Mathematics.

parametric and polar

Calculus in Polar Coordinates

Calculus II25 minDifficulty6 out of 10

You should know: polar coordinates, integral

Overview

Differentiating and integrating polar curves r = f(θ) requires adapting Cartesian calculus tools, since x and y are both functions of θ rather than one being a function of the other. The two workhorse formulas are the polar slope formula (for tangent lines) and the polar area formula (for regions swept out by a rotating radius), the latter built from thin circular-sector approximations instead of rectangles.

Formal Definition

Definition

Treating r = f(θ) as parametric with parameter θ, via x = r cos θ, y = r sin θ:

dydx=dy/dθdx/dθ=f′(θ)sin⁡θ+f(θ)cos⁡θf′(θ)cos⁡θ−f(θ)sin⁡θ\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{f'(\theta)\sin\theta + f(\theta)\cos\theta}{f'(\theta)\cos\theta - f(\theta)\sin\theta}
Polar slope formula
A=12∫αβ[f(θ)]2 dθA = \frac{1}{2}\int_{\alpha}^{\beta} [f(\theta)]^2\, d\theta

Area of the region swept by r=f(θ) between rays θ=α and θ=β

Polar area formula
L=∫αβ[f(θ)]2+[f′(θ)]2 dθL = \int_{\alpha}^{\beta} \sqrt{[f(\theta)]^2 + [f'(\theta)]^2}\, d\theta
Polar arc length formula

Worked Examples

  1. Apply the polar area formula with f(θ) = cos(2θ).

    A=12∫−π/4π/4cos⁡2(2θ) dθA = \frac{1}{2}\int_{-\pi/4}^{\pi/4} \cos^2(2\theta)\, d\theta

  2. Use the power-reduction identity cos²u = (1+cos2u)/2 with u=2θ.

    =12∫−π/4π/41+cos⁡(4θ)2 dθ=14[θ+sin⁡(4θ)4]−π/4π/4= \frac{1}{2}\int_{-\pi/4}^{\pi/4} \frac{1+\cos(4\theta)}{2}\,d\theta = \frac{1}{4}\left[\theta + \frac{\sin(4\theta)}{4}\right]_{-\pi/4}^{\pi/4}

  3. Evaluate: sin(π) = sin(−π) = 0, so only the θ term survives.

    A=14(π4−(−π4))=π8A = \frac{1}{4}\left(\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right) = \frac{\pi}{8}

Answer: A = π/8

Practice Problems

Difficulty 6/10

Find the area of one full circle r = 2sinθ, θ ∈ [0, π].

Common Mistakes

Common Mistake

Using A = ∫f(θ)dθ (forgetting the ½ and the square) as the polar area formula.

The correct formula is A = ½∫[f(θ)]²dθ — it comes from summing thin circular sectors of area ½r²dθ, not thin rectangles.

Summary

  • Polar curves are treated as parametric in θ: x=r cosθ, y=r sinθ.
  • Slope: dy/dx = (dy/dθ)/(dx/dθ), computed via the product/chain rule on x(θ), y(θ).
  • Area swept by r=f(θ) from α to β: A = ½∫[f(θ)]²dθ, derived from thin circular sectors.
  • Arc length: L = ∫√(f(θ)²+f'(θ)²)dθ.

References

  1. WebsiteWikipedia — Polar coordinate system § Calculus