Question

Please help me find all points on the surface at which the tangent plane is horizontal.
z=(x^3)(y^2)

Answer 1

\[F(x,y,z)=x^3y^2-z\]

Answer 2

Gradient of F(x,y,z):
\[<3x^2y^2,2x^3y,-1>\]

Answer 3

Do I set the gradient equal to zero and then find the values for x,y,z?

Answer 4

Or do I set the partial of x and the partial of y equal to zero respectively and find the corresponding values?

Answer 5

@dan815

Answer 6

The normal of a surface at a point (x,y,z) given implicitly
\[i.e. F(x,y,z)=0\]
is the gradient of that function evaluated at that point. Since we are trying to find every point that satisfies the condition that the tangent plane is horizontal (I read this to mean that the normal vector is <0,0,1> or <0,0,-1>), Take the gradient of the function and set it equal to this normal vector and you will have a set of equations that will define all points that satisfy your criteria.
\[grad(F(x,y,z))=grad(x^3y^2-z) = <3x^2y^2,2x^3y,-1>\]
Setting the normal vector equal to the gradient gives:
\[3x^2y^2=0\]\[2x^3y=0\]\[-1=-1\]
The only points that satisfy these equations are\[<0,y,z>\]\[<x,0,z>\] such that\[z=x^3y^2\]
So in the end our points are
\[<0,y,0>\]\[<x,0,0>\]
Or all points on the x- and y- axes