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So if we're doing this, we can interpret this as a slope, but you have to be very careful. If you're gonna interpret this as a slope, it has to be the case that you're dealing with a unit vector that the magnitude of your vector is equal to one. I mean, it doesn't have to be, you can kind of account for it later, but it makes it more easy to think about if we're just thinking of a unit vector. So when I go over here, instead of saying that it's one, one, I'm gonna say it's whatever vector points in that same direction, but it has a unit length. And in this case, that happens to be square root of two over two for each of the components. And you can kind of think about why that would be true, Pythagorean theorem and all, but this is a vector with unit length, its magnitude is one, and it points in that direction. And if we're evaluating this at a point like negative one, one, we can draw that on the graph, see where it actually is.
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Directional derivatives and slope.mp3
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So when I go over here, instead of saying that it's one, one, I'm gonna say it's whatever vector points in that same direction, but it has a unit length. And in this case, that happens to be square root of two over two for each of the components. And you can kind of think about why that would be true, Pythagorean theorem and all, but this is a vector with unit length, its magnitude is one, and it points in that direction. And if we're evaluating this at a point like negative one, one, we can draw that on the graph, see where it actually is. And in this case, it'll be, whoop, moving things about when I add a point. It'll be this point here, and you kind of look from above, and you say, okay, that is kind of negative one, negative one. And if we want the slope at that point, you're kind of thinking of the tangent line here, tangent line to that curve, and we're wondering what its slope is.
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Directional derivatives and slope.mp3
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And if we're evaluating this at a point like negative one, one, we can draw that on the graph, see where it actually is. And in this case, it'll be, whoop, moving things about when I add a point. It'll be this point here, and you kind of look from above, and you say, okay, that is kind of negative one, negative one. And if we want the slope at that point, you're kind of thinking of the tangent line here, tangent line to that curve, and we're wondering what its slope is. So the reason that the directional derivative is gonna give us this slope is because you think about this, another notation that might be kind of helpful for what this directional derivative is, some people will write partial f and then partial v. And you know, you can think about that as taking a slight nudge in the direction of v, right? So this would be a little nudge, a little partial nudge in the direction of v, and then you're saying, what change in the value of the function does that result in? You know, the height of the graph tells you the value of the function, and as this initial change approaches zero, and the resulting change approaches zero as well, that ratio, the ratio of df to, or partial f to partial v, is gonna give you the slope of this tangent line.
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Directional derivatives and slope.mp3
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And if we want the slope at that point, you're kind of thinking of the tangent line here, tangent line to that curve, and we're wondering what its slope is. So the reason that the directional derivative is gonna give us this slope is because you think about this, another notation that might be kind of helpful for what this directional derivative is, some people will write partial f and then partial v. And you know, you can think about that as taking a slight nudge in the direction of v, right? So this would be a little nudge, a little partial nudge in the direction of v, and then you're saying, what change in the value of the function does that result in? You know, the height of the graph tells you the value of the function, and as this initial change approaches zero, and the resulting change approaches zero as well, that ratio, the ratio of df to, or partial f to partial v, is gonna give you the slope of this tangent line. So conceptually, that's kind of a nicer notation, but the reason we use this other notation, this nabla sub v one, is it's very indicative of how you compute things once you need to compute it. You take the gradient of f, just the vector-valued function gradient of f, and take the dot product with your vector, with the vector. So let's actually do that, just to see what this would look like.
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Directional derivatives and slope.mp3
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You know, the height of the graph tells you the value of the function, and as this initial change approaches zero, and the resulting change approaches zero as well, that ratio, the ratio of df to, or partial f to partial v, is gonna give you the slope of this tangent line. So conceptually, that's kind of a nicer notation, but the reason we use this other notation, this nabla sub v one, is it's very indicative of how you compute things once you need to compute it. You take the gradient of f, just the vector-valued function gradient of f, and take the dot product with your vector, with the vector. So let's actually do that, just to see what this would look like. And I'll go ahead and write it over here, I'll use a different color. So the gradient of f, first of all, gradient of f, is a vector full of partial derivatives. So it'll be the partial derivative of f with respect to x, and the partial derivative of f with respect to y.
