Finding the Slope of a Polar Curve at a Given Point

[spinnaker]

Homework Statement

Find the slope to the tangent line to the polar curve r^2 = 9 sin (3θ) at the point (3, π/6)

Homework Equations

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

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.

 

[spinnaker]

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

 

[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

 

[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!

 

[BiGyElLoWhAt]

No problem