8. Curves in Rn

These lecture notes present my interpretation of Ruth Lawrence’s lec-ture notes (in Hebrew)

8.1 Definitions

If the previous chapters we dealt with functions whose domains and ranges wereeither R or subsets of R. Functions may be defined between any two sets. In manyapplications, both domain and range are vector spaces. In a later chapter we willstudy functions Rm→Rn. This chapter is devoted to functions R→Rn.

A function f ∶R→Rn gets for input a real number and returns a vector. Its action isof the form

x� f(x) =�����

f1(x)f2(x)⋮f

n

(x)�����.

Note that each f

j

is a function R→R, so that:

A function R→Rn can be represented as a column of n real-valued functions

� Example 8.1 The trajectory ( �-&-2/) of a point particle in Euclidean space,where space is parametrized with Cartesian coordinates, is an example of a functionR→R3. In this case it is customary to denote the independent variable by t, andwrite

r(t) = ���x(t)y(t)z(t)��� .

114 Curves in Rn

�� Example 8.2 The trajectory of a point trajectory in Euclidean space, where spaceis parametrized with spherical coordinates, is another example of a function R→R3.In this case,

f(t) = ���r(t)q(t)f(t)��� .

Note that if both examples refer to the same trajectory, then we know the relationbetween r(t) and f(t). �

Functions R→Rn are called paths (�;&-*2/) because their image is a one-dimensionalcurve in n-dimensional space. Since in many applications the independent variablerepresents time, it is suggestive to denote it by t; also, it is customary to denote thederivative by a dot rather than by a prime.

8.1.1 The derivative of a path

Let f ∶R→Rn be a path. Can we make sense of its derivative at t? Let’s try to followthe standard definition of the derivative:

f(t) = limh→0

f(t +h)− f(t)h

.

Does it make sense?

Let’s examine the right-hand side more carefully:

limh→0

f(t +h)− f(t)h

= limh→0

1h

�����f1(t +h)− f1(t)f2(t +h)− f2(t)⋮f

n

(t +h)− f

n

(t)�����= lim

h→0

������

f1(t+h)− f1(t)h

f2(t+h)− f2(t)h⋮

f

n

(t+h)− f

n

(t)h

������.

If each of the n-functions f

j

has a derivative at t, then as h→ 0, the j-th entry of thevector on the right-hand side tends to f

′j

(t), namely,

f(t) =�����

f1(t)f2(t)⋮f

n

(t)�����.

� Example 8.3 Consider a fly moving on the Euclidean plane parametrized withCartesian coordinates,

r(t) = �t2

t

3� .

8.1 Definitions 115

The derivative of its trajectory is

r(t) = � 2t

3t

2� .What is the meaning of this derivative? Each entry of the derivative quantifies therate of change at time t of that coordinate. The derivative r(t) is called the velocity

vector of that fly. As a vector, it has a magnitude,

v(t) = �r(t)� =√4t

2+9t

4.

This is really the magnitude of the velocity as

v(t) = limh→0

1h

�r(t +h)−r(t)�.(Check that this is indeed the case.) �� Example 8.4 Consider another motion of that same fly, but this time, Euclideanspace is parametrized by polar coordinates,

f(t) = �r(t)q(t)� = �3t� .

The derivative of this path is

f(t) = �01� .

Here too, each entry of the derivative quantifies the rate of change at time t of thatcoordinate. But recall that polar coordinates are not Euclidean vectors. We cancertainly calculate the “magnitude" of the derivative

√02+12 = 1,

but this is not the velocity of the fly! In fact, we know what the velocity is: it is

2p ⋅32p= 3.

We can also see it by switching to Euclidean coordinates:

r(t) = �3 cost

3 sint

� ,getting

r(t) = �−3 sint

3 cost

� and �r(t)� = 3.

�

116 Curves in Rn

Proposition 8.1 Let u(t),v(t),w(t) be paths in Rn, then

1. (u ⋅v)′ = u ⋅v+u ⋅ v.2. (u×v)′ = u×v+u× v (in R3).3. [u,v,w]′ = [u,v,w]+ [u, v,w]+ [u,v,w].

� Example 8.5 Let r(t) be the trajectory of a particle in Cartesian coordinates.Then, r(t) = �r(t)� is its distance from the origin. The derivative of this distance is

r(t) = d

dt

�r(t) ⋅r(t) = r(t) ⋅r(t)+r(t) ⋅ r(t)

2r(t) = r(t) ⋅r(t)r(t) .

In particular, if r(t) ⊥ r(t) then the rate of change of the distance from the origin iszero. �� Example 8.6 Let r(t) denote the trajectory of a charged particle moving underconstant electric and magnetic fields, E and B. Its velocity vector is v(t) = r(t) andits acceleration vector is a(t) = v(t) = r(t). The force f experienced by the particleis

f(t) = qE+qv(t)×B,

where q is its charge. By Newton’s second law,

m v(t) = qE+qv(t)×B,

where m is its mass. Take the scalar product of both sides with v(t). The last term isa vanishing triple-product. Hence,

m v(t) ⋅v(t) = qE ⋅ r(t).We can rewrite this as follows,

d

dt

�12

mv(t) ⋅v(t)−qE ⋅r(t)� = 0.

That is, the quantity

E(t) = 12

m �v(t)�2−qE ⋅r(t)does not change in time. �� Example 8.7 Consider the previous example, this time with E = 0. Then,

v(t) = q

m

v(t)×B = �− q

m

B�×v(t).Since the orientation of the coordinates is arbitrary, we may as well choose it so thatB is along the z-axis. Suppose that

− q

m

B = ���00w

��� .

8.2 The length of a curve in Rn 117

Then, ���v1(t)v2(t)v3(t)

��� =���

00w

������

v1(t)v2(t)v3(t)

��� =���−w v2(t)w v1(t)

0

��� .You can check directly that

���v1(t)v2(t)v3(t)

��� =���

a coswt +b sinwt

a sinw −b coswt

c

��� ,where a,b,c are constants of integration. Since v(t) = r(t), it follows that

���r1(t)r2(t)r3(t)

��� =���

a

w sinwt − b

w coswt +A− a

w cosw − b

w sinwt +B

ct +C

��� ,where A,B,C are also constants of integration. This is a helical motion (%3&1;�;*#9&"). �

8.2 The length of a curve in Rn

Consider a curve/trajectory in Rn,

r ∶ [a,b]→Rn.

If Rn represent the Euclidean space in Cartesian coordinates, then the velocity ofthe trajectory is

v(t) = �r(t)�.The length of curve/trajectory is

` =� b

a

v(t)dt =� b

a

�r(t)�dt.

� Example 8.8 Consider the curve

r(t) = ���t sint

t cost

t

��� .Its velocity is

v(t) =�����������������

t cost + sint−t sint +cost

1

�����������������=√t

2+2.

Between t = a and t = b,

` =� b

a

√t

2+2dt.

�Comment 8.1 Richardson’s measurement of the length of the coast of Britain.