1. Find unit normal to the surface point (4 ,-2 , 1 )
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i differentiate r with respect to u and achieve ( 0 , -v , 2u ) --- eqn 1
i differentiate r with respect to v and achieve ( 2v , -u , 0 ) ---- eqn 2

at point (4 ,-2 , 1 ),

( v^2 = 4 , -uv = -2 , u^2 = 1 ) ,<---- i like to ask if this step is correct?

v = 2 , u = 1

sub in the u & v values to eqn 1 & 2 and cross the 2 vector to achieve normal vector....

many thanks!

Yes, becuase if you take the negative root for V it forces u=-1, and that value is not in the domain of your function.