Derivatives for Parametric Curves

# Derivatives for Parametric Curves

Suppose we have a parametric curve defined by the functions $x = x(t)$ and $y = y(t)$, and we want to find a derivative to tell us the slope of the tangent at various points on the curve. We must come up with a new rule to find a derivative for these parametric curves.

Recall that we can obtain two derivatives, one for the rate of change of $x$ with respect to $t$, and one for the rate of change of $y$ with respect to $t$ (with $t$ being our parameter). That is we can obtain:

(1)

But we want to be able to determine the rate of change of $x$ with respect to $x$, or rather $\frac{dy}{dx}$. Recall the chain rule in Leibniz notation which says:

(2)
\begin{align} \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \end{align}

We will apply a similar property such that:

(3)
\begin{align} \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \end{align}

In a sense, this can be visualized by cancelling out the "$dt$'s", though not entirely accurate, it is a good way to memorize this for finding derivatives of parametric curves.

## Example 1

Find the derivative for the parametric curves defined by $x = 2\cos t$ and $y = 3x^2$.

Let's first find dx/dt and dy/dt:

(4)

Hence since $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$, it follows that:

(5)
\begin{align} \frac{dy}{dx} = \frac{6t}{-2 \sin t} \\ \frac{dy}{dx} = - \frac{3t}{\sin t} \end{align}

## Example 2

Find the derivative for the parametric curves defined by $x = 2x^8 - \sin t$ and $y = e^t$.

Finding $\frac{dx}{dt}$ and $\frac{dy}{dt}$ we obtain:

(6)

Thus $\frac{dy}{dx}$ is as follows:

(7)
\begin{align} \frac{dy}{dx} = \frac{e^t}{16t^7 - \cos t} \end{align}

## Example 3

Find the derivative for the parametric curves defined by $x = 3t \cos t$ and $y = \sin 2t$.

Finding $\frac{dx}{dt}$ and $\frac{dy}{dt}$ we obtain:

(8)
Thus $\frac{dy}{dx}$ is as follows: