Im studying Multivariable calculus and in this video (https://www.youtube.com/watch?v=e7Nel0UoXR8&list=PLSQl0a2vh4HC5feHa6Rc5c0wbRTx56nF7&index=45) the lecturer talks about a function that maps two dimensional input (s and t) and output a parametric surface. my question is after we find $\frac{\partial \vec{V}}{\partial s}$ and $\frac{\partial \vec{V}}{\partial t}$ can we say that $\nabla \vec{V} = [\frac{\partial \vec{V}}{\partial s} , \frac{\partial \vec{V}}{\partial t}]$ and this gradient is pointing to the direction of steepest ascent in the output space?
if it is/isn't why? and how can we interpret steepest? is it along X or Y or Z direction of the output space? please tell me step by step for beginner.
The function $\vec{V}$ gives the surface parametrically as:
$$ \vec{V}(s,t) = \left< x(s,t), \, y(s,t), \, z(s,t) \right> $$
You have the partial derivative vectors:
$$ \frac{\partial \vec{V}}{\partial s} = \left< \frac{\partial x}{\partial s}, \, \frac{\partial y}{\partial s}, \, \frac{\partial z}{\partial s} \right> $$
$$ \frac{\partial \vec{V}}{\partial t} = \left< \frac{\partial x}{\partial t}, \, \frac{\partial y}{\partial t}, \, \frac{\partial z}{\partial t} \right> $$
The $\left[ \frac{\partial V}{\partial s}, \, \frac{\partial V}{\partial t}\right]$ that you suggest wouldn't really be a vector, so I'm not sure it makes sense to say it "points" in some direction. But you can interpret it as the Jacobian matrix
$$ DV = \left( \begin{array}{cc} \frac{\partial x}{\partial s} & \frac{\partial x}{\partial t} \\ \frac{\partial y}{\partial s} & \frac{\partial y}{\partial t} \\ \frac{\partial z}{\partial s} & \frac{\partial z}{\partial t} \end{array} \right) $$
You could interpret the rows of this matrix as the gradients $\nabla x$, $\nabla y$, and $\nabla z$. Then the last row, $\nabla z$, would point in the direction (in the $s,t$-plane) of greatest increase in $z$.