Curves in three dimensions are often described by a pair of equations; say, g(x ,y, z) = 0, and h(x, y, z) = 0.

Suppose you want to find the curvature radius of curvature, center of curvature, or torsion for a curve at some point r', r' = (x',y'z'), for a curve C defined in this way. One way would be to parametrize the curve, and differentiate, but this is not always easy. There is an easy, if sometimes tedious way, to find these things, as follows

1. First find a tangent vector to the curve at r'. Since grad g and grad h are both perpendicular to the curve, their cross product is such a vector: form  T = g h

2. Normalize this vector to find a unit tangent vector, t(r'),  for C at r', by dividing this cross product  by its magnitude:
form

3. Compute the derivative, of this vector with respect to length along C by taking the directional derivative of each of its components in the direction of t: form (t t)

Note

4. The magnitude of the resulting vector is the curvature of C at r'.  Its direction, having unit vector n, is in the plane of the curve at r', and is normal to t.

5. The radius of curvature is the reciprocal of the curvature:

R_c = 1 / .

6. The center of curvature is a distance given by the radius of curvature from r' in the direction  having unit vector n.

C_c = r' + n /

7. The torsion is the magnitude of the directional derivative of n in the direction of t: