Thread: Find a tangent plane horizontal to the xy plane?

find the points on the surface f(x,y,z)=1 that have a horizontal plane (ie a tangent plane horizontal to the xy plane)
where
f(x,y,z)= x^3 +y^34 -z^3.

I'm getting stuck as I can only find one point.

I got (0,0,-1) but this is by using a random vector I set as (0,0,-3z^2).

Originally Posted by Roland25

find the points on the surface f(x,y,z)=1 that have a horizontal plane (ie a tangent plane horizontal to the xy plane)
where
f(x,y,z)= x^3 +y^34 -z^3.

I'm getting stuck as I can only find one point.

I got (0,0,-1) but this is by using a random vector I set as (0,0,-3z^2).

Surely you don't mean "horizontal to the xy plane". Either just "horizontal" or "parallel to the xy plane". Any such plane is of the form z= constant and a unit normal vector is <0, 0, 1>. A vector normal to the surface given by is . That will be parallel to <0, 0, 1> when 2x= 0, .

(Surely you didn't mean " ". I have assumed it was " and you accidently hit two keys.)

Sorry, you were right, it wasm eant to be y^4.

The question states find the points on the surface that have a horizontal tangent plane
which I'm guessing means a tangent plane parallel to the xyplane right?