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Directional derivatives and slope.mp3
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So let's actually do that, just to see what this would look like. And I'll go ahead and write it over here, I'll use a different color. So the gradient of f, first of all, gradient of f, is a vector full of partial derivatives. So it'll be the partial derivative of f with respect to x, and the partial derivative of f with respect to y. And when we actually evaluate this, we take a look, partial derivative of f with respect to x, x looks like the variable, y is just a constant, so its partial derivative is two times x times y. Two times x times y, but when we take the partial with respect to y, y now looks like a variable, x looks like a constant. Derivative of a constant times the variable is just that constant, x squared.
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Directional derivatives and slope.mp3
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So it'll be the partial derivative of f with respect to x, and the partial derivative of f with respect to y. And when we actually evaluate this, we take a look, partial derivative of f with respect to x, x looks like the variable, y is just a constant, so its partial derivative is two times x times y. Two times x times y, but when we take the partial with respect to y, y now looks like a variable, x looks like a constant. Derivative of a constant times the variable is just that constant, x squared. And if we were to evaluate this at the point negative one, negative one, now you could plug that in, two times negative one times negative one would be two, and then negative one squared would be one. So that would be our gradient at that point, which means if we want to evaluate gradient of f times v, we could go over here and say that that's two, one, because we evaluate the gradient at the point we care about, and then the dot product with v itself, in this case, root two over two, and root two over two. The answer that we'd get, we multiply the first two components together, two times root two over two is square root of two, and then here, we multiply the second components together, and that's gonna be one times root two over two, root two over two, and that would be our answer, that would be our slope.
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Directional derivatives and slope.mp3
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Derivative of a constant times the variable is just that constant, x squared. And if we were to evaluate this at the point negative one, negative one, now you could plug that in, two times negative one times negative one would be two, and then negative one squared would be one. So that would be our gradient at that point, which means if we want to evaluate gradient of f times v, we could go over here and say that that's two, one, because we evaluate the gradient at the point we care about, and then the dot product with v itself, in this case, root two over two, and root two over two. The answer that we'd get, we multiply the first two components together, two times root two over two is square root of two, and then here, we multiply the second components together, and that's gonna be one times root two over two, root two over two, and that would be our answer, that would be our slope. But this only works if your vector is a unit vector, because, and I showed this in the last video where we talked about the formal definition of the directional derivative, if you scale v by two, and I could do it here, if instead of v, you're talking about two v, so I'll go ahead and make myself some room here. If you're taking the directional derivative along two v of f, the way that we're computing that, you're still taking the gradient of f, dot product with two times your vector, and dot products, you can pull out that two, this is just gonna double the value of the entire thing, dot v of this dotted with v, it's gonna be twice the value, the derivative will become twice the value, but you don't necessarily want that, because you'd say this plane that you slice it with, if instead of doing it in the direction of v, the unit vector, you did it in the direction of two times v, it's the same plane, it's the same slice you're taking, and you'd want that same slope, so that's gonna mess everything up. So this is super important if you're thinking about things in the context of slope, one thing that you could say is your formula for the slope of a graph in the direction of v, is you'd take your directional derivative, that dot product between f and v, and you just always make sure to divide it by the magnitude of v, divide it by that magnitude, and that'll always take care of what you want, that's basically a way of making sure that really you're taking the directional derivative in the direction of a certain unit vector, some people even go so far as to define the directional derivative to be this, to be something where you normalize out the length of that vector, I don't really like that, but I think that's because they're thinking of the slope context, they're thinking of rates of change as being the slope of a graph, and one thing I'd like to emphasize as always, graphical intuition is good, and visual intuition is always great, you should always be trying to find a way to think about things visually, but with multivariable functions, the graph isn't the only way, you can kind of more generally think about just a nudge in the v direction, and in the context where v doesn't have a length one, the nudge doesn't represent an actual size, but it's a certain scaling constant times that vector, you can look at the video on the formal definition for the directional derivative if you want more details on that, but I do think this is actually a good way to get a feel for what the directional derivative is all about.
