 Welcome back to our final video for lecture 34 in our series and I want to talk about how arc length can be Interacted with a polar curve. How do you find the length of a polar curve? Well, we're gonna use the arc length formula that we saw before S equals the integral of ds where we're using the fact that ds is the square root of dx squared plus dy squared Now when we dealt with parametric functions We showed that we can modify this for a parametric setting and we do have a parametric setting x equals r cosine theta and y equals r sine theta and so if we take derivatives of these things appropriately we can factor out a d theta and So factor out the d theta you're gonna get the integral of the square root of dx over d theta squared plus dy over d theta squared That's a theta believe me on that one And then there's a d theta on the outside So we can adapt this for for a d theta We just have to take the derivative of x with respect to theta and the derivative of y with respect to theta now Remember since r is itself a function theta the derivative of r with respect to theta will be involved here And so taking the derivatives of x and y with respect to theta will give the following The derivative of x since we have r times cos I mean you use the product rule in which case We're gonna get an r prime times cosine theta plus. I guess I should say minus Minus r times sine theta That's what happens when you take dx over d theta squared Then if you do that for for dy over d theta you're gonna get an r prime sine theta And you're gonna get plus r times cosine theta Like so now this is just the derivatives of those functions with respect to theta This all sits inside of the d theta or the square root now you have to foil this things out This it might seem complicated, but this is actually a Saving grace for us when you foil up the first one you're gonna r prime Squared and like usual r prime here is short for dr over d theta So you're gonna get an r prime squared cosine squared theta you're gonna get a minus two r prime r cosine sine And then you're gonna get a positive r squared sine squared theta Sounds like a lot, but that's just the first one foiled out The next one foiled out you're gonna get some stuff that's strikingly similar. You're gonna get an r prime squared sine squared You're going to get a positive to r prime r Sine theta cosine theta and then lastly you're going to get an r an r squared cosine squared theta This is the second piece now this all sits in inside of a square root and there's a d theta associated to that Now there's gonna be some nice combinations going on here. So notice some things So if you take the first one r prime squared cosine squared r prime squared sine squared Well, I forgot to write this right the square there by the Pythagorean identity those will combine just to be an r prime Squared because cosine squared plus sine squared equal to one for the next one You have a negative to r prime r cosine theta sine theta You also have a positive to r prime r sine theta cosine theta when those add together There's gonna cancel out and then lastly you're going to get a R squared sine squared an r squared cosine squared sine square plus cosine squared is again one Those will combine together just to be an r squared This sits inside the square root you're integrating with respect to theta and Wallah the arc length formula actually simplifies very nicely for the polar form the arc length of a polar curve It's gonna be the integral of the square root of oops. Sorry. That was the original one Let's see. I think I have it hidden somewhere on the next slide perhaps. Here we go The derivative are the integral The the arc length is gonna be the integral of the square root of the derivative of r squared plus r squared d theta This is a nice formula. We derived it using parametric equations, but this one's probably worth Memorizing our writing in a very convenient place So if you want to find the circumference of a cardioid, right? Remember the cardioid is This lima being like shape the lima noid That's not a word. I just made it up. I'm sorry We want to find the circumference of a cardioid. How far around is it once? So to find the arc length We have to integrate this thing if we want to get the circumference we have to do one rotation around the cardioid So we go from 0 to 2 pi Which is what we could do right there Now for reasons that'll be a little bit clearer or later I actually don't want to do 0 to 2 pi because 0 to 2 pi would be going from this point all the way around once Which is fine? 0 and 2 pi what I actually instead want to do is I want to go from this point right here So this would be negative pi halves to 3 pi halves So that's actually the direction we're going to go It doesn't make much of a difference but it's going to lead to a slightly easier integral So a little bit of foresight can be helpful in that regard And I'll tell you in a moment just why we're doing that So by the formula we have we have to take the square root of r prime Squared so take the derivative of 1 plus sine theta with respect to theta It's gonna be a cosine theta or squaring that and then we have to square r, which is 1 plus sine theta Like so d theta So the reason that we want to go from negative pi halves to 3 pi halves is to utilize symmetry, right? If you cut along the y-axis the cardioid symmetric with respect to that So the circumference of half of it gives you Exactly