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Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

Idaho State University

00:56

Felicia S.

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

01:59

Nutan C.

01:02

Anshu R.

0:00

Simon E.

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All right, This is a trick question if I've ever seen one. So we have a function f in the grating of that function is this factor Field capital F. And we actually know exactly what f iss. So using our language that we've developed in this lecture little F lower case death is a potential function or anti derivative of the vector field capital F. Okay, so keep that in the back of your mind. Now, I want to compute the line in a girl of the backer field over this curve, see, connecting the points 000 and 0 10, 10. And the curve is given by this factor value function. That is a mess. Now, you could do this. Well, maybe you could do this. I don't know. You might run into some serious trouble. You could find the derivative you could compose. Um, let's take So what can we do? So we could actually take the Grady int of f to get our vector field capital f evaluate at, you know, the ex except e y z f t etcetera. Do all of that? That's gonna be a mess, though. This is an example of where we can use the fundamental The're, um just like when you were doing a riemann some it was just so much work. But then you learn Oh, wait, If I can find an anti derivative, it's a walk in the park. That's the same thing here. Fundamental Ethereum for line integral. So what does it say? It says that if I integrate the Grady int of a function has the line integral over a curve seat, so die with our prime a t b t. This is equal to f of starting point. So here, let me say that the curve goes from A to B. So it's just f evaluated at our of B to the end point minus f of thes starting point, which is our of a. So let's apply that here. So in our case, F is this function. Here are potential function and a zero this point here 000 and be the point we get 0 10, 10. So, actually, our line in a girl is just f of 0 10 10 minus f of zero zero Europe f of 0 10 10 is well, that's gonna be co sign of zero so one plus 10 square times tense. That's 1000 and then minus F of 000 But if I plug in 000 I have co sign of zero, which is one plus zero, so minus one. So our final answer is 1000. So this is a trick question, but it's in a very important illustration that when you have a potential function, it doesn't matter how complicated of a path your particle is moving along. It on Lee depends the value of the line integral or the work done physically. Onley depends on the end points, so I could just exactly easy as it gets.

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