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MATH2051 Geometry of Curves and
Surfaces, 201920Notes c©D.G. Harland 2019
Contact Details
Lecturer: Derek HarlandOffice: School of Mathematics, 8.06E-mail: [email protected]: 343 5152
All handouts will be made available on Minerva.
MATH2051: Assessment arrangements
There will be a 2 hour exam during the January exam period, counting 85% towardsyour final module grade.
There will also be 5 homework assignments, collectively counting 15% towardsyour final grade. These will be due in by 2pm on the following Fridays:
• Friday 18th October 2019
• Friday 1st November 2019
• Friday 15th November 2019
• Friday 29th November 2019
• Friday 13th December 2019
Homework should be submitted to your marker’s mailbox on level 8 of the Schoolof Mathematics. (If you don’t know who your marker is, consult the lists publishednear the mailboxes, or ask in the Taught Students’ Office.)
Each homework assignment will be supported by two workshop sessions. It is veryimportant that you attend these.
Model solutions to the homework assignments will automatically become availableon Minerva at 9pm on the day of the deadline. This will allow you to make a timelyself-assessment of your progress. Clearly this means that late homework cannotbe accepted. If you miss a homework deadline for a good reason (e.g. illness), youshould notify the taught student office, and the homework concerned will be ommitedfrom the calculation of your module grade.
1
0 Why study geometry?
Curves and surfaces are all around us in the natural world, and in the built environ-ment. The first step in understanding these structures is to find a mathematicallynatural way to describe them. That is the primary aim of this course, which will focusparticularly on the concept of curvature.
Figure 1: Left: nautilus shell in the shape of the Fibonacci spiral. Right: The 1972 Olympic stadium,which was modelled on a minimal surface.
As a side effect, you will develop some very useful transferable skills. Key amongthese is the ability to translate a mathematical system (a set of equations, or formulae,or inequalities) into a visual picture. Thinking about a problem geometrically oftenleads to fresh insights.
Example 0 How does the number of solutions of the pair of simultaneous equations
xy = 1
x2 + y2 = a2
depend on the constant a > 0?
x1
x2 If a is large the circle overlaps the hy-perbola and the system has four solutions.If a is small the circle doesn’t meet thehyperbola and there are no solutions. Forthe critical value a =
√2 they just meet,
and there are two solutions.. �
The above example featured a pair of curves, each associated with an algebraicequation. In fact, throughout this course we will think of curves in a rather differentway, not as a set of points satisfying an equation, but rather as the range of a suitablemapping. That is, we will deal primarily with parametrized curves.
2
1 Regularly Parametrized Curves
1.1 Basic definitions
Let I ⊆ R be an open interval,
e.g. (0, π), (−∞, 1), R etc.
Recall that a function f : I → R is smooth if all its derivatives f ′(t), f ′′(t), f ′′′(t), . . .exist for all t (shorthand: ∀ t ∈ I).
E.g. polynomials, trigonometric functions (sin, cos etc.), exponentials, logarithms,hyperbolic trig functions (sinh, cosh etc.) are all smooth.
f : R → R such that f(t) = t4
3 is not smooth. Check:
f ′(t) =4
3t1
3 ⇒ f ′′(t) =4
9t−
2
3
so f ′′(0) does not exist.
We can extend this definition of smoothness to maps γ : I → Rn where n ≥ 2.Such a map is a rule which assigns to each “time” t a vector
γ(t) = (γ1(t), γ2(t), . . . , γn(t)) ∈ Rn.
We say that the map γ is smooth if every one of its component functions γi : I → R,i = 1, 2, . . . , n is smooth in the usual sense (all derivatives exist everywhere). We maythink of γ as describing the trajectory of a point particle moving in Rn. This leadsus to:
Definition 1 A parametrized curve (PC) in Rn is a smooth map γ : I → Rn. Atime t ∈ I is a regular point of γ if γ′(t) 6= 0. If γ′(t) = 0, then t ∈ I is a singularpoint of γ. If every t ∈ I is regular then γ is said to be a regularly parametrizedcurve (RPC). In other words, a PC is a RPC if and only if
there does not exist a time t ∈ I such that γ′(t) = 0 = (0, 0, . . . , 0).
The image set of a curve γ is the range of the mapping, that is,
γ(I) = {γ(t) ∈ Rn | t ∈ I} ⊂ R
n�
Example 2 Consider the parabola x2 = x21. There are infinitely many PCs whose
image set is this parabola.
3
x1
x2 Two examples:
γ : R → R2, γ(t) = (t, t2)
δ : R → R2, δ(t) = (t3, t6).
γ is a regularly parametrized curve:
γ′(t) = (1, 2t) 6= (0, 0) ∀t because 1 6= 0
But δ is not:δ′(t) = (3t2, 6t5) = (0, 0) when t = 0
. �
So Definition 1 concerns the parametrization of the curve, not just its image setγ(I) ⊂ Rn. Why? A PC is a smooth map γ : I → Rn, but it does not necessarilyrepresent a “smooth” curve! A RPC does, however.
Example 3 γ : R → R2, γ(t) = (t2, t3) is a smooth map, hence a PC. But its imageset is not “smooth” – it has a cusp.
x1
x2 Note that γ is not a RPC:
γ′(t) = (2t, 3t2) = (0, 0) when t = 0
Note also that the nasty point in γ(I) oc-curs precisely where γ′(t) = 0, that is, atthe singular point of γ. �
Definition 4 Given a PC γ : I → Rn, its velocity is γ′ : I → Rn, its accelerationis γ′′ : I → Rn and its speed is |γ′| : I → [0,∞). �
Notes:
• |v| denotes the Euclidean norm of a vector v ∈ Rn, that is,
|v| =√
v21 + v22 + · · ·+ v2n ≥ 0.
• It’s important to distinguish between vector and scalar quantities.
Velocity is a vector . Acceleration is a vector . Speed is a scalar .
• We can rephrase definition 1 as follows:
PC γ is a RPC ⇐⇒ its velocity (or its speed) never vanishes
4
Example 5 (straight line) Simple but important. For any v ∈ Rn and x ∈ Rn onehas the PC
γ : R → Rn, γ(t) = x+ vt.
Note that γ′(t) = v, constant, so γ is a RPC unless v = 0.Note also that the direction of the straight line is determined solely by v. �
A RPC has a well-defined tangent line at each t0 ∈ I:
Definition 6 Let γ : I → Rn be a RPC. Then its tangent line at t0 ∈ I is the PC
γt0 : R → Rn, γt0(t) = γ(t0) + tγ′(t0).
Note that γ′t0(t) = 0 + γ′(t0) 6= 0 since γ is a RPC. Hence every tangent line γt0 is aRPC too. �
Example 7 Consider the PC γ : R → R2, γ(t) = (t3 − t, t2 − 1).
(A) Is it a RPC?
(B) Are any of its tangent lines vertical?
(A) Just check whether its velocity ever vanishes. Assume it does at time t. Then:
(0, 0) = γ′(t) = (3t2 − t, 2t).
The equation 2t = 0 implies that t = 0. Substituting in gives
γ′ (0) = (−1, 0) ,
so the first component of γ′ does not vanish. Therefore γ′(t) is never zero.
Hence γ is a RPC.
(B) Tangent line to γ at t0 ∈ R is
γt0(t) = (t30 − t0, t20 − 1) + t(3t20 − 1, 2t0)
The direction of the tangent line is given by its (constant) velocity vector
γ′t0(t) = (3t20 − 1, 2t0) = γ′(t0).
5
The tangent line is vertical if the horizontal component (the x1 component) of thisvector is 0. Hence γt0 is vertical if and only if
3t20 − 1 = 0 ⇐⇒ t0 = ± 1√3
so γ 1√3
and γ− 1√3
are vertical.
Here’s a picture of γ, together with the two vertical tangent lines:
x1
x2
Note that the curve intersects itself exactly once. A self-intersection point is onewhere γ(t1) = γ(t2) but t1 6= t2, which implies, in this case,
t31 − t1 = t32 − t2, and t21 − 1 = t22 − 1
⇒ t21 = t22⇒ t2 = −t1
⇒ t31 − t1 = −t31 + t1 = −(t31 − t1)
⇒ 0 = t31 − t1 = t1(t21 − 1)
⇒ t1 = 0 or t1 = ±1
If t1 = 0 then t2 = −0 = t1, so this doesn’t give a self-intersection point. Likewiset1 = 1 and t1 = −1 give the same self-intersection point, γ(1) = γ(−1) = (0, 0).
Another question: what is the length of the segment of γ from t = 0 to t = 1? �
6
1.2 Arc length
γ(t0)
γ(t1) γ(t2)
γ(tN )
Given a RPC γ : I → Rn, what is the length of the curve segment from t = a to t = bsay? Partition [a, b] into N pieces [tn−1, tn], n = 1, 2, . . . , N (with t0 = a, tN = b) ofequal length δt = b−a
N. If N is large, then δt is small. The length of the straight line
segment from γ(tn−1) to γ(tn) is
δsn = |γ(tn)− γ(tn−1)| = |γ(tn−1 + δt)− γ(tn−1)| ≈ |γ′(tn−1)|δt.
So the total length of the piecewise straight line from γ(a) to γ(b) is
sN =N∑
n=1
δsn ≈N∑
n=1
|γ′(tn−1)|δt.
