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.

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.
