Pursuit Curve of Boat in River

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Theorem

Consider a straight river $R$ whose parallel banks are aligned with the $y$-axis and the line $x = c$ of a cartesian plane.

Let the current of $R$ have a constant and uniform speed $a$ in the negative $y$ direction.

Let a boat $B$ be launched from the point $\tuple {c, 0}$ and headed directly towards the origin with speed $b$ relative to the water.


The path of $B$ is defined by the equation:

$c^k \paren {y + \sqrt {x^2 + x^2} } = x^{k + 1}$

where $k = \dfrac a b$.


Proof

The components of the velocity of $b$ are:

$\dfrac {\d x} {\d t} = - b \cos \theta$
$\dfrac {\d y} {\d t} = - a + b \sin \theta$


Hence:

\(\ds \frac {\d y} {\d x}\) \(=\) \(\ds \frac {-a + b \sin \theta} {-b \cos \theta}\)
\(\ds \) \(=\) \(\ds \frac {-a + b \paren {-\dfrac y {\sqrt {x^2 + y^2} } } } {-b \paren {\dfrac x {\sqrt {x^2 + y^2} } } }\)
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \frac {-a \sqrt {x^2 + y^2} + b y} {b x}\)


$(1)$ is a homogeneous differential equation.

Let $z = \dfrac y x$.

Then by Solution to Homogeneous Differential Equation:

\(\ds \ln x\) \(=\) \(\ds \int \frac {\d z} {\dfrac {-a \sqrt {1 + z^2} + b z} b - z} + C\)
\(\ds \) \(=\) \(\ds \int \frac {b \rd z} {-a \sqrt {1 + z^2} + b z - b z} + C\)
\(\ds \) \(=\) \(\ds \int \frac {b \rd z} {-a \sqrt {1 + z^2} } + C\)
\(\ds \) \(=\) \(\ds -\frac b a \int \frac {\d z} {\sqrt {1 + z^2} } + C\)
\(\ds \) \(=\) \(\ds -\frac b a \map \ln {z + \sqrt {1 + z^2} } + C\) Primitive of $\dfrac 1 {\sqrt{x^2 + a^2} }$

Substituting back for $z$, putting $k = \dfrac a b$, and rearranging:

$C^k \paren {y + \sqrt {x^2 + x^2} } = x^{k + 1}$

$\blacksquare$


Sources

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