The idea

Parametric differentiation

Finding dy/dx for a curve given parametrically as dy/dx = (dy/dt) / (dx/dt), using it for tangents and normals, and the second derivative of a parametric curve.

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Pure · Differentiation

Parametric differentiation

Finding dy/dx for a curve given parametrically as dy/dx = (dy/dt) / (dx/dt), using it for tangents and normals, and the second derivative of a parametric curve.

Why it works

The bridge

A parametric curve gives

x

and

y

each as a function of the parameter

t

, not of each other. To find the gradient

\frac{dy}{dx}

you need the rate of change of

y

against

x

— but all you can differentiate directly is each coordinate against

t

. The chain rule bridges the gap:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

Differentiate $y$ with respect to $t$ , differentiate $x$ with respect to $t$ , and divide.

The gradient lives on t

Take

x = t^2,\ y = t^3

. Then

\frac{dx}{dt} = 2t

and

\frac{dy}{dt} = 3t^2

, so

\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}.

The gradient comes out in terms of

t

— which is natural, since each point of the curve is labelled by its

t

-value. To get the gradient at a specific point, find the

t

for that point and substitute. (Where

\frac{dx}{dt} = 0

the tangent is vertical and

\frac{dy}{dx}

is undefined — here that is

t = 0

, the cusp at the origin.)

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