Answer
$x=3+t,\ \ \ y=2t$, $\ \ $ $z =2+4t$
Work Step by Step
Differentiate $(\displaystyle \frac{d}{dt})$ each component function. to find the tangent vector at t.
$\mathrm{r}^{\prime}(t)=\langle 1/\sqrt{t}$, $\ \ 3t^{2}-1, \ \ 3t^{2}+1\rangle$
Fot the point $(3,0,2),$
find t from the parametric equations: $\quad\left\{\begin{array}{lll}
3=1+2\sqrt{t} & \Rightarrow 2\sqrt{t}=2 & \Rightarrow t=\pm 1\\
0=t^{3}-t & \Rightarrow t\neq-1 & \Rightarrow t=1\\
2=t^{3}+t & \Rightarrow t\neq-1 & \Rightarrow t=1
\end{array}\right.$
so $\mathrm{r}(1)=(3,0,2).$
A tangent vector at $(3,0,2)$ is
$\mathrm{r}^{\prime}(1)=\langle 1,2,4\rangle$
The tangent line contains $(3, 0,2)$ and is parallel to the vector $\langle 1,2,4\rangle$.
Its parametric equations are (see sec 12-5):
$x=3+t,\ \ \ y=2t$, $\ \ $ $z =2+4t$
