Chapter A6, Problem 72E

(a)

To determine

To show: That $A\left(u\right)=\frac{1}{2}\cosh u\sinh u-{\displaystyle \underset{1}{\overset{\cosh u}{\int }}\sqrt{x^2-1}}dx$ .

Explanation of Solution

Given:

The given figures are as follows.

  

And,

  

Calculation:

Let the point located at $\left(\cosh x,0\right)$ be called $T$ then,

  $\begin{array}{c}A\left(u\right)=\text{Area}\text{of}\text{the}\Delta OTP-\text{Area}\text{under}\text{the}\text{curve}y=\sqrt{x^2-1}\text{from}A\text{to}T\\ =\frac{1}{2}\cosh u\sinh u-{\displaystyle \underset{1}{\overset{\cosh u}{\int }}\sqrt{x^2-1}dx}\end{array}$

Therefore, it is proved that $A\left(u\right)=\frac{1}{2}\cosh u\sinh u-{\displaystyle \underset{1}{\overset{\cosh u}{\int }}\sqrt{x^2-1}dx}$ .

(b)

To determine

To show: That $A\text{'}\left(u\right)=\frac{1}{2}$ .

Explanation of Solution

Given:

The equation is $A\left(u\right)=\frac{1}{2}\cosh u\sinh u-{\displaystyle \underset{1}{\overset{\cosh u}{\int }}\sqrt{x^2-1}dx}$ .

Calculation:

Differentiate both sides of the equation $A\left(u\right)=\frac{1}{2}\cosh u\sinh u-{\displaystyle \underset{1}{\overset{\cosh u}{\int }}\sqrt{x^2-1}dx}$ .

  $\begin{array}{c}A\text{'}\left(u\right)=\frac{1}{2}\left({\cos }^{2}hu+{\sin }^{2}hu\right)-\left(\sqrt{{\cos }^{2}hu-1}\right)\left(\sinh u\right)\\ =\frac{1}{2}{\cos }^{2}hu+\frac{1}{2}{\sin }^{2}hu-{\sin }^{2}hu\\ =\frac{1}{2}\left({\cos }^{2}hu-{\sin }^{2}hu\right)\\ =\frac{1}{2}\end{array}$

Therefore, it is proved that $A\text{'}\left(u\right)=\frac{1}{2}$ .

(c)

To determine

To solve: The equation in (b) for $A\left(u\right)$ , find the value of $A\left(0\right)$ , find the value of constant $C$ , and find the relationship of $u$ to $A\left(u\right)$ .

Answer to Problem 72E

The equation for $A\left(u\right)$ is $A\left(u\right)=\frac{u}{2}+C$ , the value of $A\left(0\right)$ is $A\left(0\right)=C$ , the value of $C$ is $0$ , and $u$ is twice of $A\left(u\right)$ .

Explanation of Solution

Given:

The equation is $A\left(u\right)=\frac{1}{2}\cosh u\sinh u-{\displaystyle \underset{1}{\overset{\cosh u}{\int }}\sqrt{x^2-1}dx}$ .

Calculation:

As $A\text{'}\left(u\right)=\frac{1}{2}$ . So.

  $\begin{array}{c}A\left(u\right)={\displaystyle \int \frac{1}{2}du}\\ =\frac{u}{2}+C\end{array}$

To find $A\left(0\right)$ , substitute $0$ for $u$ in the above equation.

  $\begin{array}{c}A\left(0\right)=0+C\\ =C\end{array}$

From part (a), $A\left(0\right)=0$ so $C=0$ . Then,

  $\begin{array}{c}A\left(u\right)=\frac{u}{2}\\ u=2A\left(u\right)\end{array}$

Hence, $u$ is twice of $A\left(u\right)$ .

Therefore, the equation for $A\left(u\right)$ is $A\left(u\right)=\frac{u}{2}+C$ , the value of $A\left(0\right)$ is $A\left(0\right)=C$ , the value of $C$ is $0$ , and $u$ is twice of $A\left(u\right)$ .