Second Derivatives of Parametric Equations (College Board AP® Calculus BC): Revision Note

It helps to note that the formulas for both the first and second parametric derivatives divide by $\frac{dx}{dt}$.

Common error: $\frac{d^{2}y}{d{x}^2}\ne \frac{d^{2}y/d{t}^2}{d^{2}x/d{t}^2}$. The formula requires differentiating the first derivative $\frac{dy}{dx}$ , which is itself a function of $t$, then dividing by $\frac{dx}{dt}$. Don't just differentiate the top and bottom of the first-derivative quotient separately.

You can find the sign of $\frac{d^{2}y}{d{x}^2}$ by substituting in the critical $t$-value without necessarily needing to fully simplify the algebraic expression for $\frac{d^{2}y}{d{x}^2}$.

When calculating the parametric second derivative, be prepared to use the quotient rule.

A curve is given parametrically by

$\begin{array}{rcl}x& =& t^2+t\\ y& =& t^3-3t\end{array}$

(a) Find and simplify an expression for $\frac{d^{2}y}{d{x}^2}$ in terms of $t$.

(b) Find the coordinates of any local extrema on the curve, and classify the nature of these extrema.

Answer:

(a)

Start by finding the parametric first derivative, given by $\frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}$

$\frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}=\frac{3t^2-3}{2t+1}$

The parametric second derivative is given by $\frac{d^{2}y}{d{x}^2}=\frac{{\displaystyle \frac{d}{dt}}\left({\displaystyle \frac{dy}{dx}}\right)}{{\displaystyle \frac{dx}{dt}}}$ where the numerator means differentiate $\frac{dy}{dx}$ from above, $\frac{3t^2-3}{2t+1}$, with respect to $t$

This requires the quotient rule, $\frac{u^{\text{'}}v-u{v}^{\text{'}}}{v^2}$, with $u=3t^2-3$ and $v=2t+1$

$\begin{array}{rcl}\frac{d}{dt}\left(\frac{dy}{dx}\right)& =& \frac{\left(6t\right)\left(2t+1\right)-\left(3t^2-3\right)\left(2\right)}{{\left(2t+1\right)}^2}\\ & =& \frac{12t^2+6t-6t^2+6}{{\left(2t+1\right)}^2}\\ & =& \frac{6t^2+6t+6}{{\left(2t+1\right)}^2}\end{array}$

The formula, $\frac{d^{2}y}{d{x}^2}=\frac{{\displaystyle \frac{d}{dt}}\left({\displaystyle \frac{dy}{dx}}\right)}{{\displaystyle \frac{dx}{dt}}}$, says to divide the expression above by $\frac{dx}{dt}$ and the question says to simplify the answer

$\begin{array}{rcl}\frac{d^{2}y}{d{x}^2}& =& \frac{\frac{6t^2+6t+6}{{\left(2t+1\right)}^2}}{2t+1}\\ & =& \frac{6t^2+6t+6}{{\left(2t+1\right)}^2}÷\left(2t+1\right)\\ & =& \frac{6t^2+6t+6}{{\left(2t+1\right)}^2}÷\frac{\left(2t+1\right)}{1}\\ & =& \frac{6t^2+6t+6}{{\left(2t+1\right)}^2}\times \frac{1}{\left(2t+1\right)}\\ & =& \frac{6t^2+6t+6}{{\left(2t+1\right)}^3}\end{array}$

You could also factor out a 6 from the numerator (this is an optional step)

$\frac{d^{2}y}{d{x}^2}=\frac{6\left(t^2+t+1\right)}{{\left(2t+1\right)}^3}$

(b)

Local extrema occur when $\frac{dy}{dx}=0$, so solve $\frac{dy}{dx}=0$

$\frac{3t^2-3}{2t+1}=0$

The left-hand side will be zero when the numerator is equal to zero

$\begin{array}{rcl}3t^2-3& =& 0\\ t^2& =& 1\\ t& =& \pm 1\end{array}$

To find the coordinates of the critical points, substitute $t=\pm 1$ into the parametric equations $\begin{array}{rcl}x& =& t^2+t\end{array}$ and $y=t^3-3t$

$\left(2,-2\right)$ and $\left(0,2\right)$

To determine the type of extrema, substitute $t=\pm 1$ into the parametric second derivative

$t=1$ gives $\frac{d^{2}y}{d{x}^2}=\frac{6\left(1^2+1+1\right)}{{\left(2\times 1+1\right)}^3}=\frac{2}{3}>0$, local minimum

$t=-1$ gives $\frac{d^{2}y}{d{x}^2}=\frac{6\left({\left(-1\right)}^2+\left(-1\right)+1\right)}{{\left(2\left(-1\right)+1\right)}^3}=-6<0$, local maximum

Write out the answer, relating the correct set of coordinates to the correct extremum

$\left(2,-2\right)$ is a local minimum

$\left(0,2\right)$ is a local maximum