Puzzle for you

[Zaphod]

(08-16-2018 11:29 PM)HP-Collection Wrote:  Distance must be zero as the rope half lengh (3 feet) is the difference between the full height (5 feet) and the minimal height over ground (2 feet). The rope must hang straight down and up (sorry about my english)

(08-17-2018 03:59 AM)Thomas Klemm Wrote:  We can model the rope as a catenary:

\(y=a\cosh(\frac{x}{a})\)

where \(a\) is a scaling factor.

Then with \(h\) as the difference of the heights and \(s\) as half the length of the rope we get:

\(h=a(\cosh(\frac{x}{a})-1)\)

\(s=a\sinh(\frac{x}{a})\)

With:

\(u=\frac{x}{a}\)

and

\(v=\frac{h}{s}=\frac{a(\cosh(u)-1)}{a\sinh(u)}=\tanh(\frac{u}{2})\)

Thus

\(u=2\tanh^{-1}(v)=2\tanh^{-1}(\frac{h}{s})\)

This allows us to calculate:

\(a=\frac{s}{\sinh(u)}=\frac{s}{\sinh(2\tanh^{-1}(\frac{h}{s}))}=\frac{s^2-h^2}{2h}\)

Now we plug both \(a\) and \(u\) in to calculate:

\(\begin{align*}
x=a\cdot u &=\frac{s^2-h^2}{2h}\cdot2\tanh^{-1}(\frac{h}{s}) \\
&=\frac{s^2-h^2}{h}\cdot\tanh^{-1}(\frac{h}{s}) \\
&=s\cdot(\frac{s}{h}-\frac{h}{s})\cdot\tanh^{-1}(\frac{h}{s}) \\
&=s\cdot(\frac{1}{v}-v)\cdot\tanh^{-1}(v)
\end{align*}\)

Example:

\(h=3\)

\(s=4\)

\(x=4\cdot(\frac{4}{3}-\frac{3}{4})\cdot\tanh^{-1}(\frac{3}{4})\approx 2.27022850723\)

Limit \(h\to s\)

For \(v\to1\) the value \(\tanh^{-1}(v)\to\infty\) but at the same time \(\frac{1}{v}-v\to0\).
Thus we can apply L'Hôpital's rule to find:

\(\begin{align*}
\lim_{v\to1}(\frac{1}{v}-v)\cdot\tanh^{-1}(v) &= \lim_{v\to1}\frac{1-v^2}{v}\cdot\tanh^{-1}(v) \\
&= \lim_{v\to1}\frac{\tanh^{-1}(v)}{\frac{v}{1-v^2}} \\
&= \lim_{v\to1}\frac{\frac{1}{1-v^2}}{\frac{1+v^2}{(1-v^2)^2}} \\
&= \lim_{v\to1}\frac{1-v^2}{1+v^2}=\frac{1-1}{1+1}=0
\end{align*}\)

And thus:

\(x=s\cdot0=0\)

Which we already knew.

Cheers
Thomas

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