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

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

$r=4\sin\theta$, $\theta = π/6$

Please show all work...I have tried solving this problem several times and keep on getting the wrong answer....

I know that:

$x=r\cos()$

$y=r\sin()$

So from this I get:

$x=(4-\sin())\cos()$

$y=(4-\sin())\sin()$

Then from there you find $\frac{dy}{d()}/\frac{dx}{d()}$.....

With $()=\pi/6=30$ degrees

• We have $y=4\sin^2\theta$ and $x=4\sin\theta\cos\theta$. The derivatives are not hard to find, for example $\frac{dy}{d\theta}=8\cos\theta\sin\theta$. Continue. – André Nicolas Nov 29 '15 at 0:45

$$\frac{\text d x}{\text d \theta}=\frac{\text d (r\cos\theta)}{\text d \theta}=\frac{\text d (4\sin \theta\cos\theta)}{\text d \theta}=\frac{\text d (2\sin 2\theta)}{\text d \theta}=4\cos 2\theta=4 \cos \frac{\pi}{3}=2$$ $$\frac{\text d y}{\text d \theta}=\frac{\text d (r\sin\theta)}{\text d \theta}=\frac{\text d (4\sin^2 \theta)}{\text d \theta}=\frac{\text d (2(1-\cos 2\theta))}{\text d \theta}=4\sin 2\theta=4 \sin \frac{\pi}{3}=2\sqrt3$$ $$\frac{\text d y}{\text d x}=\frac{\frac{\text d y}{\text d \theta}}{\frac{\text d x}{\text d \theta}}=\sqrt 3$$