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.
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