Pursuit Curve of Boat in River

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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 $\left({c, 0}\right)$ and headed directly towards the origin with speed $b$ relative to the water.

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

$c^k \left({y + \sqrt {x^2 + x^2} }\right) = x^{k + 1}$


The components of the velocity of $b$ are:

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


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

$(1)$ is a homogeneous differential equation.

Let $z = \dfrac y x$.

Then by Solution to Homogeneous Differential Equation:

\(\displaystyle \ln x\) \(=\) \(\displaystyle \int \frac {\mathrm d z} {\dfrac {-a \sqrt {1 + z^2} + b z} b - z} + C\)
\(\displaystyle \) \(=\) \(\displaystyle \int \frac {b \, \mathrm d z} {-a \sqrt {1 + z^2} + b z - b z} + C\)
\(\displaystyle \) \(=\) \(\displaystyle \int \frac {b \, \mathrm d z} {-a \sqrt {1 + z^2} } + C\)
\(\displaystyle \) \(=\) \(\displaystyle -\frac b a \int \frac {\mathrm d z} {\sqrt {1 + z^2} } + C\)
\(\displaystyle \) \(=\) \(\displaystyle -\frac b a \ln \left({z + \sqrt {1 + z^2} }\right) + C\) Primitive of $\dfrac 1 {\sqrt{x^2 + a^2} }$

Substituting back for $z$ and rearranging:

$C^k \left({y + \sqrt {x^2 + x^2} }\right) = x^{k + 1}$