Find the equation of the tangent line to the polar curve: $r=3-3\sin\theta$ at $\theta=\frac{3\pi}{4}$
I have the equation:
$$\frac{dy}{dx} =\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}= \frac{-3\cos\theta\sin\theta+(3-3\sin\theta)\cos\theta}{-\cos^2\theta-(3-3\sin\theta)\sin\theta}=2\sqrt{2}-3$$
which, if I did the math correctly (if I didn't could someone point it out), is the slope of the tangent line. How do I find the equation?
Best Answer
This seems to be the question that I answered earlier. $x_0 = r\cos \theta = (3 - 3\sin 3\pi/4)\cos 3\pi/4$, and similarly you can find $y_0$, and then use $y - y_0 = m(x - x_0)$ with $m = 2\sqrt{2}-3$