# Finding the equation of a plane

1. Sep 6, 2007

### Tiffw-s88

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

This isn't actually a homework question, but I'm revising for my resits and can't do this past paper question. And of course if i don't figure out how to do it it will obviously come up on the exam

A plane is tangential to the surface xexp(-x^2 - y^2) - z = 0 at the point where z takes on its maximum value. Find the equation of the plane

2. Relevant equations

3. The attempt at a solution

I think I have to find where z is at its maximum and then find the normal to the tangent at that point to enable me to write the plane equation in the form r.n=0
z will be at it's maximum when xexp(-x^2 - y^2) is at its maximum but i don't know when that will be?

2. Sep 6, 2007

### NateTG

Do you know how to find the maximum of a function?

3. Sep 6, 2007

### HallsofIvy

Staff Emeritus
As NateTG suggested, first find the maximum point by setting the gradient of z equal to 0. Of course, the tangent plane at a maximum (or minimum) is parallel to the xy-plane, just as the tangent line at a maximum (or minimum) of y= f(x), in two dimensions, is parallel to the x-axis.

4. Sep 6, 2007

### Tiffw-s88

so since z = x exp(-x^2 - y^2) if i differentiate this and set it to zero i should get my point. But what do i differentiate z with respect to since there's x and y in the function.
Since I'm trying to find the equation of the plane does that just mean that it's a plane parallel to the x-y axis?

5. Sep 6, 2007

### NateTG

Actually, it's parallel to the x-y plane.

Since you're apparently not familiar with partial derivatives, I'd suggest looking at
$$x e^{-x^2-y^2}$$
and trying to find what $y$ is at the maximum by inspection.

6. Sep 6, 2007

### Tiffw-s88

oops yeh i meant plane

i am familiar with partial derivatives.

so the partial derivative wrt x is -2x^2[exp(-x^2-y^2)] + exp(-x^2-y^2)
and wrt y is -2y^2[exp(-x^2-y^2)]

but I'm still confused as to where i'm going with this, i don't really get the bigger picture of what i'm trying to do

7. Sep 6, 2007

### Dick

If you are at a maximum point, the partial derivatives wrt x and y will both vanish.

8. Sep 6, 2007

### Tiffw-s88

ok now i'm really confused. how do you recomend i start this question as i don't seem to be getting anywhere with my current route =(

9. Sep 6, 2007

### Dick

Just set your two partial derivatives equal to zero. That's two equations in two unknowns, x and y. Once you've found your critical points test them to see if you can find which one could give you a maximum for z.