# Ask Uncle Colin: A Polar Expression

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 Irritated, Raging At Computer

Hi, DIRAC, and thank you for your question!

Let me work through it and see! I reckon we’re looking at something like a spiral here: if $r\theta = 8$, then small $\theta$s give large $r$s that get progressively smaller.

However, let’s differentiate it. We’ll figure out $\diff y \theta$ and $\diff x \theta$ separately:

Since $y = r \sin(\theta)$, $\diff y \theta = r \cos(\theta) + \diff r \theta \sin(\theta)$.

Similarly, since $x = r\cos(\theta)$, $\diff x \theta = - r \sin(\theta) + \diff r \theta \cos(\theta)$.

We also need $\diff r \theta$, which is simple enough: $r = 8\theta^{-1}$, so $\diff r \theta = -8\theta^{-2}$.

Rather than simplify and then evaluate, we can do it the other way around. If we know $\theta = \pibytwo$, then we know $r= \frac{16}{\pi}$, that $\sin(\theta)=1$ and $\cos(\theta)=0$.

Also, where $\theta = \pibytwo$, $\diff r \theta = -\frac{32}{\pi^2}$, while $\cos(\theta)=0$ and $\sin(\theta)=1$.

So, $\diff y \theta = -\frac{32}{\pi^2}$ and $\diff x \theta = - \frac{16}{\pi}$, which makes $\dydx = \frac{2}{\pi}$ - as you had for the gradient ((I suspect you differentiated with respect to $r$ rather than $\theta$, but that makes no odds.)).

Your problem, so far as I can see, is with the $y$-intercept. Because $r = \frac{16}{\pi}$ when $\theta = \frac{\pi}{2}$, your $y$-intercept should be $\frac{16}{\pi}$ as well.

The line you’re looking for is $y=\frac{2}{\pi}x + \frac{16}{\pi}$, or equivalently $y = \frac{2}{\pi}\left( x + 8 \right)$. Happily, Desmos agrees:

Hope that helps!

- Uncle Colin