Mathematical Methods for Physics and Engineering : A Comprehensive Guide

VECTOR CALCULUS

y

x

φ

ˆeφ

eˆρ

ρ

i

j

Figure 10.2 Unit basis vectors for two-dimensional Cartesian and plane polar

coordinates.

Thespeedoftheparticleatt= 1 is simply

|v(1)|=

√

42 +3^2 +6^2 =

√

61.

The acceleration of the particle is constant (i.e. independent oft), and its component in
the directionsis given by

a·ˆs=

(4i+6k)·(i+2j+k)

√

12 +2^2 +1^2

=

5

√

6

3

.

Note that in the case discussed abovei,jandkare fixed, time-independent

basis vectors. This may not be true of basis vectors in general; when we are

not using Cartesian coordinates the basis vectors themselves must also be dif-

ferentiated. We discuss basis vectors for non-Cartesian coordinate systems in

detail in section 10.10. Nevertheless, as a simple example, let us now consider

two-dimensional plane polar coordinatesρ, φ.

Referring to figure 10.2, imagine holdingφfixed and moving radially outwards,

i.e. in the direction of increasingρ. Let us denote the unit vector in this direction

byeˆρ. Similarly, imagine keepingρfixed and moving around a circle of fixed radius

in the direction of increasingφ. Let us denote the unit vector tangent to the circle

byˆeφ. The two vectorsˆeρandeˆφare the basis vectors for this two-dimensional

coordinate system, just asiandjare basis vectors for two-dimensional Cartesian

coordinates. All these basis vectors are shown in figure 10.2.

An important difference between the two sets of basis vectors is that, while

iandjare constant in magnitudeand direction, the vectorseˆρ andˆeφhave

constant magnitudes but their directions change asρandφvary. Therefore,

when calculating the derivative of a vector written in polar coordinates we must

also differentiate the basis vectors. One way of doing this is to expressˆeρandˆeφ