum up all of the dr's to figure out the total work done, and that's where the integral comes in. We will do a line integral over, I mean, you could think of it two ways. You could write just dw there, but we could say, we'll do a line integral that says along this curve C, we could call that C or along R, whatever you want to say it, of dw. That'll give us the total work.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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We want to sum up all of the dr's to figure out the total work done, and that's where the integral comes in. We will do a line integral over, I mean, you could think of it two ways. You could write just dw there, but we could say, we'll do a line integral that says along this curve C, we could call that C or along R, whatever you want to say it, of dw. That'll give us the total work. So let's say work is equal to that. Or we could also write it over the integral, over the same curve, of F dot dr. And this might seem like a really, gee, this is really abstract, Sal. How do we actually calculate something like this?
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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That'll give us the total work. So let's say work is equal to that. Or we could also write it over the integral, over the same curve, of F dot dr. And this might seem like a really, gee, this is really abstract, Sal. How do we actually calculate something like this? Especially because we have everything parameterized in terms of t. How do we get this in terms of t? And if you just think about it, what is F dot r? Or what is F dot dr?
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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How do we actually calculate something like this? Especially because we have everything parameterized in terms of t. How do we get this in terms of t? And if you just think about it, what is F dot r? Or what is F dot dr? Well, actually, to answer that, let's think about, let's remember what dr looked like. If you remember, dr dt is equal to x prime of t. I'm writing it like I could have written dx dt if I wanted to, times the i unit vector, plus y prime of t times the j unit vector. And if we just wanted dr, we could multiply both sides if we're being a little bit, maybe more hand-wavy with the differentials, not too rigorous.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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Or what is F dot dr? Well, actually, to answer that, let's think about, let's remember what dr looked like. If you remember, dr dt is equal to x prime of t. I'm writing it like I could have written dx dt if I wanted to, times the i unit vector, plus y prime of t times the j unit vector. And if we just wanted dr, we could multiply both sides if we're being a little bit, maybe more hand-wavy with the differentials, not too rigorous. We get dr is equal to x prime of t dt times the unit vector i, plus y prime of t times the differential dt times the unit vector j. So this is our dr right here. That is our dr. And remember what our vector field was.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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And if we just wanted dr, we could multiply both sides if we're being a little bit, maybe more hand-wavy with the differentials, not too rigorous. We get dr is equal to x prime of t dt times the unit vector i, plus y prime of t times the differential dt times the unit vector j. So this is our dr right here. That is our dr. And remember what our vector field was. It was this thing up here. Let me copy and paste it. And then we'll see that the dot product is actually not so crazy.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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That is our dr. And remember what our vector field was. It was this thing up here. Let me copy and paste it. And then we'll see that the dot product is actually not so crazy. So let me copy, and then let me paste it down here. So what's this integral going to look like? This integral right here that seems, gives us the total work done by the field on the particle as it moves along that path, which is super fundamental to pretty much any serious physics that you might eventually find yourself doing.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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And then we'll see that the dot product is actually not so crazy. So let me copy, and then let me paste it down here. So what's this integral going to look like? This integral right here that seems, gives us the total work done by the field on the particle as it moves along that path, which is super fundamental to pretty much any serious physics that you might eventually find yourself doing. So you could say, well gee, it's going to be the integral, well, let's just say from t is equal to a to t is equal to b. a is where we started off on the path. t is equal to a to t is equal to b. You could imagine it as being time, as a particle moving as time increases.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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This integral right here that seems, gives us the total work done by the field on the particle as it moves along that path, which is super fundamental to pretty much any serious physics that you might eventually find yourself doing. So you could say, well gee, it's going to be the integral, well, let's just say from t is equal to a to t is equal to b. a is where we started off on the path. t is equal to a to t is equal to b. You could imagine it as being time, as a particle moving as time increases. And then what is f dot dr? What is f dot dr? Well, if you remember from just what the dot product is, you can essentially just take the product of the corresponding components of your vector and add them up.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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You could imagine it as being time, as a particle moving as time increases. And then what is f dot dr? What is f dot dr? Well, if you remember from just what the dot product is, you can essentially just take the product of the corresponding components of your vector and add them up. So this is going to be the integral from t equals a to t equals b of p of x, really, instead of writing xy, it's x of t, right? x is a function of t. y is a function of t. So that's that. Times this thing right here, times this component, right?
