# 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)We will apply a similar property such that:

(3)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)## 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)## 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:

(9)