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(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.

Erik Olesen
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    You can use $\TeX$ on this site by enclosing formulas in dollar signs; single dollar signs for inline formulas and double dollar signs for displayed equations. You can see the source code for any math formatting you see on this site by right-clicking on it and selecting "Show Math As:TeX Commands". Here's a basic tutorial and quick reference. There's an "edit" link under the question. – joriki Sep 20 '12 at 04:35
  • Thank you. I wasn't sure how to do that. The link you provided greatly helped me. I fixed everything the best I could. Can you or someone else help me now? Thanks again. –  Sep 20 '12 at 15:53
  • I could really use a hint or something. –  Sep 20 '12 at 18:53
  • I don't understand -- (a) and (b) refer to Example 1.18 but talk about variables $a$ and $b$ that don't occur in Example 1.18 -- but they do occur in Example 1.21. Perhaps you could explain more about the context? – joriki Sep 20 '12 at 21:59
  • By the way, you can get the appropriate size for a pair of parentheses (or other pairs of delimiters, like brackets, braces, absolute value bars) by preceding them with \left and \right, respectively. – joriki Sep 20 '12 at 22:08
  • I fixed it. This is evidently a typo in the book. The correct problem is now stated. –  Sep 21 '12 at 03:08
  • Now it makes sense. What have you tried? Where are you stuck? Are you aware of this FAQ? (By the way, the parentheses around the fractions are still the wrong size -- did you see my comment above on how to fix that?) – joriki Sep 21 '12 at 04:38

2 Answers2

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.

MA214isKillingMe
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  • There, I changed the question. It makes more sense now, but this is still not enough for me to manage to answer the question. –  Sep 21 '12 at 03:07
  • This should have been a comment rather than an answer, but you don't have enough reputation for that. [Upvotes your answer]. There, now you do. – Rick Decker Oct 12 '12 at 17:28
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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 $y=+\infty$.

Cleyton Muto
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