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


 * $$\frac{\partial x}{\partial\theta}=r'(\theta)\cos(\theta)-r(\theta)\sin(\theta)$$


 * $$\frac{\partial y}{\partial\theta}=r'(\theta)\sin(\theta)+r(\theta)\cos(\theta)$$

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


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