# Differentiate a Polar Curve

1. Sep 10, 2014

### spinnaker

1. The problem statement, all variables and given/known data
Find the slope to the tangent line to the polar curve r^2 = 9 sin (3θ) at the point (3, π/6)

2. Relevant equations

dy/dx = (r cos θ + sin θ dr/dθ)/(-r sin θ + cos θ dr/dθ)

3. The attempt at a solution

So I have no issues with taking r^2 = 9 sin (3θ) and taking the root to get r = 3√(sin 3θ)
and then subbing that into the equation.

My problem is that it's only one case, and the derivative of a square root of a trig equation is messy. I think I'm missing something - any assistant would be appreciated.

2. Sep 10, 2014

### spinnaker

(i.e. what about the case where r = -3sqrt(sin 3theta)?

3. Sep 10, 2014

### BiGyElLoWhAt

I think this is one of those things where only one of the answers you get will make sense with the situation:

If r can equal +- (stuff) that means the function is mirrored about the center of your polar function. I didn't graph it but I think this is like a clover thing, right? Lemme draw it.

It's not very well drawn, but I think it gets the point across. Polar functions are weird lol

#### Attached Files:

• ###### pm_r_of_theta.png
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4. Sep 11, 2014

### spinnaker

You were right. I followed it through, and because the two were opposites of each other, I had one case where it was a negative divided by a positive, and the other was a positive divided by a negative, getting the same answer.

Thanks!

5. Sep 11, 2014

No problem