Angle of inclincation between tangent and plane

[filter54321]

1. The problem statement, all variables and given/known data
Find the angle of inclination of the tangent and plane to the surface at the given point.
(I we are comparing the tangent plane to the XY-plane)

2. Relevant equations
2xy-z3=0
Point: (2,2,2)

3. The attempt at a solution
This one was thrown into the textbook just to piss us off. No other problem has z to an exponent. My first thought was to rearrange the surface equation and take fx and fz rather than fx and fy as we had always been trained. I found that plane, took the <A,B,C> coefficient normal vector and found \theta between that normal vector and the K vector (Z-axis) but my answer isn't correct.


Thanks

[lanedance]

show your working...

that said, the easiest way to find the nomal to the tangent plane is as follows. consider the scalar function of 3 variables, giving each point in R3 a value
F(x,y,z) = 2xy-z3

The surface you are considering is given by the the level (constant value) surface for the above function when F(x,y,z) = 0

consider what this means, any path on the surface will not change the vaule of F(x,y,z) as it equals zero. The dierction of maximum change will be perpidicular to all level surfaces (and so tangent planes) and is given by the gradient

\nabla F(x,y,z)

this means your normal vector should be reasonably easy to find, regardless of the z exponent