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.