Anyone Good with Calculus

Find the slope of the tangent line to the given polar curve at the point specified by the value of θ.

r = 6sin(θ)
θ = π/3

I attempted this and got the answers 0 and .36397 but both were incorrect. If anyone knows how to do this please help me out and show how you can arrive at the answer. Thanks in advance!

Maybe this will help:

In order to find the slope of the tangent line for use in a linear equation you need to know the derivative in terms of y and x. You would need to know dy/dx. Remember that you can treat this derivative as a ratio of change in y over change in x:

dy/dx = [dy/dθ] / [dx/dθ]

So you must know the conversion form polar to rectangular coordinates, which you probably do:

x=rcosθ
y=rsinθ

r is given as 6sinθ

x=6sinθcosθ (this is actually just 3sin2θ)
y=6sin^2θ

Now you can find dy/dθ and dx/dθ:

dy/dθ = 12sinθcosθ = 6sin2θ
dx/dθ = 6cos2θ

Now dy/dx = dy/dθ / dx/dθ = (6sin2θ) / (6cos2θ) = tan2θ

tan(2*π/3) = - sqrt(3)

which is approximately equal to -1.7321.

If you look at the graph of this function you would know that the slope would have to be negative at θ= π/3.