Parametrized Curves
Will Rosenbaum
Updated: November 2, 2014
Department of Mathematics
University of California, Los Angeles
In this essay, we consider curves in Rn described by parametric equations. In the first
section, we define the basic terminology concerning parametrized curves, in particular the
notions of continuity, differentiability and smoothness. In the second section, we study the
geometry of curves. We define what it means for two parametrizations to be equivalent and
define notions of arc length, curvature and torsion of a curve.
1 Parametrized Curves
1.1 Basic definitions Let I ⊆ R be an interval and n a natural number. A parametrized curve
is a function
x : I −→ Rn .
Writing x(t) in terms of its coordinates in Rn , we have
x(t) = (x1 (t), x2 (t), . . . , xn (t)) for
t∈I
where
xi : I → R for
i = 1, 2, . . . , n
are the coordinate functions of x.
Example 1. Consider I = [0, 2π], n = 2 and
x(t) = (sin t, cos t).
This parametrized curve is the unit circle in R2 . For any t ∈ [0, 2π], we compute the distance
from x(t) to the origin
p
p
|x(t)| = hx(t), x(t)i = sin2 t + cos2 t = 1.
Thus, the range of x does indeed lie on the unit circle.
1.2 Continuity Recall that for a function f : I → R and t0 ∈ I, we say that limt→t0 f (t) = L
if for every ε > 0, there exists δ > 0 such that
|t − t0 | < δ =⇒ |f (t) − L| < ε.
We say f is continuous at a point t0 ∈ I if limt→t0 f (t) = f (t0 ). Informally, a function is continuous if input values which are close together map to outputs which are correspondingly close. The concepts of limits and continuity easily generalize to the context of
parametrized curves.
Definition 2. Let x : I → Rn be a parametrized curve, and t0 ∈ I. We say that the limit
as t approaches t0 of x is v and write
lim x(t) = v
t→t0
Parametrized Curves
if for every ε > 0 there exists δ > 0 such that
0 < |t − t0 | < δ =⇒ |x(t) − v| < ε.
We say that x is continuous at t0 if limt→t0 x(t) = x(t0 ). We say that x is continuous on I
if x is continuous at all t0 ∈ I.
Notice that |x(t) − v| in the above expression is a vector length. If
x(t) = (x1 (t), x2 (t), . . . , xn (t))
then
|x(t) − v| =
and
v = (v1 , v2 , . . . , vn )
p
(x1 (t) − v1 )2 + (x2 (t) − v2 )2 + · · · + (xn (t) − vn )2 .
This observation leads to the following proposition, which will be helpful in proving the
continuity of parametrized curves.
Proposition 3 (Limits are component-wise). Let x = (x1 , x2 , . . . , xn ) : I → Rn be a
parametrized curve. Then limt→t0 x(t) = v if and only if limt→t0 xi (t) = vi for all i =
1, 2, . . . , n. In particular, x is continuous if and only if the xi are continuous for all i.
Proof. ⇒: Suppose limt→t0 x(t) = v. Then given any ε > 0, there exists a δ > 0 such that
|t − t0 | < δ implies |x(t) − v| < ε. Notice that for all i = 1, 2, . . . , n,
p
|xi (t) − vi | ≤ ((x1 (t) − v1 )2 + · · · + (xi (t) − vi )2 + · · · + (xn (t) − vn )2 .
Thus, for |t − t0 | < δ, we have |xi (t) − vi | < ε. Therefore, limt→t0 xi (t) = vi .
⇐: Suppose each xi satisfies limt→t0 xi (t) = vi . Given ε > 0, there exist δ1 , δ2 , . . . , δn >
0 such that
|t − t0 | < δi =⇒ |xi (t) − vi | < ε/n.
Taking δ = min {δ1 , δ2 , . . . , δn }, |t − t0 | < δ implies
p
|x(t) − v| = (x1 (t) − v1 )2 + (x2 (t) − v2 )2 + · · · + (xn (t) − vn )2
p
< (ε/n)2 + (ε/n)2 + · · · + (ε/n)2
= ε.
Therefore, limt→t0 x(t) = v.
Example 4. We will show that the curve x : R → R2 given by
x(t) = (1 + t2 , 2t)
is continuous on R. By Proposition 3, it suffices to show that the coordinate functions
x1 (t) = 1 + t2 and x2 (t) = 2t are continuous on R. Let’s start with x2 (t).
Suppose ε > 0 is given. We compute
|x2 (t) − x2 (t0 )| = |2t − 2t0 | = 2 |t − t0 | .
Therefore, choosing δ = ε/2, we have
|t − t0 | < δ =⇒ |x2 (t) − x2 (t0 )| = 2 |t − t0 | < 2ε/2 = ε
2
Parametrized Curves
as desired.
Now for the continuity of x1 . First we compute
|x1 (t) − x1 (t0 )| = (1 + t2 ) − (1 + t20 ) = t2 − t20 = |t − t0 | |t + t0 | .
If |t − t0 | < 1, then |t + t0 | ≤ 2t0 + 1, hence for δ < 1,
|t − t0 | < δ =⇒ |x1 (t) − x1 (t0 )| < δ(2t0 + 1).
Thus, given ε > 0 and t0 ∈ R, choose
δ = min
ε
,1 .
2t0 + 1
Then the above calculation shows that for this choice of δ, |t − t0 | < δ implies |x1 (t) − x1 (t0 )| <
ε, hence x1 is continuous at t0 . Since x1 and x2 are both continuous at t0 for all t0 ∈ R, x
is continuous on R by Proposition 3.
Exercise 5. Prove that the parametrized curve x : (0, ∞) → R2 given by
1 3
,t
x(t) =
t
is continuous.
Proposition 6. Suppose x, y : I → Rn are continuous and c ∈ R. Then
(a) x + y is continuous;
(b) cx is continuous;
(c) hx, yi is continuous;
(d) if n = 3, x × y is continuous.
Exercise 7. Prove Proposition 6. You may use familiar facts about continuous functions
from I ⊆ R to R such as: the sum of continuous functions is continuous; the product of
continuous functions is continuous; the composition of continuous functions is continuous.
1.3 Differentiability Recall that a function f : I → R is differentiable at t0 ∈ I with derivative f ′ (t0 ) if
f (t) − f (t0 )
lim
= f ′ (t0 ).
t→t0
t − t0
This expression easily generalizes to parametrized curves.
Definition 8. Suppose x : I → Rn is a parametrized curve where I = [a, b] is a closed,
bounded interval in R. We say that x is differentiable at t0 ∈ (a, b) and has derivative
x′ (t0 ) if
1
lim
(x(t) − x(t0 )) = x′ (t0 ).
t→t0 t − t0
If x′ (t) exists for all t ∈ (a, b), we say that x is differentiable on (a, b). We will alternatively
denote
d
x(t) = x′ (t).
dt
3
Parametrized Curves
We may compute higher order derivatives, defined inductively by
x(k) (t) =
d (k−1)
x
(t),
dt
assuming these derivatives exist. If the first k derivatives of x exist and are continuous, we
say that x is k-times continuously differentiable.
Definition 9. Let I = [a, b] be a closed bounded interval in R. We say that a parametrized
curve x : I → Rn is smooth if x is continuous on [a, b], (once) continuously differentiable
on (a, b), and x′ (t) 6= 0 for all t ∈ (a, b).
An immediate consequence of Proposition 3 is the the following.
Proposition 10 (Differentiation is component-wise). Let x : I → Rn with
x(t) = (x1 (t), x2 (t), . . . , xn (t)).
Then x is differentiable at t0 ∈ I if and only if each xi is differentiable at t0 . Furthermore,
if x is differentiable at t0 then
x′ (t0 ) = (x′1 (t0 ), x′2 (t0 ), . . . , x′n (t0 )).
Proposition 11 (Properties of differntiable functions). Suppose x, y : I → Rn , f : I → R,
and g : J → I are differentiable. Then
(a)
d
dt (x(t)
(b)
d
dt (f (t)x(t))
(c)
d
dt x(g(t))
(d)
d
dt
+ y(t)) = x′ (t) + y′ (t)
= f ′ (t)x(t) + f (t)x′ (t)
= g ′ (t)x′ (g(t))
hx(t), y(t)i = hx′ (t), y(t)i + hx(t), y′ (t)i.
Exercise 12. Prove Proposition 11. By Proposition 10 it suffices to prove the identities componentwise. To this end, you may appeal to the familiar properties of ordinary differentiation: the
product rule and chain rule.
Exercise 13. Suppose x : I → Rn is differentiable on I. Prove that x is continuous on I.
Exercise 14. Suppose x : [a, b] → Rn is differentiable on (a, b), and that x does not pass
through the origin (x(t) 6= 0 for all t ∈ I). Suppose t0 ∈ (a, b) minimizes the distance of
x(t) to the origin. That is,
|x(t0 )| ≤ |x(t)|
for all
t ∈ I.
Show that x(t0 ) and x′ (t0 ) are orthogonal.
1.4 Tangent lines One interpretation of the derivative x′ (t0 ) of a parametrized curve x : I →
Rn is that it gives a tangent vector to x at the point x(t0 ). So long as x′ (t0 ) 6= 0, we define
the tangent line of x through x(t0 ) to be
L = {x(t0 ) + tx′ (t0 ) | t ∈ R} .
4
Parametrized Curves
Exercise 15. Find the tangent line to the helix given by
x(t) = (cos t, sin t, t)
through the point (1, 0, 0).
Problem 16. Consider the unit circle parametrized by x(t) = (cos t, sin t) for t ∈ R.
(a) Show that x(t) and x′ (t) are orthogonal for all t ∈ R.
(b) Show that every point v ∈ R2 with |v| > 1 lies on some tangent line L for x.
(Hint: for part (b), by symmetry, it suffices to consider v of the form v = (a, 0) for a ≥ 1.
It might be helpful to draw a picture.)
1.5 Physical interpretation of the derivative The derivative is indispensable to physics, and indeed much of the early development of calculus was motivated by physical problems. Classically, we think of physical space as three dimensional, modeled by R3 . A parametrized
curve x(t) : I → R3 then can represent the trajectory of some particle as a function of
time, t ∈ I. The derivative of x gives the velocity of the particle:
v(t) = x′ (t) for all
t∈I
and the derivative of velocity is acceleration
a(t) = v′ (t) = x′′ (t)
for all
t ∈ I.
The length of the velocity vector is the speed, which we denote by v(t) = |v(t)|.
Exercise 17 (Uniform circular motion). Suppose a particle experiences uniform circular motion in the plane:
x(t) = (r cos(αt), r sin(αt))
where α ∈ R is a constant and r > 0 is the radius of the circle. Show that the acceleration
a(t) is parallel to x(t) for all t ∈ I and that
|a(t)| =
v 2 (t)
.
r
Exercise 18. Suppose a particle has trajectory given by x : I → R3 , and suppose the speed
of the particle is constant: v(t) = c for all t ∈ I. Show that v(t) and a(t) are orthogonal for
all t ∈ I.
The mechanics of classical particles are governed by Newton’s second law:
F(x(t), v(t), t) = ma(t).
(1)
Here x(t) describes the trajectory of a particle, m its mass, and F the forces acting upon the
particle. Since (1) is an equation that relates x and its first and second derivatives, we call
(1) a second order differential equation. Solving such equations is generally difficult, so we
will not go into great depth here.
5
Parametrized Curves
Exercise 19. According to Newton’s universal law of gravitation, a particle of mass m orbiting
a single large mass experiences force∗
∗ ignoring some physical
constants
F(x(t), v(t), t) = −m
x(t)
|x(t)|
3.
Show that for any a > 0,
t
t
x(t) = r cos 3/2 , r sin 3/2
r
r
is a solution to (1).
Remark 20. In the previous exercise, the period of the orbit (the amount of time it takes the
particle to return to its original position) is T = 2πr3/2 , while the radius of the orbit is r.
Thus, the quantity
T2
= 4π 2
r3
is a constant, independent of r. This observation is a special case of Kepler’s third law of
planetary motion.