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Directional derivatives and slope.mp3
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Let me actually graph this. So let's say R of t, I'll draw it a little bit straighter than that. That's my y axis, that is my x axis. Let's say R of t, and this is for t is less than, let me write this. So this is for t is between a and b. So when t is equal to a, we're at this vector right here. So if you actually substitute t is equal to a here, you would get a position vector that would point to that point over there.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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Let's say R of t, and this is for t is less than, let me write this. So this is for t is between a and b. So when t is equal to a, we're at this vector right here. So if you actually substitute t is equal to a here, you would get a position vector that would point to that point over there. And then as t increases, it traces out a curve, or the endpoints of our position vectors trace out a curve that looks something like that. So when t is equal to b, we get a position vector that points to that point right there. So this defines a path, and the path is going in this upward direction, just like that.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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So if you actually substitute t is equal to a here, you would get a position vector that would point to that point over there. And then as t increases, it traces out a curve, or the endpoints of our position vectors trace out a curve that looks something like that. So when t is equal to b, we get a position vector that points to that point right there. So this defines a path, and the path is going in this upward direction, just like that. Now let's say that we have another position vector function. Let me call it R, I'll call it R, I don't know, R reverse, R sub R of t. Actually, let me just call it R of t, it'll make it more confusing if I try to differentiate. But this is a different one, it's the green R. R of t, instead of being x of t times i, it's going to be x of a plus b minus t times i.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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So this defines a path, and the path is going in this upward direction, just like that. Now let's say that we have another position vector function. Let me call it R, I'll call it R, I don't know, R reverse, R sub R of t. Actually, let me just call it R of t, it'll make it more confusing if I try to differentiate. But this is a different one, it's the green R. R of t, instead of being x of t times i, it's going to be x of a plus b minus t times i. And instead of y of t, it's going to be y of a plus b minus t times i. And we've seen this in the last two videos. The path defined by this position vector function is going to look more like this.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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But this is a different one, it's the green R. R of t, instead of being x of t times i, it's going to be x of a plus b minus t times i. And instead of y of t, it's going to be y of a plus b minus t times i. And we've seen this in the last two videos. The path defined by this position vector function is going to look more like this. Let me draw my axes. That's my y-axis, that is my x-axis, maybe I should label them y and x. This path is going to look just like this, but instead of starting here and going there, when t is equal to 1, let me make it clear, this is also true for a is less than or equal to t, which is less than b.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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The path defined by this position vector function is going to look more like this. Let me draw my axes. That's my y-axis, that is my x-axis, maybe I should label them y and x. This path is going to look just like this, but instead of starting here and going there, when t is equal to 1, let me make it clear, this is also true for a is less than or equal to t, which is less than b. So t is going to go from a to b. But here, when t is equal to a, you substitute it over here, you're going to get this vector. You're going to start over here, you're going to start over there, and as you increment t, as you make it larger and larger and larger, you're going to trace out that same path, but in the opposite direction.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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This path is going to look just like this, but instead of starting here and going there, when t is equal to 1, let me make it clear, this is also true for a is less than or equal to t, which is less than b. So t is going to go from a to b. But here, when t is equal to a, you substitute it over here, you're going to get this vector. You're going to start over here, you're going to start over there, and as you increment t, as you make it larger and larger and larger, you're going to trace out that same path, but in the opposite direction. So it's the same path in the opposite direction. And so when t is equal to b, you put that in here, you're actually going to get x of a and y of a there, the b's cancel out, and so you're going to point right like that. So these are the same, you can imagine the shape of these paths are the same, but we're going in the exact opposite direction.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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You're going to start over here, you're going to start over there, and as you increment t, as you make it larger and larger and larger, you're going to trace out that same path, but in the opposite direction. So it's the same path in the opposite direction. And so when t is equal to b, you put that in here, you're actually going to get x of a and y of a there, the b's cancel out, and so you're going to point right like that. So these are the same, you can imagine the shape of these paths are the same, but we're going in the exact opposite direction. So what we're going to do in this video is to see what happens, how, I guess you could say, if I have some vector field, let's say I have some vector field, f of x, y equals p of x, y, i plus q of x, y, j, right? This is just a vector field over the x, y plane. How the line integral of this vector field over this path compares to the line integral of the same vector field over that path.