Defining and Differentiating Parametric Equations BC only

Along a parametric curve the chain rule gives $\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}$, so $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ wherever $\frac{dx}{dt} \neq 0$; deriving it prevents the equally memorable inverted version. The result is a function of $t$ rather than of $x$, which is why these problems supply a parameter value. Tangents come from the two parts read separately: $\frac{dy}{dt} = 0$ with $\frac{dx}{dt} \neq 0$ is horizontal, $\frac{dx}{dt} = 0$ with $\frac{dy}{dt} \neq 0$ is vertical, and both zero at once means the formula says nothing and a limit is needed.

A parametrisation also carries the direction of travel, the starting point and how often each point is visited, none of which a rectangular equation records. For $x = t^{2} + 1$, $y = t^{3} - 3t$ on $[-2, 2]$ the loop is traced counterclockwise and the point $(4, 0)$ is visited twice, at $t = \pm\sqrt{3}$, with slopes that differ in sign. Eliminating the parameter destroys orientation ($x = \cos t$ with $y = \pm\sin t$ both give $x^{2} + y^{2} = 1$), the domain actually traced, and the speed of the motion, so a question about the curve may survive elimination while a question about the motion may not.