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Posted by on Wednesday, November 10, 2010 at 11:24pm.

Find the parametric equations for the tangent line to the curve with the given parametric equations at specified point.
x= e^t
y=te^t
z=te^(t^2)
(1,0,0)

  • Calc 3 - , Thursday, November 11, 2010 at 9:16am

    Try a circle with parameter t=θ
    x=cos(t)
    y=sin(t)

    dx/dt=-sin(t)
    dy/dt=cos(t)

    at t=0, (dx/dt,dy/dt)=(0,1)
    at t=π/4, (dx/dt,dy/dt)=(-√2/2,√2/2)
    ...etc

    Here:
    x= e^t
    y=te^t
    z=te^(t^2)
    (1,0,0)

    Solve for t at the point (1,0,0) gives t=0 (the only way to get y=z=0).

    dx/dt=e^t
    dy/dt=(1+t)e^t
    dz/dt=t(2+t)e^t

    Substitute t=0 and find the equation of the line passing through (1,0,0) with the given slopes.

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