half of the circumference. So we're actually going to simplify Our calculation we're going to do s is equal to two times the integral from negative pi halves to pi halves So that's sort of a simplification. We got we don't have a line of symmetry that goes through zero And which is why we stepped away from it Symmetry is a great thing when you can find it and you should use it now looking inside of the radical here We have a cosine squared foil out the 1 plus sine squared. You're going to get 1 plus 2 sine Theta plus sine squared theta And this is all d theta the cosine squared combines with the sine squared to give us a 1 So that's equal to a 1 for which we can then add that one With the one that's already there. So we get 1 plus 1, which is 2 So what we're looking at now, we have two times integral from negative pi halves to pi halves We have the square root of 2 plus 2 sine theta D theta and so now we have to make a judgment on how are we going to proceed from here Well, you could factor out the 2 That's inside of the radical you get 2 times the square root of 2 as you integrate from negative pi halves to pi halves Take the square root of 1 plus sine theta here d theta Now this kind of just kicks the can down the road a little bit We still have to deal with this square root of 1 plus sine theta and there are some options We could try some type of trigometric substitution, but it's already a trig function. That seems like you'd have limited benefit We could try some type of rationalizing substitution That would also work, but it turns out that really the best technique for us is just to use the right trig identity That's always the thing for us If we were to take the square root of 1 minus sine theta That's actually something we really want on the top you have to divide by it to compensate for it And so notice what happens here if we do this a few times by square root of 1 minus sine theta on the top on the top We end up with We foil that thing out. You're gonna get the square root of 1 minus sine squared theta d theta You get this on the bottom. How did this help at all? Well notice 1 minus sine squared is the same thing as a cosine squared And the square root of a cosine squared gives you a cosine And so what you see here this two root two integral from negative pi halves to pi halves You're gonna end up with a cosine theta d theta Over the square root of 1 minus sine theta And so the reason why this was such a nice move here is because now we're in a situation where u substitution Is going to save us here and so that again with a trig identity substitution makes us a lot cleaner Take u to be 1 minus sine theta Then du equals minus cosine theta d theta We'll put a negative sign to compensate for that Since we're doing u substitution, we'll switch the bounds theta will switch to u So you have pi halves and negative pi halves If you suck if you stick pi halves in for theta You're going to get 1 minus 1, which is a zero when you stick in negative pi halves You should get 1 minus a negative one, which is a 2 That puts things in upside down order, but since there's a negative sign We can correct that by switching the order and we're going to end up with two root two The integral from zero to two And then we end up with u to the negative one half d theta Isn't that awesome the right u substitution can always save us And so by the power rule we end up with Uh, well the power will go up by one So we're going to get u to the one half power. We have to divide by one half But really divided by one half is the same thing. It's just times in by two And then we have to go from Zero to two right here. So we end up with two times two Which is four four times a square root of two And we're going to get the square root of two minus square root of zero in the end we end up with eight And I really like this one here because you come back to the original problem We want to find the circumference of a cardioid a cardioid is kind of like a Like a like a circle, but I got stabbed right has this cusp in it Because someone destroyed the circle and while the circumference of a circle is going to be two pi r The the circumference for this cardioid we ended up with an eight. There's no pi in the answer Um, it would seem so bizarre given how there's there's there's round There's curvature. There was trigonometry. Where is the pi? I guess someone ate it I guess haha. I know that's a horrible pun. I know but Someone ate the pie that that's the final answer here And that's also brings us to the end of lecture 34 on our discussion on polar functions and calculus of polar functions So this is great. Um, what what we're going to do in the next lecture 35 We're going to start our final unit of this series, which will be out sequences and series This will correspond to chapter 11 Of james stewart's calculus textbook. It's kind of appropriate to end a semester into course with chapter 11 That's where many businesses in their lives, isn't it and we'll take a look at that next time So stay tuned for those In the meanwhile, if you have any questions on any of these videos you've ever seen Please please please post your questions in the comments below. I'd be happy to answer them If you do like this video, please, uh, you know Up click that like button for us and you know, feel free to subscribe So you get more updates about videos like this in the future and I will see you next time. See you everyone. Bye