2
Geometry of Curves
In this section, we analyze geometric properties of curves in Rn . Although we describe
and define curves via parametrizations, geometric properties of a curve are those properties
which do not depend on the particular choice of parametrization. For example, we may
consider the open half circle in R2 given by
S + = v = (v1 , v2 ) ∈ R2 |v| = 1, v2 > 0 .
We can parametrize this curve many different ways, for example
x(t) = (cos t, sin t) for
t ∈ (0, π)
and
y(t) = (−t,
p
1 − t2 ) for
t ∈ (−1, 1).
Although these parametrizations appear very different, they still describe the same geometric object, S + . Thus, if we define a formula to compute a geometric quantity—for example
arc-length or curvature—we must ensure that the formula gives the same answer for x, y,
and indeed any other parametrization for S + .
Throughout this section, we will assume that I = [a, b] is a closed and bounded interval. All parametrized curves x : I → Rn in this section will be assumed to be smooth:
continuous on I, differentiable on (a, b) and |x′ (t)| > 0 for all t ∈ (a, b).
2.1 Equivalence of parametrizations We describe what it means for two parametrizations to
be equivalent in the sense that the represent the same geometric object. As alluded to above,
we will use this notion of equivalence to ensure that geometric formulas are well-defined:
equivalent parametrizations must agree on the value of a geometric quantity.
6
Parametrized Curves
Definition 21. Let x : I → Rn and y : J → Rn be smooth parametrized curves. We say
that x and y are equivalent parametrizations and write x ∼ y if there exists a differentiable,
increasing bijection ϕ : I → J such that
x(t) = y(ϕ(t))
for all
t ∈ I.
Exercise 22. Prove that ∼ as defined above is an equivalence relation. That is, if x : I → Rn ,
y : J → Rn and z : K → Rn are parametrizations with x ∼ y and y ∼ z, then
(a) x ∼ x;
(b) y ∼ x;
(c) x ∼ z.
Exercise 23. Let S + be the open half circle in R2 described above,
x(t) = (cos t, sin t) for
and
y(t) = (−t,
Show that x ∼ y.
t ∈ (0, π),
p
1 − t2 ) for
t ∈ (−1, 1).
2.2 Arc length The first geometric property of a parametrized curve that we consider is its arc
length. Intuitively, if we were to make a model of our curve out of wire, say, the curve’s arc
length is the amount of wire we would need in order to make our model.
Definition 24. Let x : I → Rn be a parametrized curve, and t0 , t ∈ I with t0 < t. Then
the arc length of x from t0 to t is given by
L(t0 , t) = Lx (t0 , t) =
Z
t
|x′ (u)| du.
t0
Example 25. Consider the circle of radius r parametrized by x(t) = (r cos t, r sin t) for
t ∈ [0, 2π]. We can compute the arc length of x by
Z
2π
2π
=
Z
2π
=
Z
L(0, 2π) =
|x′ (u)| du
0
0
p
r2 sin2 u + r2 cos2 u du
r du
0
= 2πr.
Thus, as one would hope, the arc length (i.e., circumference) of the circle of radius r is 2πr.
Exercise 26. Consider the helix given by x(t) = (cos t, sin t, t) for t ∈ [0, 4π]. Find the arc
length of x.
In order to be sure that the formula given in Definition 24 represents a bona fide geometric quantity, we must ensure that equivalent parametrizations have the same arc length.
7
Parametrized Curves
Proposition 27. Let I = [a, b] and J = [c, d]. Suppose x : I → Rn and y : J → Rn are
equivalent parametrizations. Then x and y have the same arc length.
Proof. Since x and y are equivalent, there exists an increasing, differentiable bijection ϕ :
I → J such that x(t) = y(ϕ(t)) for all t ∈ I. Then we compute the arc length of y to be
Ly (c, d) =
Z
d
Z
ϕ(b)
Z
b
|y′ (u)| du
c
=
|y′ (u)| du
ϕ(a)
|y′ (ϕ(t))| |ϕ′ (t)| dt
Z b
d
=
dt y(ϕ(t)) dt
a
Z b
=
|x′ (t)| dt
=
(change of variables)
a
a
= Lx (a, b).
