Surface Normals and Tangent Planes to Parametric Surfaces   

If r(u,v) is a regular parameterization of a surface, then the vector r_u×r_v is perpendicular to both r_u and r_v. Thus, r_u×r_v must also be perpendicular to the tangent plane spanned by r_u and r_v. We say that the cross product r_u×r_v is normal to the surface, and the vector r_u×r_v can be used as the normal vector in determining the equation of the tangent plane at a point of the form (x₁,y₁,z₁) = r( p,q) .

EXAMPLE 3    Find the equation of the tangent plane to the torus

r = á ( 2+sin( v) ) cos(u) ,( 2+sin( v) ) sin( u) ,cos( v) ñ

at the point r( 0,0) .        

Solution: The vectors r_u and r_v are given by

ru = á -( 2+sin( v) ) sin( u) ,( 2+sin( v) ) cos( u),0 ñ rv = á cos( v) cos( u),cos( v) sin( u) ,-sin( v) ñ

so that r_u( 0,0) = á 0,2,0 ñ = 2j and r_v( 0,0) = á1,0,0 ñ = i. Thus, the normal to the plane is

ru( 0,0) ×rv(0,0) = 2j×i = -2k = á 0,0,-2 ñ

Since r( 0,0) = ( 2,0,1) , the equation of the tangent plane at r( 0,0)  is

0( x-2) +0( y-0) -2( z-1) = 0

which reduces to z = 1.

      

Since the crossproduct r_u×r_v is normal to the surface r(u,v), the unit vector

n =  ru × rv || ru × rv ||

is also normal to the surface. The vector n is thus called the unit surface normal of the surface.   It is important to note that n = n( u,v) is a function of u and v, or equivalently, that n(u,v) defines a unit normal at each point on the surface.  

EXAMPLE 4    Find the unit normal to the cylinder

r( u,v) = á 3cos( u), 3sin(u), v ñ

at r( p,2) = ( -3,0,2) .    

Solution: Since r_u = á -3sin( u), 3cos( u) ,0 ñ = -3sin( u) i + 3cos( u) j and since r_v = á0,0,1 ñ = k, their cross product is

ru×rv  =  ( -3sin( u) i + 3cos( u) j) ×k  =  -3sin( u) i×k + 3cos( u) j×k  =  3sin( u) j + 3cos( u) i

It is easy to shown that || r_u×r_v || = 3, so that the unit surface normal  is 

n =  ru×rv || ru×rv ||  =  sin( u) j + cos( u) i  =   á cos( u), sin( u), 0 ñ

The tangent plane, which is the plane through ( -3,0,2) with normal

n( p, 2) = sin( p) j+cos( p) i = á -1,0,0 ñ

has an equation of

-1( x- -3) +0( y-0) +0( z-0) = 0

That is, the equation of the tangent plane is x = -3.

LiveGraphics3d Applet click and drag the red dot to see n(u,v) in action.