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Really quick question about minimum of function

  1. Feb 5, 2008 #1
    !!!Really quick qestion about minimum of function

    I ADDED ANOTHER QUESTION IN POST #3 ! Thanks for looking :)


    I am told the a jet's follows the curve [itex]y=20(10^{-6})x^2+5000[/itex] where y AND x are in feet. Basically all I need help with is for the first part of the problem, I need to find out where the jet is at its LOWEST point on the curve.

    I am pretty sure this is at x=0. But how can I prove it? Usually I would just plug in zero for x. But this is not a function of time, it is of position, so I am not quite sure how to show that if I plug in a neg number I would not get a number less than 5000. I mean I know since it's an even function thath even numbers<0 for x will yield numbers>5000.

    I was hoping to show this with Calculus rather than analysis, though maybe the latter is more reasonable in an engineering course?

    Any thoughts are welcome, I can move on withthe problem regardless (since I could use the analysis) but I just want some input.
     
    Last edited: Feb 5, 2008
  2. jcsd
  3. Feb 5, 2008 #2

    Dick

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    In general find where dy/dx=0 and then check if that's a minimum. It doesn't matter that the equation is not a function of time. You are just finding the minimum point on a curve. In this case though you hardly even need to do that. You know x^2>0 unless x=0. So the lowest point is clearly at x=0, isn't it?
     
  4. Feb 5, 2008 #3
    Like I said, I know that it is at x=0, but I'm an idiotm so I would to prove that it is (to myself). So I see your point. 1st Derivative test does work.

    ANOTHER QUESTION:

    How about this now: I need to find the normal acceleration [itex]a_n[/itex] of a point traveling along said curve AT X=0.

    Now I know that [tex]a_n=\frac{v^2}{\rho}[/tex]

    My teacher gave the hint [itex]ds=rd\theta[/itex]. How can I find velocity from this? I would nee to write theta as a function of x?

    Any ideas?
     
  5. Feb 5, 2008 #4
    ....................
     
  6. Feb 5, 2008 #5
    :confused:
     
  7. Feb 6, 2008 #6

    Dick

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    ??? Gravity has absolutely no effect on the acceleration of the plane. It's not in free fall. It's following a fixed trajectory. You can't find the normal acceleration from the information you are given. To use v^2/R you would need to find the radius of curvature, R, of the curve at x=0 and then find v. But you have no way of finding v without knowing something about the position of the plane as a function of t. For all you know, it could just be sitting at x=0 without accelerating at all.
     
    Last edited: Feb 6, 2008
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