Thus, x and y give the same arc length.
2.3 Arc length parametrizations There are many different parametrizations that correspond
to the same geometric object. In this section, we describe how to define a unique “canonical” parametrization for a given smooth curve. This canonical representative will prove be
computationally convenient in many situations that follow.
Suppose x : I → Rn is a smooth parametrization. for some fixed t0 ∈ I, we can
compute the arc length of x from t0 to t as a function of t:
s(t) = Lx (t0 , t) =
Z
t
|x′ (u)| du.
t0
We call s as above an arc length parameter for x. Notice that since |x′ (u)| > 0 for all u ∈ I,
s is an increasing function for t ≥ t0 : t0 ≤ t1 < t2 implies that s(t1 ) < s(t2 ). Further, s is
differentiable by the fundamental theorem of calculus.
Problem 28. Suppose f : [a, b] → R is continuous on [a, b], continuously differentiable
on (a, b) and f ′ (t) > 0 for all t ∈ (a, b). Prove that f is invertible and its inverse f −1 is
continuously differentiable.
Applying the result Problem 28 to arc length parameters, we obtain the following result.
Proposition 29. For any smooth parametrization x : (a, b) → Rn , the arc length parameter
s : (t0 , b) → R is invertible and its derivative s−1 is continuously differentiable.
Definition 30. Suppose x : I → Rn is a smooth parametrization. We call x an arc length
parametrization if |x′ (s)| = 1 for all s ∈ I.
Proposition 31. A smooth parametrization x : I → Rn is an arc length parametrization if
and only if for all a, b ∈ I, a < b,
Lx (a, b) = b − a.
8
Parametrized Curves
Exercise 32. Prove Proposition 31.
Theorem 33. Let x : (a, b) → Rn be a smooth parametrization. Then for all t0 in (a, b),
there exists an arc length parametrization y : J → Rn with 0 ∈ J such that y is equivalent
to x and x(t0 ) = y(0). Further, any other arc length parametrization z : K → Rn
equivalent to x is of the form z(s) = y(s + c) for some constant c ∈ R.
Rt
Proof. The idea is to invert the arc length parameter s(t) = t0 |x′ (u)| du, so that we get
t = t(s). By Proposition 29 such a function t(s) exists and is continuously differentiable.
We will argue that y(s) = x(t(s)) is an arc length parametrization. To this end, we compute
d
′
|y (s)| = x(t(s))
ds
= |x′ (t(s))t′ (s)|
1
= |x′ (t(s))| ′
s (t)
1
= |x′ (t(s))| ′
|x (t(s))|
= 1.
(chain rule)
(FToC)
Thus, such an arc length parametrization exists. To prove the second assertion in the theorem, suppose z : K → Rn is another arc length parametrizations equivalent to x. Since
y and z are both equivalent to x, y and z are equivalent to each other, so there exists a
continuously differentiable, increasing function ϕ : J → K such that
y(s) = z(ϕ(s))
for all
s ∈ J.
Since y and z are arc length parametrizations, we compute
1 = |y′ (s)|
d
= z(ϕ(s))
ds
= |z′ (ϕ(s))ϕ′ (s)|
= |z′ (ϕ(s))| |ϕ′ (s)|
= |ϕ′ (s)| .
Since |ϕ′ (s)| = 1 for all s and ϕ′ (s) > 0, we conclude that ϕ(s) = s + c for all s and some
constant c ∈ R.
Exercise 34. Compute arc length parametrizations which are equivalent to the following
parametrized curves.
(a) The circle of radius r > 0: x(t) = (r cos t, r sin t) for t ∈ [0, 2π].
(b) The helix given by y(t) = (r cos t, r sin t, qt) where q, r ∈ R, q, r > 0.
Exercise 35. Consider a curve C of the form y = f (x) in the plane R2 where x ∈ [a, b].
Show that the arc length of C can be computed by the formula
Z bp
1 + (f ′ (x))2 dx.
