WEBVTT mathematics/multivariable-calculus/hovasapian
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Hello, and welcome back to educator.com and multivariable calculus.
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Today we are going to continue our discussion of line integrals, so, let us just jump right on in.
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We are going to talk about notation, and as it turns out, you know, the line integral notation, there are several different notations for it.
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So, I am going to introduce a pretty standard notation and it is going to be the one we end up using more and more especially when we end up talking about double integrals and Green's theorem later on, which is the fundamental theorem of calculus in two dimensions.
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So, let us introduce this notation, so that we are aware of it. Actually, you know what, I think I am going to do this lesson in blue. There we go. Okay.
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So, notation. Okay. Let f(x,y) = f and g. So, this is a function. It is a function of 2 variables, it is a vector field. This is the first coordinate function, this is the second coordinate function, but notice I have not included the f(x,y), g(x,y), but it is implicit there.
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We are going to start to use a slightly shortened notation, and c(t), the curve over which we integrate, or the line, is going to be some x of t, so x is a function of t and y is a function of t. Something like that.
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Therefore, then the integral of f along c is equal to... we write it as f dx + g dy. That is the notation right here. fdx + gdy. Okay.
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This just -- the integrand -- is just fg ⋅ dx/dy, so it actually has the advantage of sort of representing some dot product already.
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Now, it is very, very important to remember of course that when we use notations, these are short-hand, so fdx + gdy is short for the following: let me write it up here, well, no, let me write it down here. It is short for f(x(t),y(t)) × dx/dt + g(x(t),y(t)) dy/dt dt.
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When we multiply all of this out, this fdx + gdy is actually just short for that, which is equivalent to our more theoretical definition f(c(t)) ⋅ c'(t). Again, this is just a shorthand notation.
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This is what it is when it is fully expanded, so, once again I am going to write this... so, again, the integral of f along c which is equal to the integral of f(c(t)) ⋅ dc/dt, c', dc/dt, just different notations... dt, forgot that little dt there... is equal to fdx + gdy.
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This is the notation that is really, really important, but again it is just a question of notation as long as you remember what this means. This believe it or not is actually the important one. It is the definition of the line integral so that is the one that we are going to keep using.
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We will use this one more and more as we move towards Green's Theorem. Okay.
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Often, a curve will be given in a non-parameterized way, so, a non-parameterized way.
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In other words, they might say integrate this function or this vector field over the parabola as the parabola moves from this point to this point.
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Well, you are going to have to come up with a parameterization because the line integral... the definition of the line integral involves the parameterization c(t).
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So, you are going to have to find a way to take that curve -- which is expressed usually as a function of (x,y) equals some function of x -- and find a way to express it as a single parameter t.
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In general, for y = f(x), let x = t and y = f(t).
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This way, you will always get the parameterization that you need. Now that does not necessarily mean that this is the only parameterization, that is the thing about parameterizations. There is usually more than the one.
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The nice thing about it is that the integral itself is actually independent of the parameterization. Some parameterizations will make the problem easier, some will make it harder. There is more than one parameterization, so those are not unique.
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Okay. Just some general examples for the circle of radius r, x = rcos(t) and y = rsin(t). That is the parameterization for a circle. For a straight line segment between p and q, between the points p and q, the parameterization is as follow: c(t) = the point p + t × q - p.
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We will actually see an application of this a little bit later -- where t runs from 0 to 1.
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Again, the integral... this is the most important part... the integral is independent of the parameterization.
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You know what, I do not like this capital letters, I will just stick to writing it out -- is independent of the parameterization chosen.
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Profoundly important. So, two different parameterizations, same curve, same direction, you are going to get the same integral.
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Now, when you choose a parameterization, you are specifying x a direction of traversal.
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We will say more about this in the next lesson when we actually talk about reversing the path, integrating in one direction, integrating along another direction. You are specifying a direction of traversal.
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In other words, if you were to parameterize that curve moving in this direction from here to here would be one parameterization, but if you wanted to move in this direction, if you wanted to move like this from here to here, it actually changes the parameterization.
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There is a formal mathematical way of doing that and again we will see that in the next lesson when we see some examples, but for the most part what is important is that the parameterization that you use does not matter, the integral will be the same, that is what is important.
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Let us do an example. Okay, so example 1. Find the integral of the vector field f(x,y) = (xy²,x²y) on the ellipse centered at the origin with major radius 4 and minor radius 2 from θ = π/6 to θ = 3π/4.
