Last modified on 21 February 2015, at 13:00

# Calculus/Polar Differentiation

### Differential calculusEdit

We have the following formulae:

$r \tfrac{\partial}{\partial r}= x \tfrac{\partial}{\partial x} + y \tfrac{\partial}{\partial y} \,$
$\tfrac{\partial}{\partial \theta} = -y \tfrac{\partial}{\partial x} + x \tfrac{\partial}{\partial y} .$

To find the Cartesian slope of the tangent line to a polar curve r(θ) 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 θ yields

$\tfrac{\partial x}{\partial \theta}=r'(\theta)\cos\theta-r(\theta)\sin\theta \,$
$\tfrac{\partial y}{\partial \theta}=r'(\theta)\sin\theta+r(\theta)\cos\theta \,$

Dividing the second equation by the first yields the Cartesian slope of the tangent line to the curve at the point (rr(θ)):

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