TITLE: Normal vector to a surface
QUESTION [5 upvotes]: I've been reading in my calc book that the gradient vector is always orthogonal to the surface. 
So for a surface in space described by the level surface $f(x,y,z) = k$ where $k$ is a constant, $\nabla f$ is orthogonal to the surface at every point because the gradient is the normal vector of the surface at every point.
Then later I read about parametric surfaces where a surface is described by vector valued function $r(u,v) = <x(u,v), y(u,v), z(u,v)>$ and a normal vector $r_u \times r_v$ or $r_v \times r_u$
How are $r_u \times r_v$ and $\nabla f$ related here? I am referring to James Stewart's Text.
Also a last comment I want to make is, what about a normal vector to a surface that doesn't need to be described by a level surface? For example $f(x,y) = z = x^2 + y^2$? How would I go finding the normal vector at any point without rewriting it as $z - x^2 - y^2 = 0$ or parametrizing it?
A final Remark: I've been confusing the notion of the gradient vector being tangent to a surface instead of normal to it. There is this rather confusing picture I have which seems to suggests that the gradient vector really is tangent to a surface rather than normal because the gradient is formed by the vector sum of $\partial/\partial x$ and $\partial/\partial y$ and according to the picture I have, both $\partial/\partial x$ and $\partial/\partial x$ are "flat" and their sum should also be "flat" and not "pointing up"
Added PIcture

REPLY [3 votes]: These methods are just applications of two different geometric ideas to help you find a normal vector to a surface.  I'm sure that you or I could do some variable pushing and prove that they are compatible, but I don't know how enlightening that would be.  I think the most important thing is just to understand the geometry behind each of these ideas.
When you have a parametrized surface $r(u,v) = \left< x(u,v), y(u,v), z(u,v) \right>$ and a point $(u_0,v_0)$, you can consider two cross sections of that surface.  The functions $$r(u_0,v) = \left< x(u_0,v), y(u_0,v), z(u_0,v) \right>$$
$$r(u,v_0) = \left< x(u,v_0), y(u,v_0), z(u,v_0) \right>$$
define curves in three dimensions which are contained in the plane $r(u,v)$.  Convince yourself that a tangent vector to any curve contained in a surface is also tangent to the surface itself.  Therefore the vectors
$$  \frac{\partial}{\partial v} r(u_0,v) \big|_{v=v_0} $$
and
$$  \frac{\partial}{\partial u} r(u,v_0) \big|_{u=u_0} $$
are both tangent to the surface at $r(u_0,v_0)$.  Convince yourself that if these two vectors were parallel, then $r$ wouldn't look like a curve at this point, rather than a surface, so they should not be parallel.  In linear algebra terms, these vectors span the space of tangent vectors.  Their cross product will yield a vector which is normal to both of them, and therefore normal to the plane.  This is the definition you stated.
The other definition uses the fact that the gradient of a function at a point is perpendicular to the level surface at that point.  To understand this, it is helpful to think of the lower dimensional analogy.  The gradient of a function $f(x,y)$ (which defines a surface) will be perpendicular to the level curve at any point.  This is geometrically obvious if $f(x,y)$ defines a plane.  The level curve will be a horizontal line, and the gradient will point in the direction of greatest slope of the plane.  The same logic works, in fact, for $f(x,y)$ that is not a plane because the differentiability of $f$ tells us that it behaves like a plane at any given point.