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Math Help - I need help finding a position vector

  1. #1
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    I need help finding a position vector

    A particle starts at the initial position r(0)=<0,0,1> with the initial velocity v(0)=<1,0,0> and the accelerlation a(t<sin(t),e^(t),12t>. Find its position function r(t).
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  2. #2
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    Re: I need help finding a position vector

    Hello, allstar2!

    \text{A particle starts at the initial position: }\:r(0)\:=\:\langle 0,0,1\rangle
    \text{with initial velocity: }\:v(0)\:=\:\langle1,0,0\rangle
    \text{and accelerlation: }\:a(t) \:=\:\langle \sin t,\,e^t,\,12t\rangle.

    \text{Find its position function, }r(t).

    The velocity function is the integral of the acceleration function.

    v(t) \;=\;\int\langle \sin t,\;e^t,\;12t\rangle\,dt \;=\;\langle \text{-}\cos t\!+\!c_1,\; e^t\!+\!c_2,\; 6t^2\!+\!c_3\rangle


    We are told that: . v(0) \:=\:\langle 1,\,0,\,0\rangle

    So we have: . \langle \text{-}\cos0\!+\!c_1,\:e^0\!+\!c_2,\:0\!+\!c_3\rangle \;=\;\langle 1,\,0,\,0\rangle

    . . . . . . . . . . . . . . \langle \text{-}1\!+\!c_1,\;1\!+\!c_2,\;c_3\rangle \;=\;\langle 1,\,0,\,0\rangle

    Then we have: . \begin{Bmatrix}\text{-}1+c_1 \:=\:1 & \Rightarrow & c_1 \:=\:2 \\ 1+c_2 \:=\:0 & \Rightarrow & c_2 \:=\:\text{-}1 \\ c_3 \:=\:0\end{Bmatrix}

    Hence: . v(t) \;=\;\langle \text{-}\cos t\!+\!2,\;e^t\!-\!1,\; 6t^2\rangle



    The position function is the integral of the velocity function.

    r(t) \;=\;\int \langle \text{-}\cos t\!+\!2,\;e^t\!-\!1,\;6t^2\rangle\,dt \;=\;\langle -\sin t\!+\!2t\!+\!k_1,\;e^t\!-\!t\!+\!k_2,\;2t^3\!+\!k_3\rangle


    We are told: . r(0) \:=\:\langle 0,0,1\rangle

    So we have: . \langle \text{-}\sin0\!+\!0\!+\!k_1,\;e^0\!-\!0\!+\!k_2,\;0\!+\!k_3\rangle \;=\;\langle 0,0,1\rangle

    . . . . . . . . . . . . . . \langle k_1,\;1\!+\!k_2,\;k_3\rangle \;=\;\langle 0,0,1\rangle

    Then we have: . \begin{Bmatrix}k_1 \:=\:0 \\ 1+k_2 \:=\:0 & \Rightarrow & k_2 \:=\:\text{-}1 \\ k_3\:=\:1 \end{Bmatrix}


    Therefore: . r(t) \;=\;\langle \text{-}\sin t\!+\!2t,\;e^t\!-\!t\!-\!1,\;2t^3\!+\!1 \rangle

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