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Math Help - A relatively easy Vector problem from calc 3?

  1. #1
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    A relatively easy Vector problem from calc 3?

    The velocity of a particle is given by v(t)
    = (t^2) i+ (t^3 +1) j

    and the particle is at
    the point (2,1) when t=0.
    a) Where is the particle when t=2?

    This was on a sample final and gave no answers...
    thanks!
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  2. #2
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    Quote Originally Posted by iamtrojan3 View Post
    The velocity of a particle is given by v(t)
    = (t^2) i+ (t^3 +1) j

    and the particle is at
    the point (2,1) when t=0.
    a) Where is the particle when t=2?

    This was on a sample final and gave no answers...
    thanks!
    Remember that velocity is the derivative of position

    \frac{dr}{dt} = (t^2) \vec i + (t^3+1)\vec j

    So we take the integral

    \int dr = \int (t^2)dt \vec i + \int (t^3+1)dt \vec j

    \vec r(t) = (\frac{1}{3}t^3+c) \vec i + (\frac{1}{4}t^4+t+d)\vec j

    \vec r(0) = 2\vec i+\vec j=(\frac{1}{3}(0)^3+c) \vec i + (\frac{1}{4}(0)^4+0+d)\vec j

    so we get c=2 and d=1

    \vec r(t) = (\frac{1}{3}t^3+2) \vec i + (\frac{1}{4}t^4+t+1)\vec j

    \vec r(2)=...
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  3. #3
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    So this was my reasoning, not quite the same but it comes out with the same answer as you

    To get the position at some later time we must integrate the velocity over time since distance is velocity times time. So at some later time the position will be given by:
    P = A + Integral{V*dt}| from 0 to 2

    So at time t = 2 we get:
    Integral {(t^2, t^3 + 1)}dt = [(t^3/3), (t^4/4 + t)]

    P = (2,1) + (2^3/3, 2^4/4 + 2)
    P = (2,1) + (8/3, 6) = (14/3 , 7)

    So the particle is at (14/3 , 7) when t=2

    can you check if thats correct?
    sorry i don't really no the coding for this...
    And, thanks alot, really appreciate your help.
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  4. #4
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    Quote Originally Posted by iamtrojan3 View Post
    To get the position at some later time we must integrate the velocity over time since distance is velocity times time. So at some later time the position will be given by:
    P = A + Integral{V*dt}| from 0 to 2

    So at time t = 2 we get:
    Integral {(t^2, t^3 + 1)}dt = [(t^3/3), (t^4/4 + t)]

    P = (2,1) + (2^3/3, 2^4/4 + 2)
    P = (2,1) + (8/3, 6) = (14/3 , 7)

    So the particle is at (14/3 , 7) when t=2

    can you check if thats correct?

    thanks and sorry i don't really no the coding for this...
    Yes you answer is correct and here is a link for the LaTex code

    Helpisplaying a formula - Wikipedia, the free encyclopedia

    Is is really nice and not too hard to learn.

    Just hit the sigma button to open the math tags.
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