In the limit N → ∞, δt → 0 and the piecewise straight line tends to the real curveγ. So the total length of the curve segment is
s = limN→∞
sN =
∫ b
a
|γ′(t)| dt.
This motivates the following definition:
Definition 8 Let γ : I → Rn be a RPC. The arc length along γ from t = t0 tot = t1 is
s =
∫ t1
t0
|γ′(t)| dt. �
More informally, distance travelled = integral of speed.
7
Example 9 γ : (0,∞) → R2, γ(t) = (t, 23t3
2 ).What is the arc length from t = 3 to t = 15?
γ′(t) = (1, t1
2 )
|γ′(t)| =√1 + t
s =
∫ 15
3
(1 + t)1
2dt =2
3
[(1 + t)
3
2
]153
=112
3
�
WARNING! This example was cooked up to be easy. It’s usually impossible tocompute s in practice.
Example 7 (revisited) γ(t) = (t3 − t, t2 − 1). What is the arc length from t = 0 tot = 1?
γ′(t) = (3t2 − 1, 2t) so
s =
∫ 1
0
√(3t2 − 1)2 + (2t)2dt
=
∫ 1
0
√9t4 − 2t2 + 1dt
= ??? (1.36 to 2d.p.)
�
Note that we can use Definition 8 even if t1 < t0:
s =
∫ t1
t0
|γ′(t)|︸ ︷︷ ︸ dt = −∫ t0
t1
|γ′(t)| dt < 0
positive
So s can be positive or negative. For this reason it is sometimes called the signedarc length.Once we’ve chosen a base point t0 ∈ I, at every other time we can assign a uniquesigned arc length.
Definition 10 Let γ : I → Rn be a RPC. The arc length function based at t0 ∈ Iis
σt0 : I → R, σt0(t) =
∫ t
t0
|γ′(u)| du. �
8
Example 9 (revisited) The arc length function based at t0 = 1 for γ(t) = (t, 23t3
2 )is
σ1(t) =
∫ t
1
√1 + udu
=2
3
((1 + t)
3
2 − 23
2
)
Trick questions: (A) What is σ1(1)?(B) What is σ′1(t)?
�
σt0(t) is very hard (usually impossible) to compute explicitly in practice. But the factthat it exists is crucial to the theory of curves. We need to understand its properties.
Remark 11 The arc length function σt0 : I → R has the following properties:
(a) σt0(t0) =
∫ t0
t0
|γ′(q)| dq = 0.
(b) σt0 is smooth, and its first derivative is strictly positive:
σ′t0(t) =d
dt
∫ t
t0
|γ′(a)| da = |γ′(t)| > 0
for all t ∈ I since γ is a RPC.
(c) It follows, by the Mean Value Theorem, that σt0 is strictly increasing (if t2 > t1then σt0(t2) > σt0(t1)), and hence is injective (one-to-one).
(d) Let J ⊆ R denote the range of σt0 . It’s another (possibly unbounded) openinterval. Given (c), there exists an inverse function to σt0 , let’s call it τt0 : J → I,so that
τt0(σt0(t)) = t for all t ∈ I, and
σt0(τt0(s)) = s for all s ∈ J (♣)
Iσt0→ J
time arc length
t←τt0 s
Trick question: what is τt0(0)?
9
(e) The inverse function τt0 is also strictly increasing. To see this, differentiate (♣)with respect to s using the chain rule:
σ′t0(τt0(s))τ′t0(s) = 1
⇒ τ ′t0(s) =1
σ′t0(τt0(s))=
1
|γ′(τt0(s))|> 0
�
Example 9 (revisited) γ : (0,∞) → R2 γ(t) = (t, 23t3
2 ). Recall
σ1 : (0,∞) → R σ1(t) =2
3((1 + t)
3
2 − 23
2 ).
What is τ1? Domain of τ1 = range of σ1. But since σ1 is increasing, this is just theinterval J = (a, b), where
a = limt→0
σ1(t) = −2
3(2
3
2 − 1) and
b = limt→∞
σ1(t) = ∞.
To find a formula for τ1(s), we must solve s = σ1(t) to find t as a function of s:
s =2
3
[(1 + t)
3
2 − 23
2
]
⇒(3
2s+ 2
3
2
) 2
3
= 1 + t
⇒ τ1(s) =
(3
2s+ 2
3
2
) 2
3
− 1.
[Check: τ1(0) = (0 + 23
2 )2
3 − 1 = 20 − 1 = 1 as expected ]
What is τ ′1(s)?
τ ′1(s) =1
σ′1(τ1(s))=
1
|γ′(τ1(s))|=
1√1 + τ1(s)
=
(3
2s+ 2
3
2
)− 1
3
�
1.3 Reparametrization
A reparametrization of a RPC γ : I → Rn is a redefinition of “time”:
β : J → Rn, β(u) = γ(h(u)) = (γ ◦ h)(u)
for some interval J and function h : J → I.
10
Example 12 Let γ : R → R2 such that γ(t) = (t, et). Let h : (0,∞) → R such thath(u) = log u. This gives the reparametrization
β : (0,∞) → R2,
β(u) = γ(h(u)) = (log u, u)
Note that although γ and β are different functionstheir image sets are identical: β((0,∞)) = γ(R).We’ve just changed the way we label the points onthe curve.Note also that both γ and β are RPCs in this case:
x1
x2
γ′(t) = (1, et) 6= 0 ∀t ∈ R β ′(u) = (u−1, 1) 6= 0 ∀0 < u < ∞
But we can’t allow h to be any function J → R if we want β to be a RPC. Forexample, let h : R → R such that h(u) = sin u. Then
β(u) = γ(h(u)) = (sin u, esinu)
β ′(u) = (cosu, cosuesinu)
so β ′(π2) = (0, 0), and β is not regular! This is an example of a bad redefinition of time.
We want to exclude things like this from our formal definition of reparametrization.�
Definition 13 A reparametrization of a PC γ : I → Rn is a map β : J → Rn
defined by β(u) = γ(h(u)), where J is an open interval, and h : J → I is smooth,surjective and has h′(u) > 0 for all u ∈ J . �
Notes:
• We require h to be smooth so that β is smooth (by the Chain Rule), hence aPC.
• We require h to be surjective so that β(J) = γ(I). In other words, this ensuresthat β covers all of γ, not just part of it.
• Since h is strictly increasing, it is injective. Hence for each time t ∈ I there isone (h surjective) and only one (h injective) corresponding new time u ∈ J .
Lemma 14 Any reparametrization of a RPC is also a RPC.
Proof: Let β(u) = γ(h(u)) where γ is a RPC. Then
β ′(u) = γ′(h(u))h′(u) = 0 ⇒γ′(h(u)) = 0 (impossible since γ is regular)
or h′(u) = 0 (impossible since h′ > 0).
11
Hence β is a RPC. �
Note that reparametrization preserves the direction and sense of the velocity vector,but not its length:
new velocity = β ′(u) = h′(u)γ′(h(u)) = positive number × old velocity.
However:
Lemma 15 Reparametrization preserves arc length.
Proof: Let γ be a PC and s be the arc length along γ from γ(t0) to γ(t1). Let β = γ◦hbe a reparametrization of γ where h(u0) = t0 and h(u1) = t1. We must show that sis also the arc length along β from β(u0) to β(u1). In fact
s =
∫ t1
t0
|γ′(t)| dt Substitution: t = h(u), dt = h′(u)du
=
∫ h−1(t1)
h−1(t0)
|γ′(h(u))| h′(u)du
=
∫ u1
u0
|γ′(h(u))h′(u)| du since h′(u) > 0
=
∫ u1
u0
|β ′(u)| du
as required. �
Definition 16 A unit speed curve (USC) is a smooth map γ : I → Rn such that|γ′(s)| = 1 for all s ∈ I �
Notes:
• Clearly USC ⇒ RPC.
• It’s conventional to denote the “time” parameter of a unit speed curve by srather than t, because the parameter is signed arc length (up to a constant):
∫ s
s0
|γ′(u)|du =
∫ s
s0
1 du = s− s0. (♣)
Unit speed curves may seem very special, but in fact they are, in a sense, completelyuniversal:
Theorem 17 Every RPC γ : I → Rn has a unit speed reparametrization (USR)β : J → Rn. This USR is unique up to “time” translation. More precisely, ifδ : K → Rn is another USR of γ, then there exists a constant c ∈ R such that
β(s) = δ(s− c).
12
Proof:(A) Existence:Choose t0 ∈ I and let σt0 : I → J be the signed arc length function of γ based at t0.Recall that there exists a smooth, well-defined inverse function τt0 : J → I which haspositive derivative (Remark 11(d) and (e)). So h = τt0 gives a reparametrization ofγ, according to Definition 13: β : J → Rn, β(s) = γ(τt0(s)). But
β ′(s) = γ′(τt0(s))τ′t0(s) = γ′(τt0(s))
1
|γ′(τt0(s))|
by Remark 11(e). Hence |β ′(s)| = 1 for all s ∈ J , and β is a USC as required.
