Chapter 7, Problem 10P

a.

To determine

To Find: The point where the dog catches the rabbit.

Answer to Problem 10P

The Dog catches the rabbit when rabbit position is $\left(0,2/3L\right)$.

Explanation of Solution

Given:

Dog run as twice as fast as the Rabbit.

Calculation:

The speed of Dog is twice the speed of Rabbit. Both are travelled in straight line. Let the distance travelled by the dog is $$ then the position of rabbit is $\left(0,s/2\right)$.

The tangent of path joining dog and rabbit has slope is $\frac{dy}{dx}=\frac{s/2-y}{0-x}$

Therefore,

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

Let $z=\frac{dy}{dx}$then put this is above equation:

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

Integrating both side then,

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

When $x=L,z=dy/dx=0$, so

  $\begin{array}{l}\ln L=2\ln 1+C=C\\ \ln x=2\ln \left(\sqrt{1+z^2}+z\right)+\ln L\\ =\ln \left(L{\left(\sqrt{1+z^2}+z\right)}^2\right)\\ x=L{\left(\sqrt{1+z^2}+z\right)}^2\\ \sqrt{1+z^2}=\sqrt{\frac{x}{L}}-z\\ 1+z^2=\frac{x}{L}+z^2-2z\sqrt{\frac{x}{L}}\\ 2z\sqrt{\frac{x}{L}}=\frac{x}{L}-1\\ \frac{dy}{dx}=z=\frac{1}{2}\sqrt{\frac{x}{L}}-\frac{1}{2\sqrt{\frac{x}{L}}}\\ \frac{dy}{dx}=\frac{1}{2\sqrt{L}}{x}^{1/2}-\frac{\sqrt{L}}{2}{x}^{-1/2}\\ y=\frac{1}{3\sqrt{L}}{x}^{3/2}-\sqrt{L}{x}^{1/2}+C_1\end{array}$

When $x=L,y=0$, then,

  $\begin{array}{l}0=\frac{1}{3\sqrt{L}}{L}^{3/2}-\sqrt{L}{L}^{1/2}+C_1\\ =\frac{L}{3}-L+C_1\\ C_1=\frac{2}{3}L\end{array}$

Therefore,

  $y=\frac{x^{3/2}}{3\sqrt{L}}-\sqrt{L}{x}^{1/2}+\frac{2}{3}L$

As $x=0$ then the dog catches the rabbit when rabbit position is $\left(0,2/3L\right)$.

Conclusion:

The Dog catches the rabbit when rabbit position is $\left(0,2/3L\right)$.

b.

To determine

The position of dog and rabbit when they are closest.

Answer to Problem 10P

The position of dog is $\left(\frac{L}{\sqrt[4]{3}},\frac{5\sqrt[4]{3}-6}{9}L\right)$ and rabbit is $\left(0,\frac{8\sqrt[4]{3}-6}{9}L\right)$.

Explanation of Solution

Given:

Dog run as half as fast as the Rabbit.

Calculation:

The speed of Dog is half the speed of Rabbit. Both are travelled in straight line.

From the part a

  $\begin{array}{l}z=\frac{dy}{dx}=\frac{x^2}{2L^2}-\frac{L^2}{2x^2}\end{array}$

Therefore,

  $y=\frac{x^3}{6L^2}+\frac{L^2}{2x}-\frac{2}{3}L$

We have to minimize the distance$D$ from the dog $\left(x,y\right)$ and rabbit $(0,2s$so,

The tangent of path joining dog and rabbit has slope is $\frac{dy}{dx}=\frac{2s-y}{0-x}$

  $\begin{array}{l}s=\frac{1}{2}y-\frac{1}{2}x\frac{dy}{dx}\\ 2s=y-xz\\ y-2s=xz=x\left(\frac{x^2}{2L^2}-\frac{L^2}{2x^2}\right)\\ y-2s=\frac{x^3}{2L^2}-\frac{L^2}{2x}\\ D=\sqrt{{\left(x-0\right)}^2+{\left(y-2s\right)}^2}\\ =\sqrt{x^2+{\left(\frac{x^3}{2L^2}-\frac{L^2}{2x}\right)}^2}\\ =\sqrt{\frac{x^6}{4L^4}+\frac{L^4}{4x^2}+\frac{x^2}{2}}=\sqrt{{\left(\frac{x^3}{2L^2}+\frac{L^2}{2x}\right)}^2}\\ =\left(\frac{x^3}{2L^2}+\frac{L^2}{2x}\right)\end{array}$

For miminum value of $D$ then check value of $D^{\prime }=0$

  $\begin{array}{l}{D}^{\prime }=0\\ \frac{3x^2}{2L^2}-\frac{L^2}{2x^2}=0\\ \frac{3x^2}{2L^2}=\frac{L^2}{2x^2}\\ x^4=\frac{L^4}{3}\\ x=\frac{L}{\sqrt[4]{3}},x>0,L>0\end{array}$

Then

  ${D^{\prime }}^{\prime }\left(x\right)=\frac{3x}{L^2}+\frac{L^2}{x^3}>0$ for all $x>0$ so,

  $\begin{array}{l}D\left(\frac{L}{\sqrt[4]{3}}\right)=\frac{{\left(\frac{L}{\sqrt[4]{3}}\right)}^3}{2L^2}+\frac{L^2}{2\left(\frac{L}{\sqrt[4]{3}}\right)}\\ =\frac{2L}{3^{3/4}}\end{array}$

  $D=\frac{2L}{3^{3/4}}$ is the minimum distance between dog and rabbit.

The position of dog and rabbit is:

Dog: $\left(x,y\right)=\left(\frac{L}{\sqrt[4]{3}},\left(\frac{5}{3^{7/4}}-\frac{2}{3}\right)L\right)=\left(\frac{L}{\sqrt[4]{3}},\frac{5\sqrt[4]{3}-6}{9}L\right)$

Rabbit: $\left(0,2s\right)=\left(0,\frac{8\sqrt[4]{3}L}{9}-\frac{2L}{3}\right)=\left(0,\frac{8\sqrt[4]{3}-6}{9}L\right)$

Conclusion:

The position of dog is $\left(\frac{L}{\sqrt[4]{3}},\frac{5\sqrt[4]{3}-6}{9}L\right)$ and rabbit is $\left(0,\frac{8\sqrt[4]{3}-6}{9}L\right)$.