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Let's apply what we learned in the last video into a concrete

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example of the work done by a vector field on something

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going through some type of path through the field.

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So let's say that I have a vector field.

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It's defined over r2 for the x-y plane.

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So it's a function of x and y.

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It associates a vector with every point on the plane.

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And let's say my vector field is y times the unit vector i

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minus x times the unit vector j.

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And so you can imagine if we were to draw-- let's

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draw our x- and y-axes.

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I'll do it over here.

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If we were to draw our x- and y-axes, this associates a

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vector, a force vector-- let's say this is actually a

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force vector-- with every point in our x-y plane.

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So this is x and this is y.

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So if we're at the point, for example, 1, 0, what will the

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vector look like that's associated with that point?

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Well, at 1, 0, y is 0, so this will be 0, i minus 1, j.

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Minus 1, j looks like this.

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So minus 1, j will look like that.

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At x is equal to 2-- I'm just picking points at random, ones

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that'll be -- y is still 0, and now the force vector here

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would be minus 2, j.

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So it would look something like this.

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Minus 2, j.

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Something like that.

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Likewise, if we were to go here, where y is equal to 1 and

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x is equal to is 0, when y is equal to 1, we have 1, i minus

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0, j, so then our vector is going to look like

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that at this point.

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If we're to go to 2-- you could get the picture.

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You can keep plotting these points.

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You just want to get a sense of what it looks like.

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If you go here, the vector's going to look like that.

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If you go maybe at this point right here, the vector's

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going to look like that.

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I think you get the general idea.

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I could keep filling in the space for this

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entire field all over.

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You know, just to make it symmetric, if I was here,

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the vector is going to look like that.

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You get the idea.

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I could just fill in all of the points if I had to.

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Now, in that field, I have some particle moving, and let's say

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its path is described by the curve c, and the

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parameterization of it is x of t is equal to cosine of t, and

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y of t is equal to sine of t.

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And the path will occur from t-- let's say, 0 is less than

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or equal to t is less than or equal to 2pi.

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You might already recognize what this would be.

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This parameterization is essentially a

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counterclockwise circle.

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So the path that this guy is going to go is

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going to start here.

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Well, you can imagine, t in this case, you could almost

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imagine is just the angle of the circle, but you can

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also imagine t is time.

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So at time equals 0, we're going to be over here.

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Then at time of pi over 2, we're going to have traveled a

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quarter of the circle to there, so we're moving in

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that direction.

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And then at time after pi seconds, we would have

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gotten right there.

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And then all the way after 2pi seconds, we would have gotten

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all the way around the circle.

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So our path, our curve, is one counterclockwise rotation

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around the circle, so to speak.

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So what is the work done by this field on this curve?

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So the work done.

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So the work, we learned in the previous video, is equal to the

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line integral over this contour of our field, of our vector

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field, dotted with the differential of our movement,

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so dotted with the differential of our movement dr.

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Well, I haven't even defined r yet.

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I mean, I kind of have just the parameterization here, so we

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need to have a vector function.

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We need to have some r that defines this path.

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This is just a standard parameterization, but if I

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wanted to write it as a vector function of t, we would write

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that r of t is equal to x of t, which is cosine of t times i

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plus y of t times j, which is just sine of t times j.

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And likewise, this is for 0 is less than or equal to t, which

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is less than or equal to 2pi.

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And this are equivalent.

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The reason why I took the pain of doing this is so now I can

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take its vector function derivative, and can figure out

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its differential, and then I can actually take the dot

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product with this thing over here.

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So let's do all of that and actually calculate this line

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integral and figure out the work done by this field.

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One thing might already pop in your mind.

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We're going in a counterclockwise direction, but

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at every point where we're passing through, it looks like

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the field is going exactly opposite the direction

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of our motion.

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For example, here we're moving upwards.

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The field is pulling us backwards.

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Here we're moving to the top left.

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The field is moving us to the bottom right.

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Here we're moving exactly to the left.

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The field is pulling us to the right.

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So it looks like the field is always doing the exact opposite

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of what we're trying to do.

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It's hindering our ability to move.

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So I'll give you a little intuition.

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This'll probably deal with negative work.

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For example, if I lift something off the ground,

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I have to apply force to fight gravity.

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I'm doing positive work, but gravity's doing

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negative work on that.

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We're just going to do the math here just to make you

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comfortable with this idea, but it's interesting to think about

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what's exactly going on even here.

