Derivatives of variables defined by parametric equations

So far $y$ has been expressed either explicitly in terms of $x$, or implicitly via an equation directly relating $x$ and $y$. A third common situation is parametric equations: both $x$ and $y$ are given separately as functions of a third, auxiliary variable $t$,

called the parameter. As $t$ ranges over some domain $[a,b]$, the pair $(x,y)=(f(t),g(t))$ traces out a curve in the plane; this specification of the relationship between $x$ and $y$ via $t$ is described as parametric. Recovering a single direct equation connecting $x$ and $y$ alone — by eliminating $t$ — is called elimination of the parameter; the point of using a parameter in the first place is often that a genuinely two-variable relationship becomes much easier to describe and differentiate through one auxiliary variable $t$.

Worked illustration — the circle. The circle $x^2+y^2=r^2$ (centre the origin, radius $r$) has the parametric form $x=r\cos t,\ y=r\sin t$; eliminating $t$ via $\cos^2t+\sin^2t=1$ recovers $x^2+y^2=r^2$ exactly.

Differentiating a parametric pair. If $y$ is regarded as (ultimately) a function of $x$ through the parameter $t$, the chain rule gives

and symmetrically, if instead $x$ is regarded as a function of $y$,

— the two are reciprocals of each other, exactly as ordinary $dy/dx$ and $dx/dy$ are.

For the circle $x=r\cos t,\ y=r\sin t$: $\dfrac{dx}{dt}=-r\sin t$ and $\dfrac{dy}{dt}=r\cos t$, so the slope of the tangent to the circle at parameter value $t$ is

Worked illustration — a non-trivial pair. For $x=at^2,\ y=2at$ ($t\ne0$): $\dfrac{dx}{dt}=2at$ and $\dfrac{dy}{dt}=2a$, so $\dfrac{dy}{dx}=\dfrac{2a}{2at}=\dfrac1t$ — no elimination of $t$ was needed at all to get the slope in terms of the parameter. …