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Homework Help: Integration of vector valued functions

  1. Jun 13, 2010 #1
    1. The problem statement, all variables and given/known data

    Evaluate the definite integral.


    [[ti + t^2j]] dt from 3 to 0. Sorry I don't know how to add the integration sign here. the parenthesis is referring to the magnitude of the vector.


    2. Relevant equations



    3. The attempt at a solution

    I have no clue.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 13, 2010 #2

    LCKurtz

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    Do you mean this:

    [tex]\vec R(t) = ti + t^2j[/tex]

    and you want to integrate

    [tex]\int_3^0|\vec R(t)|\, dt\hbox{ ?}[/tex]

    Click on the equations to see how to write them. If that is what your question is, what exactly are you stuck on?
     
  4. Jun 13, 2010 #3
    [tex]
    \int_3^0|\vec R(t)|\, dt\hbox{ ?}
    [/tex]


    the integration is from zero to three. the other way around that I wrote it ( I have to get used to this stuff very soon)

    I'm stuck on the second thing you wrote there, and I'm stuck on the whole thing. Our instructor didn't really explain this stuff a lot, and the text book kind of assumes what we should be doing, so that's not much help also.

    So, my main problem is, do we integrate first, or we set up the magnitude, find that and then integrate it. And any help with that would be much appreciated.
     
    Last edited: Jun 13, 2010
  5. Jun 13, 2010 #4

    LCKurtz

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    Do you know how to find the magnitude (length) of a vector? Just write down its length and integrate it.
     
  6. Jun 14, 2010 #5
    would it be like this;


    It comes out of the integral like this

    ti^2/2 + t^3/3 + C

    and the length of this would be,

    (.5^2 + .33^2)^.5 =.599 ?
     
  7. Jun 14, 2010 #6

    Mark44

    Staff: Mentor

    That's not what I get. What did you get for the magnitude of the vector - |ti + t2j|? I think you skipped that part, and it looks like you just integrated t + t2, which is not a vector.
     
  8. Jun 14, 2010 #7
    |ti + t^2| isn't this simply the sum of the square root of the coefficients squared. so it would be (2)^.5 = 1.41
     
  9. Jun 14, 2010 #8

    HallsofIvy

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    What do you mean by "it" here? Grammatically, it should refer to |ti+ t^2j| but I think you mean the integral. |ti+ t^2j| is equal to [tex]\sqrt{t^2+ t^4}= t\sqrt{1+ t^2}[/itex]. That should be easy to integrate. Also, you said in your first post that you were integrating "from 3 to 0". Since 3< 0 and your integrand is positive, the integral would be negative. Or are you really integrating from 0 to 3?

    Finally, [itex]2^{.5}[/itex] is only approximately equal to 1.41. It would be better to leave it as [tex]\sqrt{2}[/tex]
     
  10. Jun 14, 2010 #9

    Mark44

    Staff: Mentor

    The coefficients are the coefficients of i and j, not the coefficients of t and t^2. t and t^2 are the coefficients.
     
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