I was doing a sample question and came across this question.
A given surface is defined by the equation: $3x^2+2y^2-z=0$. Describe the normal vector at a point (x, y, z) on the surface. Calculate the normal vector at the point $(1,-1,5)$ on the surface.
The normal vector is $(6x, 4y, -1)$
How did the answer come out to be $(6x,4y,-1)$?
And how do I calculate the unit normal vector at the point $(1,-1,5)$?
On
It is the gradient vector which is normal to the level surface.
The gradient vector is defined to be the vector of partial derivatives which in your case is $(6x, 4y, -1)$
On
Given an infinitesimal displacement to the point $(x,y,z)$ such that it remains in the surface. By differential calculus, we know that
$$f(x+dx,y+dy,z+dz)=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz=0.$$
This expression can be seen as the dot product of the gradient vector and the displacement vector. As they are orthogonal (zero dot product), the gradient vector must be normal to the surface.
Because that's the gradient of $3x^2+2y^2-z$.