V vinson24 Sep 2008 144 2 May 26, 2010 #1 Find the slope of the tangent line to the polar curve for the given value of \(\displaystyle \theta\) \(\displaystyle r=1/\theta; \theta=2\)

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@ MHF Hall of Honor Sep 2008 1,163 429 Champaign, Illinois May 26, 2010 #2 vinson24 said: Find the slope of the tangent line to the polar curve for the given value of \(\displaystyle \theta\) \(\displaystyle r=1/\theta; \theta=2\) Click to expand... \(\displaystyle \frac{dy}{dx}=\frac{r'(\theta)\sin\theta+r(\theta)\cos\theta}{r'(\theta)\cos\theta-r(\theta)\sin\theta} \) 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. Reactions: vinson24

vinson24 said: Find the slope of the tangent line to the polar curve for the given value of \(\displaystyle \theta\) \(\displaystyle r=1/\theta; \theta=2\) Click to expand... \(\displaystyle \frac{dy}{dx}=\frac{r'(\theta)\sin\theta+r(\theta)\cos\theta}{r'(\theta)\cos\theta-r(\theta)\sin\theta} \) 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.