(B) Uniqueness:Let δ : K → Rn be another USR of γ, σ be the arc length along γ from γ(t0) =β(s0) = δ(s0) to γ(t) = β(s) = δ(s). By Lemma 15 and (♣),
σ = s− s0 = s− s0
⇒ s = s− (s0 − s0)
Hence β(s) = δ(s) = δ(s− (s0 − s0)). �
Example 18 Circle γ(t) = a(cos t, sin t), or radius a > 0. To find a USR of γ wecompute the arclength function and invert it, then use τ0 to reparametrize γ:
γ′(t) = a(− sin t, cos t)
|γ′(t)| = a√
sin2 t+ cos2 t = a
σ0(t) =
∫ t
0
a du = at
τ0(s) =s
a
β(s) = γ(τ0(s)) = a(cos(sa
), sin
(sa
))
x1
x2
a
�
Although Theorem 17 ensures a USR exists, we may not be able to construct itexplicitly. For example, for the curve in example 7 we were not able to find thearclength function explicitly, so certainly wouldn’t be able to find a USR.
13
Summary
• A regularly parametrized curve (RPC) is a smooth map γ : I → Rn (whereI is an open interval) such that, for all t ∈ I, γ′(t) 6= 0.
• Given a RPC γ, its velocity is γ′, its speed is |γ′| and its acceleration is γ′′.
• The tangent line to γ at t0 ∈ I is
γt0 : R → Rn, γt0(t) = γ(t0) + γ′(t0)t.
• The arclength function based at t0 ∈ I is
σt0(t) =
∫ t
t0
|γ′(u)|du.
Geometrically, this is the arclength along γ from γ(t0) to γ(t) if t ≥ t0 (andminus the arclength if t < t0).
• A reparametrization of a curve γ : I → Rn is a curve γ ◦ h : J → Rn whereh : J → I is smooth, surjective and has strictly positive derivative. If γ is aRPC, so is every reparametrization of γ.
• Arclength is unchanged by reparametrization.
• A unit speed curve (USC) is a curve with |γ′(t)| = 1 for all t.
• Every RPC has a reparametrization which is a USC. One can construct it, inprinciple, by reparametrizing with h = σ−1t0 .
14
2 Curvature of a parametrized curve
2.1 Basic definition
In this section we will develop a measure of the curvature of a RPC.
x
x
1
2
What distinguishes a region of high curvature from one of low curvature?
High curvature: tangent lines change direction very rapidlyLow curvature: tangent lines change direction only slowlyNo curvature (straight line): tangent lines don’t change direction at all
So we require a measure of the rate of change of direction of the curve’s tangentlines. Recall that the tangent line at γ(t0) of a RPC γ is
γt0(u) = γ(t0) + u γ′(t0)︸ ︷︷ ︸determines
direction
Perhaps we can used
dt0(γ′(t0)) = γ′′(t0) to measure the curvature of γ at t = t0? Not
quite: γ′′ tells us about rate of change of length of γ′ as well as rate of change ofdirection.
Example 19 γ(t) = (log t, 2 log t + 1) is a straight line, and hence is not curved.
x1
x2
However
γ′(t) =
(1
t,2
t
)
γ′′(t) =
(−1
t,− 2
t2
)
so γ′′ is not zero. �
15
But what if γ happens to be a unit speed curve, γ(s)? Then |γ′(s)| = 1 for alls, so γ′′(s) does tell us only about the rate of change of the direction of the tangentlines.
Definition 20 (a) Let γ : I → Rn be a USC. Then the curvature vector of γ isk : I → Rn where
k(s) = γ′′(s).
Note that k is a vector quantity. We shall refer to the norm of k, |k| : I → [0,∞) asthe curvature of γ.
(b) If a RPC γ : I → Rn is not a USC, then Theorem 17 says that it has a unitspeed reparametrization β : J → Rn, β = γ ◦ h. In that case, we define the curvaturevector of γ at t = h(s) to be the curvature vector of β at s, as in part (a), that is,β ′′(s). In other words, k : I → Rn such that
k = β ′′ ◦ h−1.
Note: To make sense, this definition should be independent of the choice of unit speedreparametrization β of γ. It is. Recall, by Theorem 17, that any pair of USRs of γdiffer only by shifting the origin of the new time coordinate s. But such a shift hasno effect on the second (or indeed the first) derivative of β. �
Example 21 Helix γ : R → R3, γ(s) =1√2(cos s, sin s, s).
γ′(s) =1√2(− sin s, cos s, 1)
|γ′(s)| =1√2
(cos2 s+ sin2 s+ 1
) 1
2 = 1
k(s) = γ′′(s) =1√2(− cos s,− sin s, 0)
|k(s)| =1√2
(cos2 s+ sin2 s
) 1
2 =1√2.
�
Example 22 Circle γ : R → R2, γ(t) = (2 cos t, 2 sin t)
γ′(t) = (−2 sin t, 2 cos t)
|γ′(t)| = 2
16
so not a USC. But we can find a USR of γ. How?
Arclength function: σ0(t) =∫ t
02dt = 2t
Inverse function: τ0(s) =12s
USR: β(s) = γ(τ0(s)) =(2 cos s
2, 2 sin s
2
)
Now compute curvature vector of β:
β ′(s) =(− sin
s
2, cos
s
2
)
β ′′(s) =
(−1
2cos
s
2,−1
2sin
s
2
)
Change back to old parametrization, s = σ0(t) = 2t
k(t) = β ′′(σ0(t)) =
(−1
2cos t,−1
2sin t
).
�
In general this process is rather clumsy. In particular, it’s usually impossible towrite down β, the USR of γ, explicitly (recall Example 7 (revisted)).
So let’s calculate k once and for all for a general RPC γ, to obtain a more user-friendly version of Definition 20. Two preliminary observations:
(A) Given two vectors u, v ∈ Rn, we define their scalar product u · v ∈ R by
u · v = u1v1 + u2v2 + · · ·unvn.
In particular, the norm of v may be rewritten
|v| =√v · v.
(B) Given two vector valued functions u, v : I → Rn, we have a product rule fordifferention,
d
dt(u(t) · v(t)) = u′(t) · v(t) + u(t) · v′(t).
In particular,
d
dt|u(t)| = d
dt(u · u) 1
2 =u · u′ + u′ · u2(u · u) 1
2
=u(t) · u′(t)
|u(t)| .
Let γ(t) be a RPC and β(s) = γ(h(s)) be a USR of γ. Then
β ′(s) =d
ds(γ(t)) =
dγ
dt
dt
ds=
γ′(t)
ds/dt.
17
But |β ′(s)| = 1, so ds/dt = |γ′(t)|, and
β ′(s) =γ′(t)
|γ′(t)| .
Differentiating this equation w.r.t. s once again yields
β ′′(s) =d
ds
(γ′(t)
|γ′(t)|
)=
d
dt
(γ′(t)
|γ′(t)|
)dt
ds
=
{γ′′(t)
|γ′(t)| −γ′(t)
|γ′(t)|2d
dt|γ′(t)|
}1
|γ′(t)|
=
{γ′′(t)
|γ′(t)| −γ′(t)
|γ′(t)|2(γ′(t) · γ′′(t)
|γ′(t)|
)}1
|γ′(t)|
=1
|γ′(t)|2{γ′′(t)−
(γ′(t) · γ′′(t)|γ′(t)|2
)γ′(t)
},
where we have used observation (B) above. This leads us to
Definition 20 (*) The curvature vector of a RPC γ : I → Rn is k : I → Rn,where
k(t) =1
|γ′(t)|2{γ′′(t)−
(γ′(t) · γ′′(t)|γ′(t)|2
)γ′(t)
}. �
Example 23 Parabola γ : R → R2, γ(t) = (t, t2).
γ′(t) = (1, 2t)
|γ′(t)|2 = 1 + 4t2
γ′′(t) = (0, 2)
γ′(t) · γ′′(t) = 4t
k(t) =1
1 + 4t2
{(0, 2)− 4t
1 + 4t2(1, 2t)
}=
1
(1 + 4t2)2(−4t, 2)
x1
x2
�
18
Example 24 Ellipse γ : R → R2, γ(t) = (2 cos t, sin t).
γ′(t) = (−2 sin t, cos t)
|γ′(t)|2 = 4 sin2 t + cos2 t
γ′′(t) = (−2 cos t,− sin t)
γ′(t) · γ′′(t) = 3 sin t cos t
k(t) =1
4 sin2 t + cos2 t
{(−2 cos t,− sin t)− 3 sin t cos t
4 sin2 t + cos2 t(−2 sin t, cos t)
}
=1
(4 sin2 t + cos2 t)2(−8 sin2 t cos t− 2 cos3 t+ 6 sin2 t cos t,
−4 sin3 t− sin t cos2 t− 3 sin t cos2 t)
=1
(4 sin2 t + cos2 t)2(−2 cos t,−4 sin t)
|k(t)| =2
(4 sin2 t + cos2 t)2
√cos2 t+ 4 sin2 t =
2
(4 sin2 t + cos2 t)3
2
x1
x2
�
Definition 20(*) is much easier to use than definition 20, but it’s not so memorable.Can we improve it?
