Finding Slope of Tangent at point on polar curve

**sfgiants13** · January 26th 2009, 03:08 PM

r=cos[theta/3], theta=pi

So dy/dx is dy/d theta / dx/d theta and x=rcos theta, y=rsin theta and i found the derivatives of each which gave me:

x=1/3(-sin 2th/3)-2sin(4th/3)
y=1/3((cos(2th/3)+2cos(4th/3))

I'm kinda stumped here. I have dy/dx as a big mess but I don't know what to do from here.

**Soroban** · January 26th 2009, 05:13 PM

Hello, sfgiants13!

Find the slope of the tangent to $r\:=\;\cos\tfrac{\theta}{3}\:\text{ at }\,\theta=\pi$

You're expected know (or be able to derive) that the slope formula is:

. . . . . $\frac{dy}{dx} \:=\:\frac{r\cos\theta + r'\sin\theta}{\text{-}r\sin\theta + r'\cos\theta}$ .[1]

We have:. $r \:=\:\cos\tfrac{\theta}{3},\quad r' \:=\:-\tfrac{1}{3}\sin\tfrac{\theta}{3}$

Substitute into [1]: . $\frac{dy}{dx} \:=\:\frac{\cos\frac{\theta}{3}\cos\theta - \frac{1}{3}\sin\frac{\theta}{3}\sin\theta} {\text{-}\cos\frac{\theta}{3}\sin\theta - \frac{1}{3}sin\frac{\theta}{3}\cos\theta}$

Let $\theta = \pi\!:\;\;\frac{dy}{dx} \:=\:\frac{\cos\frac{\pi}{3}\cos\pi = \frac{1}{3}\sin\frac{\pi}{3}\sin\pi}{\text{-}\cos\frac{\pi}{3}\sin\pi - \frac{1}{3}\sin\frac{\pi}{3}\cos\pi}$ . $\;=\;\frac{(\frac{1}{2})(-1) - (\frac{1}{3})(\frac{\sqrt{3}}{2})(0)} {\text{-}(\frac{1}{2})(0) - \frac{1}{3}(\frac{\sqrt{3}}{2})(-1)} <br />$

Therefore: . $\frac{dy}{dx} \;=\;\frac{\text{-}\frac{1}{2}}{\frac{\sqrt{3}}{6}} \;=\;-\sqrt{3}$

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