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Vector Calculus

  1. Dec 4, 2005 #1
    Hello everyone,
    I found a random question regarding finding the parametric equations for a tangent line to a space curve and I'm striving to solve it, but no results. I consulted the book but there isn't anything similar.

    Find the parametric equations for the tangent line to the space curve:
    x = ln(t), y = 2*Sqrt(t), z = t^2 at the point (0,2,1)

    I would appreciate any suggesstions or hints how to solve it.
  2. jcsd
  3. Dec 4, 2005 #2


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    The time derivative of [itex]\vec x[/itex] is tangent to the curve. If [itex]\vec v = d\vec x/dt[/itex] then the line you are looking for is given by [itex]\vec x_{tangent} = \vec x_0 + \vec v(t = t_0) (t - t_0)[/itex] where [itex]t_0[/itex] is the time corresponding to the point of interest.
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