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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Well, if you remember from just what the dot product is, you can essentially just take the product of the corresponding components of your vector and add them up. So this is going to be the integral from t equals a to t equals b of p of x, really, instead of writing xy, it's x of t, right? x is a function of t. y is a function of t. So that's that. Times this thing right here, times this component, right? We're multiplying the i components. So times x prime of t dt. And then that plus, we're going to do the same thing with the q function.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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Times this thing right here, times this component, right? We're multiplying the i components. So times x prime of t dt. And then that plus, we're going to do the same thing with the q function. So this is q plus, I'll go to another line, hopefully you realize I could have just kept writing, but I'm running out of space, plus q of x of t, y of t, times the component of our dr, times the y component, or the j component, y prime of t dt. And we're done. This might still seem a little bit abstract, but we're going to see in the next video that this is actually, everything is now in terms of t, so this is just a straight up integration with respect to dt.
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Line integrals and vector fields Multivariable Calculus Khan Academy.mp3
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The last video was very abstract in general and I used f of x and g of t and h of t. What I want to do in this video is do an actual example. So let's say I have f of x, y is equal to x, y. Let's say that we have a path in the x, y plane or a curve in the x, y plane. I'm going to define my curve, say my curve is going to be defined by x being equal to cosine of t and y being equal to sine of t. We have to define what are our boundaries on our t. We're going to go from t is equal to 0 or t is going to be greater than or equal to 0 and then less than or equal to, we're going to deal in radians, pi over 2. If this was degrees, that would be 90 degrees. So that's our curve and immediately you might already know what this type of a curve looks like and I'm going to draw that really fast right here and then we'll try to visualize this. I've actually graphed it ahead of time so that we can visualize this.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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I'm going to define my curve, say my curve is going to be defined by x being equal to cosine of t and y being equal to sine of t. We have to define what are our boundaries on our t. We're going to go from t is equal to 0 or t is going to be greater than or equal to 0 and then less than or equal to, we're going to deal in radians, pi over 2. If this was degrees, that would be 90 degrees. So that's our curve and immediately you might already know what this type of a curve looks like and I'm going to draw that really fast right here and then we'll try to visualize this. I've actually graphed it ahead of time so that we can visualize this. So this curve right here, if I were to just draw it in the standard x, y plane, let me do that in a different color so we can make the curve green. Let's say that is y and this is right here x. So when t is equal to 0, x is going to be equal to cosine of 0.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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I've actually graphed it ahead of time so that we can visualize this. So this curve right here, if I were to just draw it in the standard x, y plane, let me do that in a different color so we can make the curve green. Let's say that is y and this is right here x. So when t is equal to 0, x is going to be equal to cosine of 0. Cosine of 0 is 1. Y is going to be equal to sine of 0, which is 0. So t is equal to 0, we're going to be at x is equal to 1, that's cosine of 0 and y is sine of 0 or y is going to be 0.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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So when t is equal to 0, x is going to be equal to cosine of 0. Cosine of 0 is 1. Y is going to be equal to sine of 0, which is 0. So t is equal to 0, we're going to be at x is equal to 1, that's cosine of 0 and y is sine of 0 or y is going to be 0. So we're going to be right there. That's at t is equal to 0. When t is equal to pi of 2, what's going to happen?
