I am having some trouble with this question:
Let surface S be represented by F = 0, where F(x,y,z) = $x^2 – yz^{-2}$. Find all points on S where a normal vector is parallel to the xy-plane.
Now, this is causing me some trouble because I do not know where to apply the function given to me. I am under the impression that any normal vector in the Z plane is going to be parallel to the xy-plane, yielding the answer $(0,0,z)$.
To try and use the function in question, I attempted to, like the similar question to this one regarding the tangent plane does, take the gradient of F (the $\nabla$F), which would be the normal vector. However, I got lost in making that parallel to the xy-plane.
Any help would be appreciated. The answer I am given by the test book is $(0,0,z)$.
Best Answer
The surface in question is
$$S=\{(x,y,z)\in\mathbb{R}^3:x^2-yz^{-2}=0\}$$
which can be rewritten as $$S=\{(x,y,z)\in\mathbb{R}^3:z\neq 0\ \text{and} \ x^2z^2-y=0\}$$
Put $h(x,y,z)=x^2z^2-y\implies \nabla h(x,y,z)=(2xz^2,1,2x^2z)$. Now, a vector is parallel to the $xy$ plane if and only if it is perpendicular to $(0,0,1)$, and $$\nabla h(x,y,z)\perp (0,0,1)\iff (2xz^2,1,2x^2z)\cdot(0,0,1)=0\iff 2x^2z=0.$$
Since $z\neq0$ for $(x,y,z)\in S$, it follows that $x=0$, and therefore $y=0$. Hence, any point of the form $(0,0,z)$ with $z\neq 0$ is a solution.