Find a unit normal vector at a point

Question

Let G be the graph of the function $f(x,y)=xy$. Let $p=(2,3,6)$ in G.

I'm trying to find a unit normal vector to G at point p. I'm using a formula: $N(p)=(\sigma_u(q)\times \sigma_v(q))/(\left | \sigma_u(q)\times \sigma_v(q)\right |)$

I think it has something to do with partial derivatives which I found to be $f_x =y=3$ at point p, and $f_y=x=2$ at point p. But not sure how they relate to the formula if at all, and I don't get what the q represents in this formula.

Also, would I have to translate this to $p$ by $p + T_pG$ first for the graph to meet the point?

I'm trying to do this problem with differential algebra instead of using calculus

Answer 1

You have a surface $xy - z = 0$

$\nabla(xy-z)$ will be normal to the surface.

$\nabla(xy-z) = (y,x,-1)$

at the point $(2,3,6)$ gives $(3,2,-1)$

and normalize $(\frac {3}{\sqrt {14}},\frac {2}{\sqrt {14}}, -\frac {1}{\sqrt {14}})$

Looking at what you have:

$z = xy$

$\frac {\partial z}{\partial x} = y\\ \frac {\partial z}{\partial y} = x$

$(1,0,y)$ and $(0,1,x)$ lie in the plane parallel to the surface. Taking the cross product

$(0,1,x)\times(1,0,y) = (y,x,-1)$