Problem on normal vectors to surfaces
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Use 3d level surfaces to show that a normal vector to $z=f(x,y)$ is given by $\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, -1 \rangle$.
Solution
First, we recall how to use level surfaces to find a normal vector.
Normal vector to an implicitly defined surface
Gradient is perpendicular to level curves
Level curves and surfaces
To find a normal vector to a surface, view that surface as a level set of some function $g(x,y,z)$.
$\nabla g(x,y,z)$ is perpendicular to the level surfaces of $g(x,y,z)$.
To find a normal vector, we identify the function $g(x,y,z)$, and compute its gradient.
Level curves and surfaces
We identify that $z=f(x,y)$ is the 0-level surface of $$g(x,y,z) = f(x,y)-z.$$
Definition of the gradient
We now compute that the gradient is $$\nabla g(x,y,z) = \left \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, -1\right \rangle.$$
Normal vector to an explicitly defined surface
Hence, a normal vector to $z=f(x,y)$ is $\left \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, -1 \right \rangle$.
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