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05:28

Felicia S.

If $f(x)=x+\sqrt{2-x}$ and $g(u)=u+\sqrt{2-u},$ is it true that $f=g ?$

02:07

Fangjun Z.

0:00

Jsdfio K.

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All right. So we want to compute the line integral of the specter field over these two pasts. Uh, traversing from the 20.0 toe 11 are connecting those two points where C one is gonna fall. The curve y equals XY two is going to cripple the curve. Y equals X squared. OK, so let's just draw a picture. And so I'm not gonna actually draw the vector field. So just imagine, the vector field is a bunch of arrows. So at every point, I have a vector pushing me somewhere. But I really want to draw. Is this starting point in the end point? Here's 11 and so see, one is going to be along the line. Why equals X? This is C one and then C two is gonna be along the line or the curve. Why equals X squared? So that's gonna be C two. And so this is the set up that we have in mind. I have two points. I have a bunch of different ways. I can take a path between those two points. Well, here's just two of them. Specifically, one is kind of following this line. Why equals X and one of them is kind of following this parabola. Why equals X squared? Okay, so let's just do it. Let's compute the been intervals. So let's do say see one first. Now remember, the most important part of finding a line and a girl is we need to parameter rise the curve. Okay, so I just need a parameter rise. The line segment from 00 thio 11 Well, that's pretty easy. And if you've never done this before, I think you should just watch me do it and then I'll explain kind of my reasoning. So I typically just always pick t to be in the interval 01 and then just suggest. And so the X coordinate is going from 0 to 1 linearly. The Y coordinate is going from 0 to 1 linearly. And so you see that for any as t goes from 0 to 1, my points are just slowly increasing literally up this line between 00 and 11 So here's my parameter realization of the curb and recall that we also need the derivative. So we'll put down the derivative right here and the derivative of this vector value function is just one one like that. Okay. And let's plug everything in so f is minus y squared Comma X. So we have t going from 0 to 1 f I'm going to evaluate at Exit E and F t So we have for F This is going to be minus t square and then X is just gonna be t So again, I'm just plugging in literally what x and y are and for F. And then I do the dot product with the derivative, which is just 11 and GT. All right, so this is just I do the start product. It's t minus T square T T, which is just t squared over two minus t cubed over three evaluated from 0 to 1. That's just one half minus one third, which is 1/6. Okay, so now let's do see too. We'll see. Two is going to be very similar. Except, uh, the curve is why equals X squared? So if X is just t going from 0 to 1, why is going to be t squared? So the privatization, instead of being TT is just going to be t t squared and then the derivative is one to t. All right, so then we do the integral from 0 to 1 of tea t squared dot with one to t d t which equals the integral from 0 to 1 of looks like t Hope. You know, I forgot to dio whips getting out of myself here. Okay, so the first thing I have to do is I have to evaluate the vector valued function at thes points. So why is t squared? So this is gonna be minus t to the fourth. And an ex is just gonna be t and then I have one to t. Here we go. Almost messed up. Okay, so I do this stop product and I have something like the integral from 0 to 1 of two. T squared minus t to the fourth t t, which is t minus t to the five over. Fit over five. Evaluated from 01 That's just one minus 1/5. 4/5. But notice I took two different paths connecting the same two points. And when I did the line integral over those two different paths of the Spectra field, I got two different answers. So somehow along see one. The Force field does less work than along C two, which means along See one. I probably have to having to do more work myself. So, in other words, this field can't be conservative, so f is not conservatives. That's what we can conclude about the vector function. I'm sorry. The vector Field F is that it's not conservative.

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