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# Worked example: differentiating polar functions

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.G (LO)
,
FUN‑3.G.1 (EK)
,
FUN‑3.G.2 (EK)

## Video transcript

let R be the function given by R theta is equal to three theta sine theta for theta is between zero and two pi including zero and 2pi the graph of R in polar coordinates consists of two loops as shown in the figure above so let's think about why it has two loops so as our theta when theta is zero R is zero and then as R theta increases we start tracing out this first loop all the way until when theta is equal to pi so we've traced out this first loop from theta is equal to 0 2 theta is equal to PI and then the second loop has a larger R so these are larger R's this is when we're going from PI to 2 pi and you might say well why doesn't it show up down here well between sine of PI and sine of 2 pi this part right over here is going to be negative so it flips it over the arm to this side and the magnitude of the R is larger and larger because of this 3 theta and so when we go from PI to 2 pi we trace out the larger circle fair enough that seems pretty straightforward point P is on the graph of R right over there and the y-axis find the rate of change of the x coordinate with respect to theta at the point P all right so let's think about this a little bit they don't give us X as a function of theta we have to figure out that from what they've given us so just as a bit of a polar coordinates refresher if this is our theta right over there this is our R and that would be a point on our curve for this theta now how do you convert that to X and Y's well you can construct a little bit of a right triangle right over here and we know from our basic trigonometry that the length of this base right over here this is going to be the hypotenuse let me just write that that's that's going to be our x coordinate our x coordinate right over here is going to be equal to our hypotenuse which is R times the cosine of theta if we wanted the y coordinate as a function of R and theta it'd be Y is equal to R sine of theta but they don't want us to worry about Y here just the x coordinate so we know this but we want it purely in terms of theta so how do we get there well what we can do is take this expression for our R itself as a function of theta and replace it right over there and so what we can do is we can write well X of theta is going to be equal to R which itself is three theta sine of theta times cosine of theta times cosine of theta and now we want to find the rate of change of the x coordinate with respect to theta at a point so let's just find the derivative of X with respect to theta so X prime of theta is equal to well I have the product of three expressions over here I have this first expression three theta then I have sine theta and then I have cosine theta so we can apply the product rule to find the derivative if you're using the product rule with the expression of three things you essentially just follow the same pattern when you're taking the product of two things the first term is going to be the derivative of the first of the expressions three times the other two expressions so we're going to have three times sine of theta cosine of theta plus the second term is going to be the derivative of the middle term times the other two expressions so we're gonna have three theta and then derivative of sine theta is cosine theta times another cosine theta you're gonna have cosine squared of theta or cosine of theta squared just like that and then you're going to have the derivative of the last term is going to be the derivative of cosine theta times these other two expressions well the derivative of cosine theta is negative sine theta so if you multiply negative sine theta times three theta sine theta you're going to have negative three theta sine squared theta and so we want to evaluate this at point P so what is theta at point B well point P does happen on our first pass around and so at Point P theta is to data right over here is equal to PI over 2 so PI over 2 so what we really just need to find is well what is X prime of PI over 2 well that is going to be equal to 3 times sine of PI over 2 which is 1 times cosine of PI over 2 which is 0 so this whole thing is 0 plus 3 times pi over 2 is this 3 PI over 2 times cosine squared of PI over 2 or cosine of PI over 2 squared well that's just 0 so so far we everything is 0 minus 3 times pi over 2 3 PI over 2 times sine of PI over 2 squared well what's sine of PI over 2 well that's 1 you square it you still get 1 so all of this simplified to negative 3 PI over 2 now it's always good to get a reality check does this make sense that the rate of change of X with respect to theta is negative 3 PI over 2 well think about what's happening as theta increases a little bit X is definitely going to decrease so it makes sense that we have a negative out here so right over here rate of change of X with respect to theta negative 3 PI over 2 is theta increases our X for sure is decreasing so at least it does make intuitive sense
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