L=
a
(Hint: how would you parametrize C?)
9
Parametrized Curves
2.4 Curves on a plane To further our understanding of the geometry of curves, we begin by
describing curves in the plane R2 , known as plane curves. In the previous sections we defined the arc length of any smooth parametrization in terms of a formula, and proceeded
to show that the formula is somehow independent of parametrization: equivalent curves
agree on the arc length. We then used arc length to define an arc length parametrization,
which gives a canonical choice of parametrization for each smooth curve. In this section, we
describe the curvature of plane curves. Instead of defining a curvature formula for an arbitrary parametrization, we first give a definition specifically for an arc length parametrization.
We then use this definition to find a formula that works for general parametrizations. This
approach still leads to a consistent definition of arc length for all parametrizations, as by
Theorem 33, arc length parametrization are essentially unique.
Arc length parametrizations
Let x : I → R2 be an arc length parametrization. We define the unit tangent vector for x
by
dx
T(s) =
.
ds
Since x is an arc length parametrization, we have |T(s)| = 1 for all s. Assuming T′ (s) 6= 0,
we define the unit normal vector to be
1
T′ (s).
N(s) = ′
|T (s)|
Exercise 36. Prove that hT(s), N(s)i = 0 for all s.
By the previous exercise and the definition of N, the set {T(s), N(s)} forms an orthonormal basis for R2 for all s (assuming T′ (s) 6= 0). We define the curvature κ(s) of x
at x(s) to be
dT .
κ(s) = ds Thus, by the definition of N,
T′ (s) = κ(s)N(s).
Exercise 37. Let C be the circle of radius r centered at the origin. Find an arc length parametrization of C, and use this parametrization to compute the curvature of C at each point.
Exercise 38. Suppose x, y : I → Rn satisfy hx(t), y(t)i = c for all t ∈ I for some constant
c. Show that
hx′ (t), y(t)i = − hx(t), y′ (t)i .
Since {T(s), N(s)} is an orthonormal basis for R2 for each value of s, we can compute
N′ (s) as a linear combination of T(s) and N(s) via
N′ (s) = hN′ (s), T(s)i T(s) + hN′ (s), N(s)i N(s).
Since |N(s)| = 1 is constant, the second term in the above expression is 0. On the other
hand, applying the result of the previous exercise,
N′ (s) = hN′ (s), T(s)i T(s) = − hN(s), T′ (s)i T(s) = −κ(s)T(s).
To summarize the relationship between T, N and their derivatives, we obtain the FrenetSerret formulas in R2 :
′ T
0
κ
T
=
N′
−κ 0
N
10
Parametrized Curves
General smooth parametrizations
In general, it may be inconvenient (or even impossible) to find an explicit arc length parametrization equivalent to a given smooth parametrization x : I → R2 . Thus, we would like a
formula for curvature which will work for any smooth parametrized curve. To this end,
Rt
suppose x : I → R2 is a smooth parametrization and s(t) = t0 |x′ (u)| du is an arc
length parameter. If y is an arc length parametrization for x, then y is characterized by the
expression
x = y ◦ s or equivalently x(t) = y(s(t)) for t ∈ I.
Differentiating twice with respect to t and applying the chain rule and product rule, we
obtain
x′ (t) = y′ (s(t))s′ (t) = s′ (t)T(t)
x′′ (t) = y′′ (s)(s′ (t))2 + y′ (s)s′′ (t) = (s′ (t))2 κ(t)N(t) + s′′ (t)T(t).
This equation is equivalent to the matrix equation
′ ′
x
s
0
1
=
x′′
s′′ (s′ )2
0
0
T
.
κ
N
(2)
In (2), we view x′ , x′′ , T, and N as row vectors, hence all matrices in the above equation are
2 × 2 matrices.
Problem 39. Given a 2 × 2 matrix
a11
A=
a21
a12
a22
we define the determinant of A to be
det A = a11 a22 − a21 a12 .
Prove the following facts about determinants:
(a) If A and B are 2 × 2 matrices, then
det(AB) = det(A) det(B).