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The field is-- the field I'm doing in that pink color,

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so let me stick to that.

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The field is pushing in that direction, so it's always

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going opposite the motion.

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But let's just do the math to make everything in the last

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video a little bit more concrete.

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So a good place to start is the derivative of our

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position vector function with respect to t.

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So we have a dr/dt, which we could also write

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as r prime of t.

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This is equal to the derivative of x of t with respect to t,

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which is minus sine of t times i plus the derivative of y

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of t with respect to t.

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Derivative of sine of t is just cosine of t.

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Cosine of t times j.

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And if we want the differential, we just multiply

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everything times dt, so we get to dr is equal to-- we could

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write it this way.

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We could actually even just put the d-- well,

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let me just do it.

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So it's minus sine of t dt-- I'm just multiplying each of

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these terms by dt, distributive property-- times the unit

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vector i plus cosine of t dt times the unit vector j.

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So we have this piece now.

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And now we want to take the dot product with this over here,

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but let me rewrite our vector field in terms of in

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terms of t, so to speak.

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So what's our field going to be doing at any point t?

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We don't have to worry about every point.

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We don't have to worry, for example, that over here the

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vector field is going to be doing something like that

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because that's not on our path.

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That force never had an impact on the particle.

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We only care about what happens along our path.

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So we can find a function that we can essentially substitute y

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and x for, their relative functions with respect to t,

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and then we'll have the force from the field at any

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point or any time t.

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So let's do that.

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So this guy right here, if I were to write it as a function

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of t, this is going to be equal to y of t, right? y is a

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function of t, so it's sine of t, right? that's that.

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Sine of t times i plus-- or actually minus x, or x of

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t. x is a function of t.

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So minus cosine of t times j.

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And now all of it seems a little bit more

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straightforward.

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If we want to find this line integral, this line integral is

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going to be the same thing as the integral-- let me pick

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a nice, soothing color.

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Maybe this is a nice one.

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The integral from t is equal to 0 to t is equal

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to 2pi of f dot dr.

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Now, when you take the dot product, you just multiply the

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corresponding components, and add it up.

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So we take the product of the minus sign and the sine of t--

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or the sine of t with the minus sine of t dt, I get-- you're

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going to get minus sine squared t dt, and then you're going

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to add that to-- so you're going to have that plus.

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Let me write that dt a little bit.

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That was a wacky-looking dt.

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dt, and then you're going to have that plus these two guys

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multiplied by each other.

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So that's-- well, there's a minus to sign here so plus.

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Let me just change this to a minus.

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Minus cosine squared dt.

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And if we factor out a minus sign and a dt, what is

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this going to be equal to?

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This is going to be equal to the integral from 0 to 2pi of,

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we could say, sine squared plus-- I want to put the t --

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sine squared of t plus cosine squared of t.

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And actually, let me take the minus sign out to the front.

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So if we just factor the minus sign, and put a minus

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there, make this a plus.

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So the minus sign out there, and then we factor dt out.

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I did a couple of steps in there, but I think you got it.

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Now this is just algebra at this point.

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Factoring out a minus sign, so this becomes positive.

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And then you have a dt and a dt.

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Factor that out, and you get this.

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You could multiply this out and you'd get what we

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originally have, if that confuses you at all.

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And the reason why I did that: we know what sine squared of

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anything plus cosine squared of that same anything is.

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That falls right out of the unit circle definition of

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our trig function, so this is just 1.

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So our whole integral has been reduced to the minus integral

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from 0 to 2pi of dt.

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And this is-- we have seen this before.

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We can probably say that this is of 1, if you want to

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put something there.

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Then the antiderivative of 1 is just-- so this is just going to

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be equal to minus-- and that minus sign is just the same

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minus sign that we're carrying forward.

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The antiderivative of 1 is just t, and we're going to evaluate

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it from 2pi to 0, or from 0 to 2pi, so this is equal to

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minus-- that minus sign right there-- 2pi minus t

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at 0, so minus 0.

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So this is just equal to minus 2pi.

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And there you have it.

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We figured out the work that this field did on the particle,

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or whatever, whatever thing was moving around in this

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counterclockwise fashion.

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And our intuition held up.

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We actually got a negative number for the work done.

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And that's because, at all times, the field was actually

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going exactly opposite, or was actually opposing, the movement

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of, if we think of it as a particle in its

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counterclockwise direction.

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Anyway, hopefully, you found that helpful.