3D - derivative of a point's function, is it the tangent? If I have (for instance) this formula which associates a $(x,y,z)$ point $p$ to each $u,v$ couple (on a 2D surface in 3D):
$p=f(u,v)=(u^2+v^2+4,2uv,u^2−v^2) $
and I calculate the $\frac{\partial p}{\partial u}$, what do I get? The answer should be "a vector tangent to the point $p$" but I can't understand why. Shouldn't I obtain another point?
 A: your equation determines a surface in R^3 . if you calculate this derivative, by definition , the value of the derivative is the angular coeficient of the line which pass in P ( for each pair (u,v) you have a point P) and has director vector (1,0,0) . a good calculus book do this geometrical interpretation . ( my english is horrible, sorry )
A: For simplicity, I'll just deal with $f$, since $p$ is just a synonym or alias for $f$. First, we have:
$$
f : \mathbb R^2 \mapsto \mathbb R^3\\
f(u, v) = (u^2 + v^2 + 4, \,\,2uv, \,\,u^2 - v^2)
$$
When we differentiate $f$ with respect to the input variable $u$, we are trying to see how the output of $f$ changes for small changes $\epsilon$ in $u$, while keeping $v$ constant. Specifically, we are:


*

*Taking the original collection of 3D points, $f(u, v)$.

*Producing a new collection of 3D points, $f(u + \epsilon, v)$.

*Producing a collection of 3D vectors by subtracting the original points $f(u, v)$ from the new points $f(u + \epsilon, v)$.

*Scaling those vectors by $\frac{1}{\epsilon}$ so that it represents a ratio of the change in output over the change in input.

*Seeing what the instantaneous change is by pushing $\epsilon$ to zero.


Putting this all together, we have:
$$\frac{\partial f}{\partial u}(u, v) = \lim_{\epsilon \to 0}\frac{f(u + \epsilon, v) - f(u, v)}{\epsilon}$$
A: Take a fixed location where $(u,v) = (u_0,v_0)$. Think about the mapping $u \mapsto f(u,v_0)$. This is a curve lying on your surface, which is formed by allowing $u$ to vary while $v$ is held fixed. In fact, in my business, we would say that this is an "isoparametric curve" on the surface. By definition, $\frac{\partial f}{\partial u}(u_0)$ is the first derivative vector of this curve at $u= u_0$. In other words, it's the "tangent" vector of this curve.
