Differential Equation of Perpendicular Pursuit Curve

Therorem
Let a rabbit $R$ be situated at the origin of a cartesian plane.

Let a dog $D$ be situated at the point $\left({c, 0}\right)$.

Let $R$ start running up the $y$-axis with speed $a$.

At the same instant, let $D$ start pursuing $R$ by running directly towards it at speed $b$.

Then the differential equation describing the path taken by $D$ is:
 * $\dfrac {\mathrm d y} {\mathrm d x} = \dfrac 1 2 \left({\left({\dfrac x c}\right)^k - \left({\dfrac c x}\right)^k}\right)$

where $k = \dfrac a b$.

Proof

 * DogAfterRabbit.png

Let $C$ be the curve traced out by $D$.

Let the time $t$ be measured from the instant $R$ and $D$ start running.

At $t$, $R$ will be at the point $\left({0, a t}\right)$.

Let $D$ be at $\left({x, y}\right)$ at $t$.

By the nature of the motion of $D$, the line $DR$ is tangent to $C$.

Hence:
 * $\dfrac {\mathrm d y} {\mathrm d x} = \dfrac {y - a t} x$

or:
 * $(1): \quad x y' - y = - a t$

Differentiating $(1)$ $x$:


 * $(2): \quad x y'' = - a \dfrac {\mathrm d t} {\mathrm d x}$

Let $s$ be the length of $C$ between $\left({c, 0}\right)$ and $\left({x, y}\right)$ at time $t$.

Then:
 * $\dfrac {\mathrm d s} {\mathrm d t} = b$

so:
 * $(3): \quad \dfrac {\mathrm d t} {\mathrm d x} = \dfrac {\mathrm d t} {\mathrm d s} \dfrac {\mathrm d s} {\mathrm d x} = - \dfrac 1 b \sqrt {1 + \left({y'}\right)^2}$

The minus sign is there because $s$ increases as $x$ decreases.

When $(2)$ and $(3)$ are combined, the differential equation describing $C$ is seen to be:
 * $(4): \quad x y'' = k \sqrt {1 + \left({y'}\right)^2}$

where $k = \dfrac a b$.

Let $p = y'$.

Then $(4)$ becomes:
 * $x \dfrac {\mathrm d p} {\mathrm d x} = k \sqrt {1 + p^2}$

When $x = c$ we have $p = 0$, so:

Hence:

Solving for $p$ reveals:
 * $p = \dfrac {\mathrm d y} {\mathrm d x} = \dfrac 1 2 \left({\left({\dfrac x c}\right)^k - \left({\dfrac c x}\right)^k}\right)$