Suppose I have a certain plane. I know that the plane is parallel to the plane z=0(3d coordinate system). Can I say that the partial deriatives of a certain surface equal to zero?
basically I need to find a certain plane that is tangent to a specific surface(some function) and is parallel to the z=0 plane.
Can I say that the partial deriatives equal to zero? because in the book they solve it in this way(which I don't really understand). what is the geometrcial meaning of taking a partial deriative of a certain surface and comparing it to zero?
The problem: $z=x^2y^2-2xy^2-6x^2y+12xy$ and there is some tangent plane to the function. find all the points on the plane, so that the plane is vertical to the vector (0,0,1)(there is no answer).
There are two aspects to the problem, and I'm not sure which is confusing you. The equation of the tangent plane to the surface $z=f(x,y)$ at the point $(x_0,y_0,f(x_0,y_0))$ is $$z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)\tag{1}$$ assuming the partial derivatives exist. See Paul's Online Math Notes if you are unsure about this.
The second part is what it means for the tangent plane to be normal (not "vertical") to the vector $(0,0,1)$. A normal vector to the plane $Ax+By+Cz=D$ has direction $(A,B,C)$. (If we look at the parallel plane $Ax+By+Cz=0$, it's clear that for any vector $(x,y,z)$ on the plane we have $(A,B,C)\cdot (x,y,z)=0$.)
Looking at $(1)$, we se that a normal vector is $$(-f_x(x_0,y_0),-f_y(x_0,y_0),1)$$ and if this is to be the same direction as $(0,0,1)$ the partial derivatives must vanish.