19
2.2 The unit tangent vector and normal projection
We start with a simple observation about the curvature vector:
Fact 25 The curvature vector k of a RPC γ is always orthogonal to its velocity,k(t) · γ′(t) = 0. This follows directly from Definition 20(*):
k(t)·γ′(t) = 1
|γ′(t)|2{γ′′(t) · γ′(t)−
(γ′(t) · γ′′(t)|γ′(t)|2
)γ′(t) · γ′(t)
}= 0. �
Looking back at examples 21, 22, 24 one sees several illustrations of this:
Helix γ(s) = 1√2(cos s, sin s, s) γ′(s) = 1√
2(− sin s, cos s, 1) k(s) = 1√
2(− cos s,− sin s, 0)
Circle γ(t) = (2 cos t, 2 sin t) γ′(t) = (−2 sin t, 2 cos t) k(t) = 12(− cos t,− sin t)
Ellipse γ(t) = (2 cos t, sin t) γ′(t) = (−2 sin t, cos t) k(t) = −2(1+3 sin2 t)2
(cos t, 2 sin t)
We can use this fact to give a more memorable version of Definition 20(*). In prepa-ration for this, we need:
Definition 26 Let γ : I → Rn be a RPC. Its unit tangent vector u : I → Rn is
u(t) =γ′(t)
|γ′(t)| .
Note that u is well defined because γ is a RPC (so |γ′(t)| > 0 for all t). Note alsothat |u(t)| = 1 for all t by construction.
Given any other vector valued function v : I → Rn, we define its normal projec-tion, v⊥ : I → Rn by
v⊥(t) = v(t)− [v(t) · u(t)]u(t).
We can think of v⊥ as that part of v left over after we have subtracted off thecomponent of v in the direction of u. �
20
Example 27 γ : R → R3, γ(t) = ( t2
2, sin t, cos t). What are u and γ′′⊥?
γ′(t) = (t, cos t,− sin t)
|γ′(t)| =√
t2 + cos2 t + sin2 t =√t2 + 1
u(t) =1√
t2 + 1(t, cos t,− sin t)
γ′′(t) = (1,− sin t,− cos t)
γ′′(t) · u(t) =1√
t2 + 1(t− sin t cos t + sin t cos t) =
t√t2 + 1
γ′′⊥(t) = (1,− sin t,− cos t)− t
t2 + 1(t, cos t,− sin t)
=
(1
t2 + 1,− sin t− t
t2 + 1cos t,− cos t+
t
t2 + 1sin t
)�
The normal projection of the acceleration vector γ′′⊥ : I → Rn is of particularinterest, because it occurs in Definition 20(*):
k(t) =1
|γ′(t)|2{γ′′(t)−
(γ′(t)
|γ′(t)| · γ′′(t)
)γ′(t)
|γ′(t)|
}
=1
|γ′(t)|2 {γ′′(t)− [u(t) · γ′′(t)]u(t)} =
γ′′⊥(t)
|γ′(t)|2
Definition 20 (**) The curvature vector of a RPC γ : I → Rn is k : I → Rn,
k(t) =γ′′⊥(t)
|γ′(t)|2 �
Example 27 (revisited) γ(t) = ( t2
2, sin t, cos t) has curvature vector
k(t) =1
t2 + 1
(1
t2 + 1,− sin t− t
t2 + 1cos t,− cos t +
t
t2 + 1sin t
)
�
21
Summary
• The curvature vector of a RPC measures how fast the tangent lines to the curvechange direction.
• If γ is a unit speed curve, the curvature vector is k(s) = γ′′(s).
• In general
k(t) =1
|γ′(t)|2{γ′′(t)− γ′′(t) · γ′(t)
|γ′(t)|2 γ′(t)
}.
• The unit tangent vector along a curve γ : I → Rn is
u(t) =γ′(t)
|γ′(t)| .
• The normal projection of v : I → Rn is
v⊥(t) = v(t)− [u(t) · v(t)]u(t).
• An alternative formula for the curvature vector is
k(t) =γ′′⊥(t)
|γ′(t)|2 .
22
3 Planar curves
3.1 Signed curvature of a planar curve
The theory of curvature can be developed further in the special case of planar curves,that is, RPCs γ : I → R2 in two dimensions. These are special because all the cur-vature information associated with γ may be encoded in a single real-valued functionκ : I → R, called the signed curvature. How?
Recall that k in this case is a 2-vector, so it consists of a pair of real functionsk(t) = (k1(t), k2(t)) say. However, Fact 25 states that k is always orthogonal to theunit tangent vector u. So in fact we already know the direction of k. The only extrainformation we need to provide is the length of k, and its sense: whether it points tothe left or the right of u.
x
x
1
2
Definition 28 Let γ : I → R2 be a RPC with unit tangent vector u (recall u =γ′/|γ′|). The unit normal vector of γ is n : I → R2,
n(t) = (−u2(t), u1(t)).
The signed curvature of γ is κ : I → R,
κ(t) = k(t) · n(t)where k : I → R2 is the curvature vector of γ, as in Definition 20(**). �
Notes:
(a) n is orthogonal to u (n(t)·u(t) = 0) and has unit length (|n|2 = u22+u2
1 = |u|2 = 1)by construction. In fact, n is the vector obtained by rotating u 90◦ anticlockwise.
(b) Since both n and k are orthogonal to u, they must be parallel. In fact we canre-interpret the above definition as follows: given that k(t) and n(t) are parallel,we define κ(t) to be the constant of proportionality,
k(t) = κ(t)n(t).
It follows that |κ(t)| = |k(t)|. However, κ contains more information than the(unsigned) curvature |k| — its sign tells us the “sense” of k.
(c) Recall that k = γ′′⊥/|γ′|2, so
κ(t) =γ′′⊥(t) · n(t)|γ′(t)|2 .
23
However,γ′′⊥ · n = (γ′′ − (γ′′ · u)u) · n = γ′′ · n,
so we can give a slightly more convenient definition of κ:
Definition 28 (*) Let γ : I → R2 be a RPC, u = γ′/|γ′| be its unit tangent vectorand n = (−u2, u1) be its unit normal. Then its signed curvature κ : I → R is
κ(t) =γ′′(t) · n(t)|γ′(t)|2 . �
Example 29 A sinusoidal curve γ : R → R2, γ(t) = (sin t, t)
γ′(t) = (cos t, 1)
|γ′(t)| =√1 + cos2 t
u(t) =1√
1 + cos2 t(cos t, 1)
n(t) =1√
1 + cos2 t(−1, cos t)
γ′′(t) = (− sin t, 0)
κ(t) =sin t
(1 + cos2 t)3
2
x1
x2
. �
Note that
κ > 0 ⇒ k points in same sense as n ⇒ curve is turning leftκ < 0 ⇒ k points in opposite sense to n ⇒ curve is turning right
x
x
1
2
24
Example 30 Looking at the parabolaγ(t) = (t, t2) it’s immediately clear thatκ(t) is always positive . Let’s check:
x1
x2
γ′(t) = (1, 2t)
u(t) =1√
1 + 4t2(1, 2t)
n(t) =1√
1 + 4t2(−2t, 1)
γ′′(t) = (0, 2)
κ(t) =2
(1 + 4t2)3
2
Note that this observation depends crucially on the orientation of the curve, thatis, the direction in which it is traversed. For example, for both of the parabolae below,the signed curvature is always negative :
γ(t) = (−t, t2) γ(t) = (t2, t)
x1
x2
x1
x2
. �
Points on a curve where the signed curvature changes sign have a special name:
Definition 31 Let γ : I → R2 be a RPC and κ : I → R be its signed curvature. Ifthere exists a time t∗ ∈ I such that κ(t∗) = 0 and κ changes sign at t = t∗, then γ(t∗)is an inflexion point of γ. �
25
Example 29 (revisited) What are the inflexion points of the sinusoidal curve γ(t) =(sin t, t)? Recall that its signed curvature function is
κ(t) =sin t
(1 + cos2 t)3
2
So κ(t) = 0 if and only if t = Nπ, whereN ∈ Z. Further, the sign of κ(t) changesat each such time. Hence for all N ∈ Z,
γ(Nπ) = (0, Nπ)
is an inflexion point.
x1
x2
WARNING!κ(t∗) = 0 does NOT imply that γ(t∗) is an inflexion point!
κ must CHANGE SIGN at t = t∗ too!
Counterexample 32 The quartic curve γ : R → R2, γ(t) = (t, t4) has κ(0) = 0.However, it clearly has no inflexion points (κ(t) ≥ 0, since the curve never turnsright),
γ′(t) = (1, 4t3)
γ′′(t) = (0, 12t2)
γ′′(0) = (0, 0)
κ(0) =γ′′(0) · n(0)|γ′(0)|2 = 0
x1
x2
Exercise: show that
κ(t) =12t2
(1 + 16t6)3
2
. �
3.2 Planar curves of prescribed curvature
In this section we will consider only planar unit speed curves (PUSCs) γ(s) (|γ′(s)| = 1for all s). Note this entails no loss of generality by Theorem 17.
So far, given a PUSC γ(s) we can construct its signed curvature κ(s). Can wego the other way? That is, given a function κ(s), can we reconstruct the PUSCγ(s) whose curvature is κ? Yes, provided we also specify an initial position γ(0) andtangent vector γ′(0).
26
How? Note |γ′(s)| = 1 so each velocity vector is determined by just its direction.That is, there exists smooth θ : I → R such that
γ′(s) = (cos θ(s), sin θ(s)) (A).
Note that u(s) = γ′(s) and hence the unit normal vector is
n(s) = (− sin θ(s), cos θ(s)).
Now, for a USC, curvature k(s) = γ′′(s) = (− sin θ(s), cos θ(s))θ′(s), and hence thesigned curvature is
κ(s) = n(s) · k(s) = θ′(s) (B).
