# Tangent planes and normal vectors

1. Oct 31, 2012

### Mdhiggenz

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. The attempt at a solution

2. Oct 31, 2012

### LCKurtz

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 $f_y(x,y)=0$.

3. Oct 31, 2012

### HallsofIvy

Staff Emeritus
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.

4. Oct 31, 2012

### 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?