Chapter 7, Problem 9P

(a)

To determine

To Prove: The dog’s path is the graph of the function

Answer to Problem 9P

The $c_1=0.00001>0$ and $c_2=0.00004>0$ shows that the relation is cooperative.

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

Consider the distance travelled by the dog is,

  $\begin{array}{l}S={\displaystyle {\int }_x^L\sqrt{1+{\left(\frac{dy}{dx}\right)}^{2}dx}}\\ \frac{dS}{dx}=-\frac{d}{dx}{\displaystyle {\int }_L^x\sqrt{1+{\left(\frac{dy}{dx}\right)}^{2}dx}}\\ \frac{dS}{dx}=-\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}\end{array}$

Since, both the dog and the rabbit are at the same speed and as the dog reaches $\left(x,y\right)$ the total rabbit will be at $\left(0,S\right)$ as the rabbit runs along the Y axis.

The slope of the line that joins the points is,

  $\begin{array}{l}\frac{dy}{dx}=\frac{S-y}{0-x}\\ S=y-x\frac{dy}{dx}\end{array}$

Then,

  $\frac{dS}{dx}=-x^2\frac{d^{2}x}{d{x}^2}$

Then, from the given equations,

Consider the distance travelled by the dog is,

  $\begin{array}{l}S={\displaystyle {\int }_x^L\sqrt{1+{\left(\frac{dy}{dx}\right)}^{2}dx}}\\ \frac{dS}{dx}=-\frac{d}{dx}{\displaystyle {\int }_L^x\sqrt{1+{\left(\frac{dy}{dx}\right)}^{2}dx}}\\ \frac{dS}{dx}=-\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}\end{array}$

Since, both the dog and the rabbit are at the same speed and as the dog reaches $\left(x,y\right)$ the total rabbit will be at $\left(0,S\right)$ as the rabbit runs along the Y axis.

The slope of the line that joins the points is,

  $\begin{array}{l}\frac{dy}{dx}=\frac{S-y}{0-x}\\ S=y-x\frac{dy}{dx}\end{array}$

Then,

  $\begin{array}{l}-x\frac{d^{2}x}{d{x}^2}=-\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}\\ x\frac{d^{2}x}{d{x}^2}=\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}\end{array}$

(b)

To determine

To Find: The solution of the equation in part (a) that satisfies the initial conditions $y=y^{\prime }$ that is equal to 0 and when $x=L$ .

Answer to Problem 9P

The solution of the equation is $y=\frac{x^2}{4L}-\frac{L}{4}+\frac{L}{2}\left(\ln L-\ln x\right)$ .

Explanation of Solution

Calculation:

Consider the relation is,

  $r=\frac{dy}{dx}$

Then,

  $\begin{array}{l}x\frac{dy}{dx}=\sqrt{1+r^2}\\ \frac{1}{\sqrt{1+r^2}}dr=\frac{1}{x}dx\\ {\displaystyle \int \frac{1}{\sqrt{1+r^2}}dr}={\displaystyle \int \frac{1}{x}dx}\\ \ln \left(r+\sqrt{1+r^2}\right)=\ln x+\ln c\end{array}$

Solve further as,

  $\begin{array}{l}\left(r+\sqrt{1+r^2}\right)=cx\\ \sqrt{1+r^2}=cx-r\\ r^2+1=c^2{x}^2-r^2-2cxr\end{array}$

Consider the case for $x=L$ , $\frac{dy}{dx}=0$ and $r=0$ then,

  $\begin{array}{l}1=c^2{L}^2+0-0\\ c=\pm \frac{1}{L}\end{array}$

Take $c=\frac{1}{L}$ . Then,

  $\begin{array}{l}1+r^2=\frac{1}{L^2}{x}^2-1\\ r=\frac{L}{2x}\left(\frac{1}{L^2}{x}^2-1\right)\\ r=\frac{1}{2L}x-\frac{1}{2x}\\ dy=\left(\frac{x}{2L}-\frac{L}{2x}\right)dx\end{array}$

Integrate both the sides as,

  $\begin{array}{l}{\displaystyle \int dy}={\displaystyle \int \left(\frac{x}{2L}-\frac{L}{2x}\right)dx}\\ y=\frac{x^2}{4L}-\frac{L}{2}\ln x+C_1\end{array}$

Consider for $x=L$ and $y=0$ .

  $\begin{array}{l}0=\frac{x^2}{4L}-\frac{L}{2}\ln x+C_1\\ C_1=\frac{L}{2}\ln L-\frac{L}{4}\end{array}$

Thus, the equation is,

  $y=\frac{x^2}{4L}-\frac{L}{4}+\frac{L}{2}\left(\ln L-\ln x\right)$

(c)

To determine

To Find: Whether the dog catch the rabbit.

Answer to Problem 9P

The dog cannot catch the rabbit.

Explanation of Solution

Calculation:

Consider the case when the point $\left(x,y\right)$ is very close to the origin.

  $\underset{x\to 0^{-}}{\lim }y=\underset{x\to 0^{-}}{\lim }\left(\frac{x^2-L^2}{4L}\right)+\frac{L}{2}\underset{x\to 0^{-}}{\lim }\ln \left(\frac{L}{x}\right)$

Since, $\frac{L}{x}\to \infty \text{as}x\to 0^{+}$ . Then, the value of the limit is $\underset{x\to \infty }{\text{lim}}\ln \left(\frac{L}{x}\right)\to \infty$ .

For the case when $x\to 0^{-}$ and $y\to \infty$ the dog cannot catch the rabbit.