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Math Help - Line Integral

  1. #1
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    Line Integral

    Hello! I need some help with this problem, please!

    Let f be a field and \gamma a curve such that for all t\in{\mathbb{R}} happens that f(\gamma (t))=m\gamma ''(t) (where m is a constant).
    Show that:

    \displaystyle\int_{\gamma (a)}^{\gamma (b)}fd\gamma = \displaystyle\frac{1}{2}m \left\|{\gamma ''(b)}\right\|-\displaystyle\frac{1}{2}m \left\|{\gamma ''(a)}\right\|

    Thank you!!
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Veronika View Post
    Hello! I need some help with this problem, please!

    Let f be a field and \gamma a curve such that for all t\in{\mathbb{R}} happens that f(\gamma (t))=m\gamma ''(t) (where m is a constant).
    Show that:

    \displaystyle\int_{\gamma (a)}^{\gamma (b)}fd\gamma = \displaystyle\frac{1}{2}m \left\|{\gamma ''(b)}\right\|-\displaystyle\frac{1}{2}m \left\|{\gamma ''(a)}\right\|

    Thank you!!
    I wonder whether this formula is at all valid, since \int_{\gamma(a)}^{\gamma(b)}f d\gamma=\int_a^bf(\gamma(t))\cdot \gamma'(t)\,dt=m\int_a^b\gamma''(t)\cdot \gamma'(t)\,dt and \frac{d}{dt}\big[\parallel \gamma'(t)\parallel^2\big]=\frac{d}{dt}\big[\gamma'(t)\cdot \gamma'(t)\big]=2\gamma''(t)\cdot\gamma'(t).
    So I would have expected the formula to look more like \int_{\gamma(a)}^{\gamma(b)}f d\gamma\overset{?}{=}\tfrac{1}{2}m\parallel \gamma'(b)\parallel^2-\tfrac{1}{2}m\parallel \gamma'(a)\parallel^2
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  3. #3
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    Yes, I think is a typing error from my professor.

    So, to prove the last equation should I just derive and see if the are the same? I mean, using the 1 Fundamental Theorem of Calculus.

    Thanks!
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by Veronika View Post
    Yes, I think is a typing error from my professor.

    So, to prove the last equation should I just derive and see if the are the same? I mean, using the 1 Fundamental Theorem of Calculus.

    Thanks!
    Um, well, yes, I think that's the way to go. You can really just determine the derivative of F(t) := \frac{m}{2}\parallel \gamma'(t)\parallel^2 and show, that it is equal to f(\gamma(t))\cdot \gamma'(t), given that f(\gamma(t))=m\gamma''(t).
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  5. #5
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    I was thinking about this:

    \displaystyle\int_{\gamma (a)}^{\gamma (b)}f(\gamma (t))d\gamma =m\displaystyle\int_{a}^{b}\gamma (t)''\gamma (t)'dt

    We change u(t)=(\gamma (t)')^2 so that du=2\gamma (t)'\gamma (t)'' and

    \displaystyle\frac{m}{2}\displaystyle\int_{a}^{b}d  u=\displaystyle\frac{m}{2}(u(b)-u(a))=\displaystyle\frac{m}{2}(\gamma (a)')^2-\displaystyle\frac{m}{2}(\gamma (b)')^2
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