Question

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

Short answer

The parametric equation of the given line is

\[\begin{array}{*{20}{l}}{x = 3 + t}\\{y = 2t}\\{z = 2 + 4t}\end{array}\]

Step-by-step solution

Step 1: Solution of the tangent vector of x, y and z values

Here, \[(x,y,z) = (3,0,2)\]

\[\begin{array}{l}1 + 2\sqrt t = \frac{1}{{2\sqrt t }}\\{t^3} - t = 3{t^2}\\{t^3} + t = 3{t^2}\end{array}\]

The tangent vector is \[r'(t) = \left( {\frac{1}{{2\sqrt t }},3{t^2},3{t^2}} \right)\]

Step 2: Solution of the direction vector

Putinto tangent vector line at point

\[\begin{array}{l}r'(0) = (\frac{1}{{2\sqrt 0 }},3(0),3(0))\\r'(0) = (0,0,0)\end{array}\]

Step 3: Solution of the parametric line equation 

The direction vector into thepointinto theof the parametric line equations

\[\begin{array}{l}x = {x_0} + at\\ = 3 + t\\x = 3 + t\\y = {y_0} + bt\\y = 0 + 2t\\y = 2t\\z = {z_0} + ct\\z = 2 + 4t\end{array}\]

The parametric equation of the given line is

\[\begin{array}{*{20}{l}}{x = 3 + t}\\{y = 2t}\\{z = 2 + 4t}\end{array}\]