Chapter 4, Problem 44RE

To determine

To find: The maximum value of the observer’s angle of sight $\theta$ between the runners.

Answer to Problem 44RE

The value of the observer’s angle of sight $\theta$ between the runners

That is, $\psi =\frac{\pi }{6}$

Explanation of Solution

Given:

Distance between the point $P$ from the track =1

Calculation:

Two runners start at the point $S$ and run along the track.

One runner runs three times as fast as the other.

Let the distance of $BC=3t$.

Consider the triangle $PSD,$

Let, $\angle P=\psi$

Consider the triangle $PCD$

$\angle P=\theta$

Consider the triangle $PSC$

$\angle P=\theta +\psi$

Then,

$\begin{array}{c}\tan \left(\theta +\psi \right)=\frac{3t}{1}\\ \frac{\tan \theta +\tan \psi }{1-\tan \theta \tan \psi }=3t\\ \frac{t+\tan \theta }{1-t\tan \theta }=3t\end{array}$

Simplify further,

$\begin{array}{c}3t\left(1-t\tan \theta \right)=t+\tan \theta \\ 2t=\left(1+3t^2\right)\tan \theta \\ \tan \theta =\frac{2t}{1+3t^2}\end{array}$

Let $f\left(t\right)=\tan \theta =\frac{2t}{1+3t^2}$

Differentiate with respect to t,

$\begin{array}{c}\frac{df}{dt}=\frac{2\left(1+3t^2\right)-2t\left(6t\right)}{{\left(1+3t^2\right)}^2}\\ =\frac{2\left(1-3t^2\right)}{{\left(1+3t^2\right)}^2}\end{array}$

For critical points, $\frac{df}{dt}=0$

$\begin{array}{c}\frac{2\left(1-3t^2\right)}{{\left(1+3t^2\right)}^2}=0\\ 1-3t^2=0\\ t=\pm \frac{1}{\sqrt{3}}\end{array}$

Differentiate $\frac{df}{dt}$ with respect to t,

$\begin{array}{c}\frac{d^{2}f}{d{t}^2}=\frac{{\left(1+3t^2\right)}^2\left(-12t\right)-2\left(1-3t^2\right)2\left(1+3t^2\right)}{{\left(1+3t^2\right)}^4}\\ =\frac{\left(1+3t^2\right)\left(-12t\right)-4\left(1-3t^2\right)}{{\left(1+3t^2\right)}^3}\end{array}$

Substitute $t=\frac{1}{\sqrt{3}}$ in $\frac{d^{2}f}{d{t}^2}$

$\begin{array}{c}\frac{d^{2}f}{d{t}^2}=\frac{\left(1+3\frac{1}{3}\right)\left(-12\frac{1}{3}\right)-4\left(1-3\frac{1}{3}\right)}{{\left(1+3\frac{1}{3}\right)}^3}\\ =\frac{-8}{8}\\ =-1<0\end{array}$

So. $f$ has absolute maximum when $t=\frac{1}{\sqrt{3}}$.

Substitute $t=\frac{1}{\sqrt{3}}$ in $\tan \theta =\frac{2t}{1+3t^2}$.

$\begin{array}{c}\tan \theta =\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1+3\frac{1}{3}}\\ =\frac{1}{\sqrt{3}}\\ =\tan \frac{\pi }{6}\\ \theta =\frac{\pi }{6}\end{array}$

Substitute $t=\frac{1}{\sqrt{3}}$ and $\theta =\frac{\pi }{6}$ in $\tan \left(\theta +\psi \right)=\frac{3t}{1}$

$\begin{array}{c}3\frac{1}{\sqrt{3}}=\tan \left(\psi +\frac{\pi }{6}\right)\\ \sqrt{3}=\tan \left(\psi +\frac{\pi }{6}\right)\\ \tan \frac{\pi }{3}=\tan \left(\psi +\frac{\pi }{6}\right)\end{array}$

Simplify that,

$\begin{array}{c}\frac{\pi }{3}=\psi +\frac{\pi }{6}\\ \psi =\frac{\pi }{3}-\frac{\pi }{6}\\ =\frac{\pi }{6}\end{array}$

Thus, the value of the observer’s angle of sight $\theta$ between the runners

That is $\psi =\frac{\pi }{6}$.