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Missouri State University

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

01:34

Scott N.

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

00:51

Heather Z.

01:59

Nutan C.

00:33

Dungarsinh P.

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All right, So now we have a line integral over a vector field. So we're given a vector field and were given this oriented curve given by this premature ization. And so recall that the work done by a force field on a particle moving along the path is Well, I add up all of the work elements, okay. And what is a small work element? Well, I take the component of the vector field that lies in the tangential direction or in the direction of motion, and then times a small little change along the tangential direction like that. Okay. And that's along the curve. Of course, that's given by this premature ization. And since we have a premature ization, we can actually write this out as the integral from 0 to 1 of f evaluated are eso recalled? That means I'm just going thio plug in x of T y t and Z F T for X, y and z. And then we take the dot product with the derivative and then DT cool. So let's write down with all of this is first of all, So here's our integral What is f of our team? Well, we're just gonna plug in X of T and for X y t and for y and Z t in for Z. So this is equal to you co sign of X, which is t cubed then sign of why which is t squared, then t Okay, so then why times e Why times e is t cute. Okay. And then the derivative prime a team iss What? Let's see three t squared and then to t and one. All right, so then what is f are of tea dot Our prime a t. Well, this is three t squared co sign t cubed plus two team times sign T squared and then plus ticket. So this is the function that were integrating from 0 to 1. So the work done is the integral from 0 to 1 of this function three t squared, assigned cute plus two t sign of t square plus t. Cute. All right, so we're actually really lucky here because if you notice if I want to take an anti derivative of this Well, this three t squared is the derivative of t cute. So the anti derivative of three t squared co sinti cubed is just sign of T. Cute and similarly known anti derivative here is just negative. Co sign t squared, Just noticed. Just take the derivative. It's set up nicely to do use substitution and then an anti derivative. Here is T to the fourth over four, and we're evaluating this from 0 to 1. It's a sign of one cube that's just sign of one. Okay, and then minus sign of zero is zero. Then we have minus co sign of one so minus co sign of one square than a minus minus co sign of zero. So plus co sign of zero, which is one and then t to the fourth over. Four, evaluated from 0 to 1 is just 1/4. So whatever this number is, that's the work done by this force field on the particle moving along this path

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