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Equation of Tangent Line to Curve at Point

  1. Feb 14, 2013 #1
    1. The problem statement, all variables and given/known data
    Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.

    2. Relevant equations
    [tex]x = t \\
    y = e^{-4t} \\
    z = 5t - t^5 \\
    P = (0, 1, 0)[/tex]

    3. The attempt at a solution
    [tex] \vec{r}(t) = < t, e^{-4t}, 5t - t^5 > [/tex]
    At the point (0, 1, 0), t = 0.

    [tex] \vec{r '}(t) = < 1, -4e^{-4t}, 5 - 5t^4 > [/tex]

    [tex] x' = 1 \\
    y ' = -4e^{-4t} \\
    z ' = 5 - 5t^4[/tex]
    These are the x, y, and z equations for the rate of change of the curve in the x, y, and z directions respectively (or, the slopes/direction of the tangent lines at the point when t = some number).

    At t = 0, the x, y, and z slopes of the tangent lines are:
    [tex] x' = 1 \\
    y' = -4e^{-4} \\
    z' = 4[/tex]

    A point on the single tangent line of the curve (found by adding the vectors x', y', and z') is (0, 1, 0)—since the curve touches that point, too. The direction vector of the single tangent line is < 1, -4(e-4), 4 >. The parameter is t.

    So the vector equation of the tangent line of the curve at the point (0, 1, 0) is:
    [tex] (x(t), y(t), z(t)) = (0, 1, 0) + t<1, \frac{-4}{E^{4}}, 4> = < t, 1 -\frac{4t}{E^{4}}, 4t >[/tex]

    This, according to Webassign, is incorrect.

    Their answer the same, except their y-component is 1 -4t. They... dropped the exponential in the denominator? I can't figure out why that is the correct answer, and why my answer is incorrect.
     
  2. jcsd
  3. Feb 14, 2013 #2

    Dick

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    If I put t=0 into -4e^(-4t) I get -4*e^0=(-4).
     
  4. Feb 14, 2013 #3

    LCKurtz

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    Right. So ##\vec r'(0) = \langle 1, -4, 5\rangle##. That gives you the direction vector of that tangent line for ##t=0##.

    After you correct it, you still won't agree because their last component is wrong; it should be ##5t##.

    [Edit] Actually, it's your last component that is also wrong.
     
  5. Feb 14, 2013 #4
    Oh.

    I was plugging in the components of the vector, not the parameter (so plugging 1 into the y-equation). I don't know why. Thinking and typing one thing, doing a completely different thing, and was totally unaware of it.

    -______________________-

    Thanks, guys.
     
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