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Projections quick help!

  1. Nov 29, 2012 #1
  2. jcsd
  3. Nov 29, 2012 #2


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    Generally, the derivative of a function is zero at a point where the function has a minimum.
    [Edit:Sorry, please ignore my comment. I was looking at the first problem.]
    Last edited: Nov 29, 2012
  4. Nov 29, 2012 #3


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    The solution is not differentiating nor "sets" x = 0 at that part of the solution.

    What the solution is trying to say is:

    "At whatever time it is when x happens to be zero, the y-component of velocity is c times greater than the x-component of velocity."

    Recall that in the original problem statement, the initial velocity was given to you as u(i + cj). So the slope ("rise over run", also called dy/dx) is c/1 = c.

    So at the very beginning, when the particle is launched, x = 0. (This is implied by the problem statement saying, "Relative to O, the position vector of a point on the path of P is (xi + yj) m.)

    So still in other words, the statement is saying, "At the time the particle was launched (i.e. x = 0), the velocity slope (dy/dt over dx/dt) is c."
    Last edited: Nov 29, 2012
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