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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So t is equal to 0, we're going to be at x is equal to 1, that's cosine of 0 and y is sine of 0 or y is going to be 0. So we're going to be right there. That's at t is equal to 0. When t is equal to pi of 2, what's going to happen? Cosine of pi over 2, that's the angle, cosine of pi over 2 is 0. Sine of pi over 2 is 1. So we're going to be at the point 0, 1.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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When t is equal to pi of 2, what's going to happen? Cosine of pi over 2, that's the angle, cosine of pi over 2 is 0. Sine of pi over 2 is 1. So we're going to be at the point 0, 1. So this is when we're at t is equal to pi over 2. You might recognize what we're going to draw is actually the first quadrant of the unit circle. When t is equal to pi over 4 or 45 degrees, we're going to be at square root of 2, square root of 2.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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So we're going to be at the point 0, 1. So this is when we're at t is equal to pi over 2. You might recognize what we're going to draw is actually the first quadrant of the unit circle. When t is equal to pi over 4 or 45 degrees, we're going to be at square root of 2, square root of 2. You can try it out for yourself, but we're just going to have a curve that looks like this. It's going to be the top right of a circle, of the unit circle. It's going to have radius 1 and we're going to go in that direction from t is equal to 0 to t is equal to pi over 2.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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When t is equal to pi over 4 or 45 degrees, we're going to be at square root of 2, square root of 2. You can try it out for yourself, but we're just going to have a curve that looks like this. It's going to be the top right of a circle, of the unit circle. It's going to have radius 1 and we're going to go in that direction from t is equal to 0 to t is equal to pi over 2. That's what this curve looks like. But in the last, our goal isn't here just to graph a parametric equation. What we want to do is raise a fence out of this kind of base and rise it to this surface.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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It's going to have radius 1 and we're going to go in that direction from t is equal to 0 to t is equal to pi over 2. That's what this curve looks like. But in the last, our goal isn't here just to graph a parametric equation. What we want to do is raise a fence out of this kind of base and rise it to this surface. So let's see if we can do that or at least visualize it first. And then we'll use the tools we used in the last video. So right here I've graphed this function and I've rotated it a little bit so you can see the other case.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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What we want to do is raise a fence out of this kind of base and rise it to this surface. So let's see if we can do that or at least visualize it first. And then we'll use the tools we used in the last video. So right here I've graphed this function and I've rotated it a little bit so you can see the other case. This right here, let me get some dark colors out, that right there is the x-axis. That's the x-axis. That in the back is the y-axis and the vertical axis is the z-axis.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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So right here I've graphed this function and I've rotated it a little bit so you can see the other case. This right here, let me get some dark colors out, that right there is the x-axis. That's the x-axis. That in the back is the y-axis and the vertical axis is the z-axis. And this is actually 2, this is 1 right here. y equals 1 is right there. So this is it graphed that way.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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That in the back is the y-axis and the vertical axis is the z-axis. And this is actually 2, this is 1 right here. y equals 1 is right there. So this is it graphed that way. So if I were to graph this contour on the xy-plane, it would be under this graph and it would go something like this. Let me see if I can draw it. It would look something like this.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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So this is it graphed that way. So if I were to graph this contour on the xy-plane, it would be under this graph and it would go something like this. Let me see if I can draw it. It would look something like this. This would be on the xy-plane. This is the same exact graph, f of x is equal to xy. Let me make that clear.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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It would look something like this. This would be on the xy-plane. This is the same exact graph, f of x is equal to xy. Let me make that clear. This is f of x, f of xy is equal to xy. That's both of these. I just rotated it.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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Let me make that clear. This is f of x, f of xy is equal to xy. That's both of these. I just rotated it. In this situation, this is now, that right there is now the x-axis. I rotated to the left, you can kind of imagine. That right there is the x-axis.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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I just rotated it. In this situation, this is now, that right there is now the x-axis. I rotated to the left, you can kind of imagine. That right there is the x-axis. That right there is the y-axis. It was rotated closer to me. That's the z-axis.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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That right there is the x-axis. That right there is the y-axis. It was rotated closer to me. That's the z-axis. And then this curve, if I were to draw it in this rotation, is going to look like this. When t is equal to 0, we're at x is equal to 1, y is equal to 0. That's going to form a unit circle, something like, or a quarter of a unit circle like that.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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That's the z-axis. And then this curve, if I were to draw it in this rotation, is going to look like this. When t is equal to 0, we're at x is equal to 1, y is equal to 0. That's going to form a unit circle, something like, or a quarter of a unit circle like that. And when t is equal to pi over 2, we're going to get there. What we want to do is find the area of the curtain that's defined. Let's raise a curtain from this curve up to f of xy.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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That's going to form a unit circle, something like, or a quarter of a unit circle like that. And when t is equal to pi over 2, we're going to get there. What we want to do is find the area of the curtain that's defined. Let's raise a curtain from this curve up to f of xy. If we keep raising walls from this up to x of y, we're going to have a wall that looks something like that. Let me shade it in, color it in so it looks a little bit more substantive. A wall that looks something like that.