# Tangent planes and normal vectors

## 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
fx=3x2y2
fy=2x3y
fz=-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 in to the partials.

LCKurtz
Homework Helper
Gold Member

## 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
fx=3x2y2
fy=2x3y
fz=-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 in to 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.