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Deriving the elastica

  1. May 24, 2012 #1

    Bernoulli's/Newton's classic elastica equation is of the form:

    1) curvature = dθ/ds

    The RHS reforms to the well known elastica equation:

    2) d2y/d2x/(1+(dy/dx)2)3/2

    How do I get from 1) to 2)?

  2. jcsd
  3. May 24, 2012 #2
    Hi there.

    Your function has the form y=y(x).

    Step 1. Find a formula for theta(x), the angle made by the tangent vector at point x.

    Step 2. ds is the element of arclength. The Pythagorean thereom says that ds^2=dx^2+dy^2. Now do the algebra to solve for ds / dx and don't let your Intro Calc. professor see you treating differentials as if they were numbers subject to the normal algebraic rules.

    Step 3. Use the Chain rule for differentiation.

  4. May 25, 2012 #3
    Thanks Vargo,

    Help me out here. If I follow the steps you outline I get stuck:

    1)ds= √dy2+dx2





    Now I'm stuck....
  5. May 25, 2012 #4
    You have sin(theta) and cos(theta). What is tan(theta) as a function of x? Then take the arctan and you have theta(x).
  6. May 30, 2012 #5
    Still no luck I'm afraid Vargo...
  7. May 30, 2012 #6

    Let P be a point in the plane with polar angle theta. Then tan(theta) represents the slope of the line connecting the origin to P. Similarly, the slope of any line is equal to the tangent of the polar angle of that line. In particular, if we have a curve y(x), then
    y'(x) = tan(theta), where theta is the angle of the tangent line.

    Therefore, theta = arctan(y'). We differentiate with respect to x and use the chain rule:
    d theta / dx = y'' / (1+ (y')^2)

    Now we know that ds/dx = sqrt( 1 +(y')^2 ), so using the Chain rule again, we find

    d theta / ds = [y'' / (1+ (y')^2)]*[1 / sqrt( 1 +(y')^2 )] = y'' / (1 + (y') )^(3/2).
  8. May 30, 2012 #7
    Brilliant-now I see it! Thanks a million Vargo.
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