\section*{Problem 11}

The distance from A to \(\mathrm{B}\) is \(\mathrm{d}\) kilometers. A plane \(\mathrm{P}\) is flying with constant speed, height and direction from A to B. Over a period of 1 second the angle PAB changes by \(\alpha\) degrees and the angle PBA by \(\beta\) degrees. What is the minimal speed of the plane?


\section*{Solution}

Answer: \(20 \pi \mathrm{d} \sqrt{ }(\alpha \beta)\) kilometers per hour.

Let the plane be at height \(\mathrm{h}\) and a (horizontal) distance \(\mathrm{y}\) from A. Let the angle PAB be \(\theta+\alpha\) and the angle PBA be \(\varphi\). After 1 second, the angle PAB is \(\theta\) and the angle PBA is \(\varphi+\beta\). We have immediately that:

\[
\mathrm{h} / \mathrm{y}=\tan (\theta+\alpha), \mathrm{h} /(\mathrm{d}-\mathrm{y})=\tan \varphi, \mathrm{h} /(\mathrm{y}+\mathrm{x})=\tan \theta, \mathrm{h} /(\mathrm{d}-\mathrm{y}-\mathrm{x})=\tan (\varphi+\beta)
\]

Eliminating \(\theta\), we obtain: \(\mathrm{h} / \mathrm{y}=(\tan \alpha+\tan \theta) /(1-\tan \alpha \tan \theta)=(\mathrm{a}(\mathrm{y}+\mathrm{x})+\mathrm{h}) /(\mathrm{y}+\mathrm{x}-\mathrm{ah})\), where \(a=\tan \alpha\). Hence \(x=a\left(h^{2}+y^{2}\right) /(h-a y)\). Similarly, eliminating \(\varphi\), we obtain \(x=b\left(h^{2}+(d-\right.\) \(\left.\mathrm{y})^{2}\right) /(h+(d-y) b)\).

At this point I do not see how to make further progress without approximating. But approximating seems reasonable, since \(\alpha\) and \(\beta\), are certainly small, at least when expressed in radians. For example, typical values might be \(10,000 \mathrm{ft}\) for \(\mathrm{h}\) and more than 10 miles for \(\mathrm{y}\) or \(\mathrm{d}-\mathrm{y}\) and \(500 \mathrm{mph}\) for the aircraft speed. That gives \(\mathrm{x}=0.14\) miles, so \(\mathrm{x} / \mathrm{y}=0.014\) and \(\mathrm{x} / \mathrm{h}=\) 0.07 . So, let us neglect \(a / h, b / h, a / y\) etc. Then we get the simplified expressions: \(x=a\left(h^{2}+\right.\) \(\left.y^{2}\right) / h=b\left(h^{2}+(d-y)^{2}\right) / h\).

If \(\mathrm{a}=\mathrm{b}\), then we quickly obtain \(\mathrm{y}=\mathrm{d} / 2, \mathrm{~h}=\mathrm{d} / 2, \mathrm{x}=\) ad. Assume \(\mathrm{a}>\mathrm{b}\). Then we can solve for \(\mathrm{h}\), substitute back in and obtain an expression for \(\mathrm{x}\) in terms of \(\mathrm{y}\). It is convenient to divide through by \(\mathrm{d}\) and to write \(\mathrm{X}=\mathrm{x} / \mathrm{d}, \mathrm{Y}=\mathrm{y} / \mathrm{d}\). Note that since we are assuming \(\mathrm{a}>\mathrm{b}\), we require \(\mathrm{Y}<1 /(1+\sqrt{ }(\mathrm{a} / \mathrm{b}))\). After some manipulation we obtain: \(\mathrm{X}=\mathrm{ab}(1-2 \mathrm{Y}) / \sqrt{ }(\mathrm{a}-\mathrm{b})\left(\mathrm{b}(1-\mathrm{Y})^{2}-\mathrm{a}^{2}\right)\). Differentiating, we find that there is a minimum at \(\mathrm{Y}=\mathrm{b} /(\mathrm{a}+\mathrm{b})\), which is in the allowed range, and that the minimum value of \(\mathrm{X}\) is \(\sqrt{ }(\mathrm{ab})\). By symmetry, we obtain the same result for \(\mathrm{a}<\mathrm{b}\) and we notice that it is also true for \(\mathrm{a}=\mathrm{b}\). So in all cases we have that the minimum value of \(\mathrm{x}\) is \(d \sqrt{ }(a b)\).

We are assuming \(\alpha\) and \(\beta\) are small, so we may take \(a=\alpha, b=\beta\). However, the question specified that \(\alpha\) and \(\beta\) were measured in degrees, so to obtain the final answer we must convert, giving: \(x=d(\pi / 180) \sqrt{ }(\alpha \beta)\), and hence speed \(=20 \pi \mathrm{d} \sqrt{ }(\alpha \beta)\) kilometers per hour.

