Mathematical Training of Lifeguards Resize


Optimal path

For simplicity, we will start from a particular case: the lifeguard \(A\) is exactly on the edge of the coast and starts to run along that line until reaching point \(C\), and then he swims towards the distressed swimmer \(B\). How should we choose \(C\), so that \(ACB\) is the fastest path? Let \(v_{1}\) be the running velocity and \(v_{2}\)(\(< v_{1}\)) the swimming velocity. We want \(C\) such that (see next picture), for any points in the coast line, \(D\) on the right of \(C\) and \(E\) on the left of \(C\), the paths \(ADB\) and \(AEB\) would take longer than the path \(ACB\).

\(F\) is the point on the coast \(AD\) and \(FB\) is perpendicular to \(AD\). In the dotted line, we have represented the perpendiculars of \(CB\) through \(C\) and \(D\), and its parallel \(CB\) through \(E\), with \(G\), \(H\), \(I\) points of intersection clearly identified in the picture.

The time spend in the path \(ACB\) is \(\frac{\overline{AC}}{v_{1}} + \frac{\overline{CB}}{v_{2}}\). And the time in \(ADB\) is \(\frac{\overline{AD}}{v_{1}} + \frac{\overline{DB}}{v_{2}}\) These two paths have a common subpath: \(AC\). We want to find a condition on \(C\) so that we can assure that \(CB\) is faster than \(CDB\) for every \(D\) on the right of \(AC\). Obviously, \(HB\) is faster than \(DB\): they are in the same environment, hence, the velocity is the same, and \(DB\) is the hypotenuse of a triangle from which \(HB\) is a cathetus, and consequently longer than \(HB\). If the time spent on \(CH \) (\(= \frac{\overline{CH}}{v_{2}}\)) is less than or equal to the time spent on \(CD\) (\(= \frac{\overline{CD}}{v_{1}}\)), we can conclude that the path \(CB\) is faster. But \(\frac{\overline{CH}}{v_{2}} \leq \frac{\overline{CD}}{v_{1}}\) is equivalent to \(\frac{\overline{CH}}{\overline{CD}} \leq \frac{v_{2}}{v_{1}}\). And the triangle \(CHD\) is similar to the triangle \(CFB\) (they are both rectangular and they have a common angle). This means that one can say that if the "slope" of \(CB\), taking the coast line as the reference axis, given by \(\frac{\overline{CF}}{\overline{CB}}\), is less than or equal to \(\frac{v_{2}}{v_{1}}\), then the path \(ADB\) will be slower than the path \(ACB\).

A similar approach, now applied to \(AEB\), given that the rectangular triangle \(EIC\) is also similar to \(CFB\), allows us to conclude that, if \(\frac{\overline{CF}}{\overline{CB}}\) is great than or equal to \(\frac{v_{2}}{v_{1}}\), the path \(ACB\) is faster than the path \(AEB\). We may now say that, if \(\frac{\overline{CF}}{\overline{CB}}\) is exactly \(\frac{v_{2}}{v_{1}}\)1, then the path through \(C\) is faster than the ones on the left and on the right.

1 Which is equivalent to \(\frac{v_{2}}{\small\mbox{slope of }CB}\) of being equal to \(v_{1}\). We should note that this condition seems to be asymmetric with respect to \(v_{1}\) and \(v_{2}\), but \(v_{1}\) can also be written as \(\frac{v_{1}}{\small\mbox{slope of }AC}\), since, in this particular case, in which \(A\) is the line of the water, the slope of \(AC\) is \(1\).