Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 64

Suppose that two vertical poles $P\theta$ and $ST$ are secured by a rope $PRS$ going from the top of the first pole to a point $R$ on the ground between the poles and then to the top of the second poles as in the figure. Show that the shortest length of such rope occues when $\theta_1 = \theta_2$.




The total length of the rope is $L = L_1 + L_2$, then...

$
\begin{equation}
\begin{aligned}
\sin \theta_1 &= \frac{a}{L_1} \qquad \text{and} \qquad \sin \theta_2 = \frac{b}{\theta_2}
\end{aligned}
\end{equation}
$


So, $\quad \displaystyle L = \frac{a}{\sin \theta_1} + \frac{b}{\sin \theta_2}$

If we take the derivative of $L$ with respect to $\theta$,

$
\begin{equation}
\begin{aligned}
\frac{dL}{d\theta_1} &= \frac{d}{d \theta_1} \left( \frac{a}{\sin \theta_1} \right) + \frac{d}{d\theta_2} \left( \frac{b}{\sin \theta_2} \right) \frac{d\theta_2}{d\theta_1}\\
\\
\frac{dL}{d\theta_1} &= -a \csc \theta_1 \cot \theta_1 - b \csc \theta_2 \cot \theta_2 \frac{d\theta_2}{d\theta_1} \qquad \Longleftarrow \text{(Equation 1)}
\end{aligned}
\end{equation}
$


$
\begin{equation}
\begin{aligned}
\text{We know that segment } |Q T| = \frac{a}{\tan \theta_1} + \frac{b}{\tan \theta_2}
\end{aligned}
\end{equation}
$


If we take the derivative of the segment $|QT|$ with respect to $\theta_1$, we get...

$
\begin{equation}
\begin{aligned}
\frac{d|QT|}{d\theta_1} &= \frac{d}{d \theta_1} \left( \frac{a}{\tan \theta_1} \right) + \frac{d}{d \theta_2} \left( \frac{b}{\tan \theta_2} \right) \frac{d\theta_2}{d \theta_1}\\
\\
\frac{d|QT|}{d\theta_1} &= -a \csc^2 \theta_1 - b \csc^2 \theta_2 \left( \frac{d\theta_2}{d\theta_1} \right)
\end{aligned}
\end{equation}
$


when $\displaystyle \frac{d|QT|}{d \theta_1} = 0$,

$
\begin{equation}
\begin{aligned}
0 &= -a \csc^2 \theta_1 - b \csc^2 \theta_2 \left( \frac{d\theta_2}{d\theta_1} \right)\\
\\
\frac{d\theta_2}{d\theta_1} &= \frac{-a \csc^2 \theta_1}{b \csc^2 \theta_2}
\end{aligned}
\end{equation}
$

By using Equation 1, we get...
$\displaystyle \frac{dL}{d \theta_1} = - a \csc \theta_1 \cot \theta_1 - b \csc \theta_2 \cot \theta_2 \left( \frac{-a \csc^2 \theta_1}{b \csc^2 \theta_2} \right)$

when $\displaystyle \frac{dL}{d \theta_1} = 0,$

$
\begin{equation}
\begin{aligned}
\cot \theta_1 \csc \theta_2 &= \csc \theta_1 \cot \theta_2\\
\\
\frac{\cot \theta_1}{\csc \theta_1} &= \frac{\cot \theta_2}{\csc \theta_2}\\
\\
\cos \theta_1 &= \cos \theta_2
\end{aligned}
\end{equation}
$


Therefore, we can say that shortest length of such a rope occurs hwen $\theta_1 = \theta_2$.