1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Surface Optimization

  1. Oct 2, 2005 #1
    Ok, so here's the problem:

    Two circular wire hoops of radius [itex]R[/itex] are spaced a distance [itex]2\ell[/itex] apart. Consider a soap film stretching beatween the two hoops. Due to surface tension the film's equlibrium form is a surface of minimal area. Use the calculus of variations to find this minimal surface.

    What happens when [itex]\ell/R[/itex] is small? What happens when [itex]\ell/R[/itex] becomes large? Is there a critical value of [itex]\ell/R[/itex]? If so, explain why; find in numerically; and say what happens when one exceeds it.

    Calculus of Variations is simple: To "optimize" a quantity, [itex]y(x)[/itex], (i.e. find its minima/maxima), you find the value of [itex]x[/itex] which satisfies [itex]y'(x)=0[/itex].

    If the independent is a function which optimizes a quantity (as in the case of this problem),
    [tex]A=\int F(y(x),\,y'(x);\,x)\,dx[/tex]​
    the optimized [itex]y(x)[/itex] is given by the solution to the differential equation below (Euler's Equation):
    [tex]\frac{\partial F}{\partial y}-\frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right)=0[/tex]​


    Here's what I did. It makes most sense to set this problem in cylindrical coordinates. So, the surface area of the soap film is given by

    [tex]A=\int_0^{2\ell}2\pi r(z)\sqrt{1+r'(z)^2}dz.[/tex]​

    Here, we are assuming that there is no angular dependence of [itex]r[/itex]. So in anticipation of applying the calculus of variations (Euler's Equation), I evaluated the following quantities for [itex]F(r(z),\,r'(z);\,z)=r(z)\sqrt{1+r'(z)^2}[/itex]:

    [tex]\frac{\partial F}{\partial r}=\sqrt{1+r'^2}\,,\,\,\,\,\,\,\,\,\frac{\partial F}{\partial r'}=\frac{r(z)r'(z)}{\sqrt{1-r'^2}}\,,\,\,\,\,\text{and}\,\,\,\,\,\,\frac{d}{dz}\left(\frac{\partial F}{\partial r'}\right)=\frac{1-r'^4+rr''}{(1+r'^2)^{3/2}}[/tex]​

    Therefore Euler's equation becomes,
    [tex]\frac{1+r'^2-rr''}{(1+r'^2)^{3/2}}=0.[/tex]​
    If we disallow the denominator from vanishing, we can simplify this:
    [tex]1+\left(\frac{dr}{dz}\right)^2-r(z)\frac{d^2r}{dz^2}=0.[/tex]​

    I have no clue how to approach this. Mathematica gave me some really wierd ugly answer, which when I asked it to FullSimplify, gave an unmanagable expression with hyperbolic sines and cosines. What's the solution to this differential equation, and how do I analytically apply the boundary conditions?
     
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted



Loading...
Similar Threads for Surface Optimization Date
Surface Tension Yesterday at 7:00 AM
Equipotential surface and electric field Apr 16, 2018
Ball floating on the surface of water Apr 12, 2018
Find the charge on the outer surface Apr 12, 2018
Optimal angle to guide pallets with the least friction Mar 14, 2018