 Let's take a look at the solution to question 2 for the final exam for our course math 12-20 We're asked to find the slope of the tangent line to the polar curve r equals 8 sine of theta at the point theta equals Pi thirds, so if we want to find the tangent line I'm gonna I'm gonna go a little bit to the side here So you get a little more space right out here if we're looking for the slope of the tangent line We're trying to find the derivative dy Whoops dy over dx Which if you treat a polar curve as a parametric function We have to take the derivative of y with respect to theta and divide that by the derivative of x evaluated theta where here remember y equals r sine theta and x equals r cosine theta and So taking the derivative here Well without even knowing if the function is going to be if we treat r just as a function of theta The general formula you can see very quickly from the product rule is you're going to get r prime sine theta and Then you're going to get plus r cosine theta as your numerator and then in the denominator you end up with r prime cosine theta minus r sine theta So I mean without even knowing what what r is we get the following and so we do know what r is That sounds like a horrible grammar right there right so r is 8 sine theta So we'll plug those in in just a second and we also want to plug in our theta equals by thirds So let's plug in that r equals 8 sine theta The derivative of 8 sine theta with respect to theta would be 8 cosine theta So we get 8 cosine sine and then we're gonna get 8 sine Cosine that's kind of nice that those actually match up and they're gonna double up there in the denominator We're gonna get 8 cosine squared theta. In fact minus 8 sine squared like so which are some trig identities We could use to simplify this thing and the double angle identities are in play right here In either situation though we can factor in 8 That's actually gonna cancel out 8 over 8 and that leaves behind a 2 sine theta cosine theta over Cosine squared theta minus sine squared theta Which at this point you could plug in the pi thirds if you so wanted to it's perfectly fine Or you could do you could rewrite this as just sine of 2 theta over cosine of 2 theta If you want to use those identities, whichever you prefer I'm gonna stick with this ladder approach if we stick in the pi thirds at that moment We're gonna end up with sine of 2 pi thirds And then in the denominator We end up with cosine of 2 pi thirds So this is an angle in the second quadrant it'll reference angles in the first quadrant Which is actually pi thirds again. So we get a negative Well, let's leave you more careful here You're gonna get sine of pi thirds on top and then you get negative cosines negative in the second quadrant So at pi thirds sine should be root 3 over 2 At pi thirds cosine is actually one half so you get negative one half And so that would simplify to be negative square root of 3 Which we then see is option C For which case we select that one and so on this exercise We had to know how to calculate the derivative of a polar curve and here when we talk about the slope of the tangent line We're talking about dy over dx and really just treat this like a parametric curve You would if you were asked to find the slope of the tangent line of some parametric curve We would do the exact same thing Except that the function would be given as x equals whatever and y equals whatever in many cases You can treat a polar curve as a parametric curve and you'll be just fine in that regard