Chapter 4.6, Problem 52E

To determine

To calculate: Show that the shortest length of such a rope occurs when ${\theta }_1={\theta }_2$ If two vertical poles $PQ$ and $ST$ k are secured by a rope $PRS$ going from the top of the first pole to point $R$ on the ground between the poles and then to the top of the second pole as in the figure.

Answer to Problem 52E

The shortest length of such a rope occurs when ${\theta }_1={\theta }_2$

Explanation of Solution

Given information:

Two vertical poles $PQ$ and $ST$ k are secured by a rope $PRS$ going from the top of the first pole to point $R$ on the ground between the poles and then to the top of the second pole as in the figure.

Formula used:

Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is:

  

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

Trigonometric ratio:

  $\cos \theta =\frac{b}{h}$

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

Calculation:

As per the given problem

Draw the diagram of two vertical poles $PQ$ and $ST$ k are secured by a rope $PRS$ going from the top of the first pole to point $R$ on the ground between the poles and then to the top of the second pole as in the figure

  

Let $d$ represent the distance between the poles and $x$ represent the distance between the base of the pole $P$ and tie-point $R$

Also, let $p$ and $s$ be the respective heights of poles $P$ and $S$

Finally, let $l\left(x\right)$ represent the rope- length based on the variable tie-point.

Recall that, Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is:

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

In right angled triangle $△PQR$

  $\begin{array}{l}{\left(PR\right)}^2=p^2+x^2\\ \Rightarrow PR=\sqrt{p^2+x^2}\end{array}$

Recall that, Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is:

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

In right angled triangle $△SRT$

  $\begin{array}{l}{\left(SR\right)}^2=s^2+{\left(d-x\right)}^2\\ \Rightarrow SR=\sqrt{s^2+{\left(d-x\right)}^2}\end{array}$

Now, the length of the rope is

  $\begin{array}{l}l\left(x\right)=PR+SR\\ =\sqrt{p^2+x^2}+\sqrt{s^2+{\left(d-x\right)}^2}\end{array}$

Recall that,

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

Differentiate with respect to $x$

  $\begin{array}{l}l\left(x\right)=\sqrt{p^2+x^2}+\sqrt{s^2+{\left(d-x\right)}^2}\\ l^{\prime }\left(x\right)=\frac{x}{\sqrt{p^2+x^2}}-\frac{\left(d-x\right)}{2\sqrt{s^2+{\left(d-x\right)}^2}}\\ \end{array}$

Solve for $l^{\prime }\left(x\right)=0$ , and simplified

  $\begin{array}{l}\frac{x}{\sqrt{p^2+x^2}}-\frac{\left(d-x\right)}{2\sqrt{s^2+{\left(d-x\right)}^2}}=0\\ \Rightarrow \frac{x}{\sqrt{p^2+x^2}}=\frac{\left(d-x\right)}{2\sqrt{s^2+{\left(d-x\right)}^2}}\mathrm{......}\left(1\right)\\ \end{array}$

Recall that,

Trigonometric ratio:

  $\cos \theta =\frac{b}{h}$

  $\cos {\theta }_1=\frac{x}{\sqrt{p^2+x^2}}$ and $\cos {\theta }_2=\frac{\left(d-x\right)}{\sqrt{s^2+{\left(d-x\right)}^2}}$

Therefore , from equation $\left(1\right)$

  $\begin{array}{l}\cos {\theta }_1=\cos {\theta }_2\\ \Rightarrow {\theta }_1={\theta }_2\end{array}$

Since $l^{\prime }\left(x\right)<0$ to the left of the value of $x$ where ${\theta }_1={\theta }_2$ and $l^{\prime }\left(x\right)>0$ to the right

  $l\left(x\right)$ must be a minimum at ${\theta }_1={\theta }_2$

Conclusion:

Thus the shortest length of such a rope occurs when ${\theta }_1={\theta }_2$