Solving Parametric Equations for Tangent Line to Space Curve

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To find the parametric equations for the tangent line to the space curve defined by x = ln(t), y = 2*Sqrt(t), and z = t^2 at the point (0, 2, 1), the time derivative of the position vector is essential. The tangent vector, denoted as \vec v = d\vec x/dt, provides the direction of the tangent line. The equation for the tangent line can be expressed as \vec x_{tangent} = \vec x_0 + \vec v(t = t_0) (t - t_0), where t_0 corresponds to the parameter value at the point of interest. Evaluating the derivatives at the appropriate t-value will yield the desired parametric equations. This approach effectively determines the tangent line to the given space curve.
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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.
Thanks
 
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The time derivative of \vec x is tangent to the curve. If \vec v = d\vec x/dt then the line you are looking for is given by \vec x_{tangent} = \vec x_0 + \vec v(t = t_0) (t - t_0) where t_0 is the time corresponding to the point of interest.
 
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