Find Unit Vector perpendicular to the Surface

dustydude

Find Unit Vector perpendicular to the Surface,

x³+zx=1 at the point P=(1,2,-1)

I figures that the perpendicular vector would be,

N(X)=grad(x³+zx)
= (3x²+z, 0, x)
N(P)= (3,0,1)
Then the unit vector would be,

n=N(P)/||N(P)||

n=(3/5^1/2,0,1/5^1/2)
The answer i get is not the right answer and i dont see where im going wrong.

Thanks,

Dick

Is (3x^2+z,0,x) at P=(1,2,-1) really (3,0,1)?

dustydude

Thanks for pointing that out Dick, minor error.

(2,0,1)

Its still not the right answer which is 1/27^1/2(5,1,1)

Dick

Then there's probably a typo in the problem. The surface x^3+zx=1 equation doesn't have a 'y' in it. That means the y direction is tangent to the surface. The normal vector can't possibly have a nonzero y component.