Hint: recall that the product of two matrices is given by
a11 b11 + a12 b21
b11 b12
a11 a12
=
a21 b11 + a22 b21
b21 b22
a21 a22
a11 b12 + a12 b22
.
a21 b12 + a22 b22
(b) If the rows of A are v and w with v and w orthonormal, then
det A = ±1.
Hint: First argue that if v and w are orthonormal and v = (a, b) then w = ±(−b, a).
Writing x(t) = (x1 (t), x2 (t)), taking determinants of both sides of (2) and applying the
previous exercise, we obtain the following very useful expression for curvature in the plane:
κ(t) =
|x′1 (t)x′′2 (t) − x′2 (t)x′′1 (t)|
|x′ (t)|3
.
(3)
11
Parametrized Curves
Exercise 40. Show that for a curve C given in the form y = f (x) for x ∈ I, the curvature
is given by the formula
|f ′′ (x)|
κ(x) =
.
(1 + (f ′ (x))2 )3/2
Geometric interpretation of curvature
We describe the geometric interpretation of curvature by analogy with a common geometric
interpretation of a tangent line. Given a smooth curve C parametrized by x : I → Rn , the
tangent line at x(t0 ) is
L = {x(t0 ) + tx′ (t0 ) | t ∈ R} .
∗ The name is derived from
the Latin name for this
circle, circulus
osculans—attributed to
the mathematician
Leibniz—which literally
means “kissing circle.”
We think of L as being the “best linear approximation to C” around the point x(t0 ). But
why should we restrict our approximation to boring old lines? Curvature arises if we instead
think of approximating C by circles. In particular, imagine all circles S tangent to C at x(t0 ).
One of these circles, call it So offers the closest approximation to C at x(t0 )—we call this
circle the osculating circle.∗ If ro is the radius of the osculating circle So , then we have the
relation
1
κ= .
ro
Thus, we may interpret curvature κ(t0 ) as the inverse radius of the circle that best approximates C at the point x(t0 ).
2.5 Curves in R3 We now consider curves in 3 dimensions. Let x : I → R3 be an arc length
parametrization. As before, we define
T(s) =
dx
ds
and
N(s) =
T′ (s)
.
|T′ (s)|
The vectors T(s) and N(s) are still orthonormal, but in R3 these vectors do not form a
basis. To obtain an orthonormal basis, we form the binormal vector
B(s) = T(s) × N(s) for all
s ∈ I.
Now whenever T, N and B are all nonzero, they form an orthonormal basis for R3 . By
definition, we again have
T′ (s) = κ(s)N(s).
Since N is a unit vector, its derivative N′ is orthogonal to N, hence it is a linear combination
of T and B. As in the case with R2 , the coefficient of T is −κ. We define the torsion τ of x
to be N′ ’s coefficient of B
N′ (s) = −κ(s)T(s) + τ (s)B(s).
Finally, we compute the derivative B′ . Since B is a unit vector, its derivative is orthogonal
to B, and thus a linear combination of T and N alone:
B′ = hB′ , Ti T + hB′ , Ni N
= hB, T′ i T + hB, N′ i N
= − hB, κNi T − hB, −κT + τ Bi N
= −τ N.
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Parametrized Curves
We summarize the relationship between T, N, B and their derivatives with the FrenetSerret formulas in R3 :

 
T′
0
N′  = −κ
B′
0
κ
0
−τ
 
0
T
τ  N .
0
B
Exercise 41. Consider the helix given by
x(t) = (r cos t, r sin t, at) for
t ∈ [0, 2π].
Find an arc length parametrization for x and use it to compute the curvature and torsion of
x.
Problem 42. Show that the range of a smooth curve x : I → R3 lies in a plane P ⊆ R3 if
and only if τ (t) = 0 for all t ∈ I. What is the normal vector to P ?
