Finding tangent vectors to unit circle I am working on the following problem from Tu's Introduction to Manifolds.

I have been able to do (a), but (b) is causing me some troubles. My approach so far is as follows. Let $C$ be some smooth curve in $\mathbb{R}^2$ with $C(0) = p$ and $C'(0) = \frac{\partial}{\partial \overline{x}}\Big|_p.$ Then
$$i_*\Big(\frac{\partial}{\partial \overline{x}}\Big|_p\Big) = \frac{d}{dt}\Big|_0(i \circ C)(t) = \frac{\partial i}{\partial C}\Big|_{C(0)} \frac{\partial C}{\partial t}\Big|_0 = \frac{\partial i}{\partial C}\frac{\partial}{\partial \overline{x}}\Big|_p.$$
I am unable to take this any further and it seems I am on the wrong track. Can anyone provide some hints?
 A: Unfortunately Tu's use of "smooth curve" is ambiguous.
In section 8.6 (Curves in a Manifold) he defines it as a smooth map from an open interval to a manifold. But a smooth curve $u : (a,b) \to \mathbb R^2$ is not a submanifold $C \subset \mathbb R^2$. We could try to argue that the image of a smooth curve is a submanifold, but this is in general not true. The image may have corners (this is possible at points where $u'(\tau) = 0$) or self-intersections.
In Appendix B.2 (The Implicit Function Theorem) he gives a completely different interpretation when he says

On a smooth curve $f (x,y) = 0$ in $\mathbb R^2$

Here $f$ is of course a smooth map and $C = \{(x,y) \in \mathbb R^2 \mid f(x,y) = 0 \}$.
This makes clear that we should understand a smooth curve $C$ as a one-dimensional submanifold of $\mathbb R^2$.
Now have a look to Tangent vectors to a plane curve Tu #11.2. You will see that everything works exactly as for the circle $S^1$ under the assumption that $U$ is an open neighborhood of $p \in C$ such that $\bar x = x \mid_U : U \to \mathbb R$ is a local coordinate. The only difference is that we do not exlipicitly know what $\bar y = y \mid_U : U \to \mathbb R$ looks like. But this does not play any role, we still get the same formula as for $S^1$.
Note that in general there are points $p \in C$ which do not admit an open neigborhood $U$ such that $\bar x = x \mid_U : U \to \mathbb R$ is a local coordinate. For $S^1$ these are $(-1,0)$ and $(1,0)$. In such cases we can exchange the role of $x$ and $y$.
