slope of tangent line to polar curve

Sep 2008
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Find the slope of the tangent line to the polar curve for the given value of \(\displaystyle \theta\)
\(\displaystyle r=1/\theta; \theta=2\)
 

chiph588@

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Sep 2008
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Champaign, Illinois
Find the slope of the tangent line to the polar curve for the given value of \(\displaystyle \theta\)
\(\displaystyle r=1/\theta; \theta=2\)
\(\displaystyle \frac{dy}{dx}=\frac{r'(\theta)\sin\theta+r(\theta)\cos\theta}{r'(\theta)\cos\theta-r(\theta)\sin\theta} \)

So in our case \(\displaystyle \frac{dy}{dx}=\frac{-\tfrac{1}{\theta^2}\sin\theta+\tfrac{1}{\theta}\cos\theta}{-\tfrac{1}{\theta^2}\cos\theta-\tfrac{1}{\theta}\sin\theta} \)

So just plug in \(\displaystyle \theta=2 \) to get your answer.
 
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