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Homework Help: Parameterized curve vs. Integral(Please look at my result)

  1. Feb 10, 2009 #1
    1. The problem statement, all variables and given/known data

    Hello just found this forum and have question if I have done my homework correctly

    Anyway let [tex]\theta: I \to \mathbb{R}^3[/tex] be a parameterized curve.
    Next assume that [tex][a,b] \subset I [/tex] and put [tex]\theta(a) = l[/tex](1) and [tex]\theta(b) = m[/tex]

    Then show for any constant vector [tex]\vec{u}[/tex],[tex]|\vec{u}| = 1[/tex].

    that the inequality

    [tex](m-l) \cdot \vec{u} = \int_{a}^{b} \vec{\theta(t)} \cdot \vec{u(t)} \ dt \leq \int_{a}^{b} |\vec{\theta'(t)}| dt[/tex](1)

    3. The attempt at a solution

    Solution:
    If I look at the dot-product of [tex]\frac{d}{dt}(\vec{\theta} \cdot \vec{u}}) = \vec{\theta(t)^{\prime}} \cdot \vec{u(t)}[/tex]

    Next by Fundamental theorem of Calculus leftside of (1) is true.

    The inequality however is a bit tricky for me. I seem to remember an inequality
    which states that the dot product of two vectors a and b can be written as
    [tex]\vec{a} \cdot \vec{b} \leq |\vec{a}| \cdot |\vec{b}|[/tex] which transfered to Our situation gives
    [tex]\vec{\theta'(t)} \cdot \vec{u(t)} \leq |\vec{\theta'(t)}| \cdot |\vec{u(t)}|[/tex] and since [tex]|\vec{u}| = 1[/tex], then this is equivalent to [tex]\theta'(t) \cdot \vec{u(t)} \leq |\theta'(t)| [/tex] and finally by the rules regarding our Riemann Integrals [tex]\int_{a}^{b} f \leq \int_{a}^{b} g[/tex], then the integrals [tex]\int_{a}^{b} \ldots dt \leq \int_{a}^{b} \ldots \ dt[/tex] is also true.

    There is a second part as well

    If I set [tex]\vec{u} = \frac{l-m}{|l-m|}[/tex]

    Then show that [tex]|\theta(b) - \theta(a)| \leq \int_{a}^{b} |\vec{\theta(t)^{\prime}| dt[/tex]

    If I insert [tex]\vec{u}[/tex] into lefthand side of the inequality I get [tex](l-m) \cdot \frac{l-m}{|l-m|} = |l-m|[/tex] since [tex](l-m) \cdot (l-m) = |l-m|^2[/tex]

    From this it should follow that the length of the curve from l to m is a straight line joining these points. But how exactly does it do that?
    Is it something to do with if I on my curve draw a straight line from l to m and use this line as a side in a triangle and by using pythagoras then claim that line previously mentioned always will be shorter than the arc length from l to m?

    Cheers Superboy
     
    Last edited: Feb 10, 2009
  2. jcsd
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