Integral curve definition in terms of the tangent vector: why $D \varphi_{t} (d/dt) = X_{\varphi(t)}$? I have this definition of an integral curve from these notes, page 29, on differentiable manifolds.
With a manifold $M$, an integral curve of a vector field $X$ is a smooth map $\varphi: (\alpha,\beta) \subset \mathbb{R} \to M$ such that $$
D \varphi_{t} \left(\frac{d}{dt}\right) = X_{\varphi(t)}
$$
An example is given where $M = \mathbb{R}^{2}$, and so the derivative of the smooth function $\varphi(t) = (x(t),y(t))$ is
$$
D \varphi\left(\frac{d}{dt}\right) = \frac{dx}{dt}\frac{\partial}{\partial x} + \frac{dy}{dt}\frac{\partial}{\partial y}
$$
What is going on here with the $d/dt$ in the brackets on the final line? i.e. this part
$$
D\varphi\left(\underbrace{\frac{d}{dt}}_{\text{this part here}}\right) = \dots
$$
I am assuming there is a function $f: M \to \mathbb{R}$ on which $D\varphi(d/dt)$ acts, as $D\varphi(df/dt)$, but I thought $\varphi$ takes in real numbers i.e. the time parameter, since its defined as a map $(\alpha,\beta) \subset \mathbf{R}$? Perhaps it's not a function, just a term in brackets to multiply $D\varphi$ by?
 A: You are correct in saying that $\varphi$ is a map from $\mathbf{R}$ to $M$. The differential $D\varphi_t$, however, is the pushforward; it takes tangent vectors in $T_{t} (\alpha,\beta) \cong \mathbf{R}$ over to a tangent vector in $T_{\varphi(t)} M$. In our case, the tangent vector $d/dt$ is pushed forward by $\varphi$ to get $X_{\varphi(t)} \in T_{\varphi(t)}M$. It may be useful to compute $\varphi_t$ for a few different vector fields $X$ on $\mathbf{R}^2$ or $\mathbf{R}^3$ to get a better idea of what is going on, and then try to carry this over to higher dimensions and other manifolds like $\mathbf{S}^n$.
A: Let $X$ be a smooth vector field on a manifold $M$. Recall that $X$ is a section of the tangent bundle $TM$, i.e. $X:M\rightarrow TM$. So for every point $p\in M$, $X(p)=X_p\in T_pM$ is a tangent vector.
Now, let $\varphi:(a,b)\rightarrow M$ be a curve in $M$. Recall that $(a,b)$ is a manifold with the global coordinate $t$. In particular, at a point $t_0\in (a,b)$, the vector $(d/dt)\vert_{t_0}$ spans the tangent space $T_{t_0}(a,b)$. We say that $\varphi$ is an integral curve of $X$ if the following holds. For all $t_0\in (a,b)$,
$$
D(\varphi)_{t_0}\left(\dfrac{d}{dt}\Big\vert_{t_0}\right)=X_{\varphi(t_0)}.
$$
What this means is that the map
$$
D(\varphi)_{t_0}:T_{t_0}(a,b)\rightarrow T_{\varphi(t_0)}M
$$
has the property that $(d/dt)\vert_{t_0}$ is mapped to $X_{\varphi(t_0)}$ (for all $t_0\in(a,b)$).
