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i start by changing polar coords into x and y and then find the derivatives to get the slope.

$$x=(3-3\sin\theta)\cos\theta $$

$$x=3\cos\theta -3\cos\theta \sin\theta $$

and took $x'=(-3\sin\theta +3\sin^2\theta -3\cos^2\theta )$

$$= -3(\sin\theta -\sin^2\theta +\cos^2\theta)$$

$$= -3\sin\theta +3$$

$$\:y=(3-3\sin\theta)\sin\theta $$

$$y=3\sin\theta -3\sin^2\theta $$

$$y'=3\cos\theta -3\cos^2\theta $$

Now that i got the (hopefully) correct derivatives i need to find the tangent line for $\theta =\frac{3\pi }{4}$

so i plug in $\theta =\frac{3\pi }{4}$ into $\frac{3cos\theta -3cos^2\theta }{-3sin\theta +3}$ and got $-\frac{\sqrt{2}}{2}$ for the slope.

How do i get the tangent line though?

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  • $\begingroup$ $\frac{d}{dx}3\sin^2\theta$ != $3\cos^2\theta$ $\endgroup$ – Jessica Oct 27 '14 at 4:25
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Hint: $y - y_0 = m(x - x_0)$ is the equation of the tangent line for any curve. You have $m = -\dfrac{1}{\sqrt{2}}$, and $x_0 = (3-3\sin 3\pi/4)\cos 3\pi/4$....

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