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Classical differential geometry of space-curves  33



                                              Normal
                                              plane
                                                            b
                                                                    Rectifying
                                                   n                plane
                                                                t
                                             Osculating  P
                                             plane

                        Fig. 2.1 Osculating, normal, and rectifying planes at a point P on a curve.


                        with respect to s gives
                                      db   dt          dn               dn      dn
                                         =    × n + t ×   = κn × n + t ×   = t ×   ,       (2.20)
                                      ds   ds          ds               ds      ds
                        since n × n = 0. We deduce that db/ds is a vector orthogonal to t. Moreover, since b
                        is defined as a unit vector, with b · b = 1, it follows that b · db/ds = 0, so that db/ds
                        is also orthogonal to b.
                          Hence db/ds can only be parallel to n, and we can write
                                                        db
                                                           =−τn,
                                                        ds
                        where the scalar τ is called the torsion of the curve at a point. Not all writers follow
                        the convention of including a negative sign in this equation; the significance of the sign
                        can be shown by considering the special case of the circular helix given by eqn (2.6).
                        Here we obtain

                                            1               s                s

                                 t = r = √        −ai sin √       + aj cos √      + ck ,
                                                                            2
                                                            2
                                           2
                                          a + c 2          a + c 2         a + c 2
                                                1               s               s

                                t = r = κn =          −ai cos √       − aj sin √       ,
                                               2
                                                                                2
                                              a + c 2          a + c 2         a + c 2
                                                                2
                        so that
                                                              a
                                                       κ =                                 (2.21)
                                                            2
                                                           a + c 2
                        and
                                                         s              s
                                            n =−i cos √       − j sin √     .
                                                                       2
                                                        2
                                                       a + c 2        a + c 2
                        Hence
                                               1              s               s
                                b = t × n = √        ci sin √      − cj cos √      + ak
                                                                             2
                                                             2
                                              2
                                             a + c 2        a + c 2         a + c 2
                        and
                                         c            s              s            c

                                 b =          i cos √      + j sin √       =−         n.
                                                                                2
                                       2
                                                                    2
                                      a + c 2       a + c 2        a + c 2     a + c 2
                                                     2
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