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Parametric equation for a tangent line (multi vars)

  1. Feb 7, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the parametric equation for the line tangent to the curve:

    x=t^3-1, y=t^4+1, z=t

    at the point (26, 82, 3).

    Use the variable t for your parameter.


    2. Relevant equations



    3. The attempt at a solution

    dx/dt=3t^2, dy/dt=4t^3, dz/dt=1

    I got then that

    x=26+(3*26^2)t
    y=82+(4*82^3)t
    z=3+t

    and only the z component is correct. Where am I going wrong. Any help is really appreciated!
     
  2. jcsd
  3. Feb 7, 2010 #2

    Dick

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    Look, at your given point t=3. When you are working out the d/dt parts why are you substituting x and y for t?
     
  4. Feb 7, 2010 #3
    You should not plug back your given point coordinates into "t" in your derived equations. Your value of t is constant. And technically it's given in your problem.
     
  5. Feb 7, 2010 #4
    I substituted x and y into my derivatives to find the slope of the that component of the line at that point. I then multiply it by t, and add x and y for intercepts or to start from the origin.
     
  6. Feb 7, 2010 #5

    Dick

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    t is t. x is x. y is y. z happens to be t. You got lucky on the z component because z=t. Just put t in where your equations say 't'. Don't substitute x for t just because it happens to be the 'x' component.
     
  7. Feb 7, 2010 #6
    I don't know why I'm so hung up on this question.

    For the x component for instance,

    it the curve is t^3-1, the slope of the tangent is 3t^2. If we're talking about the slope of the tangent at 26 in the curve, why wouldn't we plug in 26 for t in dx/dt?

    What is the next step I should take after finding the derivative?
     
  8. Feb 7, 2010 #7
    Oh jesus. t=3 right. That can't change.

    Wow I feel stupid. Thanks all for the help.
     
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