We may collect (A), (B) into a coupled system of 3 nonlinear ordinary differentialequations (ODEs):
(∗)
dθ
ds= κ(s) (1)
dγ1ds
= cos θ (2)
dγ2ds
= sin θ (3)
So any PUSC of curvature κ is a solution of system (*), and vice versa. Given aprescribed κ(s), we can solve (*) for the curve γ(s).
Theorem 33 Given any smooth κ : I → R, 0 ∈ I, and constants θ0 ∈ R andγ0 ∈ R2, there exists a unique USC γ : I → R2 with γ(0) = γ0, γ
′(0) = (cos θ0, sin θ0)and signed curvature κ.
Proof: We must prove existence of a unique global solution (θ(s), γ1(s), γ2(s)) ofthe initial value problem [IVP] (θ(0), γ1(0), γ2(0)) = (θ0, a, b) for system (*) [whereγ0 = (a, b)].
In fact, (*) is separable. First consider IVP (1):
dθ
ds= κ(s), θ(0) = θ0.
This has solution
θ(s) = f(s) := θ0 +
∫ s
0
κ(α)dα.
[Note that f ′(s) = κ(s) by the Fundamental Theorem of the Calculus, and f(0) = θ0,so θ = f is a solution with the right initial data.]Is this solution unique? Yes, by the Mean Value Theorem (MVT).[Assume it is not unique. Then there exists another solution θ(s) = g(s) with g(0) =θ0 but g 6= f , that is, there exists s0 ∈ I such that g(s0) 6= f(s0). Consider the
27
function F (s) = f(s)− g(s). Clearly F (0) = 0, F (s0) 6= 0 and F is differentiable onI. Hence, by the MVT, there exists s∗ between 0 and s0 such that
F ′(s∗) =F (s0)− F (0)
s0 − 0=
F (s0)
s06= 0.
But f and g both solve (1), so F ′(s) = κ(s)− κ(s) = 0 for all s, a contradiction.]Now substitute this unique solution (θ = f) into IVP (2):
dγ1ds
= cos f(s) γ1(0) = a.
This has unique solution
γ1(s) = a +
∫ s
0
cos f(α) dα
by an identical argument. Similarly, substituting θ = f into IVP (3),
dγ2ds
= sin f(s) γ2(0) = b,
one has the unique solution
γ2(s) = b+
∫ s
0
sin f(α) dα. �
Note that Theorem 34 doesn’t just prove existence and uniqueness of the curveγ(s), it also gives a formula for it:
γ(s) = γ0 + (
∫ s
0
cos θ(α) dα,
∫ s
0
sin θ(α) dα), where θ(α) = θ0 +
∫ α
0
κ(β) dβ. (C)
Example 34 Curves of constant curvature: κ(s) = κ0 6= 0, constant. Let’s alwayschoose γ(0) = (0, 0), θ(0) = 0 henceforth. Then formula (C) implies
θ(α) =
∫ α
0
κ0dβ = κ0α
⇒ γ1(s) =
∫ s
0
cos(κ0α)dα =1
κ0
sin(κ0s)
γ2(s) =
∫ s
0
sin(κ0α)dα =1
κ0(1− cos(κ0s))
⇒ γ(s) =1
κ0(sin(κ0s), 1− cos(κ0s))
This curve is a circle of radius 1/|κ0| centred on (0, 1/κ0). Question: what happenswhen κ0 = 0?
28
If we go back and use formula (C) withκ(β) ≡ 0 we find that
κ0 = 0 ⇒ γ(s) = (s, 0),
that is, the solution degenerates to a hor-izontal straight line.
-1
-0.5
0
0.5
1
-0.6-0.4-0.2 0 0.2 0.4 0.6
Example 35 κ(s) =1
1 + s2. Note κ(s) > 0 for all s, so curve always turns left and
has no inflexion points. Applying formula (C) again:
θ(α) =
∫ α
0
1
1 + β2dβ = arctanα
⇒ cos θ(α) =1
sec θ(α)=
1√1 + tan2 θ(α)
=1√
1 + α2
sin θ(α) = tan θ(α) cos θ(α) =α√
1 + α2
γ1(s) =
∫ s
0
1√1 + α2
dα = arcsinh(s)
γ2(s) =
∫ s
0
α√1 + α2
dα =√1 + s2 − 1
⇒ γ(s) =(arcsinh(s),
√1 + s2 − 1
).
In fact, using the reparametrization s = σ(t) = sinh t, this becomes
γ(t) = (γ ◦ σ)(t) = (t, cosh t− 1),
the graph of the cosh function shifted down one unit. �
It’s actually quite difficult to cook up curvature functions κ(s) for which the inte-grals in formula (C) are explicitly calculable. Even a seemingly simple choice such asκ(s) = s turns out to be intractable:
θ(α) =
∫ α
0
β dβ =1
2α2
γ1(s) =
∫ s
0
cosα2
2dα =???
γ2(s) =
∫ s
0
sinα2
2dα =???
What can we say about the geometry of this curve?
Reminder: A function f : I → R is
29
even if f(−t) = f(t) for all t ∈ Iodd if f(−t) = −f(t) for all t ∈ I.
Proposition 36 Let γ(s) be the PUSC of curvature κ(s) with γ(0) = 0, γ′(0) =(1, 0).
(a) If κ is even, γ is symmetric under reflexion in the x2 axis.
(b) If κ is odd, γ is symmetric under rotation by 180 degrees about (0, 0).
Proof: From formula (C) one sees that
θ(−α) =
∫ −α
0
κ(β)dβ = −∫ α
0
κ(−ξ)dξ
where ξ := −β. Hence κ even implies θ odd, while κ odd implies θ even. Similarly,
γ1(−s) =
∫ −s
0
cos(θ(α))dα = −∫ s
0
cos(θ(−η))dη
so if κ is odd or even (meaning θ is even or odd) then γ1 is odd. Further,
γ2(−s) =
∫ −s
0
sin(θ(α))dα = −∫ s
0
sin(θ(−η))dη
so if κ is even (θ odd) γ2 is even while if κ is odd (θ even) γ2 is odd. Summarizing:
κ even ⇒ γ(−s) ≡ (−γ1(s), γ2(s))
κ odd ⇒ γ(−s) ≡ (−γ1(s),−γ2(s))
and hence γ has the symmetry claimed. �
Applying Proposition 36 to κ(s) = s, an odd function, we see that the correspond-ing curve γ must have rotational symmetry about the origin. Also, γ has one and onlyone inflexion point: γ(0) = (0, 0). For s > 0, κ > 0 meaning the curve always turnsleftwards, and as s grows this turning gets tighter and tighter (|κ| is unbounded).The behaviour for s < 0 is determined by that for s > 0 by the symmetry property.
To get an idea of the specific shape of the curve γ, we can solve system (*) ap-proximately using an ODE solver package, in Python for example. Given a functionκ and an interval I = (s0, s1) with s0 < 0 < s1, the following Python code computesthe curve γ : I → R2 of curvature κ (γ(0) = 0, γ′(0) = (1, 0)) and then plots it. Webegin (as always) by importing some useful Python modules:
>>> import numpy as np
>>> import matplotlib.pyplot as plt
>>> from scipy.integrate import odeint
Numpy (which we abbreviate to np) gives us access to the trig functions sin andcos, and to numerical vectors; matplotlib is a plotting library; scipy.integrate has theODE solver we will use. We next define a Python function which solves the ODEsand plots the solution:
30
>>> def show_curve(kappa,s0,s1):
... Y0=[0,0,0]
... xr=np.linspace(0,s1,200)
... xl=np.linspace(0,s0,200)
... def dY_ds(Y, s):
... return [kappa(s), np.cos(Y[0]), np.sin(Y[0])]
... Yr=odeint(dY_ds,Y0,xr)
... Yl=odeint(dY_ds,Y0,xl)
... yr1=Yr[:,1]
... yr2=Yr[:,2]
... yl1=Yl[:,1]
... yl2=Yl[:,2]
... plt.plot(yr1,yr2,yl1,yl2)
... plt.axis(’equal’)
... plt.show()
...
To apply this to our case (κ(s) = s) one defines
>>> def kappa(s):
... return s
...
and then executes (for example)
>>> show_curve(kappa,-8,8)
The result is:
-2
-1
0
1
2
-2 -1 0 1 2
Note the curve has the predicted turning behaviour and symmetry. It is now straight-forward to turn the program loose on just about any curvature function. The resultscan be quite entertaining.
Example 37 Let κ(s) = s2 sin s. That is:
>>> def kappa(s):
... return s**2*np.sin(s)
...
31
Note that κ is again odd, so the corresponding curve must have rotational symme-try about the origin. Note also that γ(s) has infinitely many inflexion points, since κchanges sign at every s = mπ, wherem ∈ Z. Executing show_curve(kappa,-15,15);one obtains (below left):
-1
-0.5
0
0.5
1
-2 -1 0 1 2
0
0.5
1
1.5
2
2.5
3
-2 -1 0 1 2
Compare with κ(s) = s sin s, an even function:
>>> def kappa(s):
... return s*np.sin(s)
...
>>> show_curve(kappa,-8,8)
The corresponding curve is depicted above, right. Note the reflexion symmetry in thex2 axis. �
Example 38 Let κ(s) = s2−1 Note that κ is even , so the corresponding curvemust have reflection symmetry. Note also that γ(s) has exactly two inflexionpoints, since κ changes sign at s = ±1 . Executing
>>> def kappa(s):
... return s**2-1.0
...
