# 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)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)And so, we will now calculate the second derivative:

(3)We will have to use the quotient rule this time:

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

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

(7)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)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)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 $t > 0.5$.