Horizontal tangent planes to a surface

makarooney · #1 $Old$ November 5th, 2009, 06:10 PM

"Find all horizontal planes tangent to the surface given by $z=xye^\frac{-(x^2+y^2)}{2}$ or $f=-z + xye^\frac{-(x^2+y^2)}{2}$ ."

I already did this problem but there's something that bugs me. Here's how I solved it:

The normal vector to a curve at a point is $<f_x, f_y, f_z>$
If this curve is to be horizontal, $f_x$ and $f_y$ must be zero, and the only component is in the z direction.

So, $\frac{\partial f}{\partial x} = (xy)' e^\frac{-(x^2+y^2)}{2} +xy(e^\frac{-(x^2+y^2)}{2})' = ye^\frac{-(x^2+y^2)}{2} - x^2ye^\frac{-(x^2+y^2)}{2} =0$ and this simplifies to $y(1-x^2)=0$

Similarly, by finding $\frac{\partial f}{\partial x} =0$ I get $x(1-y^2)=0$

So now I get two equatons:
(1) $y(1-x^2)=0$
(2) $x(1-y^2)=0$

By setting y=0 in (1), x=0 in (2).
I also get four other points $(\pm1, \pm1)$

I did a few calculations and found 3 (!) tangent planes: including one at (0,0) which doesn't make sense... I even plotted the surface on Maple and saw that there cannot be a tangent plane at (0,0) because it would intersect the curve at other points.

And THAT is what I don't get... How come I get (0,0) as a point that has a horizontal tangent plane from my equations? Or should I just disregard this point when solving the problem for the same reason that I stated?
And how would I know for sure if I didn't have Maple, for example?

TriKri · #2 $Old$ November 6th, 2009, 01:36 PM

Quote:

Originally Posted by makarooney $View Post$

there cannot be a tangent plane at (0,0) because it would intersect the curve at other points

If z(x, y) has a gradient in a point, the surface that is formed also has a tangent plane in that point. It doesn't matter if the plane intersects other parts of the surface or not.