# Perpendicular vector Q

1. Feb 20, 2006

### ElDavidas

The question reads:

"Consider the surface given by the equation $$x^3 + xy^3 - z^2 = -4$$ The point p = (1,2,3) lies on this surface. Give a vector that is perpendicular to the surface at p?"

I'm not too confident about this question although there is a theorem in my notes saying:

if p exists within the level surface equal to c and the gradient of the function f is not equal to zero, then the gradient at point p is perpendicular to every path in the level surface equal to c which passes through p.

Does this apply here?

2. Feb 20, 2006

### HallsofIvy

Staff Emeritus
Well, yes, of course, it does!! Let f(x,y,z)= x2+ xy3- z2. Then this plane is a level curve of f: f(x,y,z)= -4. The gradient of f is perpendicular to that plane.

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