Normal to surface at point

Question

I have this function:

$F(x,y,z)=x^2−y^2−z^2+4$ where $z\ge 0,0\le x \le 2,0 \le y \le 2$.

How can I find the normal at some point $P=(p_x,p_y,p_z)$?

I have tried to calculate the derivatives of the axes:

$F′x=2x,F′y=-2y$

But how can I calculate the normal with this gradient?

Answer 1

The normal to a surface of the form $f=c$ is just the gradient of $f$, in your example $$\nabla f = (2x, -2y, -2z)^T =\left( \frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right)^T$$ whenever $(x,y,z)$ fulfils your equation. This is assuming the gradient does not vanish (in which case the equation does not define a surface in general). Usually one is interest in a unit normal, in which case you have to divide by the norm of the normal.

Answer 2

Given a surface of equation $$z=f(x,y)$$ you can find the normal vector $N$ at a point $P(x_0,y_0)$ as: $$N=[f_x(x_0,y_0),f_y(x_0,y_0),-1]^T$$ where: $f_x=\dfrac{\partial f}{\partial x}$ and $f_y=\dfrac{\partial f}{\partial y}$ If the equation has given in implicit form you can get for $N$: $$N=[f_x(x_0,y_0,z_0),f_y(x_0,y_0,z_0),f_z(x_0,y_0,z_0)]^T$$