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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So these are the same, you can imagine the shape of these paths are the same, but we're going in the exact opposite direction. So what we're going to do in this video is to see what happens, how, I guess you could say, if I have some vector field, let's say I have some vector field, f of x, y equals p of x, y, i plus q of x, y, j, right? This is just a vector field over the x, y plane. How the line integral of this vector field over this path compares to the line integral of the same vector field over that path. How that compares to this. We'll call this the minus curve. So this is the positive curve, we're going to call this the minus curve.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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How the line integral of this vector field over this path compares to the line integral of the same vector field over that path. How that compares to this. We'll call this the minus curve. So this is the positive curve, we're going to call this the minus curve. So how does it going over the positive curve compare to going over the minus curve, f of f dot dr. So before I break into the math, let's just think about it a little bit. Let me draw this vector field f. So maybe it looks, I'm just going to draw random stuff.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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So this is the positive curve, we're going to call this the minus curve. So how does it going over the positive curve compare to going over the minus curve, f of f dot dr. So before I break into the math, let's just think about it a little bit. Let me draw this vector field f. So maybe it looks, I'm just going to draw random stuff. So you know, on every point in the x, y plane, it has a vector, it defines or maps a vector to every point in the x, y plane. But we really care about the points that are on the curve. So maybe on the curve, this is the vector field at the points on the curve.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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Let me draw this vector field f. So maybe it looks, I'm just going to draw random stuff. So you know, on every point in the x, y plane, it has a vector, it defines or maps a vector to every point in the x, y plane. But we really care about the points that are on the curve. So maybe on the curve, this is the vector field at the points on the curve. And let me draw it over here too. So on the points on the curve where we care about, this is our vector field. That is our vector field.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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So maybe on the curve, this is the vector field at the points on the curve. And let me draw it over here too. So on the points on the curve where we care about, this is our vector field. That is our vector field. And let's just get an intuition of what's going to be going. We're summing over the dot, we're taking each point along the line. We're going to take each point along the line.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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That is our vector field. And let's just get an intuition of what's going to be going. We're summing over the dot, we're taking each point along the line. We're going to take each point along the line. And we're summing, let me start over here. Each point along the line, let me do it in a different color. And we're summing the dot product of the value of the vector field at that point, the dot product of that, with dr, or the differential of our position vector function.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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We're going to take each point along the line. And we're summing, let me start over here. Each point along the line, let me do it in a different color. And we're summing the dot product of the value of the vector field at that point, the dot product of that, with dr, or the differential of our position vector function. And dr you can kind of imagine as a small, an infinitesimally small vector going in the direction of our movement. And when we take this dot product here, it's essentially, it's going to be a scalar value, but the dot product, if you remember, it's the magnitude of f in the direction of dr times the magnitude of dr. So it's this, you can imagine it's the shadow of f onto dr. Let me zoom into that because I think it's useful.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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And we're summing the dot product of the value of the vector field at that point, the dot product of that, with dr, or the differential of our position vector function. And dr you can kind of imagine as a small, an infinitesimally small vector going in the direction of our movement. And when we take this dot product here, it's essentially, it's going to be a scalar value, but the dot product, if you remember, it's the magnitude of f in the direction of dr times the magnitude of dr. So it's this, you can imagine it's the shadow of f onto dr. Let me zoom into that because I think it's useful. So this little thing that I'm drawing right here, let's say that this is my path. This is f at that point. F at that point looks something like that.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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So it's this, you can imagine it's the shadow of f onto dr. Let me zoom into that because I think it's useful. So this little thing that I'm drawing right here, let's say that this is my path. This is f at that point. F at that point looks something like that. And then dr at this point looks something like that. Let me do it in a different color. dr looks something like that.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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F at that point looks something like that. And then dr at this point looks something like that. Let me do it in a different color. dr looks something like that. So that is f. And so the dot product of these two says, okay, how much of f is going in the same direction as dr? And you can kind of imagine it as a shadow. So you take the f that's going in the same direction as dr, the magnitude of that times the magnitude of dr, that is the dot product.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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dr looks something like that. So that is f. And so the dot product of these two says, okay, how much of f is going in the same direction as dr? And you can kind of imagine it as a shadow. So you take the f that's going in the same direction as dr, the magnitude of that times the magnitude of dr, that is the dot product. In this case, we're going to get a positive number because this length is positive, this length is positive. That's going to be a positive number. Now what if our dr was going in the opposite direction as it is in this case?
