Mathematical Methods for Physics and Engineering : A Comprehensive Guide

VECTOR CALCULUS

The order of the factors in the terms on the RHS of (10.6) is, of course, just as

important as it is in the original vector product.

Aparticleofmassmwith position vectorrrelative to some originOexperiences a force

F, which produces a torque (moment)T=r×FaboutO. The angular momentum of the

particle aboutOis given byL=r×mv,wherevis the particle’s velocity. Show that the

rate of change of angular momentum is equal to the applied torque.

The rate of change of angular momentum is given by

dL

dt

=

d

dt

(r×mv).

Using (10.6) we obtain

dL

dt

=

dr

dt

×mv+r×

d

dt

(mv)

=v×mv+r×

d

dt

(mv)

= 0 +r×F=T,

where in the last line we use Newton’s second law, namelyF=d(mv)/dt.

If a vectorais a function of a scalar variablesthat is itself a function ofu,so

thats=s(u), then the chain rule (see subsection 2.1.3) gives

da(s)

du

=

ds

du

da

ds

. (10.7)

The derivatives of more complicated vector expressions may be found by repeated

application of the above equations.

One further useful result can be derived by considering the derivative

d

du

(a·a)=2a·

da

du

;

sincea·a=a^2 ,wherea=|a|,weseethat

a·

da

du

=0 ifais constant. (10.8)

In other words, if a vectora(u) has a constant magnitude asuvaries then it is

perpendicular to the vectorda/du.

10.1.2 Differential of a vector

As a final note on the differentiation of vectors, we can also define thedifferential

of a vector, in a similar way to that of a scalar in ordinary differential calculus.

In the definition of the vector derivative (10.1), we used the notion of a small

change ∆ain a vectora(u) resulting from a small change ∆uin its argument. In

the limit ∆u→0, the change inabecomes infinitesimally small, and we denote it

by the differentialda. From (10.1) we see that the differential is given by

da=

da

du

du. (10.9)