 Welcome back everyone in this video. We're gonna start doing some calculus with polar functions That is functions of the form r equals f of theta and frankly speaking a lot of the calculus is gonna be based upon the fact That we can view polar functions as parametric functions And how one does that comes from what we discussed before We had seen previously that as we switch from Polar coordinates to Cartesian coordinates we get the following r equals r I'm sorry x equals r cosine theta and y equals r sine theta Well since r is itself a function of theta we could insert for r f of theta Like we do right here, and so we get x equals f of theta times cosine of theta and y equals f of theta times sine of theta And so that way we can view polar functions as Parametric equations associated to the parameter theta And so the reason why this is an important perspective is that as we want to talk about tangent lines of a polar curve The tangent line will still be determined by this rise over run We're looking the slope of the tangent line will still be dy over dx and so we have to express dy over dx knowing how r is a function of theta So if y and x are both functions of theta You can replace dy with dy over d theta and dx with dx over d theta by the chain rule that these two quotients will still be the same thing and if you take the derivative of Why I guess we'll do why first if you take the derivative of y with respect to theta first using the Product rule you're going to get r prime and by r prime here This will always mean r prime always mean dr over d theta and Theta prime will always mean dy over dx if we ever abbreviate that we won't deviate from those So taking the derivative of y with respect to theta remember f of theta is just r here So by the product rule you're going to get the derivative You'll get the derivative of r first times sine theta then you're going to get r times the derivative of sine theta Which is cosine Taking the derivative of x with respect to theta you're going to get r prime Cosine and then you're going to get negative r times sine theta the derivative of cosine is a negative sign and So we can actually get the derivative here And so this is a form of one can Memorize but really these yellow boxes right here is what you want to know that x equals r cosine theta y equals r sine theta And since r is a function of theta you can take the derivative with respect to theta on top and bottom And that's going to help us out here So let's do a little bit of analysis of a Corioid take the function r equals 1 plus sine of theta and let's look at some tangent lines for this curve For example, let's find the tangent line When theta equals pi thirds Well, let's focus on the slope. We can do the rest of it if we want to now pi thirds be aware it's going to be It's gonna be about this angle right here And so the tangent line we can kind of visualize would be something like this thing right here If ever you miss your dot just draw the dot bigger so you get to become the tangent line So this is roughly speaking what we're looking for we need to find the slope of this line So what we're looking for is we're trying to find y prime like we saw before is dy over d theta divided by dx over d theta and So be aware that we're going to be getting r which is 1 plus sine theta times that by sine We need to take the derivative of this on top and then we have to do the same thing for x. We have 1 plus sine theta Times cosine theta Take the derivative with respect to theta there So you're just gonna replace the r with the 1 plus sine theta we have from the function right here So by the product rule on the top you take the derivative of 1 plus sine You're gonna get a cosine and then you'll times that by sine theta and then You take the derivative well the 1 plus sine theta will stick around then you take the derivative of sine Which is cosine we'll come back to that one in a little bit on the derivative on the bottom Well again derivative 1 plus sine is gonna be cosine again, but this time you're gonna times that by cosine and Then you're gonna get a 1 plus sine theta and you're gonna times by the derivative of Cosine which is sine so I'm gonna get a sine right there And I'm also gonna replace this with a minus sign right there. So let's distribute Across the sum right here on top and bottom This is gonna give us a sine theta times cosine theta This is gonna give us a cosine theta when you distribute the cosine onto the one when you distribute the cosine onto the sign You're gonna get another sine theta. Sorry another another sine theta cosine theta So I'm just gonna put a 2 in front right there In the denominator What we end up with we're gonna get a cosine times cosine so a cosine squared When you distribute you're gonna get one on the sign, so it's gonna be a negative sign and Then when you get sine and sine that's gonna be a negative sine squared Like so and so to try to make this thing a little bit simpler I'm gonna want to utilize some trig identities notice that 2 sine theta cosine theta That's the that's half of the double angle identity so we can end up with sine of 2 theta plus cosine and Then this it's above well actually kind of kind of works out nicely here cosine squared minus sine squared That's half of the double angle identity for cosine you end up with a cosine of 2 theta Minus sine theta so there's some symmetry to there. So this right here is just the derivative y prime, right? We next need to calculate. What is the derivative when it's evaluated at? Theta equals pi thirds Now it's the angle of consideration. So plug in all of these appropriate parts in there. You're gonna get