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Parametric curve /derivative problems

  1. Apr 28, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider the parametric curve given by the equation

    x(t) = ti + t^(1/3)j

    1. Calculate x'(t). Does the vector exist at [itex]t=0[/itex]?

    2. Find a new parametrisation of the curve for which the tangent vector is well
    defined at all points. What is the value of the vector at the origin?

    3. The attempt at a solution

    For q.1, is it as simple as finding dx/dt? which would be [itex]1 + \frac{1}{3\sqrt[3]{t^{2}}}[/itex]?

    As for q.2, I'm completely stumped.

    Any help would be greatly appreciated.
    Last edited: Apr 28, 2013
  2. jcsd
  3. Apr 28, 2013 #2


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    Science Advisor

    Yes, it really is as simple as finding dx/dt. No, what you have is NOT dx/dt. x(t) is a vector and dx/dt is a vector: dx/dt= i+ [itex]\frac{1}{3}t^{-2/3}[/itex]j.

    As for (2), [itex]e^s[/itex] exists for all s so try [itex]e^s[/itex]i+ [itex]e^{s/3}[/itex]j.
    Do you see why that is also a parameterization for the curve?
  4. Apr 28, 2013 #3
    Ah yes, my mistake. So this should mean that dx/dt does not exist at t=0, since [itex]\frac{1}{3}t^{-2/3}[/itex]j becomes 1/0, giving i + ∞j.

    Yeah. dx/ds in this case would be defined at s=0, which is what we want. Is this sort of like 'moving' the origin?
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