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Problem 105 Medium Difficulty

[T] Evaluate line integral $\int_{C} \mathbf{F} \cdot d \mathbf{r},$ where $\mathbf{F}(x, y)=\left(e^{x} \sin y-y\right) \mathbf{i}+\left(e^{x} \cos y-x 2\right) \mathbf{j}, \quad$ and $C$ is the path given by $r(t)=\left[t^{3} \sin \frac{\pi t}{2}\right] \mathbf{i}-\left[\frac{\pi}{2} \cos \left(\frac{\pi t}{2}+\frac{\pi}{2}\right)\right] \mathbf{j}$ for $0 \leq t \leq 1$

Answer

$-1.9941$

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Video Transcript

Okay, This question wants us to evaluate this line into crow. So to start, we want to check if we have a conservative field. So we're looking for If Dick UT X is equal to DPD y and the extra relative of the Y component, looking at our vector field would just be eating the X co sign. Why then, minus one. Then for the wide derivative of the X component, we get eat of the axe. The sign turns into a co sign and then minus one. And I wrote that down wrong. This should be a co sign. And now we see we do indeed have a conservative field. So f is equal to the Grady INTs of some potential function. So to go about finding that we know that AF is equal to an ex derivative and a wide derivative. So we're just going to see all the information we can recover. So our first contribution to the potential will come from the anti derivative of the X component. So we'll integrate heat of the X sign y minus y with respect to X, giving us each of the X sign. Why minus x y, plus some function of why then, for our second contribution to the potential, we integrate the why component. So eat a vax co sign Why minus X minus two and this gives us e to the X sign. Why minus X Y minus two. Why plus f of X And as we can see here, we'll pick up and eat it The x sign y a minus X y and a negative two y plus some constant. So our potential up to a constant would be Phi X Y is equal to e to the X Times sign Why minus X y minus two way. So now we can use this to evaluate Are lining to go because we know the integral oversee of f dot de are If we have a Grady Int, it would just be sigh of the end minus phi of the beginning, so we should probably find these ending points and starting points. So it actually gives us the curve that were integrating over so we can just use that. So it says we start at time equals zero. So we're just going to find where our path starts, which is r of zero. So we get zero for the X component because zero times sign 00 and then we get hi over to co sign of pirate, too. So, again, zero and then we're ending when t is equal to one. So if that's the case, then we end at one time. Sign pi over two. So one and then plugging in one for the y component. We get pi over two times co sign of Hi. So we end up getting negative one. But there's a negative sign for that whole term. So this just turns into positive pi over two. So we're just going to evaluate fi of one comma pirate too, and then subtract that with Fi at 00 And if we look at our potential again, this gives us e times signed pirates, You minus pi over two, then minus pie and then subtracting are lower bounds. We get zero because signed 00 minus zero minus zero. So that whole lower bound does not matter. So our final answer would be yeah, minus high over to minus pi or e minus three pi over two. Or, if you want to simplify that a bit more, this is approximately negative. 1.9941 Neither is acceptable

University of Michigan - Ann Arbor
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