How would you find the unit normal vector to the surface $x+y^{2}+2z=4$?

Question

What is the method that would be used to find the unit normal vector to the surface given by the equation $x+y^{2}+2z=4$?

Answer 1

$$\{(x,y,z)\mid x+y^2+2z=4\}=\{u(y,z)=(y^2+4-2z, y,z)\mid y,z\in \mathbb R\}$$

I denote $u_y=\frac{\partial u}{\partial y}$ and $u_z=\frac{\partial u}{\partial z}$. A normal is given by $u_y\times u_z$.

Answer 2

if you difference $x + y^2 + 2z = 4, $ and use the fact that $(dx, dy, dz)$ is tangent vector ay $(x, y, z),$ then you get $$0=dx + 2ydy + 2dz = (1,2y, 2)^\top \cdot (dx, dy, dz)^\top.$$ that is the normal vector at the point $(x,y,z)$ to the surface $x + y^2 + 2z = 4 $ is $(1,2y,z)^\top.$