Problem
I have surface defined in implicit form. $x^3+3x^2y+y^2+2\cos(z)=13$. Now would like to compute normal vector for example in point $(1,2,0)$
attempt to solve
Now if i call this surface as $g(x,y,z)=x^3+3x^2y+y^2+2\cos(z)-13$ isn't $\nabla g(g,y,z)$ normal vector field for this surface ?
$$ \nabla g(x,y,z)= \begin{bmatrix} 3x^2+6xy \\ 3x^2 \\ -2\sin(z) \end{bmatrix} $$ $$ \nabla(1,2,0)=\begin {bmatrix} 3\cdot 1^2+6\cdot 1\cdot 2 \\ 3\cdot 1^2 \\ -2\sin(0) \end{bmatrix} $$ $$ \nabla(1,2,0)=\begin{bmatrix} 15 \\ 3 \\ -2 \end{bmatrix} $$
Now i am not sure if this is correct answer of if there is flaw. I understand that gradient is the highest increasing direction and it's length is the slope of the increase in some direction. I am not really sure how gradient behaves when surface is given in implicit from for instead of explicit one. I think i lack intuition on this subject.
Yes this is correct, the normal vector to the surface $F(x,y,z)=c\,$ is just the gradient vector $(F_x,F_y,F_z)$.
Indeed, for a geometric intepretation, if you consider the funtion $w=F(x,y,z)$, the gradient vectors is orthogonal to the level surface $w=c\,$ that is the normal to $F(x,y,z)=c$.