Second Derivative of Parametric Functions

This lesson finds the second derivative $\tfrac{d^2 y}{dx^2}$ for a curve given in parametric form, where both $x$ and $y$ are written in terms of the parameter $\theta$.

Setting up the problem

We are given the parametric equations

$x = \sin\theta, \qquad y = \cos\theta.$

Because $x$ and $y$ depend on the parameter $\theta$, we differentiate each one with respect to $\theta$ first.

Derivative of $x$

$\frac{dx}{d\theta} = \frac{d}{d\theta}(\sin\theta) = \cos\theta.$

Derivative of $y$

$\frac{dy}{d\theta} = \frac{d}{d\theta}(\cos\theta) = -\sin\theta.$

Forming the first derivative

The first derivative comes from dividing the two parameter derivatives:

$\frac{dy}{dx} = \frac{\;dy/d\theta\;}{\;dx/d\theta\;} = \frac{-\sin\theta}{\cos\theta} = -\tan\theta.$

The second derivative

To differentiate again with respect to $x$, differentiate $\tfrac{dy}{dx}$ with respect to $\theta$ and multiply by $\tfrac{d\theta}{dx}$:

$\frac{d^2 y}{dx^2} = \frac{d}{d\theta}\left(-\tan\theta\right)\cdot \frac{d\theta}{dx}.$

Differentiating gives $\tfrac{d}{d\theta}(-\tan\theta) = -\sec^2\theta$. Since $\tfrac{dx}{d\theta} = \cos\theta$, its reciprocal is

$\frac{d\theta}{dx} = \frac{1}{\cos\theta} = \sec\theta.$

Multiplying the two factors:

$\frac{d^2 y}{dx^2} = -\sec^2\theta \cdot \sec\theta = -\sec^3\theta.$

Key takeaways

• For parametric curves, find $\tfrac{dx}{d\theta}$ and $\tfrac{dy}{d\theta}$ first, then divide to get $\tfrac{dy}{dx}$.
• The second derivative is $\tfrac{d}{d\theta}\!\left(\tfrac{dy}{dx}\right)\cdot \tfrac{d\theta}{dx}$, not just differentiating $\tfrac{dy}{dx}$ a second time.
• Here the answer simplifies to $\tfrac{d^2 y}{dx^2} = -\sec^3\theta$.