How do you find the slope of a polar curve?

Aug 13, 2015

If $r = f \left(\theta\right)$ is the polar curve, then the slope at any given point on this curve with any particular polar coordinates $\left(r , \theta\right)$ is $\frac{f ' \left(\theta\right) \sin \left(\theta\right) + f \left(\theta\right) \cos \left(\theta\right)}{f ' \left(\theta\right) \cos \left(\theta\right) - f \left(\theta\right) \sin \left(\theta\right)}$

Explanation:

If $r = f \left(\theta\right)$, then $x = r \cos \left(\theta\right) = f \left(\theta\right) \cos \left(\theta\right)$ and $y = r \sin \left(\theta\right) = f \left(\theta\right) \sin \left(\theta\right)$. This implies, by the Product Rule, that $\frac{\mathrm{dx}}{d \theta} = f ' \left(\theta\right) \cos \left(\theta\right) - f \left(\theta\right) \sin \left(\theta\right)$ and $\frac{\mathrm{dy}}{d \theta} = f ' \left(\theta\right) \sin \left(\theta\right) + f \left(\theta\right) \cos \left(\theta\right)$.

Therefore $\setminus m b \otimes \left\{s l o p e\right\} = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{d \theta}}{\frac{\mathrm{dx}}{d \theta}} = \frac{f ' \left(\theta\right) \sin \left(\theta\right) + f \left(\theta\right) \cos \left(\theta\right)}{f ' \left(\theta\right) \cos \left(\theta\right) - f \left(\theta\right) \sin \left(\theta\right)}$

I tested this with the polar curve $r = f \left(\theta\right) = \theta$, which gave $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sin \left(\theta\right) + \theta \cos \left(\theta\right)}{\cos \left(\theta\right) - \theta \sin \left(\theta\right)}$ and, at the point with polar coordinates $\left(r , \theta\right) = \left(f \left(\theta\right) , \theta\right) = \left(f \left(\frac{5 \pi}{6}\right) , \frac{5 \pi}{6}\right) = \left(\frac{5 \pi}{6} , \frac{5 \pi}{6}\right) \approx \left(2.62 , 2.62\right)$ (and rectangular coordinates $\left(x , y\right) \approx \left(- 2.28 , 1.31\right)$)

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sin \left(\frac{5 \pi}{6}\right) + \frac{5 \pi}{6} \cdot \cos \left(\frac{5 \pi}{6}\right)}{\cos \left(\frac{5 \pi}{6}\right) - \frac{5 \pi}{6} \cdot \sin \left(\frac{5 \pi}{6}\right)} = \frac{\frac{1}{2} + \frac{5 \pi}{6} \cdot \left(- \frac{\sqrt{3}}{2}\right)}{- \frac{\sqrt{3}}{2} - \frac{5 \pi}{6} \cdot \frac{1}{2}} \approx 0.81252$. I graphed the polar curve along with its tangent at this point and got the following picture. It looks good.