The hanging chain
Tonight's idea
Two functions built from \(e^x\), the hyperbolic cosine and sine:
\(\cosh x = \dfrac{e^x + e^{-x}}{2}, \qquad \sinh x = \dfrac{e^x - e^{-x}}{2}.\)
They are the even and odd halves of \(e^x\), so \(\cosh x + \sinh x = e^x\). Square each and subtract. The cross terms are \(\pm 2e^xe^{-x} = \pm 2\):
\[\cosh^2 x - \sinh^2 x = \tfrac14\big[(e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x})\big] = 1.\]
Compare \(\cos^2 + \sin^2 = 1\). The trig pair walks round the circle \(x^2 + y^2 = 1\); this pair walks along the hyperbola \(x^2 - y^2 = 1\), hence the name. Derivatives are easy, since \((e^{-x})' = -e^{-x}\): \((\cosh x)' = \sinh x\) and \((\sinh x)' = \cosh x\), with no minus sign to remember.
Now hang a chain weighing \(w\) per metre. The horizontal part of its tension is one constant \(T_0\) all along it, because nothing pulls sideways. The vertical part at horizontal position \(x\) is the weight of chain between the lowest point and \(x\): \(ws(x)\), where \(s\) is arc length. The slope is their ratio, \(y' = ws/T_0\). Differentiate, using \(s' = \sqrt{1 + y'^2}\) (Pythagoras on a short piece):
\(y'' = \dfrac{1}{a}\sqrt{1 + y'^2}, \qquad a = \dfrac{T_0}{w}.\)
The solution is the catenary \(y = a\cosh(x/a)\). A parabola needs \(y''\) constant; here \(y''\) grows with the slope, so the ends bend up more sharply than any parabola.
Watch one: check the solution, then test a parabola
Take \(y = a\cosh(x/a)\). Then \(y' = \sinh(x/a)\) and \(y'' = \tfrac1a\cosh(x/a)\). On the other side, \(1 + y'^2 = 1 + \sinh^2(x/a) = \cosh^2(x/a)\), so \(\tfrac1a\sqrt{1 + y'^2} = \tfrac1a\cosh(x/a)\), since cosh is positive. The two sides agree.
Now try \(y = 1 + x^2/2\), the parabola that matches \(\cosh x\) best near 0. Its \(y''\) is 1, but \(\sqrt{1 + y'^2} = \sqrt{1 + x^2}\), which equals 1 only at \(x = 0\). The values drift too: at \(x = 2\), \(\cosh 2 = 3.76\) against 3; at \(x = 3\), \(10.07\) against \(5.5\).
Your turn
- From the definitions, show that \(\cosh(-x) = \cosh x\), \(\sinh(-x) = -\sinh x\) and \(\cosh x + \sinh x = e^x\). What does \(\cosh x\) look like for large \(x\)?
- Differentiate the definitions to confirm \((\cosh x)' = \sinh x\) and \((\sinh x)' = \cosh x\). Then show that \(\tanh x = \sinh x / \cosh x\) has derivative \(1/\cosh^2 x\).
- The parabola through \((0, 1)\) and \((\pm 3, \cosh 3)\) is \(y = 1 + kx^2\). Find \(k\) to three decimals, then compare the parabola with \(\cosh x\) at \(x = 1.5\). Which is higher, and by how much?
- A chain with \(a = 10\) m hangs between posts at \(x = \pm 15\) m. (a) Find the sag, \(y(15) - y(0)\). (b) The arc length from the lowest point is \(s = a\sinh(x/a)\). Check that \(s' = \sqrt{1 + y'^2}\), then find the chain's total length and compare it with the 30 m span.
Stretch: the height is the tension
The tension \(T\) at a point acts along the chain, at the slope angle \(\theta\) with \(\tan\theta = y'\). Its horizontal part is \(T\cos\theta = T_0\). Show that \(T = wy\): the tension anywhere on a catenary equals the weight of a length \(y\) of chain, so the curve's height reads off its own tension.
Answers
- Replacing \(x\) by \(-x\) swaps \(e^x\) and \(e^{-x}\): the sum is unchanged and the difference changes sign. Adding the two definitions cancels the \(e^{-x}\) terms and leaves \(e^x\). For large \(x\), \(e^{-x}\) is negligible and \(\cosh x \approx \sinh x \approx e^x/2\): at \(x = 3\), \(\cosh 3 = 10.068\) and \(e^3/2 = 10.043\).
- \((e^x)' = e^x\) and \((e^{-x})' = -e^{-x}\), so \((\cosh x)' = \tfrac12(e^x - e^{-x}) = \sinh x\) and \((\sinh x)' = \tfrac12(e^x + e^{-x}) = \cosh x\). Quotient rule: \((\tanh x)' = \dfrac{\cosh^2 x - \sinh^2 x}{\cosh^2 x} = \dfrac{1}{\cosh^2 x}\), by the identity.
- \(\cosh 3 = 10.0677\), so \(k = (\cosh 3 - 1)/9 = 1.008\). At \(x = 1.5\): parabola \(1 + 1.008 \times 2.25 = 3.267\); \(\cosh 1.5 = 2.352\). The parabola is higher by \(0.91\). (The figure shows this gap.)
- (a) \(y(15) - y(0) = 10(\cosh 1.5 - 1) = 10(2.3524 - 1) = 13.52\) m. (b) \(s' = \cosh(x/a) = \sqrt{1 + \sinh^2(x/a)} = \sqrt{1 + y'^2}\). Total length \(= 2 \times 10\sinh 1.5 = 20 \times 2.1293 = 42.59\) m, about 42% longer than the 30 m span.
Stretch. \(T = T_0/\cos\theta = T_0\sqrt{1 + \tan^2\theta} = T_0\sqrt{1 + y'^2} = T_0\cosh(x/a)\). Since \(T_0 = wa\), this is \(wa\cosh(x/a) = wy\). At the lowest point \(T = wa = T_0\); the tension grows with height, and at the posts it is the weight of a chain as long as the post is tall (measured from \(y = 0\), which sits a distance \(a\) below the lowest point).