Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Element of Arc Length Problem

  1. Oct 9, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]x = \frac{u^{2} + v^{2}}{2}[/tex]

    [tex]y = uv[/tex]

    [tex]z = z[/tex]

    Find the arc length given:

    [tex]u(t) = cos(t), v(t) = sin(t), z = \frac{2t^{\frac{3}{2}}}{3}[/tex]

    2. Relevant equations
    [tex]ds^{2} = dx^{2} + dy^{2} + dz^{2}[/tex]

    In curvilinear coordinates thhis becomes

    [tex]ds = \sqrt{h^{2}_{1}du^{2}_{1} + h^{2}_{2}du^{2}_{2} + h^{2}_{3}du^{2}_{3}}[/tex]

    3. The attempt at a solution
    First I need to get the scale factors, so I took the derivative of each x, y, z component.

    I came up with:
    [tex]dx = udu - vdv[/tex]
    [tex]dy = vdu + udv[/tex]
    [tex]dz = dz[/tex]

    I then found the scale factors,

    [tex]h_{1} = h_{u} = \sqrt{u^{2} + v^{2}}[/tex]
    [tex]h_{2} = h_{v} = \sqrt{u^{2} + v^{2}}[/tex]
    [tex]h_{3} = h_{z} = 1[/tex]

    Then we inject the scale factors into the element arc length formula.

    [tex]ds = \sqrt{h^{2}_{1}du^{2}_{1} + h^{2}_{2}du^{2}_{2} + h^{2}_{3}du^{2}_{3}}[/tex]

    I'm not sure what to do about du1, du2, and du3. Are they just dx, dy and dz? And if so, would this mean I have to integrate 3 times to get the arc length?
  2. jcsd
  3. Oct 9, 2008 #2
    I'm not entirely sure about what all this stuff about scale factors is, but it seems that your solution is correct ( though I haven't checked everything). In general, the 3 dimensional Euclidean line element is [itex] ds^2 = dx \wedge dx + dy \wedge dy + dz \wedge dz [/itex]. However, since you are considering zero curvature space, the wedges can just be dropped, so that [itex] ds^2 = dx^2 + dy^2 + dz^2 [/itex].

    Now you've calculated each of dx, dy, and dz in terms of du, dv, dz. To find the line element in the new system, simply plug those expressions directly into the above equation. That is

    [tex] ds^2 = dx^2 + dy^2 + dz^2 [/tex]
    [tex] = (u du - v dv)^2 + (udv + v du) + dz^2 [/tex]
    [tex] = (u^2+v^2) du^2 + (u^2 + v^2) dv^2 + dz^2 [/tex]

    which is exactly what you have.

    Now for finding the arclength, you want to integrate ds, that is [itex] s = \displaystyle \int ds [/itex]. You could indeed integrate three times, though this would be very inefficient. Instead, consider the following.

    For simplicity of notation, let the line element be [itex] ds = \sqrt{ dx^2 + dy^2 + dz^2} [/itex]. Now lets multiply by one, in the form of [itex] \frac{dx}{dx} [/itex]. That is

    [tex] \sqrt{ dx^2 + dy^2 + dz^2} \frac{dx}{dx} = \sqrt{ \frac{dx^2}{dx^2} + \frac{dy^2}{dx^2} + \frac{dz^2}{dx^2} } dx [/tex]
    [tex] = \displaystyle \sqrt{ 1 + \left( \frac{dy}{dx} \right)^2 + \left( \frac{dz}{dx} \right)^2 } dx [/tex]

    It's easy enough to calculate dy/dx and dz/dx, and now you need only do one integration.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook