Show that $f(t)=(a_{1}t^{2}+b_{1}t+c_{1}, a_{2}t^{2}+b_{2}t+c_{2}, a_{3}t^{2}+b_{3}t+c_{3})$, $t \in \mathbb{R}$ is a plane curve.

Question

A curve $f:I \subseteq \mathbb{R} \rightarrow \mathbb{R^n}$ is said to be flat, if there is a plane $\beta=\{(x,y,z)\in\mathbb{R^3}:ax+by+cz=d\}$ such that $f(t) \in \beta$, for all $t \in I$. Show that $f(t)=(a_{1}t^{2}+b_{1}t+c_{1}, a_{2}t^{2}+b_{2}t+c_{2}, a_{3}t^{2}+b_{3}t+c_{3})$, $t \in \mathbb{R}$ is a plane curve.

Answer 1

You have $\vec{f}=\vec{c}+t\,\vec{b}+t^2 \, \vec{a}$ with constant vectors $\vec{a},\vec{b},\vec{c}.$ Take $\vec{k} = \vec{a} \times \vec{b}$ to be the cross product of $\vec{b},\vec{a},$ so that $\vec{k} \cdot \vec{a} = 0$ and $\vec{k} \cdot \vec{b} = 0.$ Then $$(\vec{f}-\vec{c}) \cdot \vec{k} = 0 $$ is your plane

Answer 2

Torsion for a general parametric curve can be computed using a determinant of the curve's first, second and third derivatives: $$\tau=\frac{\det(r',r'',r''')}{\|r'×r''\|^2}$$ Since all components of this curve are quadratic, the third derivative $r'''$ is the zero vector, so the determinant and hence the torsion are zero.