Differentiation of vectors

10
Vector calculus
In chapter 7 we discussed the algebra of vectors, and in chapter 8 we considered
how to transform one vector into another using a linear operator. In this chapter
and the next we discuss the calculus of vectors, i.e. the differentiation and
integration both of vectors describing particular bodies, such as the velocity of
a particle, and of vector fields, in which a vector is defined as a function of the
coordinates throughout some volume (one-, two- or three-dimensional). Since the
aim of this chapter is to develop methods for handling multi-dimensional physical
situations, we will assume throughout that the functions with which we have to
deal have sufficiently amenable mathematical properties, in particular that they
are continuous and differentiable.
10.1 Differentiation of vectors
Let us consider a vector a that is a function of a scalar variable u. By this
we mean that with each value of u we associate a vector a(u). For example, in
Cartesian coordinates a(u) = ax (u)i + ay (u)j + az (u)k, where ax (u), ay (u) and az (u)
are scalar functions of u and are the components of the vector a(u) in the x-, yand z- directions respectively. We note that if a(u) is continuous at some point
u = u0 then this implies that each of the Cartesian components ax (u), ay (u) and
az (u) is also continuous there.
Let us consider the derivative of the vector function a(u) with respect to u.
The derivative of a vector function is defined in a similar manner to the ordinary
derivative of a scalar function f(x) given in chapter 2. The small change in
the vector a(u) resulting from a small change ∆u in the value of u is given by
∆a = a(u + ∆u) − a(u) (see figure 10.1). The derivative of a(u) with respect to u is
defined to be
a(u + ∆u) − a(u)
da
= lim
,
du ∆u→0
∆u
334
(10.1)
10.1 DIFFERENTIATION OF VECTORS
∆a = a(u + ∆u) − a(u)
a(u + ∆u)
a(u)
Figure 10.1 A small change in a vector a(u) resulting from a small change
in u.
assuming that the limit exists, in which case a(u) is said to be differentiable at
that point. Note that da/du is also a vector, which is not, in general, parallel to
a(u). In Cartesian coordinates, the derivative of the vector a(u) = ax i + ay j + az k
is given by
dax
day
daz
da
=
i+
j+
k.
du
du
du
du
Perhaps the simplest application of the above is to finding the velocity and
acceleration of a particle in classical mechanics. If the time-dependent position
vector of the particle with respect to the origin in Cartesian coordinates is given
by r(t) = x(t)i + y(t)j + z(t)k then the velocity of the particle is given by the vector
v(t) =
dx
dy
dz
dr
=
i + j + k.
dt
dt
dt
dt
The direction of the velocity vector is along the tangent to the path r(t) at the
instantaneous position of the particle, and its magnitude |v(t)| is equal to the
speed of the particle. The acceleration of the particle is given in a similar manner
by
a(t) =
d2 x
d2 y
d2 z
dv
= 2 i + 2 j + 2 k.
dt
dt
dt
dt
The position vector of a particle at time t in Cartesian coordinates is given by r(t) =
2t2 i + (3t − 2)j + (3t2 − 1)k. Find the speed of the particle at t = 1 and the component of
its acceleration in the direction s = i + 2j + k.
The velocity and acceleration of the particle are given by
dr
= 4ti + 3j + 6tk,
dt
dv
= 4i + 6k.
a(t) =
dt
v(t) =
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VECTOR CALCULUS
y
êφ
j
êρ
i
ρ
φ
x
Figure 10.2 Unit basis vectors for two-dimensional Cartesian and plane polar
coordinates.
The speed of the particle at t = 1 is simply
|v(1)| =
42 + 32 + 62 =
√
61.
The acceleration of the particle is constant (i.e. independent of t), and its component in
the direction s is given by
a · ŝ =
√
(4i + 6k) · (i + 2j + k)
5 6
√
=
.
3
12 + 22 + 12
Note that in the case discussed above i, j and k are 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 differentiated. 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 coordinates ρ, φ.
Referring to figure 10.2, imagine holding φ fixed and moving radially outwards,
i.e. in the direction of increasing ρ. Let us denote the unit vector in this direction
by êρ . Similarly, imagine keeping ρ fixed and moving around a circle of fixed radius
in the direction of increasing φ. Let us denote the unit vector tangent to the circle
by êφ . The two vectors êρ and êφ are the basis vectors for this two-dimensional
coordinate system, just as i and j are 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
i and j are constant in magnitude and direction, the vectors êρ and êφ have
constant magnitudes but their directions change as ρ and φ vary. 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 express êρ and êφ
336
10.1 DIFFERENTIATION OF VECTORS
in terms of i and j. From figure 10.2, we see that
êρ = cos φ i + sin φ j,
êφ = − sin φ i + cos φ j.
Since i and j are constant vectors, we find that the derivatives of the basis vectors
êρ and êφ with respect to t are given by
dφ
dφ
dêρ
= − sin φ
i + cos φ
j = φ̇ êφ ,
dt
dt
dt
dφ
dφ
dêφ
= − cos φ
i − sin φ
j = −φ̇ êρ ,
dt
dt
dt
(10.2)
(10.3)
where the overdot is the conventional notation for differentiation with respect to
time.
The position vector of a particle in plane polar coordinates is r(t) = ρ(t)êρ . Find expressions for the velocity and acceleration of the particle in these coordinates.
Using result (10.4) below, the velocity of the particle is given by
v(t) = ṙ(t) = ρ̇ êρ + ρ ˙êρ = ρ̇ êρ + ρφ̇ êφ ,
where we have used (10.2). In a similar way its acceleration is given by
d
(ρ̇ êρ + ρφ̇ êφ )
dt
= ρ̈ êρ + ρ̇ ˙êρ + ρφ̇ ˙êφ + ρφ̈ êφ + ρ̇φ̇ êφ
a(t) =
= ρ̈ êρ + ρ̇(φ̇ êφ ) + ρφ̇(−φ̇ êρ ) + ρφ̈ êφ + ρ̇φ̇ êφ
= (ρ̈ − ρφ̇2 ) êρ + (ρφ̈ + 2ρ̇φ̇) êφ . Here we have used (10.2) and (10.3).
10.1.1 Differentiation of composite vector expressions
In composite vector expressions each of the vectors or scalars involved may be
a function of some scalar variable u, as we have seen. The derivatives of such
expressions are easily found using the definition (10.1) and the rules of ordinary
differential calculus. They may be summarised by the following, in which we
assume that a and b are differentiable vector functions of a scalar u and that φ
is a differentiable scalar function of u:
da dφ
d
(φa) = φ +
a,
du
du
du
d
db da
(a · b) = a ·
+
· b,
du
du du
d
db da
(a × b) = a ×
+
× b.
du
du du
337
(10.4)
(10.5)
(10.6)
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.
A particle of mass m with position vector r relative to some origin O experiences a force
F, which produces a torque (moment) T = r × F about O. The angular momentum of the
particle about O is given by L = r × mv, where v is 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
d
dL
= (r × mv).
dt
dt
Using (10.6) we obtain
dr
d
dL
=
× mv + r × (mv)
dt
dt
dt
d
= v × mv + r × (mv)
dt
= 0 + r × F = T,
where in the last line we use Newton’s second law, namely F = d(mv)/dt. If a vector a is a function of a scalar variable s that is itself a function of u, so
that s = s(u), then the chain rule (see subsection 2.1.3) gives
da(s)
ds da
=
.
(10.7)
du
du ds
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
da
d
(a · a) = 2a · ;
du
du
since a · a = a2 , where a = |a|, we see that
da
= 0 if a is constant.
(10.8)
du
In other words, if a vector a(u) has a constant magnitude as u varies then it is
perpendicular to the vector da/du.
a·
10.1.2 Differential of a vector
As a final note on the differentiation of vectors, we can also define the differential
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 ∆a in a vector a(u) resulting from a small change ∆u in its argument. In
the limit ∆u → 0, the change in a becomes infinitesimally small, and we denote it
by the differential da. From (10.1) we see that the differential is given by
da =
da
du.
du
338
(10.9)