# Ask Uncle Colin: A Polar Expression | Colin

Dear Uncle Colin,

I was asked to find the tangent to the curve $r=\frac{8}{\theta}$ at the point where $\theta = \frac{\pi}{2}$.

I worked out $\dydx = \frac{ \frac{8 \left(\theta \cos(\theta)-\sin(\theta)\right)}{\theta^2}}{\frac{-8\left(\theta \sin(\theta)+\cos(\theta)\right)}{\theta^2} }$, which simplifies to $-\frac{\theta \cos(\theta)-\sin(\theta)} {\theta \sin(\theta)-\cos(\theta)}$. Evaluated at $\theta = \frac{\pi}{2}$, that gives $\dydx=\frac{2}{\pi}$ and a line of $y = \frac{2}{\pi}x +\frac{2}{\pi}$.

However, it’s been marked wrong by the computer. How come?

– Doubly…

http://ift.tt/2u88DQ4