Chapter 8.4, Problem 9E

a.

To determine

To Find: The length of the curve.

Answer to Problem 9E

The length of the curve $\approx 2.057$ .

Explanation of Solution

Given information: curve $y=\sec x,\frac{-\pi }{3}\le x\le \frac{\pi }{3}$

Formula used: if a smooth curve begins at $\left(a,c\right)$ and ends at $\left(b,d\right)$

  $a<b,c>d,$ then the length of the curve is

  $L={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}}dx$

If y is a smooth function of x on $\left[a,b\right]$

  $L={\displaystyle \underset{c}{\overset{d}{\int }}\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}}dy$

If x is a smooth function of y on $\left[c,d\right]$

Calculation:

curve $y=\sec x,\frac{-\pi }{3}\le x\le \frac{\pi }{3}$

  $\frac{dy}{dx}=\sec x\tan x$

Which is continuous on $\left[-\frac{\pi }{3},\frac{\pi }{3}\right]$ . Therefore

  $\begin{array}{l}\frac{dy}{dx}=\sec x\tan x\\ L={\displaystyle \underset{\frac{-\pi }{3}}{\overset{\frac{\pi }{3}}{\int }}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}}dx\\ L={\displaystyle \underset{\frac{-\pi }{3}}{\overset{\frac{\pi }{3}}{\int }}\sqrt{1+{\left(\sec x\tan x\right)}^2}}dx\\ L\approx 3.14\end{array}$

b.

To determine

To find the graph of the curve $y=\sec x$ .

Explanation of Solution

Given information: curve $y=\sec x,\frac{-\pi }{3}\le x\le \frac{\pi }{3}$

Graph:

  

c.

To determine

To Find: Use NINT to find the length of the curve.

Answer to Problem 9E

The length of the curve $\approx 3.14$ .

Explanation of Solution

Given information: curve $y=\sec x,\frac{-\pi }{3}\le x\le \frac{\pi }{3}$

Formula used: if a smooth curve begins at $\left(a,c\right)$ and ends at $\left(b,d\right)$

  $a<b,c>d,$ then the length of the curve is

  $L={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}}dx$

If y is a smooth function of x on $\left[a,b\right]$

  $L={\displaystyle \underset{c}{\overset{d}{\int }}\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}}dy$

If x is a smooth function of y on $\left[c,d\right]$

Calculation:

curve $y=\sec x,\frac{-\pi }{3}\le x\le \frac{\pi }{3}$

  $\begin{array}{l}L={\displaystyle \underset{\frac{-\pi }{3}}{\overset{\frac{\pi }{3}}{\int }}\sqrt{1+{\left(\sec x\tan x\right)}^2}}dx\\ L\approx 3.14\text{By NINT}\end{array}$