How to find the second derivative of a parametric curve

Formula for the second derivative of a parametric curve

To find the second derivative of a parametric curve, we need to find its first derivative $dy/dx$, using the formula

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

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and then plug it into this formula for the second derivative:

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

where $d^2y/dx^2$ is the second derivative of the parametric curve, $dy/dx$ is its first derivative and $dx/dt$ is the first derivative of the equation for $x$. The $d/dt$ is notation that tells us to take the derivative of $dy/dx$ with respect to $t$.

How to calculate the second derivative of a parametric curve

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Calculating the second derivative of the parametric equation

Example

Find the second derivative of the parametric curve.

$x=5t^3+6t$

$y=t^4-3$

To find the first derivative, we’ll solve for $dx/dt$ and $dy/dt$.

This means we will have to solve for $dy/dt$, $dx/dt$ and $dy/dx$ first. Let’s start with $dy/dt$.

$x=5t^3+6t$

$\frac{dx}{dt}=15t^2+6$

and

$y=t^4-3$

$\frac{dy}{dt}=4t^3$

Plugging these two derivatives into the formula for the first derivative,

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

we get

$\frac{dy}{dx}=\frac{4t^3}{15t^2+6}$

The d/dt is notation that tells us to take the derivative of dy/dx with respect to t.

Now plugging the first derivative $dy/dx$ and the value we found earlier for $dx/dt$ into the formula for the second derivative,

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

we get

$\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{4t^3}{15t^2+6}\right)}{15t^2+6}$

We’ll use quotient rule to take the derivative of $dy/dx$ with respect to $t$.

$\frac{d^2y}{dx^2}=\frac{\frac{\left(12t^2\right)\left(15t^2+6\right)-\left(4t^3\right)\left(30t\right)}{\left(15t^2+6\right)^2}}{15t^2+6}$

$\frac{d^2y}{dx^2}=\frac{\left(12t^2\right)\left(15t^2+6\right)-\left(4t^3\right)\left(30t\right)}{\left(15t^2+6\right)^2}\cdot\frac{1}{15t^2+6}$

$\frac{d^2y}{dx^2}=\frac{180t^4+72t^2-120t^4}{\left(15t^2+6\right)^3}$

$\frac{d^2y}{dx^2}=\frac{60t^4+72t^2}{\left(15t^2+6\right)^3}$

$\frac{d^2y}{dx^2}=\frac{12t^2\left(5t^2+6\right)}{\left(15t^2+6\right)^3}$

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