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:



So from this I get:



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

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

  • $\begingroup$ 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. $\endgroup$ – André Nicolas Nov 29 '15 at 0:45

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


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