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Okay, so find the integral of the function on the ellipse, so we are integrating along the ellipse, centered at the origin, the major radius is 4 the minor radius is 2 and we want to go from the point on the ellipse represented by π/6 to 3π/4.
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Let us go ahead and draw what this looks like, and then we will do the parameterization of the ellipse.
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We have that, major radius is 4 so this distance from there to there is 4, minor radius is 2, so something like this -- sort of like that, I am not the best ellipse drawer but something like this -- and θ = π/6, that means this point and 3π/4 is the 45 degree angle to this point.
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What we are doing is we are integrating from here to here, so let me do this in red. We are integrating along this path. Here to here. This function.
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Let us go ahead and write our, so we have f(x), let us go ahead and write our c(t). As it turns out, you parameterize an ellipse sort of the same way you parameterize a circle.
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You have rcos(t),rsin(t) but now the two radii are different, so it is just going to be this radius cos(t), and this radius sin(t).
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What we have is 4cos(t) and 2sin(t), now, this particular parameterization, by choosing this parameterization I am actually moving in this direction.
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This is what we mean when we say any time you choose a parameterization you are specifying a direction. I am specifying a counterclockwise movement along this path.
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If I had gone, let us say this way, let us say I wanted to parameterize from this point to this point, the parameterization would actually have to change.
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So, when you choose a parameterization, you are actually specifying a particular direction and it is very important that you know what that direction is.
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You have to make sure that you understand all of the particulars of the objects that you are working with. Okay, so let us go ahead and flip the page. I am going to rewrite the parameterization of the function and then I am going to start working with it and integrate it.
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Let me go back to blue here. So, we have f(x,y) = (xy²,x²y), and c(t) = 4cos(t)2sin(t), and t is going to run from π/6 to 3π/4.
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Okay. Let us form, as always, we form f(c(t)) ⋅ dc/dt. O
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Okay, so, f(c(t)) = let us put -- this is x, this is y -- we put those in here, we get 4cos(t),sin²(t)... uh, no, sorry, that is not right... 4cos(t) × 2sin(t)², and we get 4cos(t)² × 2sin(t).
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This is going to equal 2 × 2 is 4, this is going to be 16 × cos(t) sin²(t), and this is going to be 32cos²(t)sin(t).
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Okay. Now we have that. Now let us form dc/dt. dc/dt = the derivative of this thing equals -4 × sin(t) and we have 2 × cos(t).
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Now, when we form the f(c(t)) ⋅ dc/dt, it is this × that + this × that. We end up with -64cos(t)sin³(t) + this × that which is 64cos³(t)sin(t).
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So, our integral of f over c is equal to the integral from π/6 to 3π/4 of this expression, -64cos(t)sin³(t) + 64cos³(t)sin(t) dt.
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When we use our mathematical software, we end up with the number 2. That is it. I am certainly hoping that you are not trying to do this manually, it is not worth it.
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One of the nice things about having -- in fact, the best hting about having math software is now you are no longer confined to examples that simply for logistical reasons are simple so that you can get through them.
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Now you can choose any function that you want, any parameterization, and boom software gives you the answer. This is what is important, being able to set this up. This a computer can do. This is what you want to understand.
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So let us go ahead and do another example here. A very, very important example.
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So, example 2. Let us see here. Find the integral of the vector field f(x,y) = -y/x² + y² x/x² + y², around the circle of radius 4 from (4,0) to the point (2,2sqrt(3)).
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In this particular case, they are actually giving us two specific points, from this point to this point.
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So, what we are going to have to do is, in order to find the parameterization, we are going to have to find the lower and upper limits of integration.
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We are going to have to convert this point and this point to t values based on the parameterization that we choose. Okay.
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Let us go ahead and do a little geometry here just to see what this looks like, so that we make sure we understand what we are doing.
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So, we have this circle of radius 4, right? This is 4, that is 4, we know how to parameterize that, that is going to be 4cos(t), 4sin(t).
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What we are doing is we are going from this point, the point (4,0) all the way to this point right here (2,2sqrt(3)) and we are moving in this direction, that direction, that is the parameterization.
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Our c(t) is going to be the 4cos(t), 4sin(t)... now, when we do the integration, we need values of t because we are integrating along the curve so the t values are going to be the lower and upper limits of integration.
