Problem A surface that is described explicitly by an equation of the form $z= f(x,y)$ can be thought as a level surface of the scalar field F defined by the equation $ F(x,y,z)= f(x,y)-z$ . Then $\nabla F = \frac{\partial f}{\partial x} i+ \frac{\partial f}{\partial y}j-k$.
Doubt $z= f(x,y)$ implies $ F(x,y,z)= f(x,y)-z=0$. Then gradient is always zero because if $F$ is identically zero , $\nabla F$ is also zero for all points on the surface.
Please help.
On
I don’t see that’s a problem.
A surface can be described in two different ways, i.e. $z=f(x,y)$ or $F(x,y,z)=0$. Let’s look at the second form. If a point $(x,y,z)$ is on the surface, then $F(x,y,z)$, of course, is zero. Otherwise, this condition does not necessarily hold. Thus, the partial derivatives may not be $0$.
Consider the following example: the plane $z=f(x,y):=x+y$ can also be written as $F(x,y,z):=x+y-z=0$. All gradients $\nabla F = ({\partial F\over \partial x}, {\partial F\over \partial y}, {\partial F\over \partial z}) = (1,1,-1)\ne 0$.
Its not that the function $F(x,y,z) $ is identically zero. Instead we have a function
$$F: \mathbb R^3 \to \mathbb R$$ or $F: U\subset \mathbb R^3 \to \mathbb R$, and the surface is the set of special points $(x,y,z)$ where $F(x,y,z) = 0$.