Concavity of Parametric Curves

Recall that when we have a function $f$, we could determine intervals where $f$ was concave up and concave down by looking at the second derivative of $f$. The same sort of intuition can be applied to a parametric curve $C$ defined by the equations $x = x(t)$ and $y = y(t)$. Recall that the first derivative of the curve $C$ can be calculated by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. If we take the second derivative of $C$, then we can now calculate intervals where $C$ is concave up or concave down.

(1)

\begin{align} \frac{d^2y}{dx^2} = \frac{d}{dx} \left ( \frac{dy}{dx} \right) = \frac{\frac{d}{dt} \left (\frac{dy}{dx} \right)}{\frac{dx}{dt}} \end{align}

Now let's look at some examples of calculating the second derivative of parametric curves.

Example 1

Determine the second derivative of the parametric curve defined by $x = 3t^2 - 2t$ and $y = 4t^3 - 2t$.

Let's first find the first derivative $\frac{dy}{dx}$:

(2)

\begin{align} \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{12t^2 - 2}{6t - 2} \end{align}

And so, we will now calculate the second derivative:

(3)

\begin{align} \frac{\frac{d}{dt} \left (\frac{dy}{dx} \right)}{\frac{dx}{dt}} = \frac{\frac{d}{dt} \left (\frac{12t^2 - 2}{6t - 2} \right)}{\frac{dx}{dt}} = \frac{\frac{d}{dt} \left (\frac{12t^2 - 2}{6t - 2} \right)}{6t - 2} \end{align}

We will have to use the quotient rule this time:

(4)

\begin{align} \frac{d^2y}{dx^2} = \frac{(6t - 2)(24t) - (12t^2 - 2)(6)}{6t - 2} \\ \frac{d^2y}{dx^2} = \frac{144t^2 -48t - 72t^2 +12}{6t - 2} \\ \frac{d^2y}{dx^2} = \frac{72t^2 - 48t + 12}{6t - 2} \end{align}

Example 2

Let $C$ be a parametric curve defined by the equations $x = t^2 + t$ and $y = 2t^3 - 1$. Find intervals when $C$ is concave up and concave down.

We will first calculate the first derivative, $\frac{dy}{dx}$ as follows:

(5)

\begin{align} \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{6t^2}{2t + 1} \end{align}

Now let's calculate the second derivative by first differentiating $\frac{dy}{dx}$ with respect to $t$:

(6)

\begin{align} \frac{d}{dt} \cdot \frac{dy}{dx} = \frac{(2t + 1)(12t) - 2(6t^2)}{(2t + 1)^2} \end{align}

Therefore, since $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\frac{dy}{dt}}{\frac{dx}{dt}}$, it follows that:

(7)

\begin{align} \frac{d^2y}{dx^2} = \frac{\frac{(2t + 1)(12t) - 2(6t^2)}{(2t + 1)^2}}{2t + 1} \\ \frac{d^2y}{dx^2} = \frac{(2t + 1)(12t) - 2(6t^2)}{(2t + 1)^3} \\ \frac{d^2y}{dx^2} = \frac{24t^2 + 12t - 12t^2}{(2t + 1)^3} \\ \frac{d^2y}{dx^2} = \frac{12t(t + 1)}{(2t + 1)^3} \\ \end{align}

We now want to know when the second derivative is positive and when the second derivative is negative. Note that the numerator is positive when $t < -1$ or $t > 0$ and negative when $-1 < t < 0$. Meanwhile, the denominator is positive when $t > -0.5$ and negative when $t < -0.5$. The diagram below makes this illustration:

When $-1 < t < -0.5$ or $0 < t$, then $C$ is concave up.

When $t < -1$ or $-0.5 < t < 0$, then $C$ is concave down.

Example 3

Let $C$ be the parametric curve defined by the equations $x = t^2 + t$ and $y = 2t - 1$. Determine intervals where $C$ is concave up or concave down.

We first calculate the first derivative for $C$, that is:

(8)

\begin{align} \frac{dy}{dx} = \frac{2}{2t + 1} \end{align}

Now we must compute the second derivative for $C$. Note that $\frac{d}{dt} \frac{dy}{dx} = \frac{(2t + 1)(0) - 2(2)}{(2t + 1)^2}$ by the quotient rule. Therefore:

(9)

\begin{align} \frac{d^2y}{dx^2} = \frac{\frac{d}{dt} \frac{dy}{dx}}{\frac{dx}{dt}} = \frac{-4}{(2t + 1)^2} \cdot \frac{1}{2t + 1} \\ \frac{d^2y}{dx^2} = -\frac{4}{(2t + 1)^3} \end{align}

We note that when $t < -0.5$, our second derivative is positive, while when $t > 0.5$, our second derivative is negative. Hence, $C$ is concave up for $t < -0.5$ and concave down for $t > 0.5$.