I am studying gradients right now and how they relate to tangent planes of a graph.
I came across how you can get a tangent curve of $z=f(x,y)$ if you define it as a level curve at $0$ for the function $F(x,y,z) = f(x,y) - z = 0$.
I came across this problem:
Find the tangent plane for $2z-x^2=0$ at $P_0(2,0,2)$
If we say that $F(x,y,z) = 2z-x^2$, then the gradient would be $\nabla{F} = -4i+2k$
Evaluating the the dot product, the tangent plane is $-4(x-2)+2(z-2)=0$ which is $2z-4x+4=0$.
Now say that $f(x,z) = y = 2z-x^2$. Then the gradient evaluated at $P_0(2,0,2)$ is $\nabla{f = -4i-j+2k}$. As a result, the tangent line is $-4(x-2) -(y-0)+2(z-2)=0$ which is $2z - 4x + 4= y$
EDIT I forgot to mention that if $f(x,z) = y$, we can define $F(x,y,z) = 2z-x^2 - f(x,y)$ for the level curve $F(x,y,z)=0$.
We then have:
$$2z-4x+4=0$$ $$2z - 4x + 4= y$$
While the first only allows $z$ and $x$ to be certain values with a free range $y$, the second one allows both $z$ and $x$ to be any value.
How can this be so?