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Math Help - Parametric tangent line

  1. #1
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    Parametric tangent line

    Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.

    x=ln(t), y=2(sqrt(t)), z=t^2; (0,2,1)

    My question is, do I simply solve for t (t=1) and then take a derivative of each? I'm confused on what it is asking for.
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  2. #2
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    Quote Originally Posted by Rhode963 View Post
    Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.

    x=ln(t), y=2(sqrt(t)), z=t^2; (0,2,1)

    My question is, do I simply solve for t (t=1) and then take a derivative of each? I'm confused on what it is asking for.
    Well, it is asking for parametric equations of the tangent line!

    You should know that if you are given a curve as (x(t), y(t), z(t)) then the derivative vector <x'(t), y'(t), z'(t)> is tangent to the curve. You should also know that parametric equations for a line through (x_0, y_0, z_0) with "direction vector" <A, B, C> are x= At+ x_0, y= Bt+ y_0, and z= Ct+ z_0.

    You are given the point and the "direction vector" of a tangent line is a tangent vector at that point. Differentiate x, y, and z with respect to t, evaluate at t= 1, and use that as the "direction vector" of the line.
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  3. #3
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    So x'=1/t, y'=t^(-1/2), z'=2t and evaluated at t=1, x'=1, y'=1, z'=2, meaning the parametric tangent line is... x=1+ln(t), y=1+2(sqrt(t)), z=2+t^2 ?
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