 In this video, we provide the solution to question number seven for practice exam number three for math 1220 We have to set up and simplify the integral that will measure the length of the parametric curve x equals e to the t cosine of t y equals e to the t sine of t as t ranges from Zero to pi. We have to set up the integral. We do not have to evaluate it. So we're looking for arc length So we're going to use our standard arc length formula the integral of ds Now in this situation because we have our parametric function, we can adapt it Because the general formula right is you take the square root of dx squared plus dy squared For which this actually suggests as we modify this into Parametric form we're going to take the derivative of x there. So e to the t cosine t Then we have to square it of course We also are going to take the derivative of the y e to the t sine of t Squared this all sits inside of the square root We're going to integrate with respect to t and then told us the bounds are going to go from zero to pi in that situation So we've now set up the integral but the instructions do specifically say we have to simplify it So my strategy is going to um, well I have a I have a I have to take the derivative of these things here by the product rule You're going to get a sum of a couple things. Let's take a look at that with the first one We take the derivative of e t cosine of t By the product rule you're going to get two terms there the derivative of e t is itself So you get e t cosine of t Then when you take the derivative of cosine you're going to get a negative sign So you get negative e to the t sine of t We have to square both of these things Now when you take the derivative of e t sine you're going to get something very very similar The derivative of e to the t is again e to the t Uh, then you're going to get the derivative when you take a sign you're going to get e to the t cosine of t Square that this all sits inside of the Square root dt. So we're going to foil each of those things Uh, so foiling the first one, um, you're going to end up with an e to the 2t cosine squared t Uh, you're going to get negative e to the 2t Uh, cosine of t sine of t And then for the next one, they're going to get a positive e to the 2t sine squared t Uh, for which, well, we're not quite done yet I do want to note here that we have a cosine squared and a sine squared That's going to combine together to give you just a cosine squared plus sine squared equal one So since they both are divisible by e to the 2t or I'm just going to e to the 2t that e to the 2t there Uh, but then for the next one when you foil that out, you're actually going to get something very similar You're going to e to the 2t uh sine squared You're going to get a plus uh two times and I forgot to write a two over here There should be two there. Um, you're going to get a two plus two e to the 2t Sign of t cosine of t and then you're going to get a e to the 2t cosine squared Of t this thing looks huge and so all of this is under the square root It looks massive, but it's going to simplify super nice So also notice we have another e to the 2t sine squared We have an e to the 2t cosine squared those will likewise combine together When you add them together, it's sine squared plus cosine squared is equal to one I also want to point your attention to the fact that we have a negative two e to the 2t cosine t sine t But we have a positive two e to the 2t sine t cosine t those things are opposite But uh, but negative one's positive one's negative. So they're actually going to cancel each other out And so when you simplify all of this stuff Right, this thing that's going to come I'll put it up here. I'm going to have space It looks way more scary than it really is you're going to end up with a e to the 2t from the this one right here The blue parts cancelled out then you're going to get an e to the 2t Right here dt For which if you combine those together You're going to get the square root of 2 times e to the 2t And notice, of course e to the 2t is just e to the t squared And so when you simplify this thing you actually end up with the square root of 2 times the integral from 0 to pi Of e to the t dt you do not need to evaluate the integral But it's like that's not a hard evaluation You just have to set it up and simplify it and as crazy as it looked it actually simplifies Very very nicely so much that I almost want to evaluate it But we won't because we won't get any credit for doing that here