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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Let's raise a curtain from this curve up to f of xy. If we keep raising walls from this up to x of y, we're going to have a wall that looks something like that. Let me shade it in, color it in so it looks a little bit more substantive. A wall that looks something like that. If I were to try to do it here, this would be under the ceiling, but the wall would look something like that. We want to find the area of that. We want to find the area of this right here, where the base is defined by this curve, and then the ceiling is defined by this surface here, xy, which I graphed and rotated in two situations.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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A wall that looks something like that. If I were to try to do it here, this would be under the ceiling, but the wall would look something like that. We want to find the area of that. We want to find the area of this right here, where the base is defined by this curve, and then the ceiling is defined by this surface here, xy, which I graphed and rotated in two situations. In the last video, we came up with a, well, you could argue whether it's simple, but the idea is, well, let's just take small arc lengths, change in arc lengths, and multiply them by the height at that point. Those small change in arc lengths, we call them ds, and then the height is just f of xy at that point. We'll take an infinite sum of these from t is equal to 0 to t equal to pi over 2, and then that should give us the area of this wall right there.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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We want to find the area of this right here, where the base is defined by this curve, and then the ceiling is defined by this surface here, xy, which I graphed and rotated in two situations. In the last video, we came up with a, well, you could argue whether it's simple, but the idea is, well, let's just take small arc lengths, change in arc lengths, and multiply them by the height at that point. Those small change in arc lengths, we call them ds, and then the height is just f of xy at that point. We'll take an infinite sum of these from t is equal to 0 to t equal to pi over 2, and then that should give us the area of this wall right there. What we said is, well, to figure out the area of that, we're just going to take the integral from t is equal to 0 to t is equal to pi over 2. It doesn't make a lot of sense when I write it like this, of f of xy times, or let me even better, instead of writing f of xy, let me just write the actual function. Let's get a little bit more concrete.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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We'll take an infinite sum of these from t is equal to 0 to t equal to pi over 2, and then that should give us the area of this wall right there. What we said is, well, to figure out the area of that, we're just going to take the integral from t is equal to 0 to t is equal to pi over 2. It doesn't make a lot of sense when I write it like this, of f of xy times, or let me even better, instead of writing f of xy, let me just write the actual function. Let's get a little bit more concrete. f of xy is xy times, so the particular xy times the little change in our arc length at that point. I'm going to be very hand wavy here. This is all a little bit of review of the last video.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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Let's get a little bit more concrete. f of xy is xy times, so the particular xy times the little change in our arc length at that point. I'm going to be very hand wavy here. This is all a little bit of review of the last video. We figured out in the last video, this change in arc length right here, this change in arc length, ds, we figured out that we could rewrite that as the square root of dx versus, or the derivative of x with respect to t squared plus the derivative of y with respect to t squared, and then all of that times dt. I'm just rebuilding the formula that we got in the last video. This expression can be rewritten as the integral from t is equal to 0 to t is equal to pi over 2 times xy.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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This is all a little bit of review of the last video. We figured out in the last video, this change in arc length right here, this change in arc length, ds, we figured out that we could rewrite that as the square root of dx versus, or the derivative of x with respect to t squared plus the derivative of y with respect to t squared, and then all of that times dt. I'm just rebuilding the formula that we got in the last video. This expression can be rewritten as the integral from t is equal to 0 to t is equal to pi over 2 times xy. You know what? Right from the get-go, we want everything eventually to be in terms of t. Instead of writing x times y, let's substitute the parametric form. Instead of x, let's write cosine of t. Let me write cosine of t. That is x. x is equal to cosine of t on this curve.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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This expression can be rewritten as the integral from t is equal to 0 to t is equal to pi over 2 times xy. You know what? Right from the get-go, we want everything eventually to be in terms of t. Instead of writing x times y, let's substitute the parametric form. Instead of x, let's write cosine of t. Let me write cosine of t. That is x. x is equal to cosine of t on this curve. That's how we define x in terms of the parameter t. Then times y, which we're saying is sine of t. That's our y. All I did is rewrote xy in terms of t times ds. ds is this.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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Instead of x, let's write cosine of t. Let me write cosine of t. That is x. x is equal to cosine of t on this curve. That's how we define x in terms of the parameter t. Then times y, which we're saying is sine of t. That's our y. All I did is rewrote xy in terms of t times ds. ds is this. It's the square root of the derivative of x with respect to t squared plus the derivative of y with respect to t squared. All of that times dt. Now we just have to find these two derivatives.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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ds is this. It's the square root of the derivative of x with respect to t squared plus the derivative of y with respect to t squared. All of that times dt. Now we just have to find these two derivatives. It might seem really hard, but it's very easy for us to find the derivative of x with respect to t and the derivative of y with respect to t. I can do it right down here. We'll lose our graphs for a little bit. We know that the derivative of x with respect to t is just going to be what's the derivative of cosine of t?
