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Curve of intersection of 2 functions

  1. Jan 30, 2012 #1
    1. The problem statement, all variables and given/known data
    A particle moves along the curve of intersection of shapes y = -x2 and z = x2 in the direction in which x increases. At the instant when the particle is at the point P(1,-1,1), its speed is 9cm/s and that speed is increasing at a rate of 3cm/s2. Find the velocity and acceleration of the particle at that instant


    2. Relevant equations
    ||[itex]\vec{v}[/itex]|| = 9

    Derivative of ||[itex]\vec{v}[/itex]|| = 3

    [itex]\vec{v}[/itex] = [itex]\vec{R'}[/itex] where [itex]\vec{R}[/itex] is the curve.

    3. The attempt at a solution
    I tried Googling it but I could not find a definite answer to my question. What I want to know is how do I find the [itex]\vec{R}[/itex] (curve defined by intersections). I have never encountered a similar problem before and I don't really know how to approach it. If you could hint me the right direction, it would be greatly appreciated. :smile:
     
    Last edited: Jan 30, 2012
  2. jcsd
  3. Jan 30, 2012 #2

    Char. Limit

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    Gold Member

    A parametric representation such that x^2 = -y and x^2 = z should work. I'd recommend x=t, which gives you a simple parametric representation. This gives you the vector function. I'm sure you can go from there.
     
  4. Jan 30, 2012 #3
    I tried that,

    then R(t) = t i - t2j + t2k

    R'(t) = i - 2t j + 2t k

    and at given point P(1,-1,1) from equation of R(t) I get t = 1

    So R'(1) = i - 2j + 2k , which is also velocity

    But then speed, ||R'(1)|| = sqrt{1 + 4 + 4} = 3.. And according to the problem description, it should be 9.
    It seems like R'(1) is actually giving the value of the change in speed (as defined in description) instead of the actual speed.

    Am I missing something?

    Thanks!
     
  5. Jan 30, 2012 #4
    anyone?
     
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