# Parametric derivative

In calculus, a parametric derivative is a derivative of a dependent variable y with respect to an independent variable x that is taken when both variables depend on an independent third variable t, usually thought of as "time" (that is, when x and y are given by parametric equations in t ).

## First derivative

Let ${\displaystyle x(t)\,}$ and ${\displaystyle y(t)\,}$ be the coordinates of the points of the curve expressed as functions of a variable t:

${\displaystyle y=y(t),\quad x=x(t).}$

The first derivative implied by these parametric equations is

${\displaystyle {\frac {dy}{dx}}={\frac {dy/dt}{dx/dt}}={\frac {{\dot {y}}(t)}{{\dot {x}}(t)}},}$

where the notation ${\displaystyle {\dot {x}}(t)}$ denotes the derivative of x with respect to t, for example. This can be derived using the chain rule for derivatives:

${\displaystyle {\frac {dy}{dt}}={\frac {dy}{dx}}\cdot {\frac {dx}{dt}}}$

and dividing both sides by ${\displaystyle {\frac {dx}{dt}}}$ to give the equation above.

In general all of these derivatives — dy / dt, dx / dt, and dy / dx — are themselves functions of t and so can be written more explicitly as, for example, ${\displaystyle {\tfrac {dy}{dx}}(t).}$

## Second derivative

The second derivative implied by a parametric equation is given by

 ${\displaystyle {\frac {d^{2}y}{dx^{2}}}}$ ${\displaystyle ={\frac {d}{dx}}\left({\frac {dy}{dx}}\right)}$ ${\displaystyle ={\frac {d}{dt}}\left({\frac {dy}{dx}}\right)\cdot {\frac {dt}{dx}}}$ ${\displaystyle ={\frac {d}{dt}}\left({\frac {\dot {y}}{\dot {x}}}\right){\frac {1}{\dot {x}}}}$ ${\displaystyle ={\frac {{\dot {x}}{\ddot {y}}-{\dot {y}}{\ddot {x}}}{{\dot {x}}^{3}}}}$

by making use of the quotient rule for derivatives. The latter result is useful in the computation of curvature.

## Example

For example, consider the set of functions where:

${\displaystyle x(t)=4t^{2}\,}$

and

${\displaystyle y(t)=3t.\,}$

Differentiating both functions with respect to t leads to

${\displaystyle {\frac {dx}{dt}}=8t}$

and

${\displaystyle {\frac {dy}{dt}}=3,}$

respectively. Substituting these into the formula for the parametric derivative, we obtain

${\displaystyle {\frac {dy}{dx}}={\frac {\dot {y}}{\dot {x}}}={\frac {3}{8t}},}$

where ${\displaystyle {\dot {x}}}$ and ${\displaystyle {\dot {y}}}$ are understood to be functions of t.