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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Now we just have to find these two derivatives. It might seem really hard, but it's very easy for us to find the derivative of x with respect to t and the derivative of y with respect to t. I can do it right down here. We'll lose our graphs for a little bit. We know that the derivative of x with respect to t is just going to be what's the derivative of cosine of t? That's minus sine of t. The derivative of y with respect to t, the derivative of sine of anything is the cosine of that anything. It's cosine of t. We could substitute these back into this equation. Remember, we're just trying to find the area of this curtain.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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We know that the derivative of x with respect to t is just going to be what's the derivative of cosine of t? That's minus sine of t. The derivative of y with respect to t, the derivative of sine of anything is the cosine of that anything. It's cosine of t. We could substitute these back into this equation. Remember, we're just trying to find the area of this curtain. It has our curve here as kind of its base and has this function, this surface, as its ceiling. We go back down here. Let me rewrite this whole thing.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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Remember, we're just trying to find the area of this curtain. It has our curve here as kind of its base and has this function, this surface, as its ceiling. We go back down here. Let me rewrite this whole thing. This becomes the integral from t is equal to 0 to t is equal to pi over 2. I don't like this color. Cosine times sine, that's just the xy, times ds, which is this expression right here.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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Let me rewrite this whole thing. This becomes the integral from t is equal to 0 to t is equal to pi over 2. I don't like this color. Cosine times sine, that's just the xy, times ds, which is this expression right here. Now we can write this as, I'll go switch back to that color I don't like. The square root of, the derivative of x with respect to t is minus sine of t. We're going to square it. Plus derivative of y with respect to t. That's cosine of t. We're going to square it.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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Cosine times sine, that's just the xy, times ds, which is this expression right here. Now we can write this as, I'll go switch back to that color I don't like. The square root of, the derivative of x with respect to t is minus sine of t. We're going to square it. Plus derivative of y with respect to t. That's cosine of t. We're going to square it. Let me make my radic a little bit bigger. Then all of that times dt. This still might seem like a really hard integral until you realize that this right here, when you take a negative number and you square it, this is the same thing.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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Plus derivative of y with respect to t. That's cosine of t. We're going to square it. Let me make my radic a little bit bigger. Then all of that times dt. This still might seem like a really hard integral until you realize that this right here, when you take a negative number and you square it, this is the same thing. Let me rewrite. Minus, let me do this in the side right here. Minus sine of t squared plus cosine of t squared.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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This still might seem like a really hard integral until you realize that this right here, when you take a negative number and you square it, this is the same thing. Let me rewrite. Minus, let me do this in the side right here. Minus sine of t squared plus cosine of t squared. This is equivalent to, this is the same thing as sine of t squared plus cosine of t squared. You lose the sine information when you square something. It just becomes a positive.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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Minus sine of t squared plus cosine of t squared. This is equivalent to, this is the same thing as sine of t squared plus cosine of t squared. You lose the sine information when you square something. It just becomes a positive. These two things are equivalent. This is the most basic trig identity. This comes straight out of the unit circle definition.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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It just becomes a positive. These two things are equivalent. This is the most basic trig identity. This comes straight out of the unit circle definition. Sine squared plus cosine squared, this is just equal to 1. All this stuff under the radical sign is just equal to 1. We're getting the square root of 1, which is just 1.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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This comes straight out of the unit circle definition. Sine squared plus cosine squared, this is just equal to 1. All this stuff under the radical sign is just equal to 1. We're getting the square root of 1, which is just 1. All of this stuff right here will just become 1. This whole crazy integral simplifies a good bit. It just equals the square root of t equals 0 to t is equal to pi over 2.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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We're getting the square root of 1, which is just 1. All of this stuff right here will just become 1. This whole crazy integral simplifies a good bit. It just equals the square root of t equals 0 to t is equal to pi over 2. I'm going to switch these around just because it will make it a little easier in the next step. Sine of t times cosine of t dt. All I did, this whole thing equals 1.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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It just equals the square root of t equals 0 to t is equal to pi over 2. I'm going to switch these around just because it will make it a little easier in the next step. Sine of t times cosine of t dt. All I did, this whole thing equals 1. Got rid of it and I just switched the order of that. It will make the next step a little bit easier to explain. This integral, you say sine times cosine, what's the antiderivative of that?
