I was solving the following problem: prove that every regular curve in $\mathbb{R^3}$ whose entries are polynomials with degree less than or equal to 2 is a plane curve. If we set $\alpha(t) = (at^2 + bt + c, dt^2 + et + f, gt^2 + ht + q )$, it is easy to find the equation of the plane that contains them by considering $\alpha(0), \alpha(1), \alpha(-1)$ and then proving $\alpha$ is contained in that plane. This, however, is quite a lot of work and not at all an elegant solution.
A curiosity question asked at the end of the exercise was: "Is every differentiable curve in $\mathbb{R^n}$ whose entries are polynomials of degree $\leq k$ contained in a subspace of $\mathbb{R^n}$ with dimension $\leq k$? Depending on how you solved the previous problem, the answer might be obvious".
I see you're studying by the exercises I gave in that DG class in 2016. The answer to the question I made in the end boils down to one word: "Taylor". If all the entries of the curve are polynomials of degree at most $k$, then $$\alpha(t) - \alpha(0) = \sum_{i=1}^k \frac{t^i}{i!}\alpha^{(i)}(0),$$and so the trace of $\alpha$ is contained in the affine variety $\alpha(0) + {\rm span}(\alpha'(0),\ldots,\alpha^{(k)}(0))$. I said there "dimension at most $k$" because we do not know anything about the linear independence of the derivatives of $\alpha$ at $0$.