(Solved): Determine whether the space curve given by r(t) = (t, t3, t + 6) intersects the xy-plane. Consider ...

Determine whether the space curve given by $\mathbf{r}(t)=⟨t,t^3,t^2+6⟩$ intersects the $xy$-plane. Consider the following method for identifying the point of intersection. Solve the system of equations $\begin{array}{c}\{\begin{array}{l}x(t)=0\\ y(t)=0\end{array}\\ \{\begin{array}{ll}t& =0\\ t^3& =0\end{array}\end{array}$ Therefore, $t=0$. So solve for $\mathbf{r}(0)$ to find the point of intersection. $\begin{array}{l}\mathbf{r}(0)=⟨(0),(0{)}^3,(0{)}^2+6⟩\\ \mathbf{r}(0)=(0,0,6)\end{array}$ Is the point of intersection $(0,0,6)$ ? If not, explain the misconception in the demonstrated method. Is the point of intersection $(0,0,6)$ ? If not, explain the misconception in the demonstrated method. A curve intersects the $xy$-plane when the $y$ and $z$ coordinate of $\mathbf{r}(t)$ is 0 . A curve intersects the $xy$-plane when the $x$ and $z$ coordinate of $\mathrm{r}(t)$ is 0 . The point of intersection is $(0,0,6)$. A curve intersects the $xy$-plane when the $z$ coordinate of $\mathbf{r}(t)$ is 0 . State the point of intersection, if it exists. (Use symbolic notation and fractions where needed. Give your answer as the coordinates of a point in the form $(\ast ,\cdots$,$).Enter$ NO SOLUTION if the curve does not intersect the $x$-axis.) point coordinates: