Find a $T_pM$ and $N_pM$ I have the following Submanifold:
$$M = \{(x,e^{x})\in \mathbb{R^{2}:x \in \mathbb{R}}\}$$
to which I have to find a tangent and normal space $T_{p}M$ and $N_{p}M$ at the point $p \in M$.
This  is one of the examples that I have but in the same  excercise I have like 10 different submanifolds to which i'll have to find $T_{p}M$ and $N_{p}M$. I am clear with their idea visually and we also had a Theorem about the Basis of $T_{p}M$ and $N_{p}M$.
if somebody could help me out with the practical solutions in these cases (since the excercises has like 10 different cases with different functions i assume that there might be a practical way to see them)
i would be very thankful
 A: Let $$M=\{(x,e^x): x\in \mathbb{R}\}$$ be a submanifold of $\mathbb{R^2}$. Let $\phi: \mathbb{R^2}\to \mathbb{R}$ defined by $\phi(x,y)= e^x-y$ on some open subset $U$ of $\mathbb{R^2}$. Notice $M$ is the level set $\phi^{-1}(0)$. Then since $T_p(M) = (\nabla f\vert p)^{\perp}$, then we can find the tangent space this way: $(\nabla f)=(e^x, -1)$. Therefore the complement is generated by $(1,e^x)$.
Notice that all tangent lines to $M$ in $\mathbb{R}^2$ are precisely $$span\{(1,e^x):x\in \mathbb{R}\}$$
If you have 50 similar exercises, I would follow this procedure.
A: You can draw a picture and see what we expect:

Here is a rigorous calculation, using curves and derivations: Let $p=(x_0,e^{x_0})\in M$. Define $\gamma:[-\epsilon, \epsilon]\to M$ by $\gamma(t)=(t+x_0,e^{t+x_0}).$ Then, $\gamma(0)=p$ and $\gamma'(0)=a\frac{\partial}{\partial x}+b\frac{\partial }{\partial y}$ is an arbitrary tangent vector at $p$. Operating with $\gamma'(0)$ on the projections $r_1(x,y)=x$ and $r_2(x,y)=y,$ respectively, we get $\gamma'(0)r_1=a$  and $\gamma'(0)r_2=b$.
Now, by definition of $\gamma'(0)$, we have
$a=\gamma'(0)r_1=\frac{d (r_1\circ \gamma)}{dt}|_{t=0}=\frac{d (t+x_0)}{dt}|_{t=0}=1$ and $b=\gamma'(0)r_2=\frac{d (r_2\circ \gamma)}{dt}|_{t=0}=\frac{d (e^{t+x_0})}{dt}|_{t=0}=e^{x_0}$.
Therefore, in terms of components, $v\in T_pM$ is given by vectors in  $\operatorname {span}\{(1,e^{x_0})\}.$
And then $N_pM$ is just the orthogonal complement, so we need to solve $A\cdot 1+B\cdot e^{x_0}=0.$ We know that this subspace has dimension $1$ (because $\mathbb R^2$ has dimension $2$) so all we need is one vector satisfying this condition. Insepction gives the vector $(e^{x_0},-1)$ and so we may take $N_pM=\operatorname {span}\{(e^{x_0},-1)\}.$
