Originally Posted by

**FelixHelix** Hi All,

I'm having problems with this question and need someone to look at my workings and give a gentle push in the right direction. If anyone can help I'd be most grateful.

Calculate$\displaystyle \frac{dy}{dx}$for the curve whose equation in polar coordinates is $\displaystyle r=1+sin^2\theta$

So my understanding is that polar coords are given as $\displaystyle r,\theta$ so not quite sure what to do with this. Would I rearrange to find $\displaystyle \theta$ such that $\displaystyle \theta = sin^{-1}(\pm\sqrt{r-1})$ and keep r as it is????

Or is it using r and $\displaystyle \theta$ with $\displaystyle x = rcos \theta $ & $\displaystyle y=rsin \theta$

Then find $\displaystyle \frac{dx}{d\theta}$ and hence $\displaystyle \frac{d\theta}{dx}$ and $\displaystyle \frac{dy}{d\theta}$ and finally get the result $\displaystyle \frac{dy}{dx}$$\displaystyle -cot\theta$

Best regards, Felix