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AP Calculus BC · Unit 9: Parametric, Polar & Vector Functions

9.1 Defining and Differentiating Parametric Equations

What are Parametric Equations?

A parametric equation defines both x and y in terms of a third variable, the parameter (usually t): x = f(t), y = g(t). As t varies, the point (x, y) traces a curve. This lets us describe motion and curves that aren't functions of x alone, like circles or loops.

Finding dy/dx Parametrically

To find the slope of a parametric curve, use the chain rule: dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0. This gives the rate of change of y with respect to x at a given t-value, letting you find tangent lines just like with regular functions.

Second Derivatives Parametrically

Finding d²y/dx² requires care: it is NOT simply the derivative of dy/dx with respect to t. Instead, take the derivative of dy/dx (as a function of t) with respect to t, then divide by dx/dt again: d²y/dx² = [d/dt(dy/dx)] / (dx/dt). This is used to analyze concavity.

Key terms

Parametric equations
Equations x = f(t), y = g(t) that define x and y in terms of a third variable, the parameter t.
dy/dx formula (parametric)
dy/dx = (dy/dt) / (dx/dt), valid when dx/dt ≠ 0.
d²y/dx² formula (parametric)
d²y/dx² = [d/dt(dy/dx)] / (dx/dt) — differentiate dy/dx with respect to t, then divide by dx/dt again.
Vertical tangent (parametric)
Occurs when dx/dt = 0 and dy/dt ≠ 0, making dy/dx undefined.
Horizontal tangent (parametric)
Occurs when dy/dt = 0 and dx/dt ≠ 0, making dy/dx = 0.

Try an AP-style question

x = t², y = t³. What is dy/dx at t = 2?

  1. 4/3
  2. 8
  3. 3
  4. 12
Show the answer

C. dy/dx = 3t²/(2t) = 3t/2 = 3.

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