Topic 9.1 Defining and Differentiating Parametric Equations: define a curve parametrically and find dy/dx as the ratio of the parametric derivatives (BC).

What this topic is asking

The College Board (Topic 9.1, BC only) opens Unit 9 with parametric equations, a way of describing a curve by giving $x$ and $y$ each as a function of a third variable, the parameter $t$ (often time). Instead of $y = f(x)$, you have $x = x(t)$ and $y = y(t)$. The first calculus skill is finding the slope $\frac{dy}{dx}$ of such a curve without eliminating the parameter.

Parametric curves and the slope formula

A worked slope and point

Horizontal and vertical tangents

Parametric form makes special tangents easy to spot from the two derivatives separately. A horizontal tangent occurs where $\frac{dy}{dt} = 0$ while $\frac{dx}{dt}\neq 0$, since then $\frac{dy}{dx} = 0$. A vertical tangent occurs where $\frac{dx}{dt} = 0$ while $\frac{dy}{dt}\neq 0$, since the slope formula has a zero denominator and the slope is undefined. When both derivatives are zero at the same $t$, the point may be a cusp or corner and requires more care, but on the AP exam you usually just identify horizontal and vertical tangents by checking each numerator and denominator. This is cleaner than the $y = f(x)$ approach, where vertical tangents are awkward to detect.

A worked tangent line

Eliminating the parameter (and when not to)

Sometimes you can recover the Cartesian equation by eliminating $t$: solve one equation for $t$ and substitute, or use an identity. For $x = \cos t$, $y = \sin t$, squaring and adding gives $x^2 + y^2 = 1$, the unit circle. This is useful for recognizing the shape, but it is usually unnecessary and often impossible for differentiation, because the slope formula $\frac{dy/dt}{dx/dt}$ works directly. The exam rewards using the parametric slope formula rather than converting, especially since conversion can lose information about direction and the range of $t$. Keep elimination as a tool for identifying the curve, not as a prerequisite for calculus on it.