>>> show_curve(kappa,-6,6)
one obtains:
-2
-1
0
1
2
-2 -1 0 1 2
32
It’s not hard to identify the inflexion points on the curve above.
Trick question: What’s the arc length along the curve from one inflexion point tothe next?Answer: �
Go forth and experiment
You are encouraged to experiment in Python with the function show_curve. To thisend, you will find a copy of its defining program linked from the course Minerva page.You should be able to cut and paste the program directly into a terminal runningPython, or a Python notebook. If you experience problems, please let me know.
The value of experimenting with show_curve is that it will help develop your intu-ition about the geometric relationship between the function κ and the correspondingcurve γ. Of course, in the final exam for this module, Python will not be availableto you. Hopefully the intuition you’ve developed will. In particular, relying heavilyon show_curve to complete Problem Sheet 2 would be a mistake. You have beenwarned...
33
Summary
• For a planar curve γ : I → R2 we can define the unit normal vector
n(t) = (−u2(t), u1(t)),
where u is the unit tangent vector.
• Since the curvature vector is parallel to n, there is a scalar function κ : I → R
called the signed curvature, such that
k(t) = κ(t)n(t).
• A convenient formula for κ(t) is
κ(t) =γ′′(t) · n(t)|γ′(t)|2 .
• If κ(t) > 0, the curve is turning to the left. If κ(t) < 0, the curve is turning tothe right.
• Given a function κ(s), there is a planar USC γ(s) whose signed curvature is κ(s).This curve is unique up to rigid motions.
• Symmetries of κ imply symmetries of γ (and vice versa).
34
4 New plane curves from old: evolutes, involutes and paral-
lels
Given one planar curve, there are a number of geometrically interesting ways togenerate new curves from it. In this section we will study three of these, showinghow their geometric properties are realted to those of the parent curve, and how thecurves themselves are related to one another.
4.1 The evolute of a plane curve
Throughout this section, γ : I → R2 will denote a planar RPC, not necessarily unitspeed, u : I → R2 will denote its unit tangent vector, n : I → R2 its unit normalvector and κ : I → R its signed curvature. First we need:
Definition 39 Given t0 ∈ I, the centre of curvature of γ at t = t0 is
c(t0) = γ(t0) +1
κ(t0)n(t0).
Note this definition only makes sense if κ(t0) 6= 0. �
Example 40 We saw in example 30 thatthe parabola γ(t) = (t, t2) has, at t = 0,
n(0) = (0, 1)
κ(0) = 2
⇒ c(0) =
(0,
1
2
)x1
x2
cH0L
. �
We can give a nice geometric interpretation of the centre of curvature in terms of thebehaviour of the normal lines to the curve γ.
Theorem 41 For t0 ∈ I fixed and t ∈ I variable, consider the normal lines to γthrough γ(t0) and γ(t). If κ(t0) 6= 0 and |t − t0| is sufficiently small then thesenormals intersect at some point α(t) ∈ R2. Then
limt→t0
α(t) = c(t0),
the centre of curvature of γ at t0. .
γ(t)
γ(t0)α(t)
c(t0)
35
To prove this, we’ll need a useful lemma:
Lemma 42 For all t ∈ I,
(a) n′(t) = −κ(t)γ′(t), (b) u′(t) = κ(t)|γ′(t)|n(t).
Proof: Since [u(t), n(t)] is an orthonormal pair of vectors, they form a (t dependent)basis for R2. So we can always express any R2 valued function as a linear combinationof u(t) and n(t). Applying this idea to n′(t), we see that there must exist smoothfunctions λ : I → R and µ : I → R such that
n′(t) = λ(t)u(t) + µ(t)n(t).
Taking the scalar product of both sides of this equation with u(t) gives
u · n′ = λ u · u︸︷︷︸ +µ u · n︸︷︷︸ = λ
1 0 (orthonormality)
Nowu · n = 0 ⇒ u′ · n+ u · n′ = 0 ⇒ u · n′ = −n · u′.
Hence
λ = −n · ddt
(γ′
|γ′|
)= −n ·
(γ′′
|γ′| − γ′d|γ′|−1dt
)= −n · γ′′
|γ′| = −κ|γ′|
Similarly,n · n′ = λn · u︸︷︷︸ +µn · n︸︷︷︸ = µ
0 1 (orthonormality)
andn · n = 1 ⇒ n′ · n + n · n′ = 0 ⇒ n · n′ = 0
so it follows that µ(t) = 0. Hence,
n′(t) = −κ(t)|γ′(t)|u(t) = −κ(t)γ′(t)
which proves part (a).Part (b) follows from similar reasoning, so we shall be more brief.
u′ = (u · u′)u+ (n · u′)n = 0u− (u · n′)n = κ|γ′|n
by part (a). �
We may now give a Proof of Theorem 41: We may give the normals through γ(t),γ(t0) the (unit speed) parametrizations (with parameters s and s0 respectively)
γ(t) + sn(t), γ(t0) + s0n(t0).
36
γ(t)
γ(t0)α(t)
c(t0)
Hence, their intersection point, which will depend on t, is
α(t) = γ(t) + s(t)n(t) = γ(t0) + s0(t)n(t0) (♣)
where s(t) and s0(t) are two unknown functions of t. Differentiating (♣) with respectto t and taking the limit t → t0, we find that
γ′(t0) + s′(t0)n(t0) + s(t0)n′(t0) = s′0(t0)n(t0). (♠)
Taking the scalar product of (♠) with u(t0) yields
|γ′(t0)|+ 0 + s(t0)n′(t0) · u(t0) = 0
⇒ |γ′(t0)| − s(t0)κ(t0)γ′(t0) · u(t0) = 0
⇒ s(t0) =1
κ(t0)
by Lemma 42. Hence
limt→t0
α(t) = γ(t0) + s(t0)n(t0) = γ(t0) +1
κ(t0)n(t0)
as was to be proved. �
Definition 43 The evolute of the planar curve γ is Eγ : I → R2, defined such that
Eγ(t) = γ(t) +1
κ(t)n(t),
in other words, it is the curve of centres of curvature of the curve γ. �
Notes:
• Eγ is only well defined (on the whole of I) provided γ has nonvanishing signedcurvature κ. In particular, γ must have no inflexion points, or else Eγ “escapesto infinity.”
• Even if κ(t) 6= 0 for all t ∈ I, the evolute may not be a RPC, as we will nowshow.
37
E ′γ(t) = γ′(t)− κ′(t)
κ(t)2n(t) +
1
κ(t)n′(t)
= γ′(t)− κ′(t)
κ(t)2n(t)− 1
κ(t)κ(t)γ′(t)
= − κ′(t)
κ(t)2n(t) ♦
by Lemma 42. So E ′γ(t) = (0, 0) if and only if κ′(t) = 0. Hence, if κ has criticalpoints, then Eγ is not regular. At such points, Eγ may exhibit “cusps”.
Example 44 (ellipse) γ(t) = (a cos t, sin t)
γ′(t) = (−a sin t, cos t)
n(t) =1√
cos2 t+ a2 sin2 t(− cos t,−a sin t)
γ′′(t) = (−a cos t,− sin t)
κ(t) =a
(cos2 t+ a2 sin2 t)3
2
Eγ(t) = (a cos t, sin t)− cos2 t+ a2 sin2 t
a(cos t, a sin t)
= (a cos t− a sin2 t cos t− 1
acos3 t, sin t− cos2 t sin t− a2 sin3 t)
= (a cos3 t− 1
acos3 t, sin3 t− a2 sin3 t)
=(a2 − 1
)(1
acos3 t,− sin3 t
)
a = 2 a = 1.3 a = 1.1
-3
-2
-1
0
1
2
3
-2 -1 0 1 2
-1
-0.5
0
0.5
1
-1 -0.5 0 0.5 1
-1
-0.5
0
0.5
1
-1 -0.5 0 0.5 1
38
In the limit a → 1, the ellipse degenerates to a circle, and its evolute collapses to asingle point (0, 0). �
We can relate the geometric quantities associated with Eγ to those of γ.
Theorem 45 Let uE, nE, σEt0
denote the unit tangent, unit normal and arclengthfunction of Eγ respectively. If κ′(t) < 0 for all t ∈ I, then
(a) uE(t) = n(t), (b) nE(t) = −u(t), (c) σEt0(t) =
1
κ(t)− 1
κ(t0).
Proof: (a) From (♦), we know that
E ′γ(t) = − κ′(t)
κ(t)2n(t)
so if κ′(t) < 0 then
|E ′γ(t)| = − κ′(t)
κ(t)2.
Hence uE = E ′γ/|E ′γ| = n.
(b) nE = (−uE2 , u
E1 ) = (−n2, n1) = (−u1,−u2) by part (a).
(c) Arclength is the integral of speed, which by part (a) is
σEt0(t) = −
∫ t
t0
κ′(q)
κ(q)2dq =
∫ t
t0
(1
κ
)′(q)dq =
1
κ(t)− 1
κ(t0)
which was to be proved. �
4.2 Involutes and parallels of a planar curve
Once again, let γ : I → R2 be a RPC, and let u, n, κ denote its unit tangent, unitnormal and signed curvature respectively. In this section we describe a different wayto generate new planar curves from γ which is in some sense the inverse of taking theevolute.
