# 7.2 Calculus of parametric curves  (Page 5/6)

 Page 5 / 6
$\frac{d}{dx}{\int }_{a}^{x}f\left(u\right)\phantom{\rule{0.2em}{0ex}}du=f\left(x\right).$

Therefore

$\begin{array}{cc}\hfill {s}^{\prime }\left(t\right)& =\frac{d}{dt}\left[s\left(t\right)\right]\hfill \\ & =\frac{d}{dt}\left[{\int }_{0}^{t}\sqrt{{140}^{2}+{\left(-32v+2\right)}^{2}}dv\right]\hfill \\ & =\sqrt{{140}^{2}+{\left(-32t+2\right)}^{2}}\hfill \\ & =\sqrt{1024{t}^{2}-128t+19604}\hfill \\ & =2\sqrt{256{t}^{2}-32t+4901}.\hfill \end{array}$

One third of a second after the ball leaves the pitcher’s hand, the distance it travels is equal to

$\begin{array}{cc}\hfill s\left(\frac{1}{3}\right)& =\left(\frac{1\text{/}3}{2}-\frac{1}{32}\right)\sqrt{1024{\left(\frac{1}{3}\right)}^{2}-128\left(\frac{1}{3}\right)+19604}\hfill \\ & \phantom{\rule{0.6em}{0ex}}-\frac{1225}{4}\text{ln}|\left(-32\left(\frac{1}{3}\right)+2\right)+\sqrt{1024{\left(\frac{1}{3}\right)}^{2}-128\left(\frac{1}{3}\right)+19604}|\hfill \\ & \phantom{\rule{0.6em}{0ex}}+\frac{\sqrt{19604}}{32}+\frac{1225}{4}\text{ln}\left(2+\sqrt{19604}\right)\hfill \\ & \approx 46.69\phantom{\rule{0.2em}{0ex}}\text{feet}.\hfill \end{array}$

This value is just over three quarters of the way to home plate. The speed of the ball is

${s}^{\prime }\left(\frac{1}{3}\right)=2\sqrt{256{\left(\frac{1}{3}\right)}^{2}-16\left(\frac{1}{3}\right)+4901}\approx 140.34\phantom{\rule{0.2em}{0ex}}\text{ft/s}.$

This speed translates to approximately 95 mph—a major-league fastball.

## Surface area generated by a parametric curve

Recall the problem of finding the surface area of a volume of revolution. In Curve Length and Surface Area , we derived a formula for finding the surface area of a volume generated by a function $y=f\left(x\right)$ from $x=a$ to $x=b,$ revolved around the x -axis:

$S=2\pi {\int }_{a}^{b}f\left(x\right)\sqrt{1+{\left({f}^{\prime }\left(x\right)\right)}^{2}}dx.$

We now consider a volume of revolution generated by revolving a parametrically defined curve $x=x\left(t\right),y=y\left(t\right),a\le t\le b$ around the x -axis as shown in the following figure.

The analogous formula for a parametrically defined curve is

$S=2\pi {\int }_{a}^{b}y\left(t\right)\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}}dt$

provided that $y\left(t\right)$ is not negative on $\left[a,b\right].$

## Finding surface area

Find the surface area of a sphere of radius r centered at the origin.

$x\left(t\right)=r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t,\phantom{\rule{1em}{0ex}}y\left(t\right)=r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t,\phantom{\rule{1em}{0ex}}0\le t\le \pi .$

This generates an upper semicircle of radius r centered at the origin as shown in the following graph.

When this curve is revolved around the x -axis, it generates a sphere of radius r . To calculate the surface area of the sphere, we use [link] :

$\begin{array}{cc}\hfill S& =2\pi {\int }_{a}^{b}y\left(t\right)\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}}dt\hfill \\ & =2\pi {\int }_{0}^{\pi }r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t\sqrt{{\left(\text{−}r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t\right)}^{2}+{\left(r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t\right)}^{2}}dt\hfill \\ & =2\pi {\int }_{0}^{\pi }r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t\sqrt{{r}^{2}{\text{sin}}^{2}t+{r}^{2}{\text{cos}}^{2}t}\phantom{\rule{0.2em}{0ex}}dt\hfill \\ & =2\pi {\int }_{0}^{\pi }r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t\sqrt{{r}^{2}\left({\text{sin}}^{2}t+{\text{cos}}^{2}t\right)}dt\hfill \\ & =2\pi {\int }_{0}^{\pi }{r}^{2}\text{sin}\phantom{\rule{0.2em}{0ex}}t\phantom{\rule{0.2em}{0ex}}dt\hfill \\ & =2\pi {r}^{2}\left({\text{−}\text{cos}\phantom{\rule{0.2em}{0ex}}t|}_{0}^{\pi }\right)\hfill \\ & =2\pi {r}^{2}\left(\text{−}\text{cos}\phantom{\rule{0.2em}{0ex}}\pi +\text{cos}\phantom{\rule{0.2em}{0ex}}0\right)\hfill \\ & =4\pi {r}^{2}.\hfill \end{array}$

This is, in fact, the formula for the surface area of a sphere.

Find the surface area generated when the plane curve defined by the equations

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

is revolved around the x -axis.

$A=\frac{\pi \left(494\sqrt{13}+128\right)}{1215}$

## Key concepts

• The derivative of the parametrically defined curve $x=x\left(t\right)$ and $y=y\left(t\right)$ can be calculated using the formula $\frac{dy}{dx}=\frac{{y}^{\prime }\left(t\right)}{{x}^{\prime }\left(t\right)}.$ Using the derivative, we can find the equation of a tangent line to a parametric curve.
• The area between a parametric curve and the x -axis can be determined by using the formula $A={\int }_{{t}_{1}}^{{t}_{2}}y\left(t\right){x}^{\prime }\left(t\right)\phantom{\rule{0.2em}{0ex}}dt.$
• The arc length of a parametric curve can be calculated by using the formula $s={\int }_{{t}_{1}}^{{t}_{2}}\sqrt{{\left(\frac{dx}{dt}\right)}^{2}+{\left(\frac{dy}{dt}\right)}^{2}}dt.$
• The surface area of a volume of revolution revolved around the x -axis is given by $S=2\pi {\int }_{a}^{b}y\left(t\right)\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}}dt.$ If the curve is revolved around the y -axis, then the formula is $S=2\pi {\int }_{a}^{b}x\left(t\right)\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}}dt.$

## Key equations

• Derivative of parametric equations
$\frac{dy}{dx}=\frac{dy\text{/}dt}{dx\text{/}dt}=\frac{{y}^{\prime }\left(t\right)}{{x}^{\prime }\left(t\right)}$
• Second-order derivative of parametric equations
$\frac{{d}^{2}y}{d{x}^{2}}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{\left(d\text{/}dt\right)\left(dy\text{/}dx\right)}{dx\text{/}dt}$
• Area under a parametric curve
$A={\int }_{a}^{b}y\left(t\right){x}^{\prime }\left(t\right)\phantom{\rule{0.2em}{0ex}}dt$
• Arc length of a parametric curve
$s={\int }_{{t}_{1}}^{{t}_{2}}\sqrt{{\left(\frac{dx}{dt}\right)}^{2}+{\left(\frac{dy}{dt}\right)}^{2}}dt$
• Surface area generated by a parametric curve
$S=2\pi {\int }_{a}^{b}y\left(t\right)\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}}dt$

For the following exercises, each set of parametric equations represents a line. Without eliminating the parameter, find the slope of each line.

how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
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