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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So you take the f that's going in the same direction as dr, the magnitude of that times the magnitude of dr, that is the dot product. In this case, we're going to get a positive number because this length is positive, this length is positive. That's going to be a positive number. Now what if our dr was going in the opposite direction as it is in this case? So let me draw maybe that same part of the curve. We have our f. Our f will look something like that. That is our f. I'm drawing this exact same part of the curve.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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Now what if our dr was going in the opposite direction as it is in this case? So let me draw maybe that same part of the curve. We have our f. Our f will look something like that. That is our f. I'm drawing this exact same part of the curve. But now our dr isn't going in that direction. Our dr at this point is going to be going in the other direction. We're tracing the curve in the opposite direction.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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That is our f. I'm drawing this exact same part of the curve. But now our dr isn't going in that direction. Our dr at this point is going to be going in the other direction. We're tracing the curve in the opposite direction. So our dr is now going to be going in that direction. So this is our dr. So if you do f dot dr, you're taking the shadow, or how much of f is going in the direction of dr, you take the shadow down here, it's going in the opposite direction of dr.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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We're tracing the curve in the opposite direction. So our dr is now going to be going in that direction. So this is our dr. So if you do f dot dr, you're taking the shadow, or how much of f is going in the direction of dr, you take the shadow down here, it's going in the opposite direction of dr. So you can imagine that when you multiply the magnitude, you should get a negative number. Our direction is now opposite. The shadow of f onto the same direction as dr is going in the opposite direction as dr.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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So if you do f dot dr, you're taking the shadow, or how much of f is going in the direction of dr, you take the shadow down here, it's going in the opposite direction of dr. So you can imagine that when you multiply the magnitude, you should get a negative number. Our direction is now opposite. The shadow of f onto the same direction as dr is going in the opposite direction as dr. So in this case, it's going in the same direction as dr. So the intuition is that maybe these two things are the negative of each other. And now we can do some math and try to see if that is definitely the case.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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The shadow of f onto the same direction as dr is going in the opposite direction as dr. So in this case, it's going in the same direction as dr. So the intuition is that maybe these two things are the negative of each other. And now we can do some math and try to see if that is definitely the case. So let us first figure out, let's write an expression for the differential dr. So in this case, dr dt is going to be equal to x prime of t times i plus y prime of t times j. In this other example, in the reverse case, our dr dt is going to be, what's it going to be equal to?
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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And now we can do some math and try to see if that is definitely the case. So let us first figure out, let's write an expression for the differential dr. So in this case, dr dt is going to be equal to x prime of t times i plus y prime of t times j. In this other example, in the reverse case, our dr dt is going to be, what's it going to be equal to? It's the derivative of x with respect to t. The derivative of this term with respect to t, that's the derivative of the inside, which is minus 1, or minus, times the derivative of the outside with respect to the inside. So the derivative of the outside is minus 1 times the derivative of the outside with respect to the inside. x prime of a plus b minus t times i. of y of this term, the derivative of the inside, which is minus 1, times the derivative of the outside with respect to the inside, which is y prime of a plus b minus t. So this is going to be the derivative of the inside times y prime of a plus b minus tj.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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In this other example, in the reverse case, our dr dt is going to be, what's it going to be equal to? It's the derivative of x with respect to t. The derivative of this term with respect to t, that's the derivative of the inside, which is minus 1, or minus, times the derivative of the outside with respect to the inside. So the derivative of the outside is minus 1 times the derivative of the outside with respect to the inside. x prime of a plus b minus t times i. of y of this term, the derivative of the inside, which is minus 1, times the derivative of the outside with respect to the inside, which is y prime of a plus b minus t. So this is going to be the derivative of the inside times y prime of a plus b minus tj. So this is dr dt in this case. This is dr dt in that case. And then if we wanted to write the differential dr in the forward curve example, it's going to be equal to x prime of t times i plus y prime of t times j times the scalar dt.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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x prime of a plus b minus t times i. of y of this term, the derivative of the inside, which is minus 1, times the derivative of the outside with respect to the inside, which is y prime of a plus b minus t. So this is going to be the derivative of the inside times y prime of a plus b minus tj. So this is dr dt in this case. This is dr dt in that case. And then if we wanted to write the differential dr in the forward curve example, it's going to be equal to x prime of t times i plus y prime of t times j times the scalar dt. I could multiply it down into each of these terms, but it keeps it simple just leaving it on the outside. Same logic over here. dr is equal to minus x. I changed my shade of green, but at least it's still green.