Tangent planes and normal vectors

[Mdhiggenz]

Homework Statement

Find all points on the surface at which the tangent plane is horizontal

z=x³y²

Things I know:
Tangent plane is horizontal then therefore the normal must be vertical in order to be perpendicular.

Dot product of the tangent plane with normal is = 0

Normal plane is given by the partial with respect to x,y,z evaluated at some points P(x₀,y₀,z₀)

Putting that together I get
f(x,y,z)=x³y²-z
f_x=3x²y²
f_y=2x³y
f_z=-1

Here is where I am completely stuck I know that I still don't have a normal vector because I do not have any points to plug into the partials.

Homework Equations

The Attempt at a Solution

 

[LCKurtz]

Homework Statement

Find all points on the surface at which the tangent plane is horizontal

z=x³y²

Things I know:
Tangent plane is horizontal then therefore the normal must be vertical in order to be perpendicular.

Dot product of the tangent plane with normal is = 0

Normal plane is given by the partial with respect to x,y,z evaluated at some points P(x₀,y₀,z₀)

Putting that together I get
f(x,y,z)=x³y²-z
f_x=3x²y²
f_y=2x³y
f_z=-1

Here is where I am completely stuck I know that I still don't have a normal vector because I do not have any points to plug into the partials.

Your original equation is of the form $z = f(x,y)$. The tangent plane will be horizontal at any point $(x,y,z)$ where $f_x(x,y)=0$ and $f_y(x,y)=0$.

 

[HallsofIvy]

Or, same thing, The vector you got, $<3x^2y^2, 2x^3y, -1>$ must be of the form <0, 0, a> to be vertical. Obviously we can take a= -1 but we need $3x^2y^2= 0$ and [itex]2x^3y= 0[/tex]. That will be true if x= 0 or if y= 0. In other words above the x and y axes.[/itex]

 

[Mdhiggenz]

hmm, so I can equal them together, and get

x=(3/2)y and y=2/3x

Don't really understand what the values represent?