Differential Equation of Perpendicular Pursuit Curve

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Therorem

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

Let a dog $D$ be situated at the point $\tuple {c, 0}$.

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:

$\dfrac {\d y} {\d x} = \dfrac 1 2 \paren {\paren {\dfrac x c}^k - \paren {\dfrac c x}^k}$

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 $\tuple {0, a t}$.

Let $D$ be at $\tuple {x, y}$ at $t$.

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

Hence:

$\dfrac {\d y} {\d x} = \dfrac {y - a t} x$

or:

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

Differentiating $(1)$ with respect to $x$:

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

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

Then:

$\dfrac {\d s} {\d t} = b$

so:

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

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:

$(4): \quad x y'' = k \sqrt {1 + \paren {y'}^2}$

where $k = \dfrac a b$.

Let $p = y'$.

Then $(4)$ becomes:

$x \dfrac {\d p} {\d x} = k \sqrt {1 + p^2}$
\(\displaystyle x \dfrac {\d p} {\d x}\) \(=\) \(\displaystyle k \sqrt {1 + p^2}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle k \int \frac {\d x} x\) \(=\) \(\displaystyle \int \frac {\d p} {\sqrt {1 + p^2} }\) Separation of Variables
\(\displaystyle \leadsto \ \ \) \(\displaystyle k \ln x\) \(=\) \(\displaystyle \map \ln {p + \sqrt {1 + p^2} } + C\) Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$

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

\(\displaystyle k \ln x\) \(=\) \(\displaystyle \map \ln {p + \sqrt {1 + p^2} } + C\) Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$
\(\displaystyle \leadsto \ \ \) \(\displaystyle k \ln c\) \(=\) \(\displaystyle \map \ln {0 + \sqrt {1 + 0} } + C\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \ln c^k\) \(=\) \(\displaystyle 0 + C\)

Hence:

\(\displaystyle k \ln x\) \(=\) \(\displaystyle \map \ln {p + \sqrt {1 + p^2} } + C\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle k \ln x - k \ln c\) \(=\) \(\displaystyle \map \ln {p + \sqrt {1 + p^2} }\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map \ln {\frac x c}^k\) \(=\) \(\displaystyle \map \ln {p + \sqrt {1 + p^2} }\)

Solving for $p$ reveals:

$p = \dfrac {\d y} {\d x} = \dfrac 1 2 \paren {\paren {\dfrac x c}^k - \paren {\dfrac c x}^k}$

$\blacksquare$


Sources