Parametric and Polar Equations for AP Calculus BC: A Complete Guide | RevisionDojo

Parametric and polar equations appear on the AP Calculus BC exam in both multiple-choice and free-response sections. They often involve derivatives, integrals, arc length, and area calculations — sometimes in combination with other topics like vector motion.

In this RevisionDojo guide, we’ll cover:

• Key formulas for parametric equations
• Key formulas for polar equations
• Area and arc length problems
• AP-style worked examples

📚 Parametric Equations – The Basics

A parametric equation expresses xx and yy in terms of a third variable, tt (the parameter).

Example:

x(t)=t2,y(t)=3tx(t) = t^2, \quad y(t) = 3t

🔹 Derivatives for Parametric Equations

• First derivative:

dydx=dydtdxdt\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}

• Second derivative:

d2ydx2=ddt(dydx)dxdt\frac{d^2y}{dx^2} = \frac{\frac{d}{dt} \left( \frac{dy}{dx} \right)}{\frac{dx}{dt}}

🔹 Arc Length for Parametric Equations

L=∫ab(dxdt)2+(dydt)2 dtL = \int_a^b \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt

📚 Polar Equations – The Basics

A polar equation represents points using radius rr and angle θ\theta.

Example:

r(θ)=2+cos⁡θr(\theta) = 2 + \cos \theta

🔹 Slope for Polar Curves

1. Convert to parametric form:

x=r(θ)cos⁡θ,y=r(θ)sin⁡θx = r(\theta) \cos \theta, \quad y = r(\theta) \sin \theta

1. Then:

dydx=dydθdxdθ\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}

🔹 Area for Polar Curves

A=12∫θ1θ2[r(θ)]2dθA = \frac{1}{2} \int_{\theta_1}^{\theta_2} \left[ r(\theta) \right]^2 d\theta

🔹 Arc Length for Polar Curves

L=∫θ1θ2(r(θ))2+(drdθ)2 dθL = \int_{\theta_1}^{\theta_2} \sqrt{\left( r(\theta) \right)^2 + \left( \frac{dr}{d\theta} \right)^2} \, d\theta

📝 AP-Style Worked Example

Problem: Find the area enclosed by one loop of r=2sin⁡θr = 2\sin\theta.

Solution:

1. Symmetry: one loop occurs from θ=0\theta = 0 to θ=π\theta = \pi
2. Area formula:

A=12∫0π[2sin⁡θ]2dθA = \frac{1}{2} \int_0^\pi \left[ 2\sin\theta \right]^2 d\theta

1. Simplify:

A=2∫0πsin⁡2θ dθA = 2 \int_0^\pi \sin^2\theta \, d\theta

1. Use identity: sin⁡2θ=1−cos⁡(2θ)2\sin^2\theta = \frac{1 - \cos(2\theta)}{2}
2. Integrate:

A=∫0π1−cos⁡(2θ) dθ=[θ−sin⁡(2θ)2]0πA = \int_0^\pi 1 - \cos(2\theta) \, d\theta = \left[ \theta - \frac{\sin(2\theta)}{2} \right]_0^\pi

1. Evaluate: A=π−0=πA = \pi - 0 = \pi

⚠️ Common Mistakes to Avoid

• Forgetting to check intervals of integration for polar area problems
• Dropping 12\frac{1}{2} in polar area formula
• Mixing up dydx\frac{dy}{dx} formulas for parametric and polar curves
• Forgetting to square r(θ)r(\theta) when finding polar area

📊 Practice Strategy from RevisionDojo

• Memorize all arc length and area formulas for both parametric and polar
• Practice converting between polar and rectangular forms
• Drill one parametric derivative problem and one polar area problem weekly

🧭 Final Advice from RevisionDojo

Parametric and polar equation questions are predictable in structure — the AP exam often tests the same formulas in different contexts. If you can quickly recall the correct form and set up the integral properly, you’ll secure these high-value points.