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So, this point is represented by t = 0, so at the point (4,0), t = 0, so that is our lower limit of integration. Well, how about here, what is t going to be when we get to here?
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Well, at the point (2, 2sqrt(3)), well, let us go ahead and solve, 4cos(t) is the x value here, this is the y value, let us just go ahead and solve for t.
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So, 4cos(t) = 2, cos(t) = 2/4 = 1/2, and since we are in the first quadrant, t = π/3.
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So, we are going to be integrating from 0 to π/3, that is our lower limit of integration, that is our upper limit of integration.
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Notice that in this case they did not give us the parameterization and the t values that we are going to traverse, they gave us 2 points and they gave us just the geometric on this unit circle.
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We came up with a parameterization, that should not be a problem. We need to find out what the t value is so we have to evaluate one of these to find the t values.
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t goes to 0 as t moves up to π/3, that covers that segment of the circle.
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Okay. So, t is going to run from 0 all the way to π/3. Alright, let us go ahead and solve the integral. Well, f(c(t)) when I put this parameterization into there, I end up with the following.
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I end up with -4sin(t)/16cos²(t) + 16sin²(t), and I end up with 4cos(t)/16cos²(t) + 16sin²(t).
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Okay. This of course is equal to -4sin(t)/16, cos²(t), sin²(t) = 1, so that factors out, and I have 4cos(t)/16. Let us get rid of these random little lines that tend to show up at the bottom of the page.
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This is going to equal sin(t) - sin(t)/4, and... I am sorry, -4... oh, this is ridiculous, let us see... -4sin(t)/4, and 4... I have already cancelled, now I am the one that is messing up arithmetic. 4 and 16, 4 and 16, so I get -sin(t)/4 and I get cos(t)/4.
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How is that? That is good. Okay. Now, let us do dc/dt.
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So, dc/dt, or c'(t), that is going to equal -4sin(t) and 4cos(t), so, f(c(t)) ⋅ dc/dt is going to equal -- let me rewrite the f(c(t)) again -- f(c(t)) is going to equal (-sin(t)/4,cos(t)/4), so it is the dot product of this thing and this thing.
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When I do that, I am going to get sin²(t) + cos²(t) and that is going to equal 1, so our final integral, the integral of f over this particular curve is going to be the integral from 0 to π/3 of just plain old dt, 1dt.
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Of course, that equals π/3.
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Here we are busy trying to solve this line integral, and we end up with notice, 0 to π/3 and we end up with an answer for this integral is π/3. This is not a coincidence.
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Okay. This is a very, very important result and a very important vector field, especially for those of you in physics and engineering. You are going to see this vector field over and over and over again.
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Let us just discuss this real quickly. So, this is a very important result and a very important vector field.
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Here is the result. Here is why it is important: if you ever integrate this vector field along any curve, along any... and this one I definitely will put in capitals... along any curve, from point p1 to p2, and op1 and op2 make angles of θ1 and θ2, respectively, with the x-axis. I will draw this out in just a minute so that you see what happens... respectively with the x-axis, then the integral of f along that particular curve is equal to θ2 - θ1, or Δθ.
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Here is what we mean by that. So, let us say we have this coordinate plane and let us say we have this curve that is going like this, okay?
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If this is point p1 and this is point p2, p1 and p2, if this is θ1 and this is -- this angle is θ1 and this angle is θ2, remember? we always measure our θs from the positive x-axis moving in a counterclockwise direction.
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Then the integral of this particular vector field, f(x,y) -y/x² + y², x/x² + y², the integral of this vector field over any curve, no matter what curve you trace, whether it is like this or this or it could be all whacky and end up here, the difference in the angle, the integral of this vector field is always going to be the difference between θ2 and θ1, no matter what it is.
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It does not have to start at the origin by the way, notice here this is one angle, this is the other angle. In our example, we start from the origin and we integrate up to π/3.
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The angle difference was π/3, the integral was π/3. Here, same thing. If this were 3π/4 and this were π/4, well, we would have 3π/4 - π/4, 2π/4, the integral would be π/2.
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This is a profoundly important vector field. You will see it over and over and over again.
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It will save you a lot of time if you recognize that as you integrate this vector field over absolutely any path, where this angle arrangement is involved, you can just take the different of the angles and that is the integral.
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Thank you for joining us here at educator.com, we will see you next time for a further discussion of line integrals. Take care, bye-bye.