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Calculus BC · Unit 9: Parametric Equations, Polar Coordinates, and Vector-Valued Functions · 14 min read · Updated 2026-05-11

Defining and differentiating parametric equations — AP Calculus BC

AP Calculus BC · Unit 9: Parametric Equations, Polar Coordinates, and Vector-Valued Functions · 14 min read

1. Defining Parametric Curves and Eliminating the Parameter ★★☆☆☆ ⏱ 4 min

Instead of defining a curve in the $xy$-plane as a direct relation between $x$ and $y$, parametric equations express both coordinates as separate functions of a third independent variable called a parameter, most commonly $t$ (often representing time or angle).

This notation lets us describe curves that fail the vertical line test (such as circles, ellipses, and cycloids) and track the position of moving objects over time, which is impossible with a single Cartesian function $y=f(x)$.

Eliminating the parameter converts the parametric pair to a single Cartesian relation to identify the curve's shape. For non-trigonometric parametric equations, solve one equation for $t$ and substitute into the other. For trigonometric equations, use Pythagorean identities, and always restrict the domain of $x$ and $y$ to match the original parameter interval.

2. First Derivative $\frac{dy}{dx}$ for Parametric Equations ★★★☆☆ ⏱ 4 min

To find the slope of the tangent line to a parametric curve at a given value of $t$, we use the chain rule to relate the derivatives with respect to $t$ to the derivative of $y$ with respect to $x$. Starting from $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$, rearrange to solve for $\frac{dy}{dx}$:

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}, \quad \frac{dx}{dt} \neq 0

Special cases: If $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \neq 0$, the tangent line is horizontal. If $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} \neq 0$, the tangent line is vertical (slope undefined). This formula works even when you cannot eliminate the parameter, making it extremely versatile.

3. Second Derivative $\frac{d^2y}{dx^2}$ and Concavity ★★★★☆ ⏱ 4 min

We calculate the second derivative to analyze concavity of parametric curves, just like we do for Cartesian curves. The most common student mistake is differentiating $\frac{dy}{dx}$ with respect to $t$ and calling that the second derivative — this is incorrect. The second derivative is the derivative of the first derivative *with respect to $x$*, not $t$.

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

The sign of $\frac{d^2y}{dx^2}$ follows the same rules as Cartesian curves: positive = concave up, negative = concave down.

4. AP-Style Concept Check ★★★☆☆ ⏱ 2 min

Common Pitfalls

Why: You forget the original parameter interval restricts $x$ to only a portion of the full Cartesian curve.

Why: You mixed up the order of the ratio when recalling the formula from memory.

Why: You incorrectly generalized the first derivative ratio pattern to second derivatives.

Why: You mixed up the conditions for horizontal and vertical tangents.

Why: You forgot $\frac{dx}{dt}$ can be negative, which flips the sign of the final second derivative.

Why: You assume two curves with the same points are identical, but orientation matters for motion problems.

Quick Reference Cheatsheet

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