Calculus/Polar Differentiation

Differential calculus

We have the following formulae:

$r\frac{\partial }{\partial r}=x\frac{\partial }{\partial x}+y\frac{\partial }{\partial y}$

$\frac{\partial }{\partial \theta }=-y\frac{\partial }{\partial x}+x\frac{\partial }{\partial y}$

To find the Cartesian slope of the tangent line to a polar curve $r(\theta )$ at any given point, the curve is first expressed as a system of parametric equations.

$x=r(\theta )\cos (\theta )$

$y=r(\theta )\sin (\theta )$

Differentiating both equations with respect to $\theta$ yields

$\frac{\partial x}{\partial \theta }=r^{\prime }(\theta )\cos (\theta )-r(\theta )\sin (\theta )$

$\frac{\partial y}{\partial \theta }=r^{\prime }(\theta )\sin (\theta )+r(\theta )\cos (\theta )$

Dividing the second equation by the first yields the Cartesian slope of the tangent line to the curve at the point $(r,r(\theta ))$ :

$\frac{dy}{dx}=\frac{r^{\prime }(\theta )\sin (\theta )+r(\theta )\cos (\theta )}{r^{\prime }(\theta )\cos (\theta )-r(\theta )\sin (\theta )}$