Page 186 - Schaum's Outline of Theory and Problems of Advanced Calculus
P. 186

CHAP. 7]                             VECTORS                                    177


                           (b)From the result in part (a), we have
                                                            1
                                                                     1
                                                   2 1   d  2      2 d      2  2

                                                     r   dr  r  r dr r   r  r
                                                  r    ¼  2    þ       ¼  3     3  ¼ 0
                              showing that 1=r is a solution of Laplace’s equation.

                     7.45. A particle moves along a space curve r ¼ rðtÞ, where t is the time measured from some initial time.
                           If v ¼jdr=dtj¼ ds=dt is the magnitude of the velocity of the particle (s is the arc length along the
                           space curve measured from the initial position), prove that the acceleration a of the particle is
                           given by

                                                              dv    v 2
                                                                      N
                                                              dt
                                                           a ¼  T þ
                           where T and N are unit tangent and normal vectors to the space curve and

                                                         8                     9  1=2
                                                      1        !      !      !
                                                                2      2       2
                                                            d x    d y     d z
                                                     2    <  2      2       2  =
                                                     d r
                                                    2         2      2       2
                                                ¼      ¼         þ      þ
                                                    ds    :  ds     ds     ds  ;
                              The velocity of the particle is given by v ¼ vT.  Then the acceleration is given by
                                             dv  d      dv    dT   dv    dT ds  dv   2 dT
                                                          T þ v     T þ v        T þ v
                                          a ¼  ¼   ðvTÞ¼         ¼           ¼                        ð1Þ
                                             dt  dt     dt     dt  dt    ds dt  dt    ds
                              Since T has a unit magnitude, we have T   T ¼ 1.  Then differentiating with respect to s,
                                              dT   dT            dT              dT
                                                       T ¼ 0;       ¼ 0   or       ¼ 0
                                               ds  ds             ds             ds
                                            T    þ            2T               T
                           from which it follows that dT=ds is perpendicular to T.Denoting by N the unit vector in the direction of
                           dT=ds, and called the principal normal to the space curve, we have

                                                              dT
                                                                ¼  N
                                                              ds                                      ð2Þ
                           where   is the magnitude of dT=ds.  Now since T ¼ dr=ds [see equation (7), Page 157], we have
                                      2
                                  2
                           dT=ds ¼ d r=ds .  Hence
                                                          8                   9 1=2
                                                               ! 2   ! 2    ! 2

                                                            d x    d y    d z
                                                       2    <  2    2      2  =
                                                       d r
                                                      2       2      2     2
                                                   ¼      ¼      þ     þ
                                                       ds    :  ds  ds    ds  ;
                              Defining   ¼ 1= ,(2)becomes dT=ds ¼ N= .  Thus from (1)we have, as required,
                                                               dv   v 2
                                                                     N
                                                           a ¼  T þ
                                                               dt
                                                    2
                              The components dv=dt and v =  in the direction of T and N are called the tangential and normal
                           components of the acceleration, the latter being sometimes called the centripetal acceleration. The quantities
                             and   are respectively the radius of curvature and curvature of the space curve.
   181   182   183   184   185   186   187   188   189   190   191