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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

Convert to parametric form:

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

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:

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

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

Simplify:

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

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

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.

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