Chapter 7, Problem 8P

To determine

The rate of the removal.

Answer to Problem 8P

The dog never catches the rabbit.

Explanation of Solution

Given information:

Snow began to fall during the morning of February 2 and continued steadily into the afternoon. At noon a snow plow began removing snow from a road at a constant rate. The plow traveled 6 km from noon to 1 PM out only 3 km from 1 PM to 2 PM. When did the snow begin to fall?

Formula used:

  $s={\displaystyle {\int }_{\infty }^L\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}}dx$

Calculation:

The distance travelled by the dog from point (L, 0) to (0, s) is given by the following equation.

  $\begin{array}{l}s={\displaystyle {\int }_{\infty }^L\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}}dx\\ =-{\displaystyle {\int }_L^{\infty }\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}}dx\\ \frac{ds}{dx}=-\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}\end{array}$

The speed of the dog is twice the speed of the rabbit it means position of the rabbit when dog has travelled the distance s is $\left(0,\frac{s}{2}\right)$

Here the dog runs in a Straight line to catch the rabbit, so the rate of change of y with respect to x is the slope of the line joining the points $\left(0,\frac{s}{2}\right),\left(x,y\right)$

  $\begin{array}{l}\frac{dy}{dx}=\frac{\frac{s}{2}-y}{0-x}\\ -x\frac{dy}{dx}=\frac{s}{2}-y\\ s=2y-2x\frac{dy}{dx}\\ \frac{ds}{dx}=2\frac{dy}{dx}-2\frac{dy}{dx}-2x\frac{d^{2}y}{d{x}^2}\\ \frac{ds}{dx}=-2\frac{d^{2}y}{d{x}^2}\end{array}$

Equate the expressions for $\frac{ds}{dx}$

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

Hence the desired differential equation is $\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}=2x\frac{d^{2}y}{d{x}^2}$

Solve the differential equation $\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}=2x\frac{d^{2}y}{d{x}^2}$

Let $\frac{dy}{dx}$ be z then $\frac{d^{2}y}{d{x}^2}$ will be $\frac{dz}{dx}$

Now put these values in the differential equation $\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}=2x\frac{d^{2}y}{d{x}^2}$

  $\sqrt{1+z^2}=2x\frac{dz}{dx}$

This equation can be solved by separating the variables and then by integrating it

  $\begin{array}{l}\frac{dz}{\sqrt{1+z^2}}=\frac{dx}{2x}\\ {\displaystyle \int \frac{dz}{\sqrt{1+z^2}}}={\displaystyle \int \frac{dx}{2x}}\end{array}$

  $\begin{array}{l}\ln \left(z+\sqrt{1+z^2}\right)=\frac{1}{2}\ln x+\frac{1}{2}\ln C\\ \ln {\left(z+\sqrt{1+z^2}\right)}^2=\ln xC\\ {\left(z+\sqrt{1+z^2}\right)}^2=xC\end{array}$

When $x=L,z=0$ this implies $\left(0+\sqrt{1+0^2}\right)=LC\Rightarrow C=\frac{1}{L}$

Put $C=\frac{1}{L}$ into the equation ${\left(z+\sqrt{1+z^2}\right)}^2=xC$

  $\begin{array}{l}{\left(z+\sqrt{1+z^2}\right)}^2=xC\\ {\left(z+\sqrt{1+z^2}\right)}^2=\frac{x}{L}\end{array}$

Now solve this equation for Z.

  $\begin{array}{l}{\left(z+\sqrt{1+z^2}\right)}^2=\frac{x}{L}\\ 1+2z\left(z+\sqrt{1+z^2}\right)=\frac{x}{L}\\ 1+2z\left(\sqrt{\frac{x}{L}}\right)=\frac{x}{L}\end{array}$

Since,

  $z+\sqrt{1+z^2}=\pm \sqrt{\frac{x}{L}}$

  $z=\frac{\frac{x}{L}-1}{2\sqrt{\frac{x}{L}}}\text{or}z=\frac{1+\frac{x}{L}}{2\sqrt{\frac{x}{L}}}$

Now solve these equations one by one separately.

