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# Minimal Surface

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

Solution

You should familiarize yourself with this identity first.

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

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

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

Check out my set Classic Demonstrations.

Note by Steven Zheng
3 years, 1 month ago

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