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?

  • $\begingroup$ Where "do you have" that equation from? $\endgroup$ – Timbuc Oct 27 '14 at 3:35
  • $\begingroup$ because $x=r\cos\theta$, $y=r\sin\theta$ $\endgroup$ – Jessica Oct 27 '14 at 4:30
  • $\begingroup$ Sorry, I still don't understand. $\endgroup$ – Timbuc Oct 27 '14 at 4:35

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 $m = 2\sqrt{2}-3$

  • $\begingroup$ Just looked at your answered question, and it is the same. Although, I did get a different slope. $\endgroup$ – Jessica Oct 27 '14 at 3:42
  • $\begingroup$ so it would just be $y-3\sin\theta-3\sin^2\theta = (2\sqrt{2}-3)(x-3\cos\theta-3\sin\theta\cos\theta)$? $\endgroup$ – Jessica Oct 27 '14 at 4:36
  • $\begingroup$ I got $y = (1-\sqrt{2})x+2\frac{3\sqrt{2}-3}{2}$ $\endgroup$ – Jessica Oct 27 '14 at 5:41
  • $\begingroup$ oh yeah, I edited it above. Is that correct? $\endgroup$ – Jessica Oct 27 '14 at 5:47
  • $\begingroup$ Cancel the two $2$. $\endgroup$ – DeepSea Oct 27 '14 at 5:51

Ok, the formula is correct (look at the comments), so that indeed


But at $\;\theta=\frac{3\pi}4\;$ :







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