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Frenet-Serret equations

  1. Dec 26, 2008 #1
    1. The problem statement, all variables and given/known data
    For the twisted curve [tex]y^3 + 27 axz - 81a^{2}y = 0[/tex], given parametrically by

    [tex]x=au(3-u^2)[/tex], [tex]y=3au^2[/tex], [tex]z=au(3+u^2)[/tex]

    show that the following hold

    [tex]\frac{ds}{du} = 3 \sqrt{2} a(1+u^2)[/tex], where s is the distance along the curve from the origin

    the length of curve from the origin to (2a, 3a, 4a) is [tex]4\sqrt{2}a[/tex]

    the radius of curvature at the point with parameter u is [tex]3a(1+u^2)^2[/tex]


    3. The attempt at a solution

    The first two parts are relatively painless. First part, evaluate [tex]\frac{d\bold{r}}{du}\cdot\frac{d\bold{r}}{du} [/tex]. Second part is to integrate its square root between s=1 and s=0.

    The last part is a bit troublesome though. First, I get [tex]\hat{\bold{t}} = \frac{dr}{ds} = \frac{dr}{du} \cdot \frac{du}{ds}[/tex], where du/ds can be gotten from above.

    then [tex]\hat{t} = \sqrt{2}(1+u^2)^{-1} [(1-u^2)\bold{i} + 2u\bold{j} + (1+u^2)\bold{k}][/tex].

    Then evaluating, x component of dt/du: [tex]-4\sqrt{2} \frac{u}{(1+u^2)^2}[/tex]

    y component of dt/du: [tex]2\sqrt{2} \frac{1 - u^2}{(1+u^2)^2}[/tex]

    z component is zero.

    Evaluate dt/ds = dt/du (dot) du/ds,

    [tex]\frac{d\hat{t}}{ds}= \frac{-4u}{3a(1+u^2)^3} \bold{i} + \frac{2(1-u^2)}{3a(1+u^2)^3} \bold{j}[/tex]

    Then taking magnitudes and applying the inverse, [tex]\rho = \frac{3}{2}a(1+u^2)^2[/tex]

    Which is pretty weird! The answer does not have the 1/2 coefficient. I'm not sure where I went wrong here. Guidance will be gratefully accepted...
     
  2. jcsd
  3. Dec 26, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Check your unit normal t-hat. I've got the sqrt(2) in the denominator. Not the numerator.
     
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