## Derivatives of Parametric Equations

### Learning Outcomes

• Determine derivatives and equations of tangents for parametric curves

We start by asking how to calculate the slope of a line tangent to a parametric curve at a point. Consider the plane curve defined by the parametric equations

$x\left(t\right)=2t+3,y\left(t\right)=3t - 4,-2\le t\le 3$.

The graph of this curve appears in Figure 1. It is a line segment starting at $\left(-1,-10\right)$ and ending at $\left(9,5\right)$.

We can eliminate the parameter by first solving the equation $x\left(t\right)=2t+3$ for t:

$\begin{array}{ccc}\hfill x\left(t\right)& =\hfill & 2t+3\hfill \\ \hfill x - 3& =\hfill & 2t\hfill \\ \hfill t& =\hfill & \frac{x - 3}{2}.\hfill \end{array}$

Substituting this into $y\left(t\right)$, we obtain

$\begin{array}{ccc}\hfill y\left(t\right)& =\hfill & 3t - 4\hfill \\ \hfill y& =\hfill & 3\left(\frac{x - 3}{2}\right)-4\hfill \\ \hfill y& =\hfill & \frac{3x}{2}-\frac{9}{2}-4\hfill \\ \hfill y& =\hfill & \frac{3x}{2}-\frac{17}{2}.\hfill \end{array}$

The slope of this line is given by $\frac{dy}{dx}=\frac{3}{2}$. Next we calculate ${x}^{\prime }\left(t\right)$ and ${y}^{\prime }\left(t\right)$. This gives ${x}^{\prime }\left(t\right)=2$ and ${y}^{\prime }\left(t\right)=3$. Notice that $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{3}{2}$. This is no coincidence, as outlined in the following theorem.

### theorem: Derivative of Parametric Equations

Consider the plane curve defined by the parametric equations $x=x\left(t\right)$ and $y=y\left(t\right)$. Suppose that ${x}^{\prime }\left(t\right)$ and ${y}^{\prime }\left(t\right)$ exist, and assume that ${x}^{\prime }\left(t\right)\ne 0$. Then the derivative $\frac{dy}{dx}$ is given by

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{{y}^{\prime }\left(t\right)}{{x}^{\prime }\left(t\right)}$.

#### Proof

This theorem can be proven using the Chain Rule. In particular, assume that the parameter t can be eliminated, yielding a differentiable function $y=F\left(x\right)$. Then $y\left(t\right)=F\left(x\left(t\right)\right)$. Differentiating both sides of this equation using the Chain Rule yields

${y}^{\prime }\left(t\right)={F}^{\prime }\left(x\left(t\right)\right){x}^{\prime }\left(t\right)$,

so

${F}^{\prime }\left(x\left(t\right)\right)=\frac{{y}^{\prime }\left(t\right)}{{x}^{\prime }\left(t\right)}$.

But ${F}^{\prime }\left(x\left(t\right)\right)=\frac{dy}{dx}$, which proves the theorem.

$_\blacksquare$

The theorem can be used to calculate derivatives of plane curves, as well as critical points. Recall that a critical point of a differentiable function $y=f\left(x\right)$ is any point $x={x}_{0}$ such that either ${f}^{\prime }\left({x}_{0}\right)=0$ or ${f}^{\prime }\left({x}_{0}\right)$ does not exist. The theorem gives a formula for the slope of a tangent line to a curve defined parametrically regardless of whether the curve can be described by a function $y=f\left(x\right)$ or not.

### Example: Finding the Derivative of a Parametric Curve

Calculate the derivative $\frac{dy}{dx}$ for each of the following parametrically defined plane curves, and locate any critical points on their respective graphs.

1. $x\left(t\right)={t}^{2}-3,y\left(t\right)=2t - 1,-3\le t\le 4$
2. $x\left(t\right)=2t+1,y\left(t\right)={t}^{3}-3t+4,-2\le t\le 5$
3. $x\left(t\right)=5\cos{t},y\left(t\right)=5\sin{t},0\le t\le 2\pi$

Watch the following video to see the worked solution to Example: Finding the Derivative of a Parametric Curve.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

### try it

Calculate the derivative $\frac{dy}{dx}$ for the plane curve defined by the equations

$x\left(t\right)={t}^{2}-4t,y\left(t\right)=2{t}^{3}-6t,-2\le t\le 3$

and locate any critical points on its graph.

### Example: Finding a Tangent Line

Find the equation of the tangent line to the curve defined by the equations

$x\left(t\right)={t}^{2}-3,y\left(t\right)=2t - 1,-3\le t\le 4\text{ when }t=2$.

Watch the following video to see the worked solution to Example: Finding a Tangent Line.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

### try it

Find the equation of the tangent line to the curve defined by the equations

$x\left(t\right)={t}^{2}-4t,y\left(t\right)=2{t}^{3}-6t,-2\le t\le 3\text{when}t=5$.

## Second-Order Derivatives

Our next goal is to see how to take the second derivative of a function defined parametrically. The second derivative of a function $y=f\left(x\right)$ is defined to be the derivative of the first derivative; that is,

$\frac{{d}^{2}y}{d{x}^{2}}=\frac{d}{dx}\left[\frac{dy}{dx}\right]$.

Since $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$, we can replace the $y$ on both sides of this equation with $\frac{dy}{dx}$. This gives us

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

If we know $\frac{dy}{dx}$ as a function of t, then this formula is straightforward to apply.

### Example: Finding a Second Derivative

Calculate the second derivative $\frac{{d}^{2}y}{d{x}^{2}}$ for the plane curve defined by the parametric equations $x\left(t\right)={t}^{2}-3,y\left(t\right)=2t - 1,-3\le t\le 4$.

Watch the following video to see the worked solution to Example: Finding a Second Derivative.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

### try it

Calculate the second derivative $\frac{{d}^{2}y}{d{x}^{2}}$ for the plane curve defined by the equations

$x\left(t\right)={t}^{2}-4t,y\left(t\right)=2{t}^{3}-6t,-2\le t\le 3$

and locate any critical points on its graph.