Minimal Surface

Prove that the minimum surface of revolution formed by a curve joining two points x1,y1{x}_{1},{y}_{1} and x1,y1{x}_{1},{y}_{1} is described by revolving the catenary.

Solution

You should familiarize yourself with this identity first.

Given the arclength formula dS=1+y2dxdS = \sqrt{1+{y'}^{2}} dx, the surface area is thus 2πx1x2y1+y2dx.2 \pi \int _{ { x }_{ 1 } }^{ { x }_{ 2 } }{ y\sqrt { 1+{ y' }^{ 2 } } } dx.

Since the integrand is independent of xx, we can apply the above Euler-Lagrange identity and show that y1+y2y[y2(1+y2)1/22y]=Ay\sqrt { 1+{ y' }^{ 2 } } - y' \left[\frac{y}{2} {(1+{y'}^{2})}^{-1/2} 2y' \right] = A for some arbitrary constant AA.

With a little algebra, y=y2A2A.y' = \frac{\sqrt{{y}^{2} - {A}^{2}}}{A}. Solving this differential equation we find y=Acosh(x+kA)y= A \cosh {\left(\frac{x+k}{A}\right)} where kk is another arbitrary constant. This curve is the famous catenary.

Check out my set Classic Demonstrations.

Note by Steven Zheng
4 years, 11 months ago

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