I find the derivation of the shape of the catenary curve by Jeremy Tatum in the libretexts version of his book quite elegant.

I reuse an image from an earler catenary question:

Catenary curve, force vectors

Tatum defines:

$\mu$ mass per unit of length of chain
$g$ gravitational force per unit of mass
$s$ length from midpoint to current coordinate
$\psi$ angle at current coordinate
$T_0$ Force component in the horizontal direction


$$ \tan \psi = \frac{\mu g s}{T_0} \tag 1 $$

for brevity a quantity $a$ is defined:

$$ a = \frac{T_0}{\mu g} \tag 2 $$


$$ \tan \psi = \frac{dy}{dx} = \frac{s}{a} \tag 3 $$

from which, by derivation and rearranging:

$$ \frac{ds}{dx} = a \frac{d^2y}{dx^2} \tag 4 $$

Then Tatum uses $ds = (1 + \tfrac{dy}{dx}^2)^\tfrac{1}{2}dx$ to arrive at:

$$ \left(1 + \tfrac{dy}{dx}^2\right)^\tfrac{1}{2} = a \frac{d^2y}{dx^2} \tag 5 $$

Then Tatum capitalizes on the relation $\cosh^2 = \sinh^2 + 1$
With the condition that $\tfrac{dy}{dx} = 0$ where $x=0$

$$ \frac{dy}{dx} = \sinh (\frac{x}{a}) \tag 6 $$


$$ y = a \cosh(\frac{x}{a}) + C \tag 7 $$

Tatum calls the move from (5) to (6) 'integration', but it seems to me that that step isn't actually integrating, but a substitution.

But: if I'm substituting anyway, then I can just as well go straight from (5) to (7)

I very much like the idea of capitalizing on the fact that $\sinh$ is the derivative of $\cosh$, and vice versa. The differential equation enforces the $\cosh$ as solution anyway, so as soon as you recognize that the differential equation is solved.

So I would love to learn about a way of going from (5) to (7) straigh away, explicitly motivated.

(The earlier question about the derivation of the catenary curves here on math.stackexchange gives in effect the same derivation as Tatum gives.)


1 Answer 1


If you allow $z(x)=y'(x)$, then (5) is $$ az'(x) = \sqrt{1+z^2}. $$

By separating variables: $$ \frac{dz}{\sqrt{1+z^2}} = \frac{dx}{a}, $$

which can be integrated to $$ \mathop{\mathrm{arcsinh}} z = \frac{x-x_0}a. $$

  • $\begingroup$ Thank you for demonstrating that indeed the transition involves an integration step. Incidentally, my overall goal is to show how differential calculus and calculus of variations are related. Finding the shape of the catenary curve can be done both with differential calculus and calculus of variations. That offers a window onto the relation between the two. (That is, for my purpose it isn't necessary to present an exhaustive derivation using differential calculus. I just need to show that it can be done with differential calculus.) $\endgroup$
    – Cleonis
    Jan 7, 2022 at 2:20

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