Finding the unit normal field of a surface A friend sent me a task which i am trying to solve but unfortunately i am not too sure whether or not my approach is correct.
Let $S = \{(x,y,z) \in \mathbb{R}^3 \vert z = xy+7e^{x-1}\}$ be an orientable Surface and $P=(1,0,7)^T \in S$.
a) Find a basis $(\mathcal{V}_1,\mathcal{V_2})$ of the tangent space $TpS$ in the point $P$ of $S$
b) Find a unit normal field $N:S\to \mathbb{R}^3$ of $S$ in the point $P$
My attempts:
Let $S$ be parameterized such that $(u,v)\mapsto (u,v,uv^2+7e^{u-1})^T = S(u,v)$
We find $$S_u(u,v) = (1,0,v^2+7e^{u-1})^T$$ and $$S_v(u,v) = (0,1,2uv)^T$$
Then we already have a basis $$(\mathcal{V}_1,\mathcal{V}_2) = (S_u(u,v),S_v(u,v))$$
The normal field is given by
$$N(u,v) = S_u(u,v) \times S_v(u,v)$$
thus
$$N(u,v) = (-v^2-7e^{u-1},-2uv,1)$$
dividing $N(u,v)$ by its norm $\vert N(u,v)\vert$ gives us the unit normal field of $S$ in $P$.
Would someone be so kind and tell me whether my attemps are generally correct or whether i might have messed up at some point.
I'd also like to know if a unit normal field is the same as the unit normal vector to a surface since both terms are for some reason being used equivalent in my documents and i was a bit confused, to say the least.
 A: Your parametrisation does not match the graph $z=f(x,y)=xy + 7e^{x-1}$ - you have $z(u,v) = uv^2 + 7e^{u-1}$ - which is the correct function? If it is the latter, then your derivation of $N = S_u \wedge S_v / \left| S_u \wedge S_v \right|$ is correct. 
It is really important to remember that $S$ is a surface (an object) which may have many parametrisations (functions which 'draw' the object), so it is better to say that $S$ is parametrised by a function $X(u,v)$ rather than abusing the notation by setting the 'name' of your parametrisation to $S(u,v)$. The reason for that is that you might have 2 parametrisations for your surface, and you want to compare which is 'better' for your purposes so it can become confusing if you denote the parametrisation $S$. With that in mind, from here on I will refer to the parametrisation of $S$ as $X(u,v) = \left( u,v,uv^2 + 7e^{u-1} \right)$. 
Do you know about the first fundamental form coefficients? There is a handy trick you can use to find $\left| X_u \wedge X_v \right|$:
\begin{align}
E &= X_u \cdot X_u = 1 + v^2 + 14 v e^{u-1} + 49 e^{2u-2}; \\ 
F &= X_u \cdot X_v = 2uv^3+14uve^{u-1}; \\
G &= X_v \cdot X_v = 1 + 4u^2 v^2.
\end{align}
These are the coefficients of the first fundamental form, such that
$$\mathrm{d}s^2 = E \mathrm{d}u^2 + 2F \mathrm{d}u \mathrm{d}v + G \mathrm{d}v^2.$$ This is useful when you have a unit-speed curve on the surface, so $\mathrm{d}s/\mathrm{d}t = 1$ implying that $$1 = E (u'(t))^2 + 2F u'(t) v'(t) + G (v'(t))^2$$. Moreover, $$\left| X_u \wedge X_v \right| = \sqrt{EG-F^2},$$ which can be particularly useful when $F = 0$ and $E=G = \lambda(u,v)>0$, such that $\left| X_u \wedge X_v \right| = \lambda(u,v)$ for some function $\lambda(u,v)$.
The unit normal field is the $N$ you found plus any other $N$ you may find using a different parametrisation of $S$, whereas the unit normal vector is $N$ evaluated at some point on the surface. So $$N(u,v) = \frac{-v^2 - 7e^{u-1},-2uv,1}{\sqrt{EG - F^2}}$$ is the unit normal field, and at the point $P = (1,0,7)$ on $S$ we have that $(u,v) = (1,0)$, so the unit normal vector at $P$ is $$N(1,0) = \frac{(-7,-1,1)}{\left.\sqrt{EG-F^2}\right|_{(u,v)=(1,0)}}, $$ where $\sqrt{EG-F^2}$ is evaluated at $(u,v)=(1,0)$.
