# Calculus 4. Pursuit Curve. Dog Chases Rabbit.

Gummy Bear

## Homework Statement

(a) In Example 1.18, 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?

## Homework Equations

N/A
There are many integration techniques that could be used.

## The Attempt at a Solution

Example 1.18:

$y''+k^2y=0$ Where $k$ is an unknown real constant)

We notice that the independent variable is missing. So we let:

$$y'=p, y''=p\frac{dp}{dy}$$

$$p\frac{dp}{dy}+k^2y=0$$

$$pdp=-k^2ydy$$

$$\frac{p^2}{2}=-k^2\frac{y^2}{2}+c$$

$$p=\pm k\sqrt{E-y^2}$$

Now re-substitute $p=\frac{dy}{dx}$ to obtain:

$$\frac{dy}{dx}= \pm k\sqrt{E-y^2}$$

$$\frac{dy}{\sqrt{E-y^2}}= \pm kdx$$

$$\sin^{-1}({\frac{y}{\sqrt{E}})}= \pm kx+F$$

$$\frac{y}{\sqrt{E}}=\sin{( \pm kx+F)}$$

$$y=\sqrt{E}\sin( \pm kx+F)$$

Now we apply the sun formula for sine to rewrite the last expression as:

$$y=\sqrt{E}\cos(F)\sin( \pm kx)+\sqrt{E}\sin(F)\cos( \pm kx)$$

We may write a general solution as $y=A\sin(kx)+B\cos(kx)$

There is another example in the book that I feel relates to this problem:

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 $y''=\frac{dp}{dx}$

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

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

Now, solve for $p$:

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

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:

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

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

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