Mathematical Methods for Physics and Engineering : A Comprehensive Guide

VECTOR CALCULUS

x

z

y

C

O

ˆb

ˆt

nˆ

P

r(u)

Figure 10.3 The unit tangentˆt, normalˆnand binormalbˆto the space curveC

at a particular pointP.

since the first term is zero by (10.10), and the second is zero because it is the vector product
of two parallel (in this case identical) vectors. Integrating, we obtain the required result

r×

dr

dt

=c, (10.11)

wherecis a constant vector.
As a further point of interest we may note that in an infinitesimal timedtthe change
in the position vector of the small mass isdrand the element of area swept out by the
position vector of the particle is simplydA=^12 |r×dr|. Dividing both sides of this equation
bydt, we conclude that

dA

dt

=

1

2

∣∣

∣

∣r×

dr

dt

∣∣

∣

∣=

|c|

2

,

and that the physical interpretation of the above result (10.11) is that the position vectorr
of the small mass sweeps out equal areas in equal times. This result is in fact valid for
motion under any force that acts along the line joining the two particles.

10.3 Space curves

In the previous section we mentioned that the velocity vector of a particle is a

tangent to the curve in space along which the particle moves. We now give a more

complete discussion of curves in space and also a discussion of the geometrical

interpretation of the vector derivative.

AcurveCin space can be described by the vectorr(u) joining the originOof

a coordinate system to a point on the curve (see figure 10.3). As the parameteru

varies, the end-point of the vector moves along the curve. In Cartesian coordinates,

r(u)=x(u)i+y(u)j+z(u)k,

wherex=x(u),y=y(u)andz=z(u)aretheparametricequations of the curve.