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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And then if we wanted to write the differential dr in the forward curve example, it's going to be equal to x prime of t times i plus y prime of t times j times the scalar dt. I could multiply it down into each of these terms, but it keeps it simple just leaving it on the outside. Same logic over here. dr is equal to minus x. I changed my shade of green, but at least it's still green. a plus b minus ti minus y prime a plus b minus tj. And I'm multiplying both sides by dt. Now we're ready to express this as a function of t. So this curve right here, I'll do it in pink.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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dr is equal to minus x. I changed my shade of green, but at least it's still green. a plus b minus ti minus y prime a plus b minus tj. And I'm multiplying both sides by dt. Now we're ready to express this as a function of t. So this curve right here, I'll do it in pink. The pink one is going to be equal to the integral from t is equal to a to t is equal to b of f of x of t, y of t, dot this thing over here, which is, I'll just write it out here, I could simplify it later. x prime of ti plus y prime of tj. And then all of that times the scalar dt.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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Now we're ready to express this as a function of t. So this curve right here, I'll do it in pink. The pink one is going to be equal to the integral from t is equal to a to t is equal to b of f of x of t, y of t, dot this thing over here, which is, I'll just write it out here, I could simplify it later. x prime of ti plus y prime of tj. And then all of that times the scalar dt. This will be a scalar value, and then we'll have another scalar value dt over there. Now what is this going to be equal to? If I were to take this reverse integral, the reverse integral is going to be the integral from, I'm going to need a little more space, from a to b, of f of not x of t, but x of a plus b minus t, y of a plus b minus t. I'm writing it small, so I have some space.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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And then all of that times the scalar dt. This will be a scalar value, and then we'll have another scalar value dt over there. Now what is this going to be equal to? If I were to take this reverse integral, the reverse integral is going to be the integral from, I'm going to need a little more space, from a to b, of f of not x of t, but x of a plus b minus t, y of a plus b minus t. I'm writing it small, so I have some space. Dot, this is a vector, so dot this guy right here, dot dr. Dot minus x prime of a plus b minus ti minus y prime of, I'm using up too much space. Let me scroll, go back a little bit.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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If I were to take this reverse integral, the reverse integral is going to be the integral from, I'm going to need a little more space, from a to b, of f of not x of t, but x of a plus b minus t, y of a plus b minus t. I'm writing it small, so I have some space. Dot, this is a vector, so dot this guy right here, dot dr. Dot minus x prime of a plus b minus ti minus y prime of, I'm using up too much space. Let me scroll, go back a little bit. Actually, let me make it even simpler. Let me take this minus sign out of it. Let me put a plus, and I'll put the minus sign out front.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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Let me scroll, go back a little bit. Actually, let me make it even simpler. Let me take this minus sign out of it. Let me put a plus, and I'll put the minus sign out front. So the minus sign is just a scalar value. So we could put that minus sign out, when you take a dot product, and if you multiply a scalar times a dot product, you could just take the scalar out. That's all I'm saying.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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Let me put a plus, and I'll put the minus sign out front. So the minus sign is just a scalar value. So we could put that minus sign out, when you take a dot product, and if you multiply a scalar times a dot product, you could just take the scalar out. That's all I'm saying. So we could take that minus sign out to this part right here, and then you have x prime of a plus b minus t, i plus y prime of a plus b minus t, let me scroll over a little bit, tj, dt. So this is the forward. This is when we're following it along the forward curve.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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That's all I'm saying. So we could take that minus sign out to this part right here, and then you have x prime of a plus b minus t, i plus y prime of a plus b minus t, let me scroll over a little bit, tj, dt. So this is the forward. This is when we're following it along the forward curve. This is when we're following it along the reverse curve. Now, like we did with the scalar example, let's make a substitution. I want to make it very clear what I did.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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This is when we're following it along the forward curve. This is when we're following it along the reverse curve. Now, like we did with the scalar example, let's make a substitution. I want to make it very clear what I did. All I did here is I just took the dot product, but this negative sign, I just took it out. I just said this is the same thing as negative 1 times this thing, or negative 1 times this thing is the same thing as that. So let's make a substitution on this side, because I really just want to show you that this is the negative of that right there, because that's what our intuition was going for, so let me just focus on that side.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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I want to make it very clear what I did. All I did here is I just took the dot product, but this negative sign, I just took it out. I just said this is the same thing as negative 1 times this thing, or negative 1 times this thing is the same thing as that. So let's make a substitution on this side, because I really just want to show you that this is the negative of that right there, because that's what our intuition was going for, so let me just focus on that side. So let me make a substitution. u is equal to a plus b minus t. Then we get du is equal to minus dt. Just take the derivative of both sides, or you get dt is equal to minus du.