On the problem of finding the normal vector of curved surface Surface s: z = Z (x, y), take differential element DS on the surface, point P (x, y, Z (x, y)) ∈ DS Then the normal vector of the surface s at this point is n = ± (＠ Z / ＠ x, ＠ Z / ＠ y, - 1) (＠ represents partial derivative) How is this worked out?

On the problem of finding the normal vector of curved surface Surface s: z = Z (x, y), take differential element DS on the surface, point P (x, y, Z (x, y)) ∈ DS Then the normal vector of the surface s at this point is n = ± (＠ Z / ＠ x, ＠ Z / ＠ y, - 1) (＠ represents partial derivative) How is this worked out?

The surface equation can be written as an implicit function f (x, y, Z (x, y)) = 0. Its normal vector n = (@ f)/@ x,@F/@y ,@F/@z ）From the knowledge of multivariable calculus, @ Z / @ x = - (@ f / @ x) / (@ f / @ z) ,@z/@y=- (@ f / @ y) / (@ f / @ z), then we substitute the expression of n to get n = ± (＠ Z / ＠ x, ＠ Z / ＠ y, - 1). Of course, the result is poor

How to judge the direction of the normal vector of a surface?

Basic idea:
Find out the two tangent vectors of a certain point on the surface, and then make the outer product of the tangent vector (cross product) to be the normal vector (pay attention to the order of cross multiplication, otherwise the direction will be opposite)
therefore
If you already know the two tangent vectors of a certain point, you can directly find them
If you know the expression of the surface, you can calculate the partial derivative in two directions, and then calculate the outer product

Direction cosine of surface normal vector
cosa=(-fx)/[(1+fx^2+fy^2)^(1/2)]
cosb=(-fy)/[(1+fx^2+fy^2)^(1/2)]
cosv=(1)/[(1+fx^2+fy^2)^(1/2)]
Why is there a minus sign, x-axis, Y-axis

A normal vector of the surface equation f (x, y, z) = 0 can be n = {&; F / &; X, &; F / &; y, &; F / &; Z}. In particular, if the surface equation can be expressed as f (x, y, z) = z-f (x, y) = 0, then the normal vector can be n = ± {- &; F / &; X, &; F / &; Z}