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?
- 4/3
- 8
- 3
- 12
Show the answer
C. dy/dx = 3t²/(2t) = 3t/2 = 3.
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