The question is as follows:
Let $\vec{a}, \vec{b} \in \mathbb{R}^3$ be given,
Consider the curve $\vec{r}(t) = \vec{OP_{0}} + t\vec{a} + t^2\vec{b}$, $t \in \mathbb{R}$
Find the curvature of the curve at $\vec{r}(0)$.
I have just started learning about arc-lengths and paremetrizations, and I have no clue how to approach this question. I've tried splitting each vector in the function to its components (e.g. $\vec{a} \rightarrow (a_1, a_2, a_3))$, but I can't progress any further. I'm wondering how to incorporate the curvature formula for this question;
$$\frac{\vec{r}'(t) \times \vec{r}''(t)}{\left|\vec{r(t)}\right|^3}$$
Any help would be greatly appreciated.
The formula is $$\kappa=\frac{||\dot r \times \ddot r||}{(\dot r \bullet \dot r)^{3/2}}$$ Your notation is really unfortunate, since the vector $a$ usually denotes the acceleration vector, i.e. $\ddot r,$ but I'll stick with that bad notation. $$\dot r=a+2tb$$ $$\ddot r=2b$$ $$\text {At }t=0, \dot r=a,$$ $$\kappa=\frac {2||a \times b||}{(a \bullet a)^{3/2}}.$$