sine of 2 pi thirds Plus cosine of pi thirds This sits above cosine of 2 pi thirds Minus sine of pi thirds If it helps of course pi thirds Would be 60 degrees 2 pi thirds would be 120 degrees now 2 pi thirds references to pi thirds It's just in the second quadrant now sine and the second quadrant is identical Just the angle doesn't make a difference. So you just get the first one right here. You could replace You could just replace it with a sine of pi thirds. It would be the exact same thing And so sine of pi thirds is gonna be root 3 over 2 Or adding to that cosine of pi thirds, which is one half in the denominator We're gonna get cosine of 2 pi thirds which 2 pi thirds It references pi thirds, but for cosine that I'll actually make it negative. So you're gonna get a Negative one half in the denominator minus pi thirds Which is going to give you the square root of 3 over 2 So despite as complicated this might look notice in the bottom in the top and bottom you get 3 over 2 Plus one half factor out the negative sign You're gonna get 3 over 2 plus one half This will simplify just to be a negative one and that's then the slope of This line that we had above here And you know if you look at that line as we drew it It's not perfectly drawn to scale, but negative one actually does seem to be a about a Slope of that tangent line. So I feel like our calculation worked out really well so another thing to consider is Where are the horizontal and Vertical tangent lines of this graph and on the picture we can see them There's looks like there's a horizontal right there at pi halves. There's gonna be two horizontals In the third and fourth quadrants. We'll have to come back and figure out those are there's gonna be two vertical tangent lines in the first and Second quadrant and then also we're gonna see that this cusp here at 3 pi thirds is gonna be kind of suspicious For our derivative here when we investigate it So let's find these points for where the tangent line is vertical or horizontal coming back to our formula, right? dy over dx Remember this was sine of 2 theta plus cosine theta Over cosine of 2 theta minus sine of theta. That's what we saw on the previous slide Did we not I think we did Let me just double-check. What was it? Yep. That's what we had So we want to figure out when this thing is horizontal that when the tangent line is gonna be horizontal and vertical so just as sort of a quick reminder here the The tangent line is gonna be a horizontal tangent line when we get that our dy is Zero, but dx is not zero and the vertical tangent lines are gonna correspond when the dy is not zero But the dx is zero now Of course if they both are zero at the same time you have the indeterminate form zero over zero And you might need low p-tolls rule to help us determine what's going on there All right, and so let's start off with the horizontal Well, let's actually solve all these things when they're equal to zero, right? Sine of 2 theta plus cosine of theta when this is equal to zero It'll probably be better to go back Sine of 2 theta will replace with 2 sine theta cosine theta I think the calculation will be a little bit easier when all of the angles are the same You notice you can factor out a cosine theta from this that leaves behind 2 sine theta plus 1 and So by the zero product property this implies that cosine theta equals zero or That 2 sine theta plus 1 equals zero so for cosine Cosine is equal to zero This will happen at the top of the unit circle and at the bottom of the unit circle So we're looking at pi halves and 3 pi halves and of course angles which are coterminal to those ones Now for 2 sine theta plus 1 if we solve for sine here, you can subtract 1 divide by 2 We're trying to solve the equation when does sine theta equal negative one half now We do know a 30-degree angle sine is equal to one half or that is sine at pi 6th will give us a one Sorry a sine of pi 6 will give us a one half so using reference angles that can help us out a lot here We are gonna get that theta Would since if sine is negative It's gonna be in the third quadrant or the fourth quadrant, but it should reference to pi 6th now That's gonna be exactly 7 pi 6th and 11 pi 6 now admittedly there are angles coterminal to that but As the whole cardioid is graphed by from 0 to 2 pi. We only have to find theta between 0 and 2 pi here So this tells us when the denominator when the numerator goes to 0 here on in contrast When does the denominator go to 0 cosine of 2 theta? minus sine theta is equal to 0 Likewise, we want to put you know, we want to compare apples to apples We need to have theta and theta not a 2 theta here So we could switch cosine of 2 theta back to cosine squared minus sine squared But I actually think a different Trigonometric identity would be useful here, right? So we could also use the fact That cosine of 2 theta. Oh boy. Let's see if I can remember it We we we use this identity all the time for trigonometric identities If you don't remember it, we'll just just kind of do this cosine squared minus sine squared Right here minus sine Now because of the pythagorean identity, we do know that cosine squared theta minus or plus sine squared theta Equals one if you solve for cosine squared here, you're going to get cosine squared equal to 1 minus sine squared So just make that substitution 1 minus sine squared In which case you end up with 1 minus 2 sine squared theta minus sine So what you kind of see right here It's just want to point out that if you don't remember a trig identity Your your first goal doesn't always have to be go look it up You can often derive them from the trig identities. You do know here. So this is cosine of 2 theta The reason I prefer this format right here is because now you have a sine squared You have a sine you have a constant you can treat this like a quadratic equation in terms of sine If like if you replace sine with just a simple x you'd have 1 minus 2x squared minus x you could factor this thing And this thing should factor off as let's see 1 plus sine theta Times 1 minus 2 sine theta Just by usual factoring technique right there and you can foil this thing out to double check 1 times 1 is 1 you get sine times negative 2 sine which gives you negative 2 sine squared And then you'll get 1 and negative 2 sine that's that's that's negative 2 sine And then you're going to get a sine with just one sign. So that combines to give you a single sign So there's reverse foil going on there kind of skip some of the details. I hope that's okay with everyone Uh, and so which case if we continue with this, we're going to get 1 plus sine theta equals 0 Or we get 1 minus 2 sine theta Equal 0. So now for the first one Sine theta equals negative 1 when is sine equal to negative 1 that happens at the bottom of the hour That is the bottom of the unit circle that'll happen at 3 pi halves And then for the other one if we solve for sine we're looking for when sine theta equals 1 half And like we said before that'll happen at pi 6th But also whenever sine is positive it references to pi 6th, which would also happen at 5 pi 6th So we see that the numerator of the derivative will be 0 At pi halves 3 pi halves 7 pi 6 and 11 pi 6 It'll be The denominator will be 0 at 3 pi halves pi 6 and 5 pi 6th And you'll notice here that There is a little bit of overlap 3 pi halves shows up in both situations We're going to have to investigate this one a little bit further because in this situation our derivative dy over dx Looks like 0 over 0 If we go back to the picture That we saw before this actually might agree with what we were looking at right So when does when does the when does the numerator go to 0? Pi halves that's the top of the circle at the top of the cardioid and then at 7 pi 6 and 11 pi 6 So what we're seeing right here Is this one happens at pi halves? This one right here. It's happening at 7 pi 6th And this one right here is happening at 11 pi 6 just like we calculated In terms of vertical tangent lines well 3 pi halves will come back to that one You have pi 6 and 5 pi 6th. So we can see that for these ones The vertical tangent on the right happens at pi 6th And the one on the left happens at 5 pi 6th And so by symmetry those all seem to match up right And so the suspicious thing that's left over is at 3 pi halves, right? That's where the cusp is on our cardioid So it's not surprising that 3 pi halves gives you this indeterminate form Funky things are happening at 3 pi halves for our cardioid so What we've seen so far is that dy over dx You're getting this indeterminate form. So we're going to apply l'Hopital's rule in this context We're going to take the derivative of the top and bottom yet again. So if we take Sine of 2 theta plus cosine theta We're going to take the derivative that again with respect to theta for the bottom. We have cosine of 2 theta That is a 2 minus sine theta Take again the derivative with respect to theta Because that's how we try to resolve this indeterminate form using l'Hopital's rule So l'Hopital's rule will give us 2 cosine 2 theta minus sine theta And then on the denominator, we're going to get negative 2 sine of 2 theta minus cosine theta Like so and so then Continuing on from there if we were to plug in 3 pi halves because we really only care about this thing as theta As theta is approaching 3 pi halves. So let's try that this time around You would get 2 cosine of 3 pi. I like that minus sine of 3 pi halves And then in the denominator, we're going to get negative 2 sine of 3 pi minus cosine of 3 pi halves So now we have the battle the clash of the titans who's stronger numerator denominator who wants to win this fight So going through this some things to note Cosine of 3 pi is the same thing as cosine of pi. That's going to be a negative 1 Sine of 3 pi is going to be a 0 Sine of negative Sine of 3 pi over 2 that's going to also be a negative 1 And then cosine of 3 pi over 2 that's going to be a 0 So you can see what happens here is in the numerator We end up with negative 2 plus 1 The denominator we're going to end up with 0 plus 0 So looks like we end up with like a negative 1 over 0. Assuming we didn't make any mistakes along the way But this seems to say that the winner winner is the denominator here So this thing wants to be a vertical tangent line A vertical tangent And if we come back to the picture of the carbioid that does seem to agree with what we were seeing right there, right? As this thing bends upward and bends upward This cusp is coming to a sharp point and thus having this vertical tangent that sits in between them So we can analyze Polar functions using the same type of tangent line consideration we had in the past Just make sure that we take the derivative that you are taking the derivative Treating the polar function as a parametric function Then we can use derivatives and tangent lines to measure things about monotonicity We could do the second derivative measure things about concavity and all of that business For these polar functions, just like other functions as well. And so that brings us to the end of lecture 33 Where we talked about some polar derivatives here in the next video for polar or for lecture 34 We're going to talk about uh, how integrals are affected when we start looking at polar functions. So stay tuned for that one