Definition 46 The involute of γ starting at t0 ∈ I is Iγ : I → R2, defined by
Iγ(t) = γ(t)− σt0(t)u(t).
Note that if γ(t) is a USC then σt0(t) = t− t0 so
Iγ(t) = γ(t)− (t− t0)γ′(t).
�
γ(t)
γ(t0)
γ(t)− σt0(t)u(t)
39
In the picture, the dashed line is tangent to γ at t. The point Iγ(t) = γ(t)−σt0(t)u(t)is the point on this line at distance σt0(t) from γ(t).
As an example of an involute, imagine that you have a spool with cotton threadwrapped around. Imagine tying a pencil to the end of the thread, and unwinding thethread, with the thread pulled taut at all times. The curve drawn by the end of thepencil is the involute of the circle.
Example 47 A circle γ(t) = (cos t, sin t).This is a USC, so the involute of γ basedat t = 0 is
Iγ(t) = (cos t, sin t)− t(− sin t, cos t)
= (cos t+ t sin t, sin t− t cos t)
. �
Note that Iγ has a cusp at t = 0 in this case. Accident? No: Iγ is never a RPC.
Iγ(t) = γ(t)− σt0(t)u(t)
⇒ I ′γ(t) = γ′(t)− σ′t0(t)u(t)− σt0(t)u′(t)
= γ′(t)− |γ′(t)|u(t)− σt0(t)κ(t)|γ′(t)|n(t)
= −σt0(t)κ(t)|γ′(t)|n(t) (♠)
where we have used Lemma 42. Hence I ′γ(t) = (0, 0) if and only if σt0(t) = 0 orκ(t) = 0 (since γ is a RPC, |γ′(t)| is never 0). So I ′γ(t) = (0, 0) when t = t0 andwhenever κ(t) = 0 (for example, where the contact point is an inflexion point of γ).
A useful analogy:
function f(x)derivative ւ ց integral
f ′(x)∫ x
x0f(q)dq
curve γevolute ւ ց involuteEγ Iγ
Fundamental Theorem of Calculus: derivative of integral is the original function f .
40
Theorem 48 Let Iγ be an involute of γ. Then the evolute EI of Iγ is γ.
The strategy of the proof is simple: we construct the evolute of Iγ, the involute ofγ starting at t0. To do this, we will need the unit normal and signed curvature of Iγ,so we begin by proving:
Lemma 49 Let γ be a unit speed curve and I be its involute based at time t0. ThenI has unit normal
nI(t) =(t− t0)κ
|t− t0||κ(t)|u(t),
and signed curvature
κI(t) =κ(t)
|κ(t)|1
|t− t0|,
where κ, u are the signed curvature and unit tangent vector of γ
Proof: Since γ has unit speed,
I ′(t) = −(t− t0)κ(t)n(t) = −(t− t0)k(t) (♣)
by equation (♠). Hence I has unit tangent
uI =I ′
|I ′| = − (t− t0)κ
|t− t0||κ|n
and so, rotating this 90 degrees anticlockwise yields
nI = (−uI2, u
I1) =
(t− t0)κ
|t− t0||κ|u.
Differentiating (♣) givesI ′′ = −k − (t− t0)k
′,
whence the signed curvature of I is
κI =I ′′ · nI
|I ′γ|2= − (t− t0)κ
|t− t0|3|κ|3[k + (t− t0)k
′] · u
= − κ
|t− t0||κ|3k′ · u
since k · u = 0 by Fact 25. But
k · u = 0 ⇒ k′ · u+ k · u′ = 0 ⇒ k′ · u = −k · u′ = −|k|2 = −κ2
since γ is a USC (so u = γ′ and u′ = γ′′ = k). Hence
κI =κ
|κ|1
|t− t0|,
as was claimed. �
Theorem 48 now follows immediately:
41
Proof of Theorem 48: By Theorem 17 we may assume without loss of generality thatγ is a USC. Then its involute I based at t0 has evolute
EI(t) = I(t) +1
κI(t)nI(t)
= [γ(t)− (t− t0)u(t)] +
[ |κ(t)|κ(t)
|t− t0|]×[(t− t0)κ(t)
|t− t0||κ(t)|u(t)
]
= γ(t)
by Lemma 49. �
So taking the evolute “undoes” the involute, just as differentiation “undoes” integra-tion. What about the converse? If we first differentiate f(x), then take a definiteintegral of f ′, we don’t necessarily get the same function f(x) back again – it couldbe shifted by a constant c. For example:
f(x) = cosx ⇒ f ′(x) = − sin x
⇒∫ x
0
f ′(q)dq = −∫ x
0
sin q dq = cosx− 1 = f(x)− 1
Thinking of the graphs of the functions, we could say that integrating the derivativein general gives a shifted, or “parallel”, function. An analogous statement holds forcurves too, i.e. if we take an involute of the evolute of γ, the result is a curve “parallel”to γ, in the following precise sense:
Definition 50 Given a RPC γ : I → R2 and a constant λ ∈ R, the curve γλ : I → R2
defined byγλ(t) = γ(t) + λn(t)
is a parallel curve to γ. �
Note the similarity between γλ and the evolute of γ,
Eγ(t) = γ(t) + κ(t)−1n(t).
Under what circumstances is γλ regular?
Lemma 51 The parallel γλ to γ is a RPC if and only if κ(t) 6= 1/λ for all t ∈ I. Inother words, 1/λ must lie outside the range of the function κ.
Proof:γ′λ(t) = γ′(t) + λn′(t) = γ′(t)− λκ(t)γ′(t)
by Lemma 42. Hence γ′λ(t) = (0, 0) if and only if κ(t) = 1/λ. �
42
Example 52 Let’s construct the general parallel curve to the parabola γ(t) = (t, t2):
γ′(t) = (1, 2t)
n(t) =1√
1 + 4t2(−2t, 1)
γλ(t) = (t, t2) +λ√
1 + 4t2(−2t, 1)
0
1
2
3
4
-2 -1 0 1 2
Which values of the constant λ give regular parallels? Example 30 ⇒
κ(t) =2
(1 + 4t2)3
2
.
Hence κ has range (0, 2], so γλ is regular if and only if 1/λ < 0 (note 1/λ cannot equal0) or 1/λ > 2. That is, γλ is regular if and only if λ < 1
2.
�
Theorem 53 Let γ have evolute Eγ. Then every involute IE of Eγ is a parallel curveto γ.
Proof: We simply construct the involute of Eγ starting at t0 ∈ I. To do this we willneed the arclength function σE
t0(t) and the unit tangent uE(t) of Eγ. We shall assumethat κ′(t) < 0 (a similar argument works for the case κ′(t) > 0). Recall that in thiscase,
σt0(t) =1
κ(t)− 1
κ(t0)and uE(t) = n(t)
by Theorem 45, parts (c) and (a). Hence,
IE(t) = Eγ(t)− σE(t)uE(t)
= γ(t) +1
κ(t)n(t)−
(1
κ(t)− 1
κ(t0)
)n(t)
= γ(t) +1
κ(t0)n(t).
But this is just γλ where λ = 1/κ(t0). �
43
Summary
• The curve obtained from γ by tracing out the locus of its centres of curvature iscalled the evolute of γ. Explicitly
Eγ(t) = γ(t) +1
κ(t)n(t).
• The involute of γ based at t0 ∈ I is
Iγ(t) = γ(t)− σt0(t)u(t).
• A parallel to γ is a curve
γλ(t) = γ(t) + λn(t)
where λ ∈ R is a constant.
• The evolute of an involute of γ is γ. Every involute of the evolute of γ is aparallel to γ.
• The regularity properties of evolutes and parallels can be analyzed in terms ofthe curvature properties of γ.
44
5 Curves in R3 and the Frenet frame
5.1 The Frenet frame
Planar curves (γ : I → R2) are special because given one vector in R2, the unittangent vector u = γ′/|γ′|) say, one can uniquely determine an orthogonal one byrotating 90◦ anticlockwise, the unit normal n in this case.Curves in R3 (γ : I → R3) are also special: given an ordered pair of vectors in R3,e.g. u = γ′/|γ′|, k = γ′′⊥/|γ′|2, one can uniquely determine a third orthogonal to bothby using the vector product (u× k).
Reminder 54 Given vectors u = (u1, u2, u3), v = (v1, v2, v3) ∈ R3, their vectorproduct is
u× v = (u2v3 − u3v2, u3v1 − u1v3, u1v2 − u2v1).
It has the following properties:
(a) v × u = −u× v
(b) For all λ, µ ∈ R, (λu+ µv)× w = λ(u× w) + µ(v × w)
(c) If u is parallel to v (i.e. u = λv) then u× v = 0 [follows from (a), (b)]
(d) |u× v|2 = |u|2|v|2 − (u · v)2.(e) u · (v × w) = w · (u× v) = v · (w × u)
(f) u× v is orthogonal to both u and v [follows from (e)] �
Definition 55 Let γ : I → R3 be a RPC whose curvature never vanishes (i.e. for allt ∈ I, |k(t)| 6= 0). Then in addition to the unit tangent vector
u(t) :=γ′(t)
|γ′(t)|one defines the principal unit normal vector
n(t) :=k(t)
|k(t)|and the binormal vector
b(t) := u(t)× n(t).
