Tangent Vectors to Parametric Surfaces

If q is a constant, then r( u,q) is a function of u only and is therefore a curve. As a result, the velocity vector

ru = á fu,gu,hu ñ

is tangent to the curve. Since the curve lies on the surface parameterized by r( u,v) = á f( u,v) ,g(u,v) ,h( u,v) ñ , it follows that r_u( p,q) is tangent to the surface at the point with coordinates ( p,q) . Likewise, we define r_v = á f_v,g_v,h_v ñ and similarly it follows that r_v( p,q) is also tangent to the surface at the point corresponding to ( p,q) .

Moreover, we say that the parameterization is regular at a point P on the surface if r_u and r_v are nonzero and non-parallel at P.  Otherwise, the parameterization is said to be singular at P.       

EXAMPLE 4    Find r_u and r_v for the sphere

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

Solution: To find r_u, we apply ¶_u to r to obtain

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

Likewise, to find r_v, we apply ¶_v to r to obtain

rv = á cos( u) cos( v) ,sin( u) cos( v) ,-sin( v) ñ

Graphic of Directional Derivative

       

Notice in example 4 that

ru·rv = -sin( u) sin(v) cos( u) cos( v) +cos( u)sin( v) sin( u) cos( v) = 0

That is, the vectors r_u and r_v are perpendicular for all ( u,v) . Parameterizations in which r_u and r_v are orthogonal for all ( u,v) are important in applications because they introduce an orthogonal coordinate system in each tangent plane:

If r_u · r_v = 0 for all ( u,v) , then we say that the parameterization r( u,v) is orthogonal.       

EXAMPLE 5    Find r_u and r_v and determine if the following parameterization is orthogonal:

r = á u2-v2,2uv,u2+v2 ñ

Solution: To find r_u, we apply ¶_u to r to obtain

ru = á 2u,2v,2u ñ

Likewise, to find r_v, we apply ¶_v to r to obtain

rv = á 2v,2u,-2v ñ

The dot product of r_u with r_v is given by

ru·rv = 4uv+4uv-4uv = 4uv

Since r_u·r_v ¹ 0, the parameterization is not orthogonal.