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Indefinite integral of vector function

  1. Feb 2, 2009 #1
    1. The problem statement, all variables and given/known data

    If [tex]{\vec{V}(t)[/tex] is a vector function of [tex] t [/tex], find the indefinite integral:

    [tex] \int (\vec{V}\times \frac{d^2\vec{V}}{dt^2}) \,dt [/tex]

    2. Relevant equations



    3. The attempt at a solution

    I have solved it by decomposing and integrating each terms of vector [tex]\vec{V}\times \frac{d^2t}{dt^2}[/tex], so I have:

    [tex]
    \vec{V}\times \frac{d^2\vec{V}}{dt^2}= \hat i (V_y * \frac{d^2V_z}{dt^2} - V_z * \frac{d^2V_y}{dt^2}) - \hat j (V_x * \frac{d^2V_z}{dt^2} - V_z * \frac{d^2V_x}{dt^2}) + \hat k (V_x * \frac{d^2V_y}{dt^2} - V_y * \frac{d^2V_x}{dt^2})
    [/tex]

    and the integral will be:

    [tex]

    \int (\vec{V}\times \frac{d^2\vec{V}}{dt^2}) \,dt = \hat i (\int (V_y * \frac{d^2V_z}{dt^2})\,dt - \int (V_z * \frac{d^2V_y}{dt^2})\,dt) - \hat j (\int (V_x * \frac{d^2V_z}{dt^2})\,dt - \int (V_z * \frac{d^2V_x}{dt^2}))\,dt + \hat k (\int (V_x * \frac{d^2V_y}{dt^2})\,dt - \int (V_y * \frac{d^2V_x}{dt^2}),dt)

    [/tex]


    where [tex]\hat i[/tex],[tex]\hat j[/tex] and [tex]\hat k[/tex] are the three unit vectors with respect to frame of reference [tex](x,y,z)[/tex]


    My questions is: that's the right way or there exist some other way to resolve it in a more faster manner?
    Thanks to all who will answer me ;)
     
    Last edited: Feb 2, 2009
  2. jcsd
  3. Feb 2, 2009 #2

    Mark44

    Staff: Mentor

    Why do you have the cross product with
    [tex]\frac{d^2t}{dt^2}?[/tex]

    Surely this is a mistake.
     
  4. Feb 2, 2009 #3
    Yeah, sorry. Really the integral was

    [tex]
    \int (\vec{V}\times \frac{d^2\vec{V}}{dt^2}) \,dt
    [/tex]

    In fact it includes a vector as a result of the cross product whose components are function of t. Does anybody have any idea?
     
  5. Feb 2, 2009 #4

    Mark44

    Staff: Mentor

    OK, that makes more sense. For your question, it might be that there is a simpler manner to approach this, but I don't know it. What you're doing is what I would do.
     
  6. Feb 2, 2009 #5

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    erm :redface:

    Hint: what is the derivative of V x dV/dt? :wink:
     
  7. Feb 2, 2009 #6

    [tex]

    \frac{d(\vec{V}\times \frac{d\vec{V}}{dt})}{dt} = \vec{V}\times \frac{d^2\vec{V}}{dt^2} + \frac{d\vec{V}}{dt} \times \frac{d\vec{V}}{dt} = \vec{V}\times \frac{d^2\vec{V}}{dt^2}

    [/tex]

    so result of the integral is [tex]\vec{V}\times \frac{d\vec{V}}{dt}[/tex], is it exact? :)
     
  8. Feb 2, 2009 #7

    tiny-tim

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    well, plus a constant vector!

    why? … is something worrying you? :wink:
     
  9. Feb 2, 2009 #8
    Thanks for the help.... ;)
     
  10. Mar 8, 2011 #9
    wizard85 is there anyway you can expand on how you took the derivative of the cross product of V and dV/dt ? I am doing the same problem and can not bridge that gap....
     
  11. Mar 8, 2011 #10

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi selms05! :smile:

    the product rule applies to a cross product (or a dot product) in exactly the same way as to an ordinary product :wink:
     
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