In my class, we defined the tangent space $T_pS$ of a regular curve $S$ at a point $p\in S$ as the set $\gamma'(0)$ such that $\gamma$ is a regular curve in $S$ and $\gamma(0)=p$.
My particular problem involves a torus $S$ parametrized as $\sigma(\theta,t)=((2+\cos{t})\cos{\theta},(2+\cos{t})\sin{\theta},\sin{t})$, and we want to describe the tangent plane at an arbitrary point of the $t=0$ parameter.
Solution attempt
Since we already have a parametrized surface, I assume we can treat $\sigma$ as a coordinate chart $U\to S$ for some $U\subset\mathbb{R}^2$. Let $p=\sigma(\theta,0)\in S$. We know that $\sigma$ is a diffeomorphism, so $q=\sigma^{-1}(p)=(\theta,0)$. Then $T_pS$ is spanned by the vectors $\sigma_\theta(q)$ and $\sigma_t(q)$. Now we compute these vectors...
$\sigma_\theta(\theta,0)=(-3\sin\theta,3\cos\theta,0)$
$\sigma_t(\theta,0)=(0,0,1)$
I want to believe these computations are correct because the span will give us a two-dimensional subspace of $\mathbb{R}^3$, which is in-line with a lemma stated in our text about the dimension of $T_pS$. But how exactly would we "describe" this span? Visually speaking, I know the first partial derivative is giving us a circle with radius 3 whereas the second one gives us a fixed point. But it feels like I am simply getting my torus back when I consider linear combinations of these two vectors.
