AP Calculus BC · Lesson 4 of 4
Parametric and polar reasoning
Treat x and y as functions of another variable, and compute slopes or areas without forcing y to be a function of x.
Parametric curves
A parametric curve gives x(t) and y(t). The slope dy/dx is (dy/dt) / (dx/dt), provided dx/dt is not zero. A horizontal tangent occurs when dy/dt = 0 and dx/dt ≠ 0. A vertical tangent can occur when dx/dt = 0 and dy/dt ≠ 0.
Arc length and speed use both derivatives: speed is the square root of (dx/dt)² + (dy/dt)². Do not use only dy/dx if the question asks how fast the particle is moving along the curve.
Polar curves
In polar coordinates, a point is a distance from the origin and an angle. r = 2 cos θ is a circle, not a cosine wave drawn in the Cartesian plane. Convert to x and y when you need a familiar Cartesian equation, using x = r cos θ and y = r sin θ.
Area in polar coordinates uses (1/2)∫ r² dθ over the given interval of θ. The factor one-half and the square of r are part of the formula. Identify the interval of θ that traces the region once.
Worked example
x = t² and y = t³. Find dy/dx at t = 2.
- dx/dt = 2t and dy/dt = 3t².
- dy/dx = (3t²) / (2t) = 3t/2 for t ≠ 0.
- At t = 2, the slope is 3.
Why this works. Differentiate with respect to the parameter, then divide. Do not treat t as y.
Check your understanding
- Compute dy/dx from the two parametric derivatives.
- Use speed, not slope, when the question asks how fast a particle moves.
- Square r and include 1/2 in a polar area integral.