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Thread: curvature

  1. #1
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    curvature

    find the curvature K of the curve....

    r(t) = 4t i + 3 cos t j + 3 sin t k

    i am confused after i find r'(t)...
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by chris25 View Post
    find the curvature K of the curve....

    r(t) = 4t i + 3 cos t j + 3 sin t k

    i am confused after i find r'(t)...
    Here is the formula

    $\displaystyle \kappa=\frac{|\frac{dr}{dt} \times (\frac{dr}{dt})^2|}{|\frac{dr}{dt}|^3}$

    I have to go to work. I will finish later if no one else has.

    Good luck.
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  3. #3
    Super Member wingless's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    Here is the formula

    $\displaystyle \kappa=\frac{|\frac{dr}{dt} \times \color{red}(\frac{dr}{dt})^2\color{black}|}{|\frac {dr}{dt}|^3}$

    I have to go to work. I will finish later if no one else has.

    Good luck.
    As far as I know, it's $\displaystyle \kappa = \frac{\left | \frac{dr}{dt} \times \frac{d^2r}{dt^2}\right |}{\left | \frac{dr}{dt} \right |^3} = \frac{\left | \dot {r} \times \ddot {r} \right |}{\left | r \right |^3}$

    See here if you wonder how to get the formula:
    Curvature -- from Wolfram MathWorld
    Last edited by wingless; Jun 10th 2008 at 12:37 PM.
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by wingless View Post
    As far as I know, it's $\displaystyle \kappa = \frac{\left | \frac{dr}{dt} \times \frac{d^2r}{dt^2}\right |}{\left | \frac{dr}{dt} \right |^3} = \frac{\left | \dot {r} \times \ddot {r} \right |}{\left | r \right |^3}$

    See here if you wonder how to get the formula:
    Curvature -- from Wolfram MathWorld
    @ wingless You are correct Thanks that is what I get for trying to do it in a hurry.

    $\displaystyle r(t)=<4t,3\cos(t),3\sin(t)$

    $\displaystyle \frac{dr}{dt}=<4,-3\sin(t),3\cos(t)>$

    $\displaystyle \frac{d^2r}{dt^2}=<0,-3\cos(t),-3\sin(t)>$

    Now we take the cross product to get

    $\displaystyle \begin{vmatrix}
    i & j & k \\
    4 & -3\sin(t) & 3\cos(t) \\
    0 & -3\cos(t) & -3\sin(t) \\
    \end{vmatrix}=(9\sin^2(t)+9\cos(t))\vec i-(-12\sin(t)-0)\vec j+(-12\cos(t)-0)\vec k$

    Simplifying and taking the magnitude we get

    $\displaystyle =|<9,12\sin(t),-12\cos(t)>|=\sqrt{(9)^2+(12\sin(t))^2+(-12\cos(t))^2}$
    $\displaystyle =\sqrt{81+144(\sin^{2}(t)+\cos^{2}(t))}=\sqrt{225} =15$

    Now we need the magnitude of

    $\displaystyle \bigg| \frac{dr}{dt}\bigg|=\sqrt{4^2+(-3\sin(t))^2+(3\cos(t))^2}=\sqrt{16+9(\sin^2(t)+\co s^2(t))}=\sqrt{25}=5$

    Finally we get

    $\displaystyle \kappa=\frac{15}{5^3}=\frac{3}{25}$

    Yeah.

    Sorry about the typo before.
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