Pursuit Curve. Dog Chases Rabbit. Calculus 4.

(a) In Example 1.21, assume that $a$ is less than $b$ (so that $k$ is less than $1$) and find $y$ as a function of $x$. How far does the rabbit run before the dog catches him?

(b) Assume now that $a=b$, and find $y$ as a function of $x$. How close does the dog come to the rabbit?

Example 1.21

A rabbit begins at the origin and runs up the $y-axis$ with speed $a$ feet per second. At the same time, a dog runs at speed $b$ from the point $(c,0)$ in pursuit of the rabbit. What is the path of the dog?

Solution: At time $t$, measured from the instant both the rabbit and the dog start, the rabbit will be at the point $R=(0,at)$ and the dog at $D=(x,y)$. We wish to solve for $y$ as a function of $x$.

$$\frac{dy}{dx}=\frac{y-at}{x}$$

$$xy'-y=-at$$

$$xy''=-a\frac{dt}{dx}$$

Since the $s$ is a arc length along the path of the dog, it follows that $\frac{ds}{dt}=b$. Hence,

$$\frac{dt}{dx}=\frac{dt}{ds}\frac{ds}{dx}=\frac{-1}{b}\sqrt{1+(y')^2}$$

$$xy''=\frac{a}{b}\sqrt{1+(y')^2}$$

For convenience, we set $k=\frac{a}{b}$, $y'=p$, and $y''=\frac{dp}{dx}$

$$\frac{dp}{\sqrt{1+p^2}}=k\frac{dx}{x}$$

$$\ln\left({p+\sqrt{1+p^2}}\right)=\ln\left(\frac{x}{c}\right)^k$$

Now, solve for $p$:

$$\frac{dy}{dx}=p=\frac{1}{2}\Bigg(\left(\frac{x}{c}\right)^k-\left(\frac{c}{x}\right)^k\Bigg)$$

In order to continue the analysis, we need to know something about the relative sizes of $a$ and $b$. Suppose, for example, that $a \lt$ $b$ (so $k\lt$ $1$), meaning that the dog will certainly catch the rabbit. Then we can integrate the last equation to obtain:

$$y(x)=\frac{1}{2}\Bigg\{\frac{c}{k+1}\left(\frac{x}{c}\right)^{k+1}-\frac{c}{1-k}\left(\frac{c}{x}\right)^{k-1}\Bigg\}+D$$

Again, this is all I have to go on. I need to answer questions (a) and (b) stated at the top.

Solution 1:

The book from which this question comes contains a typo. 1.18 is a typo. It should read 1.21. example 1.18 is irrelevant for this problem.

Solution 2:

According to this paper about pursuit curves: http://sections.maa.org/okar/papers/2006/lloyd.pdf, it considers $k=\dfrac{b}{a}$, but it does not affect the formulas when $x=0$.

(a) When $k\neq1$, the rabbit runs $y\left(0\right)=D=\dfrac{ck}{k^{2}-1}$ , because $x=0$.

(b) Now that $a=b$, it means $k=1$, then the dog will never catch the rabbit, since their distance's difference tends to $\dfrac{c}{2}$. There is also a special case of $y$ when $k=1$: \begin{equation*} y=\dfrac{1}{2}\left[\dfrac{x^{2}-c^{2}}{2c}-c\ln{\dfrac{x}{c}}\right] \end{equation*} making $x=0$, $-c\ln{\dfrac{x}{c}}$ tends do $y=+\infty$.