> e.g. f(x, y) = x ^ 2 * y
>
> What is the normal vector at (1, 1) ?
>
> I still remember from university that f'x(x, y) = 2 * x * y and
> f'y(x, y) = x ^ 2. At (1, 1) this would give me f'x = 2 and f'y = 1 - but
> from there on I'm lost..

as you say:
f(x, y) = x ^ 2 * y

f'x(x, y) = 2 * x * y, f'y(x, y) = x ^ 2

the two derivative functions give you slope functions in the x and y
directions respectively. Since the slope is a gradient (i.e., rise over run)
you can write the 2 functions as the following two vectors:

f'x(x, y) = 2 * x * y : <1,0,2*x*y>
f'y(x, y) = x ^ 2       : <0,1,x^2>

these represent the slope in th x and y directions, take the cross product
and you should have the normal.

-tgq

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