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Homework Help: Parametrized curve

  1. Apr 7, 2005 #1
    Find the length of parametrized curve given by
    [tex]x(t) = 0t^3 +12t^2 - 24t[/tex],
    [tex]y(t) = -4t^3 +12t^2+0t[/tex],
    where t goes from zero to one.
    Hint: The speed is a quadratic polynomial with integer coefficients.

    it's an arclength question right?

    x' = 24*t-24
    y' = -12*t^2+24*t

    [tex]\int_{0}^{1} \sqrt{(24*t-24)^2 + (-12*t^2+24*t)^2}[/tex]

    i get 6.49 when i use a math program to integrate it which is incorrect. anyone know where i went wrong?
     
  2. jcsd
  3. Apr 7, 2005 #2

    xanthym

    User Avatar
    Science Advisor

    Note that:

    [tex] 1: \ \ \ \ \int_{0}^{1} \sqrt{(24*t-24)^2 + (-12*t^2+24*t)^2} \, dt \ = \ \int_{0}^{1} \sqrt{ \left ( 12t^{2} \ - \ 24t \ + \ 24 \right )^{2} } \, dt \ = [/tex]

    [tex] 2: \ \ \ \ = \ \int_{0}^{1} \left | \left ( 12t^{2} \ - \ 24t \ + \ 24 \right ) \right | \, dt \ \color{red} = \ \mathbf{\left (16 \right )} [/tex]


    ~~
     
    Last edited: Apr 7, 2005
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