Continuity and Differentiability

Derivatives of Parametric Functions

Understand

A curve may be given by \(x=f(t)\), \(y=g(t)\) with a parameter \(t\). Then

\[\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{g'(t)}{f\,'(t)},\quad f\,'(t)\neq0.\]

Example: \(x=a\cos t,\ y=a\sin t\) gives \(\frac{dy}{dx}=\frac{a\cos t}{-a\sin t}=-\cot t\). The circle’s slope at the point with parameter \(t\) is \(-\cot t\).

For the second derivative, differentiate \(\frac{dy}{dx}\) with respect to \(t\) and divide again by \(\frac{dx}{dt}\): \(\frac{d^2y}{dx^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\Big/\frac{dx}{dt}\).

Key Concepts

dy/dx = (dy/dt)/(dx/dt).
d²y/dx² needs one more division by dx/dt.
Substitute the parameter value at the end.

Formula Bank

Parametric derivative

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

Common Mistakes

Second derivative in parametric form

\(\frac{d^2y}{dx^2}\neq\frac{d^2y/dt^2}{d^2x/dt^2}\). Differentiate \(\frac{dy}{dx}\) with respect to \(t\), then divide by \(\frac{dx}{dt}\).

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Topic Summary

Differentiate both coordinates with respect to the parameter and divide.

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