## Expressions for curvature of parametric surfaces

If we define a curve in parameter space, (and thus on the surface) and restrict the normal field to the curve, then we get a function of normals along the curve . The derivative is precisely , where . Futhermore, given a parameterization of the surface , we have a basis for vectors in the tangent plane and , can be represented in this basis:

 (2.41) (2.42)

Labouring on, we have

 (2.43) (2.44)

or in vector form,

 (2.45)

So in the basis , is the linear map given by the . This is convienient, and we now want to find expressions for the in terms of the parameterization and its derivatives.

The shape operator can be expressed in terms of the parameterization by invoking it on the tangent to a curve :

 (2.46) (2.47) (2.48) (2.49)

where we have defined

 (2.50) (2.51) (2.52)

The transformations producing the second derivatives of are generated by noticing that since, say, is in the tangent plane, it must be true that , hence .

Substituting (2.41)-(2.42) into (2.50)-(2.51) generates a set of relations between , , and , for example, . When solved, these give

 (2.53)

Hence, given a parameterization of a surface, we can write down the differential or shape operator in the basis .