  $\begin{array}{l}z=\frac{\frac{x}{L}-1}{2\sqrt{\frac{x}{L}}}\\ \frac{dy}{dx}=\frac{\frac{x}{L}-1}{2\sqrt{\frac{x}{L}}}\\ {\displaystyle \int dy}={\displaystyle \int \frac{\frac{x}{L}-1}{2\sqrt{\frac{x}{L}}}dx}\end{array}$

Here $y=0$ when $x=L$

So,

  $\begin{array}{l}0=\frac{L\sqrt{L}}{3\sqrt{L}}-\sqrt{L.L}+C_1\\ C_1=\frac{2L}{3}\end{array}$

Put this value into the equation $y=\frac{x\sqrt{x}}{3\sqrt{L}}-\sqrt{Lx}+C_1$

  $y=\frac{x\sqrt{x}}{3\sqrt{L}}-\sqrt{Lx}+\frac{2L}{3}$

Hence the solution to the differential equation $z=\frac{\frac{x}{L}-1}{2\sqrt{\frac{x}{L}}}$ is $y=\frac{x\sqrt{x}}{3\sqrt{L}}-\sqrt{Lx}+\frac{2L}{3}$

Solve the second equation $z=\frac{1-\frac{x}{L}}{2\sqrt{\frac{x}{L}}}$

It can be rewritten as $z=-\left(\frac{\frac{x}{L}-1}{2\sqrt{\frac{x}{L}}}\right)$

Hence solution to equation will be $y=-\left(\frac{x\sqrt{x}}{3\sqrt{L}}-\sqrt{Lx}+C_2\right)$

Here $y=0$ when $x=L$ so $0=-\left(\frac{L\sqrt{L}}{3\sqrt{L}}-\sqrt{L.L}+C_1\right)\Rightarrow C_2=\frac{2L}{3}$

Put this value into the equation $y=-\left(\frac{x\sqrt{x}}{3\sqrt{L}}-\sqrt{Lx}+C_2\right)$

  $y=-\left(\frac{x\sqrt{x}}{3\sqrt{L}}-\sqrt{Lx}+\frac{2L}{3}\right)$

Hence the solution to the differential equation $z=\frac{1-\frac{x}{L}}{2\sqrt{\frac{x}{L}}}\text{is}y=-\left(\frac{x\sqrt{x}}{3\sqrt{L}}-\sqrt{Lx}+\frac{2L}{3}\right)$

Evaluate the $\underset{x\to 0}{\lim }y$ tofind the distance at which the dog catches the rabbit.

Use either of the equations $y=-\left(\frac{x\sqrt{x}}{3\sqrt{L}}-\sqrt{Lx}+\frac{2L}{3}\right)$ and $y=\frac{x\sqrt{x}}{3\sqrt{L}}-\sqrt{Lx}+\frac{2L}{3}$ to evaluate $\underset{x\to 0}{\lim }y$

  $\begin{array}{l}\underset{x\to 0}{\lim }y=\underset{x\to 0}{\lim }\left(\frac{x\sqrt{x}}{3\sqrt{L}}-\sqrt{Lx}+\frac{2L}{3}\right)\\ =0-0+\frac{2L}{3}\\ \underset{x\to 0}{\lim }y=\frac{2L}{3}\end{array}$

Hence dog catches the rabbit at distance of $\underset{x\to 0}{\lim }y=\frac{2L}{3}$ units

The distance travelled by the dog from point (L, 0) to (0, 5) is given by the following equation.

  $\begin{array}{l}s={\displaystyle {\int }_{\infty }^L\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx}\\ =-{\displaystyle \underset{L}{\overset{\infty }{\int }}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx}\\ \frac{ds}{dx}=-\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}\end{array}$

The speed of the dog is half the speed of the rabbit it means position of the rabbit when dog has travelled the distance sis (0,2s).

Here the dog runs in a straight line to catch the rabbit, so the rate of change of y with respect to x is the slope of the line joining the points (0, 2s), (x, y).

  $\begin{array}{l}\frac{dy}{dx}=\frac{2s-y}{0-x}\\ -x\frac{dy}{dx}=2s-y\\ s=\frac{y}{2}-\frac{x}{2}\frac{dy}{dx}\\ \frac{ds}{dx}=\frac{1}{2}\frac{dy}{dx}-\frac{1}{2}\frac{dy}{dx}-\frac{1}{2}x\frac{d^{2}y}{d{x}^2}\\ \frac{ds}{dx}=-\frac{1}{2}x\frac{d^{2}y}{d{x}^2}\end{array}$

Equate the expressions for $\frac{ds}{dx}$

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

Hence the desired differential equation is $\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}=\frac{x}{2}\frac{d^{2}y}{d{x}^2}$

Solve the differential equation $\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}=\frac{x}{2}\frac{d^{2}y}{d{x}^2}$

Let $\frac{dy}{dx}$ be z then $\frac{d^{2}y}{d{x}^2}$ will be $\frac{dz}{dx}$

Now put these values in the differential equation $\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}=\frac{x}{2}\frac{d^{2}y}{d{x}^2}$

  $\sqrt{1+z^2}=\frac{x}{2}\frac{dz}{dx}$

This equation can be solved by separating the variables and then by integrating it

  $\begin{array}{l}\frac{dz}{\sqrt{1+z^2}}=\frac{2dx}{x}\\ {\displaystyle \int \frac{dz}{\sqrt{1+z^2}}=2{\displaystyle \int \frac{dx}{x}}}\\ \ln \left(z+\sqrt{1+z^2}\right)=2\ln x+2\ln C\\ \ln \left(z+\sqrt{1+z^2}\right)=\ln xC\\ \left(z+\sqrt{1+z^2}\right)={\left(xC\right)}^2\end{array}$

When $x=L,z=0$ this implies $\left(0+\sqrt{1+0^2}\right)={\left(CL\right)}^2\Rightarrow C=\frac{1}{L}$

Put $C=\frac{1}{L}$ into the equation $\left(z+\sqrt{1+z^2}\right)={\left(xC\right)}^2$

  $\begin{array}{l}\left(z+\sqrt{1+z^2}\right)={\left(xC\right)}^2\\ \left(z+\sqrt{1+z^2}\right)={\left(\frac{x}{L}\right)}^2\end{array}$

Now solve this equation for Z.

  $\begin{array}{l}{\left(z+\sqrt{1+z^2}\right)}^2={\left(\frac{x}{L}\right)}^4\\ 1+2z\left(z+\sqrt{1+z^2}\right)={\left(\frac{x}{L}\right)}^4\\ 1+2z{\left(\frac{x}{L}\right)}^2={\left(\frac{x}{L}\right)}^4\end{array}$

Since,

  $\begin{array}{l}{\left(z+\sqrt{1+2z}\right)}^2={\left(\frac{x}{L}\right)}^4\\ z=\frac{{\left(\frac{x}{L}\right)}^4-1}{2{\left(\frac{x}{L}\right)}^2}\\ \end{array}$

  $\begin{array}{l}\frac{dy}{dx}=\frac{{\left(\frac{x}{L}\right)}^4-1}{2{\left(\frac{x}{L}\right)}^2}\\ {\displaystyle \int dy={\displaystyle \int \frac{{\left(\frac{x}{L}\right)}^4-1}{2{\left(\frac{x}{L}\right)}^2}dx}}\\ {\displaystyle \int dy=\frac{1}{2L^2}{\displaystyle \int x^{2}dx}}-\frac{L^2}{2}{\displaystyle \int x^{-2}dx+C_1}\\ y=\frac{1}{2\sqrt{L}}\frac{x^3}{3}+\frac{L^2}{2x}+C_1\\ y=\frac{x^3}{6\sqrt{L}}+\frac{L^2}{2x}+C_1\end{array}$

Here $y=0$ when $x=L$

So

  $\begin{array}{l}0=\frac{L^3}{6\sqrt{L}}+\frac{L^2}{2L}+C_1\\ C_1=-\frac{L}{2}-\frac{L^{\frac{5}{2}}}{6}\end{array}$

Hence the solution to the differential equation $z=\frac{{\left(\frac{x}{L}\right)}^4-1}{2{\left(\frac{x}{L}\right)}^2}\text{is}y=\frac{x^3}{6\sqrt{L}}+\frac{L^2}{2x}-\frac{L}{2}-\frac{L^{\frac{5}{2}}}{6}$

Evaluate the limit $\underset{x\to 0}{\lim }y$ to find the distance at which the dog catches the rabbit.

Use the equation $y=\frac{x^3}{6\sqrt{L}}+\frac{L^2}{2x}-\frac{L}{2}-\frac{L^{\frac{5}{2}}}{6}$ to evaluate $\underset{x\to 0}{\lim }y$

  $\begin{array}{l}\underset{x\to 0}{\lim }y=\underset{x\to 0}{\lim }\left(\frac{x^3}{6\sqrt{L}}+\frac{L^2}{2x}-\frac{L}{2}-\frac{L^{\frac{5}{2}}}{6}\right)\\ \underset{x\to 0}{\lim }y=\infty \end{array}$

Hence dog never catches the rabbit.

Conclusion:

The dog never catches the rabbit.