# 7.2 Calculus of parametric curves  (Page 2/6)

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

$x\left(t\right)={t}^{2}-4t,\phantom{\rule{1em}{0ex}}y\left(t\right)=2{t}^{3}-6t,\phantom{\rule{1em}{0ex}}-2\le t\le 3$

and locate any critical points on its graph.

${x}^{\prime }\left(t\right)=2t-4$ 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 $t=±1.$

## Finding a tangent line

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

$x\left(t\right)={t}^{2}-3,\phantom{\rule{1em}{0ex}}y\left(t\right)=2t-1,\phantom{\rule{1em}{0ex}}-3\le t\le 4\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}t=2.$

First find the slope of the tangent line using [link] , 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:

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

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

$x\left(2\right)={\left(2\right)}^{2}-3=1\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}y\left(2\right)=2\left(2\right)-1=3,$

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

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

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

$x\left(t\right)={t}^{2}-4t,\phantom{\rule{1em}{0ex}}y\left(t\right)=2{t}^{3}-6t,\phantom{\rule{1em}{0ex}}-2\le t\le 3\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}t=5.$

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

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

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

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

## Finding a second derivative

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

From [link] we know that $\frac{dy}{dx}=\frac{2}{2t}=\frac{1}{t}.$ Using [link] , we obtain

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

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

$x\left(t\right)={t}^{2}-4t,\phantom{\rule{1em}{0ex}}y\left(t\right)=2{t}^{3}-6t,\phantom{\rule{1em}{0ex}}-2\le t\le 3$

and locate any critical points on its graph.

$\frac{{d}^{2}y}{d{x}^{2}}=\frac{3{t}^{2}-12t+3}{2{\left(t-2\right)}^{3}}.$ Critical points $\left(5,4\right),\left(-3,-4\right),\text{and}\phantom{\rule{0.2em}{0ex}}\left(-4,6\right).$

## Integrals involving parametric equations

Now that we have seen how to calculate the derivative of a plane curve, the next question is this: How do we find the area under a curve defined parametrically? Recall the cycloid defined by the equations $x\left(t\right)=t-\text{sin}\phantom{\rule{0.2em}{0ex}}t,\phantom{\rule{1em}{0ex}}y\left(t\right)=1-\text{cos}\phantom{\rule{0.2em}{0ex}}t.$ Suppose we want to find the area of the shaded region in the following graph.

To derive a formula for the area under the curve defined by the functions

$x=x\left(t\right),\phantom{\rule{1em}{0ex}}y=y\left(t\right),\phantom{\rule{1em}{0ex}}a\le t\le b,$

we assume that $x\left(t\right)$ is differentiable and start with an equal partition of the interval $a\le t\le b.$ Suppose ${t}_{0}=a<{t}_{1}<{t}_{2}<\text{⋯}<{t}_{n}=b$ and consider the following graph.

We use rectangles to approximate the area under the curve. The height of a typical rectangle in this parametrization is $y\left(x\left({\stackrel{–}{t}}_{i}\right)\right)$ for some value ${\stackrel{–}{t}}_{i}$ in the i th subinterval, and the width can be calculated as $x\left({t}_{i}\right)-x\left({t}_{i-1}\right).$ Thus the area of the i th rectangle is given by

${A}_{i}=y\left(x\left({\stackrel{–}{t}}_{i}\right)\right)\phantom{\rule{0.2em}{0ex}}\left(x\left({t}_{i}\right)-x\left({t}_{i-1}\right)\right).$

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