Once again, we have defined curvature and torsion specifically in terms of arc length
parametrizations. Now, we derive formulas for these quantities for general smooth parametrizations. If x : I → R3 is a smooth parametrization and s : I → R an arc-length parameter,
a routine but somewhat tedious calculation (along with the Frenet-Serret equations) shows
that
 ′  ′

 
x
s
0
0
1
0 0
T
 x′′  =  s′′ (s′ )2
0  0
κ 0  N
(4)
x′′′
s′′′ 3s′ s′′ (s′ )3
−κ2 κ′ κτ
B
We can express the first two rows of this equation as
x′
x′′
=
s′ T
.
s′′ T + (s′ )2 κN
Taking the cross product of the first two rows of both sides and taking lengths gives
κ=
|x′ × x′′ |
|x′ |
3
.
(5)
Taking determinants of both sides of (4) gives
τ=
det(x′ , x′′ , x′′′ )
|x′ × x′′ |
2
(6)
Exercise 43. Use equations (5) and (6) to compute the curvature and torsion of the curve
given by
x(t) = (t, t2 , t3 ).
Exercise 44. Show that (3) is a special case of (5). Specifically, suppose x : I → R3 is of the
form
x(t) = (x1 (t), x2 (t), 0) for all t ∈ I.
Show that (5) gives the same curvature as (3).
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Parametrized Curves
Geometric interpretation
For a smooth curve C parametrized by x : I → R3 and a point t0 ∈ I, the plane Po (t0 )
determined by T and N is called the osculating plane. Specifically,
Po (t0 ) = {x(t0 ) + sT(t0 ) + tN(t0 ) | s, t ∈ R} .
As with plane curves (i.e., curves in R2 ), we may interpret the curvature κ as the reciprocal
of the radius of the osculating circle for x. The osculating circle is contained in the osculating plane. Notice that the binormal vector B = T × N is a (unit) normal vector for the
osculating plane.
The Frenet-Serret formulas show that the torsion τ of C measures how much the binormal vector B changes with respect to the normal vector N as we traverse C. Since B is
the normal vector for the osculating plane, the torsion therefore measures how quickly the
orientation of the osculating plane changes as we traverse C.
It is a nontrivial (and useful) fact that curvature and torsion completely determine the
geometry of a curve in R3 . Specifically, if one knows the curvature κ(s) and torsion τ (s)
for all values of s, one can in principle determine the curve C, up to translation, reflection,
and rotation. Although we will not prove this general fact here, the motivated reader might
attempt to prove the following characterization of certain curves in R3 :
(a) A curve C with zero curvature and zero torsion is a line.
(b) A curve C with constant (nonzero) curvature and zero torsion is a circle. In fact, the
radius r of the circle is r = 1/κ.
(c) A curve C with constant (nonzero) curvature and constant (nonzero) torsion is a helix.
The radius of the helix is given by r = 1/κ while the pitch of the helix is proportional
to the torsion τ .
2.6 Applications to physics Suppose x : I → R3 represents the trajectory of a physical particle. Recall that v = x′ is the velocity and a = x′′ is the acceleration of the particle. One can
prove a generalization of the result of Exercise 17 to general (not uniform circular) motion.
Define the linear and centripetal components of acceleration by
aℓ = ha, Ti
and
ac = ha, Ni
respectively. Since a lies in the span of T and N (why?) and the latter two vectors are
orthonormal, we can write
a = aℓ T + ac N.
Recall that the speed of x is defined to be v(t) = |v(t)| for all t.
Proposition 45. Let x : I → R3 be the (smooth) trajectory of a particle, and let aℓ and ac
be the linear and centripetal components (respectively) of a. Then
aℓ = v ′ (t)
and
ac = (v(t))2 κ(t).
Notice that in the case of uniform circular motion, v ′ (t) = 0 and κ(t) = 1/r where
r is the (constant) radius of the circle. Thus, Proposition 45 is indeed a generalization of
Exercise 17. This proposition gives a very concrete interpretation of the two components of
acceleration. Essentially there are two ways to accelerate: linear acceleration corresponds to
changing the speed of the particle, while centripetal acceleration corresponds to changing
the direction of the particle. Proposition 45 shows that these two modes of acceleration
completely determine the acceleration.
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Parametrized Curves
Problem 46. Prove Proposition 45. (Hint: start by writing v(t) = v(t)T(t) and differentiating both sides with respect to t.)
15