Chapter 7.4, Problem 2E

(a)

To determine

To set: An integral for the length of the curve $y=\tan x,-\pi /3\le x\le 0$ needs to be determined.

Answer to Problem 2E

The length of the curve is $L={\displaystyle \underset{-\pi /3}{\overset{0}{\int }}\sqrt{1+{\sec }^{2}x}dx}$ .

Explanation of Solution

Given information: A curve is given as $y=\tan x,-\pi /3\le x\le 0$ .

Formula used: The arc length of curve is given by $L={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{1+{\left(\frac{dy}{dx}\right)}^{2}dx}}$ where y is a smooth function of x on $\left[a,b\right]$ .

Calculation:

The given curve $y=\tan x$ is a smooth function of x on interval $\left[-\pi /3,0\right]$

so using the formula for arc length of curve one gets as:

  $\begin{array}{l}L={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{1+{\left(\frac{dy}{dx}\right)}^{2}dx}}\\ L={\displaystyle \underset{-\pi /3}{\overset{0}{\int }}\sqrt{1+\frac{d}{dx}{\left(\tan x\right)}^{2}dx}}\\ L={\displaystyle \underset{-\pi /3}{\overset{0}{\int }}\sqrt{1+{\left({\sec }^{2}x\right)}^{2}dx}};\left(\frac{d}{dx}\left(\tan x\right)={\sec }^{2}x\right)\end{array}$

Hence the length of the curve is $L={\displaystyle \underset{-\pi /3}{\overset{0}{\int }}\sqrt{1+{\sec }^{2}x}dx}$ .

(b)

To determine

To graph: The graph of the given curve needs to be determined.

Answer to Problem 2E

The graph of given curve is plotted below using Desmos graphing calculator.

  

Explanation of Solution

Given information: A curve is given as $y=\tan x,-\pi /3\le x\le 0$ .

Calculation: The graph is plotted below:

  

(c)

To determine

To find: The length of the curve needs to be determined using NINT.

Answer to Problem 2E

The length of the curve using NINT is $\text{NINT}\left(\sqrt{1+{\left({\sec }^{2}x\right)}^2},x,-\frac{\pi }{3},0\right)\approx 2.057$ .

Explanation of Solution

Given information: A curve is given as $y=\tan x,-\pi /3\le x\le 0$ .

Calculation: The length of the curve is found using NINT with the help of calculator as $\text{NINT}\left(\sqrt{1+{\left({\sec }^{2}x\right)}^2},x,-\frac{\pi }{3},0\right)\approx 2.057$ .

Hence the answer is $\text{NINT}\left(\sqrt{1+{\left({\sec }^{2}x\right)}^2},x,-\frac{\pi }{3},0\right)\approx 2.057$ .