Vector functions. Find r(t). How and why?

Given r'(t)= (t^2)i + tj + (e^(2t))k and r(0) = -j+k, find r(t)

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  • Matt
    Lv 5
    9 years ago
    Favorite Answer

    Given r'(t) = t² i + t j + e^(2t) k

    I am first going to just use exp(x) = e^x since it makes my math easier to follow, secondly, I am going to use math vector notation since it is easier to follow than physics notation.

    So the above is the same as:

    r'(t) = <t², t, exp(2t)>

    r(t) = ∫ r'(t) dt

    r(t) = ∫ <t², t, exp(2t)> dt

    I think perhaps you've never integrated a vector function before. You just integrate each component.

    r(t) = <∫ t² dt,∫ t dt,∫ exp(2t) dt >

    r(t) = <⅓ t³, .5 t², .5 exp(2t) > + C

    Where C is a VECTOR constant.

    Impose our initial condition:

    r(0) = <0, 0, .5> + C = <0, -1, 1>

    C = <0, -1, .5>

    In math notation:

    r(t) = <⅓ t³, .5 t² - 1, .5(exp(2t) + 1)>

    I am switching back to physics notation.

    r(t) = (⅓ t³) i + (.5 t² - 1)j + .5(exp(2t) + 1) k

  • 9 years ago

    Here,

    r'(t)= (t^2)i + tj + (e^(2t))k and r(0) = -j+k, find r(t)

    r(t) = (t^3/3)+(t^2/2)j +(e^2t/2)k + C,

    for C, r(0) = -j+k, SO,

    r(0) = 0 +0 +k/2 +C = -j+k,

    C = -j +k/2,

    ,

    r(t) = (t^3)i/3 +(t^2)-1)j/2 + (1+e^2t)k/2 >======< ANSWER

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