# Finding the Equation of a Tangent Line to a Polar Curve

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?

• Where "do you have" that equation from? – Timbuc Oct 27 '14 at 3:35
• because $x=r\cos\theta$, $y=r\sin\theta$ – Jessica Oct 27 '14 at 4:30
• Sorry, I still don't understand. – 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 $m = 2\sqrt{2}-3$

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

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

$$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}$$

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

$$\frac{dr}{d\theta}_{\theta=\frac{3\pi}4}=-3\cos\theta_{\theta=\frac{3\pi}4}=\frac3{\sqrt2}$$

$$r\cos\theta=(3-3\sin\theta)\cos\theta\stackrel{\theta=\frac{3\pi}4}\longrightarrow\;\;=\left(3-\frac3{\sqrt2}\right)\left(-\frac1{\sqrt2}\right)=\frac32-\frac3{\sqrt2}$$

$$r\sin\theta=(3-3\sin\theta)\sin\theta\stackrel{\theta=\frac{3\pi}4}\longrightarrow\;\;=\left(3-\frac3{\sqrt2}\right)\left(\frac1{\sqrt2}\right)=\frac3{\sqrt2}-\frac32$$

So

$$\frac{dy}{dx}|_{\theta=\frac{3\pi}4}=\frac{\frac3{\sqrt2}\frac1{\sqrt2}+\frac32-\frac3{\sqrt2}}{\frac3{\sqrt2}\left(-\frac1{\sqrt2}\right)-\frac3{\sqrt2}+\frac32}=\frac{3-\frac3{\sqrt2}}{-\frac3{\sqrt2}}=-\frac{1-\frac1{\sqrt2}}{\frac1{\sqrt2}}=1-\sqrt2$$