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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So let's make a substitution on this side, because I really just want to show you that this is the negative of that right there, because that's what our intuition was going for, so let me just focus on that side. So let me make a substitution. u is equal to a plus b minus t. Then we get du is equal to minus dt. Just take the derivative of both sides, or you get dt is equal to minus du. And then you get when t is equal to a, u is equal to a plus b minus a. So then u is equal to b. And then when t is equal to b, u is equal to a.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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Just take the derivative of both sides, or you get dt is equal to minus du. And then you get when t is equal to a, u is equal to a plus b minus a. So then u is equal to b. And then when t is equal to b, u is equal to a. When t is equal to b, a plus b minus b is a. u is equal to a. So this thing, using that substitution, simplifies to this is the whole point, that simplifies to minus the integral from when t is a, u is b. From b, when t is b, u is a.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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And then when t is equal to b, u is equal to a. When t is equal to b, a plus b minus b is a. u is equal to a. So this thing, using that substitution, simplifies to this is the whole point, that simplifies to minus the integral from when t is a, u is b. From b, when t is b, u is a. The integral from u is equal to b to u is equal to a of f of x of u, y of u, dot x prime of u times i, that's u right there, plus y prime of u times j. And then instead of a dt, I need to put a du. dt is equal to minus du.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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From b, when t is b, u is a. The integral from u is equal to b to u is equal to a of f of x of u, y of u, dot x prime of u times i, that's u right there, plus y prime of u times j. And then instead of a dt, I need to put a du. dt is equal to minus du. So I could write minus du here, or just to not make it confusing, I'll put the du here. And take the minus out front. I already have a minus out there, so they cancel out.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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dt is equal to minus du. So I could write minus du here, or just to not make it confusing, I'll put the du here. And take the minus out front. I already have a minus out there, so they cancel out. They will cancel out just like that. And so you might say, hey Sal, these two things look pretty similar to each other. They don't look like they're negative of each other.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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I already have a minus out there, so they cancel out. They will cancel out just like that. And so you might say, hey Sal, these two things look pretty similar to each other. They don't look like they're negative of each other. And I'd say, well, you're almost right, except this guy's limits of integration are reversed from this guy. So this thing right here, if we reverse the limits of integration, we have to then make it negative. So this is equal to minus the integral from a to b of f dot, or the vector f of x of u, y of u, dot x prime of u i plus y prime of u j du.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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They don't look like they're negative of each other. And I'd say, well, you're almost right, except this guy's limits of integration are reversed from this guy. So this thing right here, if we reverse the limits of integration, we have to then make it negative. So this is equal to minus the integral from a to b of f dot, or the vector f of x of u, y of u, dot x prime of u i plus y prime of u j du. And now this is identical, this integral, this definite integral, is identical to that definite integral. We just have a different variable. We're doing dt here, we're having du here, but we're going to get the same exact number for any a or b, and given this vector f and the position vector path r of t. So just to summarize everything up, when you're dealing with line integrals over vector fields, the direction matters.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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So this is equal to minus the integral from a to b of f dot, or the vector f of x of u, y of u, dot x prime of u i plus y prime of u j du. And now this is identical, this integral, this definite integral, is identical to that definite integral. We just have a different variable. We're doing dt here, we're having du here, but we're going to get the same exact number for any a or b, and given this vector f and the position vector path r of t. So just to summarize everything up, when you're dealing with line integrals over vector fields, the direction matters. If you go in the reverse direction, you're going to get the negative version of that. And that's because at any point we take the dot product, you're going in the opposite direction, so it'll be the negative of each other. But when you're dealing with a scalar field, we saw in the last video, that it doesn't matter which direction that you traverse the path in.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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We're doing dt here, we're having du here, but we're going to get the same exact number for any a or b, and given this vector f and the position vector path r of t. So just to summarize everything up, when you're dealing with line integrals over vector fields, the direction matters. If you go in the reverse direction, you're going to get the negative version of that. And that's because at any point we take the dot product, you're going in the opposite direction, so it'll be the negative of each other. But when you're dealing with a scalar field, we saw in the last video, that it doesn't matter which direction that you traverse the path in. That the positive path has the same value as the negative path, and that's just because we're just trying to find the area of that curtain. Hopefully you found that mildly amusing.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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But when you're dealing with a scalar field, we saw in the last video, that it doesn't matter which direction that you traverse the path in. That the positive path has the same value as the negative path, and that's just because we're just trying to find the area of that curtain. Hopefully you found that mildly amusing.
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Vector field line integrals dependent on path direction Multivariable Calculus Khan Academy.mp3
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