Parametric Parabola in 3D Space

The vertex of the parabola will occur when the curvature is maximum. The curvature is given by $\kappa = \dfrac{\lVert \vec{r''}(t) \times \vec{r'}(t) \rVert}{\lVert \vec{r'}(t) \rVert^3}$ .

In this case we have: $\vec{r'}(t)=(1+2t, 2+2t, 3+2t)$ and $\vec{r''}(t)=(2,2,2)$ . So the curvature is $\kappa = \dfrac{2\sqrt{6}}{(12t^2+24t+14)^{3/2}}$ . We must minimize $12t^2+24t+14$ which occurs at $t=-1$ .

Finally, the vertex is $V=\vec{r}(-1)=\boxed{(1,3,7)}$ .

For the parabola given by $\vec{r}(t) = \vec{v_0} + \vec{v_1} t + \vec{v_2} t^2$ , it can be shown that the vertex occurs when $\dot{r} \cdot \ddot{r} = 0$
We have $\dot{r}(t) = \vec{v_1} + 2 \vec{v_2} t$ and $\ddot{r}(t) = 2 \vec{v_2}$
Therefore, the given condition is equivalent to
$\vec{v_1} \cdot \vec{v_2} + 2 \vec{v_2} \cdot \vec{v_2} t = 0$ .
Hence, $t^* = - \dfrac{\vec{v_1} \cdot \vec{v_2}}{2 \vec{v_2} \cdot \vec{v_2}} = - \dfrac{ (1, 2, 3) \cdot (1,1,1)}{2 (1,1,1) \cdot (1,1,1)} = -1$ . Therefore, the vertex is given by, $\vec{r}^* = (1, 4, 9) + (1, 2, 3) (-1) + (1, 1, 1)(-1)^2 = (1, 3, 7)$