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 (t) = 2 t + 3, y (t) = 3 t - 4, - 2 \leq t \leq 3

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

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

A straight line from (−1, −10) to (9, 5). The point (−1, −10) is marked t = −2, the point (3, −4) is marked t = 0, and the point (9, 5) is marked t = 3. There are three equations marked: x(t) = 2t + 3, y(t) = 3t – 4, and −2 ≤ t ≤ 3

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

\begin{matrix} x (t) & = & 2 t + 3 x - 3 & = & 2 t t & = & \frac{x - 3}{2} . \end{matrix}

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

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

\begin{matrix} y (t) & = & 3 t - 4 y & = & 3 (\frac{x - 3}{2}) - 4 y & = & \frac{3 x}{2} - \frac{9}{2} - 4 y & = & \frac{3 x}{2} - \frac{17}{2} . \end{matrix}

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

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

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

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

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.

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

Show Solution

To apply the theorem , first calculate ${x}^{\prime }\left(t\right)$ and ${y}^{\prime }\left(t\right)\text{:}$

$\begin{array}{c}{x}^{\prime }\left(t\right)=2t\hfill \\ {y}^{\prime }\left(t\right)=2.\hfill \end{array}$

Next substitute these into the equation:

$\begin{array}{c}\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\hfill \\ \frac{dy}{dx}=\frac{2}{2t}\hfill \\ \frac{dy}{dx}=\frac{1}{t}.\hfill \end{array}$

This derivative is undefined when $t=0$ . Calculating $x\left(0\right)$ and $y\left(0\right)$ gives $x\left(0\right)={\left(0\right)}^{2}-3=-3$ and $y\left(0\right)=2\left(0\right)-1=-1$ , which corresponds to the point $\left(-3,-1\right)$ on the graph. The graph of this curve is a parabola opening to the right, and the point $\left(-3,-1\right)$ is its vertex as shown.

Figure 2. Graph of the parabola described by parametric equations in part a.
To apply the theorem, first calculate ${x}^{\prime }\left(t\right)$ and ${y}^{\prime }\left(t\right)\text{:}$

$\begin{array}{c}{x}^{\prime }\left(t\right)=2\hfill \\ {y}^{\prime }\left(t\right)=3{t}^{2}-3.\hfill \end{array}$

Next substitute these into the equation:

$\begin{array}{c}\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\hfill \\ \frac{dy}{dx}=\frac{3{t}^{2}-3}{2}.\hfill \end{array}$

This derivative is zero when $t=\pm 1$ . When $t=-1$ we have

$x\left(-1\right)=2\left(-1\right)+1=-1\text{and}y\left(-1\right)={\left(-1\right)}^{3}-3\left(-1\right)+4=-1+3+4=6$ ,

which corresponds to the point $\left(-1,6\right)$ on the graph. When $t=1$ we have

$x\left(1\right)=2\left(1\right)+1=3\text{and}y\left(1\right)={\left(1\right)}^{3}-3\left(1\right)+4=1 - 3+4=2$ ,

which corresponds to the point $\left(3,2\right)$ on the graph. The point $\left(3,2\right)$ is a relative minimum and the point $\left(-1,6\right)$ is a relative maximum, as seen in the following graph.

Figure 3. Graph of the curve described by parametric equations in part b.
To apply the theorem, first calculate ${x}^{\prime }\left(t\right)$ and ${y}^{\prime }\left(t\right)\text{:}$

$\begin{array}{c}{x}^{\prime }\left(t\right)=-5\sin{t}\hfill \\ {y}^{\prime }\left(t\right)=5\cos{t}.\hfill \end{array}$

Next substitute these into the equation:

$\begin{array}{c}\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\hfill \\ \frac{dy}{dx}=\frac{5\cos{t}}{-5\sin{t}}\hfill \\ \frac{dy}{dx}=-\cot{t}.\hfill \end{array}$

This derivative is zero when $\cos{t}=0$ and is undefined when $\sin{t}=0$ . This gives $t=0,\frac{\pi }{2},\pi ,\frac{3\pi }{2},\text{and}2\pi$ as critical points for t. Substituting each of these into $x\left(t\right)$ and $y\left(t\right)$ , we obtain

$t$ $x\left(t\right)$ $y\left(t\right)$

0 5 0

$\frac{\pi }{2}$ 0 5

$\pi$ −5 0

$\frac{3\pi }{2}$ 0 −5

$2\pi$ 5 0

These points correspond to the sides, top, and bottom of the circle that is represented by the parametric equations (Figure 4). On the left and right edges of the circle, the derivative is undefined, and on the top and bottom, the derivative equals zero.

Figure 4. Graph of the curve described by parametric equations in part c.

$t$	$x\left(t\right)$	$y\left(t\right)$
0	5	0
$\frac{\pi }{2}$	0	5
$\pi$	−5	0
$\frac{3\pi }{2}$	0	−5
$2\pi$	5	0

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.

You can view the transcript for this segmented clip of “7.2 Calculus of Parametric Curves” here (opens in new window).

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.

Hint

Calculate ${x}^{\prime }\left(t\right)$ and ${y}^{\prime }\left(t\right)$ and use the theorem.

Show Solution

${x}^{\prime }\left(t\right)=2t - 4$ and ${y}^{\prime }\left(t\right)=6{t}^{2}-6$ , so $\frac{dy}{dx}=\frac{6{t}^{2}-6}{2t - 4}=\frac{3{t}^{2}-3}{t - 2}$ .
This expression is undefined when $t=2$ and equal to zero when $t=\pm 1$ .

A curve going from (12, −4) through the origin and (−4, 0) to (−3, 36) with arrows in that order. The point (12, −4) is marked t = −2 and the point (−3, 36) is marked t = 3. On the graph there are also written three equations: x(t) = t2 – 4t, y(t) = 2t3 – 6t, and −2 ≤ t ≤ 3.

Figure 5.

Example: Finding a Tangent Line

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

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

Show Solution

First find the slope of the tangent line using the theorem, which means calculating ${x}^{\prime }\left(t\right)$ and ${y}^{\prime }\left(t\right)\text{:}$

$\begin{array}{c}{x}^{\prime }\left(t\right)=2t\hfill \\ {y}^{\prime }\left(t\right)=2.\hfill \end{array}$

Next substitute these into the equation:

$\begin{array}{c}\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\hfill \\ \frac{dy}{dx}=\frac{2}{2t}\hfill \\ \frac{dy}{dx}=\frac{1}{t}.\hfill \end{array}$

When $t=2$ , $\frac{dy}{dx}=\frac{1}{2}$ , so this is the slope of the tangent line. Calculating $x\left(2\right)$ and $y\left(2\right)$ gives

$x\left(2\right)={\left(2\right)}^{2}-3=1\text{ and }y\left(2\right)=2\left(2\right)-1=3$ ,

which corresponds to the point $\left(1,3\right)$ on the graph (Figure 6). Now use the point-slope form of the equation of a line to find the equation of the tangent line:

$\begin{array}{ccc}\hfill y-{y}_{0}& =\hfill & m\left(x-{x}_{0}\right)\hfill \\ \hfill y - 3& =\hfill & \frac{1}{2}\left(x - 1\right)\hfill \\ \hfill y - 3& =\hfill & \frac{1}{2}x-\frac{1}{2}\hfill \\ \hfill y& =\hfill & \frac{1}{2}x+\frac{5}{2}.\hfill \end{array}$

A curved line going from (6, −7) through (−3, −1) to (13, 7) with arrow pointing in that order. The point (6, −7) is marked t = −3, the point (−3, −1) is marked t = 0, and the point (13, 7) is marked t = 4. On the graph there are also written three equations: x(t) = t2 − 3, y(t) = 2t − 1, and −3 ≤ t ≤ 4. At the point (1, 3), which is marked t = 2, there is a tangent line with equation y = x/2 + 5/2.

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

You can view the transcript for this segmented clip of “7.2 Calculus of Parametric Curves” here (opens in new window).

try it

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

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

Hint

Calculate ${x}^{\prime }\left(t\right)$ and ${y}^{\prime }\left(t\right)$ and use the theorem.

Show Solution

The equation of the tangent line is $y=24x+100$ .

Try It

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

Show Solution

From the example: Finding the Derivative of a Parametric Curve we know that $\frac{dy}{dx}=\frac{2}{2t}=\frac{1}{t}$ . Using our above equation, we obtain

$\frac{{d}^{2}y}{d{x}^{2}}=\frac{\left(\frac{d}{dt}\right)\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}=\frac{\left(\frac{d}{dt}\right)\left(\frac{1}{t}\right)}{2t}=\frac{-{t}^{-2}}{2t}=-\frac{1}{2{t}^{3}}$ .

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

You can view the transcript for this segmented clip of “7.2 Calculus of Parametric Curves” here (opens in new window).

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.

Hint

Show Solution

$\frac{{d}^{2}y}{d{x}^{2}}=\frac{3{t}^{2}-12t+3}{2{\left(t - 2\right)}^{3}}$ . Critical points $\left(5,4\right),\left(-3,-4\right),\text{and}\left(-4,6\right)$ .

Module 7: Parametric Equations and Polar Coordinates