Calculus/Polar Differentiation

Differential calculusEdit

We have the following formulas:

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}
Last modified on 11 July 2009, at 20:03