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A chain hanging freely under gravity between two fixed points $(x, y)$ = $(\pm \:x_0, 0)$ (where $x_0>0$) adopts the shape given by $y=\frac{1}{k}\left(\cosh (kx) - \cosh (kx_0) \right)$ for $\lvert x \rvert < x_0$, where $k>0$ is a constant. Using this expression of the curve, find the length L of the chain between $(\pm \: x_0, 0)$ in terms of $k$.

Can anyone please show me the steps of solving this question? If someone can, please show the steps and name the method used in each step. I would be grateful. Thank you very much for helping.

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There are various formulas for the arclength of a curve. The first one mentioned in a typical calculus course says that the arclength of $y=f(x)$ from $x=a$ to $x=b$ is $$\int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx.$$ In our case $\frac{dy}{dx}=\sinh kx$. Here we have used the Chain Rule and the fact that the derivative of $\cosh t$ is $\sinh t$.

Thus we find that $1+\left(\frac{dy}{dx}\right)^2=1+\sinh^2 kx=\cosh^2 kx$. Here we have used the identity $1+\sinh^2 t=\cosh^2 t$.

I expect that you can take over from here.

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  • $\begingroup$ How do we only get $sinh\:\left(kx\right)$ from the derivative, shouldn't we get $sinh\:\left(kx\right)-sinh\:\left(kx_0\right)$ ? and another question, in my case what is are the limits of the integration a and b ? $\endgroup$
    – ASD123
    Commented Nov 18, 2014 at 13:45
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    $\begingroup$ The $-\cosh kx_0$ part is a constant, its derivative is $0$. We can integrate from $-x_0$ to $x_0$. I prefer to use symmetry, integrate from $0$ to $x_0$ and multiply the result by $2$. $\endgroup$
    – André Nicolas
    Commented Nov 18, 2014 at 13:50
  • $\begingroup$ I think I am a bit lost, if it's possible for you to solve it on a paper and post a picture of it here that would be great. Thank you very much for helping. Just a reminder I need the find the Length which is L and find the constant which is k $\endgroup$
    – ASD123
    Commented Nov 18, 2014 at 14:04
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    $\begingroup$ The answer you posted is basically right. You need to do the plugging in and get $\frac{2}{k}\sinh(kx_0)$. That's the end, we can do no more since we are not told the values of $k$ and $x_0$. $\endgroup$
    – André Nicolas
    Commented Nov 18, 2014 at 14:18
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    $\begingroup$ Just pick $x_0$ arbitrary, like $1$, and calculate for a few values of $k$. Or treat it as a function of $k$ and use calculus tools. $\endgroup$
    – André Nicolas
    Commented Nov 18, 2014 at 14:26

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