Check if a curve is contained in a plane

Question

Let $$\gamma :\mathbb{R}\rightarrow\mathbb{R}^{3},$$ $$\gamma(t)=(t,t^{2},\ln(1+t^{2})).$$ be a parametrization of a curve. Establish if $\gamma$ is contained in a plane.

I have a problem here. First of all, I know that a curve $\alpha$ with curvature $k(t)\neq 0, \ \forall t\in Dom(\alpha)$ is contained in a plane if and only if $B(t)=\frac{\alpha '(t)\wedge\alpha ''(t)}{\mid\mid\alpha '(t)\wedge\alpha ''(t)\mid\mid }$ is constant. This curve seems to respect this condition but it's easy to see that $\gamma '(t)\mid_{t=0}=\underline{0}$ so in the origin it doesn't make sense to speak about curvature. I though to proceed in this way: I consider two curves $\gamma_{1}:(-\infty,0)\rightarrow\mathbb{R}^{3}$ and $\gamma_{2}:(0,+\infty)\rightarrow\mathbb{R}^{3}$. I see that these two curves are contained on a plane so I check if the planes are the same. Is it the right way to proceed?

Answer 1

Here is a better way: If the curve lines in a plane and $(a,b,c)$ is a unit normal vector to the plane then $at+bt^{2}+c \ln (1+t^{2})=0$ for all $t$. To get a contradiction easily by differentiating twice and setting $t=0$.

[First differentiation gives $a=0$ and second differentiation gives $b+c=0$ But then you get $t^{2}=-\ln (1+t^{2})$ for all $t$ which is false when $t=1$].

Answer 2

If this curve were contained in a plane, that plane would have to pass through $\gamma(0)=(0,0,0)$ and be spanned by $\gamma'(0)=(1,0,0)$ and $\gamma''(0)=(0,2,2)$. So the plane equation (again, if this were the case) would be $y=z$. But $t^2$ is not equal to $\ln(1+t^2)$ for all $t$, so here's your contradiction.