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Calculus 3 vector analysis question (newton's 2nd problem)

  1. Feb 6, 2009 #1
    This problem was discussed in my calculus 3 class, and there is one step that I don't understand.

    1. The problem statement, all variables and given/known data
    An object of mass m travels along the parabola y = x[tex]^{2}[/tex] with a constant speed of 10 units/sec. What is the force on the object due to its acceleration at (0,0) (write the answer in terms of unit vectors i,j,k)? (this problem is from Thomas' calculus 11th edition, section 13.5 #20)


    2. Relevant equations
    f(x) = x[tex]^{2}[/tex]
    a = atT + anN
    at = d(|v|)/dt
    an = [tex]kappa[/tex]|v|[tex]^{2}[/tex]
    T = v/|v|
    [tex]kappa[/tex] = |f''(x)|/((1+f'(x)[tex]^{2}[/tex])[tex]^{3/2}[/tex])
    r = ti + t[tex]^{2}[/tex]j
    => v = i+2tj
    => |v| = (1+4t[tex]^{2}[/tex])[tex]^{1/2}[/tex]
    N = (d(T)/dt)/(|d(T)/dt|)


    3. The attempt at a solution**** this was the solution presented in class:
    at = 0 because d(10)/dt = 0
    an = 100[tex]kappa[/tex]
    [tex]kappa[/tex] = 2 at the point (0,0)
    *******T = 1/((1+4t[tex]^{2}[/tex])[tex]^{1/2}[/tex]) (i+2tj)
    T = i at the point (0,0)
    N = j at the point (0,0)

    => F = ma = m(200)j


    The step that I don't understand is marked with stars (Deriving T). Since the speed is given as 10 units/sec, shouldn't that be used as |v| rather than (1+4t[tex]^{2}[/tex])[tex]^{1/2}[/tex]? And if |v| = (1+4t[tex]^{2}[/tex])[tex]^{1/2}[/tex], doesn't this contradict the problem statement, which says that speed is constant? This makes little sense to me because we used different values for |v| throughout the problem
     
    Last edited: Feb 6, 2009
  2. jcsd
  3. Feb 6, 2009 #2

    D H

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    Staff Emeritus
    Science Advisor

    The source of your problem is here:
    This parameterization conflicts with the given fact that the speed is constant. Since you are given that y=x2, you do know that
    [tex]\boldsymbol r = x \hat{\boldsymbol i} + x^2\hat{\boldsymbol j}[/tex]
    See where this leads you.
     
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