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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All I did, this whole thing equals 1. Got rid of it and I just switched the order of that. It will make the next step a little bit easier to explain. This integral, you say sine times cosine, what's the antiderivative of that? The first thing you should recognize is I have a function or an expression here. I have its derivative. The derivative of sine is cosine of t. You might be able to do u substitution in your head.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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This integral, you say sine times cosine, what's the antiderivative of that? The first thing you should recognize is I have a function or an expression here. I have its derivative. The derivative of sine is cosine of t. You might be able to do u substitution in your head. It's a good skill to be able to do in your head, but I'll do it very explicitly here. If you have something as derivative, you define that something is u. You say u is equal to sine of t and then du dt, the derivative of u with respect to t, is equal to cosine of t. Or if you multiply both sides by the differential dt, if we're not going to be too rigorous, you get du is equal to cosine of t dt.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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The derivative of sine is cosine of t. You might be able to do u substitution in your head. It's a good skill to be able to do in your head, but I'll do it very explicitly here. If you have something as derivative, you define that something is u. You say u is equal to sine of t and then du dt, the derivative of u with respect to t, is equal to cosine of t. Or if you multiply both sides by the differential dt, if we're not going to be too rigorous, you get du is equal to cosine of t dt. Notice right here I have a u and then cosine of t dt, this thing right here, that thing is equal to d of u. Then we just have to redefine the boundaries. When t is equal to 0, this thing is going to turn into the integral.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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You say u is equal to sine of t and then du dt, the derivative of u with respect to t, is equal to cosine of t. Or if you multiply both sides by the differential dt, if we're not going to be too rigorous, you get du is equal to cosine of t dt. Notice right here I have a u and then cosine of t dt, this thing right here, that thing is equal to d of u. Then we just have to redefine the boundaries. When t is equal to 0, this thing is going to turn into the integral. Instead of t is equal to 0, when t is equal to 0, what is u equal to? Sine of 0 is 0. This goes from u is equal to 0.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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When t is equal to 0, this thing is going to turn into the integral. Instead of t is equal to 0, when t is equal to 0, what is u equal to? Sine of 0 is 0. This goes from u is equal to 0. When t is pi over 2, sine of pi over 2 is 1. When t is pi over 2, u is equal to 1. From u is equal to 0 to u is equal to 1.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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This goes from u is equal to 0. When t is pi over 2, sine of pi over 2 is 1. When t is pi over 2, u is equal to 1. From u is equal to 0 to u is equal to 1. Just redid the boundaries in terms of u. Then we have instead of sine of t, I'm going to write u. Instead of cosine of t dt, I'm just going to write du.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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From u is equal to 0 to u is equal to 1. Just redid the boundaries in terms of u. Then we have instead of sine of t, I'm going to write u. Instead of cosine of t dt, I'm just going to write du. This is a super easy integral in terms of u. This is just equal to the antiderivative of u is u 1 half times u squared. We just raised the exponent and then divided by that raised exponent.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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Instead of cosine of t dt, I'm just going to write du. This is a super easy integral in terms of u. This is just equal to the antiderivative of u is u 1 half times u squared. We just raised the exponent and then divided by that raised exponent. 1 half u squared. We're going to evaluate it from 0 to 1. This is going to be equal to 1 half times 1 squared minus 1 half times 0 squared.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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We just raised the exponent and then divided by that raised exponent. 1 half u squared. We're going to evaluate it from 0 to 1. This is going to be equal to 1 half times 1 squared minus 1 half times 0 squared. Which is equal to 1 half times 1 minus 0, which is equal to 1 half. We did all that work and we got a nice simple answer. The area of this curtain, we just performed a line integral.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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This is going to be equal to 1 half times 1 squared minus 1 half times 0 squared. Which is equal to 1 half times 1 minus 0, which is equal to 1 half. We did all that work and we got a nice simple answer. The area of this curtain, we just performed a line integral. The area of this curtain along this curve right here is 1 half. If this was in centimeters, it would be 1 half centimeter squared. I think that was a pretty neat application of the line integral.
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Line integral example 1 Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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So first of all, this idea of a critical point basically means anywhere where the gradient equals zero. So you're looking for places where the gradient of your function at some kind of input, some specified input x and y that you're solving for is equal to zero. And as I've talked about in the last couple videos, the reason you might want to do this is because you're hoping to maximize the function or to maybe minimize the function. And now the second requirement of classifying those points, that's what the second derivative test is all about. Once you find something where the gradient equals zero, you want to be able to determine, is it a local maximum, is it a local minimum, or is it a saddle point? So let's go ahead and work through this example. The first thing we're gonna need to do if we're solving for when the gradient equals zero, and remember, when we say equals zero, we really mean the zero vector, but it's just a convenient way of putting it all on one line.
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Second partial derivative test example, part 1.mp3
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