Step 1
Given:
\(x = 3\ \ln (t)\) ......(1)
\(y = 4t^{1/2}\) ......(2)
\(z = t^{3}\) ......(3)
\((x_{0},\ y_{0},\ z_{0}) = (0,\ 4,\ 1)\)
For the specified points,
Substitute 1 for z in equation 3.
\(z = t^{3}\)

\(1 = t^{3}\)

\(t = 1\) The function r(t) of these equations is, \)r(t) =\ <\ x,\ y,\ z\ >\) Substitute the values in the above equation. \(r(t)=\ <\ 3\ \ln(t), 4t^{\frac{1}{2}},\ t^{3}\ >\) Step 2 Derivative of the vector r(t) is, \(r'(t)=\ <\ \frac{3}{t},\ \frac{2}{\sqrt{t}},\ 3t^{2}\ >\) Substitute 1 for t in the above vector. \(r'(1)=\ <\ 3,\ 2,\ 3\ >\) Step 3 The parametric equations for the tangent line are, \(X = x_{0}\ +\ (r'(1))_{x} t\)

\(Y = y_{0}\ +\ (r'(1))_{y} t\)

\(Z = z_{0}\ +\ (r'(1))_{z} t\) Substitute the values in the above equation. \(X=3t\)

\(Y=4\ +\ 2t\)

\(Z=1\ +\ 3t\) Thus, the parametric equations for the tangent line are \(X = 3t,\ Y = 4\ +\ 2t,\ and\ z = 1\ +\ 3t.\)

\(1 = t^{3}\)

\(t = 1\) The function r(t) of these equations is, \)r(t) =\ <\ x,\ y,\ z\ >\) Substitute the values in the above equation. \(r(t)=\ <\ 3\ \ln(t), 4t^{\frac{1}{2}},\ t^{3}\ >\) Step 2 Derivative of the vector r(t) is, \(r'(t)=\ <\ \frac{3}{t},\ \frac{2}{\sqrt{t}},\ 3t^{2}\ >\) Substitute 1 for t in the above vector. \(r'(1)=\ <\ 3,\ 2,\ 3\ >\) Step 3 The parametric equations for the tangent line are, \(X = x_{0}\ +\ (r'(1))_{x} t\)

\(Y = y_{0}\ +\ (r'(1))_{y} t\)

\(Z = z_{0}\ +\ (r'(1))_{z} t\) Substitute the values in the above equation. \(X=3t\)

\(Y=4\ +\ 2t\)

\(Z=1\ +\ 3t\) Thus, the parametric equations for the tangent line are \(X = 3t,\ Y = 4\ +\ 2t,\ and\ z = 1\ +\ 3t.\)