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Trajectory and parametric equation

  1. Jan 23, 2009 #1
    1. The problem statement, all variables and given/known data
    a particle is on a trajectory defined by the parametric equations x(t)= 2t^2,y(t)= t^2-4t,
    z(t)= 3t-5, where t is the time.Find the components of its velocity and acceleration at time =1, in the direcion i-3j+2k.


    2. Relevant equations
    what i thought is r(t)=x(t) i +y(t) j +z(t) k, then take first derivative and get velocity vector. take the second derivative and get the acceleration vector.
    However, the question provides a direction vector and i have no idea how to deal with the direction vector.

    3. The attempt at a solution
    r(t)'=4t i + (2t-4) j+(3) k.
    r(t)''= (4) i + (2)j +(0) k
     
  2. jcsd
  3. Jan 23, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    Find the direction cosines between velocity and acceleration vectors and given direction vector.
     
  4. Jan 23, 2009 #3
    You are on the correct path to solving the problem. Once you find r'(1) and r''(1) you will project the components of these two vectors on to the vector given by i-3j+2k. The dot product gives the component of one vector onto another. Since this component is multiplied by the magnitude of other vector, you must divide by the magnitude of the vector i.e.

    Let A = i-3j+2k

    then

    [tex]\mbox{component}=\frac{r'(1)\bullet\mbox{A}}{\mid\mbox{A}\mid}[/tex]

    Use the same approach for r''(1).
     
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