9.7 Defining Polar Coordinates and Differentiating in Polar Form

Syllabus
2020
Topic
9.7
Level

Learning objectives

Treat a Polar Curve as a Parametric Curve

For a polar curve r=f(θ)r=f(\theta), write x=rcosθx=r\cos\theta and y=rsinθy=r\sin\theta. Both coordinates depend on θ\theta, so differentiate each with the product rule and then use the parametric derivative rule.

\frac{dx}{d\theta}=\frac{dr}{d\theta}\cos\theta-r\sin\theta,\qquad \frac{dy}{d\theta}=\frac{dr}{d\theta}\sin\theta+r\cos\theta

\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta},\qquad \frac{d^2y}{dx^2}=\frac{d}{d\theta}\left(\frac{dy}{dx}\right)\bigg/\frac{dx}{d\theta}

  1. Find dr/dθdr/d\theta.
  2. Substitute rr and dr/dθdr/d\theta into the formulas for dx/dθdx/d\theta and dy/dθdy/d\theta.
  3. Divide to find dy/dxdy/dx and simplify before evaluating a requested angle.
  4. For d2y/dx2d^2y/dx^2, differentiate the entire slope with respect to θ\theta, then divide by dx/dθdx/d\theta again.

Example: let r=2cosθr=2\cos\theta. Then dr/dθ=2sinθdr/d\theta=-2\sin\theta, dx/dθ=2sin(2θ)dx/d\theta=-2\sin(2\theta), and dy/dθ=2cos(2θ)dy/d\theta=2\cos(2\theta). Therefore dy/dx=cot(2θ)dy/dx=-\cot(2\theta). At θ=π/4\theta=\pi/4, the slope is 00. Also,
d2ydx2=2csc2(2θ)2sin(2θ)=csc3(2θ),\frac{d^2y}{dx^2}=\frac{2\csc^2(2\theta)}{-2\sin(2\theta)}=-\csc^3(2\theta),
so at θ=π/4\theta=\pi/4 the second derivative is 1-1, indicating local concave-down behavior with respect to xx.

Do not use dr/dθdr/d\theta as the Cartesian slope: rr measures radial distance, not vertical position. The quotient for dy/dxdy/dx requires dx/dθ0dx/d\theta\ne0; when dx/dθ=0dx/d\theta=0, analyze the component derivatives separately because the tangent may be vertical or the point may require further investigation.