A woman stands in a field at a distance of \(a\,\mathrm{m}\) from the straight bank of a river which flows with negligible speed. She sees her frightened child clinging to a tree stump standing in the river \(b\,\mathrm{m}\) downstream from where she stands and \(c\,\mathrm{m}\) from the bank. She runs at a speed of \(u\,\mathrm{ms}^{-1}\) and swims at \(v\,\mathrm{ms}^{-1}\) in straight lines. Find an equation to be satisfied by \(x,\) where \(x\,\mathrm{m}\) is the distance upstream from the stump at which she should enter the river if she is to reach the child in the shortest possible time.
Suppose now that the river flows with speed \(v\) ms\(^{-1}\) and the stump remains fixed. Show that, in this case, \(x\) must satisfy the equation
\[
2vx^{2}(b-x)=u(x^{2}-c^{2})[a^{2}+(b-x)^{2}]^{\frac{1}{2}}.
\]
For this second case, draw sketches of the woman's path for the three possibilities \(b>c,\) \(b=c\) and \(b< c\).
Solution:
The distance to where she enters the water is \(\sqrt{a^2+(b-x)^2}\) and the distance through the water is \(\sqrt{x^2+c^2}\).
The total time will be \(\frac{\sqrt{a^2+(b-x)^2}}{u}+\frac{\sqrt{x^2+c^2}}{v}\).
To minimise this, we can differentiate.
\begin{align*}
\frac{\d}{\d x}: && \frac{-(b-x)}{u\sqrt{a^2+(b-x)^2}} + \frac{x}{v \sqrt{x^2+c^2}} &= 0 \\
\Rightarrow && v(b-x)(x^2+c^2)^{\frac12} &= xu(a^2+(b-x)^2)^{\frac12}
\end{align*}
When she is in the water, she can will move with velocity \(\begin{pmatrix} v \cos \theta \\ v \sin \theta -v \end{pmatrix}\). She needs to travel a distance \(\begin{pmatrix} c \\ -x \end{pmatrix}\), so we must have that
\begin{align*}
&& \frac{x}{c} &= \frac{1-\sin \theta}{\cos \theta} \\ \Rightarrow && \sec \theta - \tan \theta &= \frac{x}{c} \\
\Rightarrow && \sec \theta &= \tan \theta + \frac{x}{c} \\
\Rightarrow && \sec^2 \theta &= \tan^2 \theta + 2 \tan \theta \frac{x}{c} + \frac{x^2}{c^2} \\
\Rightarrow && 1 + \tan^2 \theta &= \tan^2 \theta + 2 \tan \theta \frac{x}{c} + \frac{x^2}{c^2} \\
\Rightarrow && \tan \theta &=\frac{c^2-x^2}{2xc} \\
\Rightarrow && \sin \theta &= \frac{c^2-x^2}{c^2+x^2} \\
&& \cos \theta &= \frac{2xc}{c^2+x^2} \\
\end{align*}
(where we have taken the positive value for \(\cos \theta\) since we must be heading towards the child).
Since \(v \cos \theta t = c\) the time taken to reach the child in the water is \(\frac{c}{v} \frac{c^2+x^2}{2xc} = \frac{c^2+x^2}{2xv}\).
So the total time is: \(\frac{\sqrt{a^2+(b-x)^2}}{u}+\frac{c^2+x^2}{2xv}\).
To minimise this, we can differentiate.
\begin{align*}
\frac{\d}{\d x}: && \frac{-(b-x)}{u\sqrt{a^2+(b-x)^2}} -\frac{c^2}{2vx^2} + \frac{x^2}{2vx^2}&= 0 \\
\Rightarrow && u(x^2-c^2)\sqrt{a^2+(b-x)^2}&= 2vx^2(b-x)
\end{align*}
as required.
When \(b = c\), the shortest path will be running directly to the bank (there's no quicker way to get to the bank) then swimming directly out (and letting the current take you downstream exactly as far as you need)). Therefore the path will be:
If \(b > c\) then she should run a little downstream first.
and if \(c > b\) she should actually run a little upstream to take advantage of the current:
