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