Chapter 8.4, Problem 21E

To determine

To Find: The length of the curve.

Answer to Problem 21E

The length of the curve $=1$ .

Explanation of Solution

Given information: curve $y={\displaystyle \underset{0}{\overset{x}{\int }}\sqrt{\cos 2t}dt,}\text{ From }x=0\text{ to }x=\frac{\pi }{4}$ .

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 $y={\displaystyle \underset{0}{\overset{x}{\int }}\sqrt{\cos 2t}dt,}\text{ From }x=0\text{ to }x=\frac{\pi }{4}$

  $\begin{array}{l}\frac{dy}{dx}=\sqrt{\cos 2x}\\ L={\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}}dx\\ L={\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\sqrt{1+{\left(\sqrt{\cos 2x}\right)}^2}}dx\\ L={\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\sqrt{1+\cos 2x}}dx\\ \because \cos 2x=2{\cos }^{2}x-1\\ L={\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\sqrt{1+2{\cos }^{2}x-1}}dx\\ L={\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\sqrt{2{\cos }^{2}x}}dx\\ L=\sqrt{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\cos x}dx\\ L=1\end{array}$