The ordered triplet [u(t), n(t), b(t)] is called the Frenet frame of the curve. γ �
Note: we call n(t) the principal unit normal to distinguish it from the infinitelymany other unit vectors lying in the plane orthogonal to u(t). This is only possibleif k(t) 6= 0.
Lemma 56 The Frenet frame is orthonormal (the vectors u, n, b are mutually orthog-onal and each has unit length).
45
Proof: |u(t)| = 1 and |n(t)| = 1 for all t by definition. Also n(t) is parallel to k(t)which is orthogonal to u(t) by Fact 25. It remains to show that (i) |b(t)| = 1 and (ii)b(t) is orthogonal to u(t) and n(t). But (i) follows from Reminder 54(d),
|b|2 = |u× n|2 = |u|2|n|2 − (u · n)2 = 1× 1− 02 = 1,
and (ii) follows directly from Reminder 54(f). �
So given any regularly parametrized curve of nonvanishing curvature (RPCNVC)γ : I → R3, the Frenet frame [u, n, b] forms an orthonormal basis for the vector spaceR3.
Example 57 Construct the Frenet frame at t = 0 for the curve γ : R → R3,γ(t) = ( t
2
2, t
3
3, t). Note: this is only possible provided k(0) 6= 0!
γ′(t) = (t, t2, 1)
γ′′(t) = (1, 2t, 0)
γ′(0) = (0, 0, 1)
γ′′(0) = (1, 0, 0)
γ′′⊥(0) = (1, 0, 0)
k(0) =γ′′⊥(0)
|γ′(0)|2 = (1, 0, 0)
0
0.5
1
1.5
2
2.5
-3
-2
-1
0
1
2
3
-2
-1
0
1
2
which is nonzero, so the Frenet frame is well-defined. So
u(0) =γ′(0)
|γ′(0)| = (0, 0, 1)
n(0) =k(0)
|k(0)| = (1, 0, 0)
b(0) = u(0)× n(0)
= (0, 1, 0)
�
46
5.2 Torsion of a unit speed curve in R3
Our aim is to describe the geometry of curves in R3 by analysing the time dependenceof the Frenet frame. This process simplifies greatly if the curve under consideration isa unit speed curve, so henceforth, we will consider only USCs. Note that this entailsno loss of generality by Theorem 17: every RPC has a unit speed reparametrization,unique up to time translation.
The construction of the Frenet frame for a USC γ : I → R3 simplifies somewhatbecause γ′ is already a unit vector, and the curvature k = γ′′. Hence,
u(s) = γ′(s), n(s) =u′(s)
|u′(s)| , b(s) = u(s)× n(s).
It is conventional in the context of curves in R3 to denote the scalar curvature byκ(s) rather than |k(s)|. This should not be confused with the signed curvature of acurve in R2. Here κ(s) just means the length of the curvature vector k(s), which ofcourse is never negative. Noting that k(s) = u′(s) for a USC, we may re-write thedefinition of the principal unit normal as
u′(s) = κ(s)n(s) (1)
Since the Frenet frame [u(s), n(s), b(s)] spans R3 we should be able to find similarformulae for n′(s) and b′(s). The coefficient functions we extract should tell us aboutthe geometry of γ, just as κ does. We start with b′(s). Since it’s a 3-vector, theremust exist smooth functions λ, µ, ν : I → R such that
b′(s) = λ(s)u(s) + µ(s)n(s) + ν(s)b(s)
Taking the scalar product of both sides with b(s) gives:
b · b′ = λ b · u︸︷︷︸ +µ b · n︸︷︷︸ +ν b · b︸︷︷︸ = ν
0 0 1 (orthonormality)
Butb · b = 1 ⇒ b′ · b+ b · b′ = 0 ⇒ b′ · b = 0
and hence ν = 0. Similarly,
u · b′ = λ u · u︸︷︷︸ +µ u · n︸︷︷︸ +ν u · b︸︷︷︸ = λ
1 0 0 (orthonormality)
But
u · b = 0 ⇒ u′ · b+ u · b′ = 0 ⇒ u · b′ = −u′ · b= −κn · b (by eqn. (1))
= 0.
Hence λ = 0 also. It follows that
b′(s) = µ(s)n(s)
for some function µ. We call −µ(s) the torsion of the curve.
47
Definition 58 Let γ : I → R3 be a USCNVC. The torsion of the curve is thatfunction τ : I → R defined by the equation
b′(s) = −τ(s)n(s) (2)
Alternatively, τ(s) = −b′(s) · n(s). �
Example 59 Construct the Frenet frame and torsion for the helix γ : R → R3,γ(s) = (cos s√
2, sin s√
2, s√
2).
First check it’s a USCNVC:
γ′(s) =
(− 1√
2sin
s√2,1√2cos
s√2,1√2
)
|γ′(s)| =
(1
2cos2
s√2+
1
2sin2 s√
2+
1
2
) 1
2
= 1
k(s) = γ′′(s) =
(−1
2cos
s√2,−1
2sin
s√2, 0
)
OK. Now compute [u, n, b]:
u(s) = γ′(s) =1√2
(− sin
s√2, cos
s√2, 1
)
n(s) =u′(s)
|u′(s)| =(− cos
s√2,− sin
s√2, 0
)
b(s) = u(s)× n(s) =1√2
(sin
s√2,− cos
s√2, 1
)
To find τ(s), we compute b′(s) and com-pare with n(s):
b′(s) =1
2
(cos
s√2, sin
s√2, 0
)= −1
2n(s)
τ(s) =1
2�
-1
-0.5
0
0.5
1
-1-0.5
00.5
1
-6
-4
-2
0
2
4
6
8
The rest of the material in section 5.2 is non-examinable, and included purely foryour delight and edification.
What is the geometric meaning of torsion? Recall that the curvature κ = |u′|measures the rate of change of the direction of the tangent line to the curve. Torsionhas a similar interpretation, but in terms of planes rather than lines.
At the point γ(s0) we define the osculating plane of the curve to be that planethrough γ(s0) spanned by the orthonormal pair [u(s0), n(s0)], or equivalently, by the
48
velocity γ′(s0) and acceleration γ′′(s0) of the curve. The orientation of this planeis uniquely determined by any vector normal to it, for example the binormal vectorb(s0) = u(s0) × n(s0). Consider the Taylor expansion of the curve γ(s) based at thetime s0:
γ(s) = γ(s0) + γ′(s0)(s− s0) +1
2γ′′(s0)(s− s0)
2
︸ ︷︷ ︸stays in osculating plane
+1
6γ′′′(s0)(s− s0)
3 + · · ·
The failure of γ(s) to stay in the osculating plane is controlled (locally) by γ′′′(s0).The osculating plane divides R3 in half: we can classify any vector v not in theplane as positive if v · b(s0) > 0 and negative if v · b(s0) < 0. Thinking of theosculating plane as horizontal, with b(s0) pointing “up”, positive vectors point onthe upside of the plane, negative vectors on the downside. So if γ′′′(s0) · b(s0) > 0,the curve punctures the osculating plane upwards as it passes through γ(s0), while ifγ′′′(s0) · b(s0) < 0, it punctures the osculating plane downwards. What does this haveto do with torsion?
Proposition 60 Let γ : I → R3 be a USCNVC. Then τ(s) = (γ′′′(s) · b(s))/κ(s).Proof: Recall b = u× n = γ′ × (γ′′/|γ′′|). Hence
b′ = γ′′ ×(
γ′′
|γ′′|
)+ γ′ ×
[γ′′′
|γ′′| −γ′′′ · γ′′|γ′′|3 γ′′
]
= 0 + u×[γ′′′
κ− γ′′′ · n
κn
].
But b′ = −τn, so
τ = −n · b′ = −n ·[u× γ′′′
κ
]=
γ′′′
κ· (u× n) =
γ′′′
κ· b.
�
So if τ(s0) > 0, the curve punctures its osculating plane at γ(s0) upwards, and ifτ(s0) < 0 it punctures downwards.
u
nb
u
nb
τ > 0 τ < 0To test your understanding, see if you can tell whether the torsion of the following
curves is positive, negative or zero. (Just ask me if you want to know the answers.)
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0
0.5
1
1.5
2
2.5
-3
-2
-1
0
1
2
3
-2
-1
0
1
2
0
0.5
1
1.5
2
2.5
-3
-2
-1
0
1
2
3
-2
-1
0
1
2
Note: the sign of τ does not depend on the orientation of the curve, that is, thedirection in which it is traversed.
-2
-1
0
1
2
-3-2
-10
12
3
-3
-2
-1
0
1
2
3
-1
-0.5
0
0.5
1
-1
-0.5
0
0.5
1
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5.3 The Frenet formulae
Recall that the curvature κ and torsion τ of a USCNVC in R3 can be extracted bydecomposing u′(s) and b′(s) relative to the Frenet frame [u(s), n(s), b(s)]:
u′(s) = κ(s)n(s) — (1) b′(s) = −τ(s)n(s) — (2)
What about n′(s)? Does this give us yet another scalar quantity analogous to κ andτ? In fact, it does not, as we shall now show.
As before, since n′ : I → R3 is a smooth, vector valued function, there must existsmooth scalar functions λ, µ, ν : I → R such that
n′(s) = λ(s)u(s) + µ(s)n(s) + ν(s)b(s),
and, since [u, n, b] is orthonormal (Lemma 56), we may extract the coefficients bytaking scalar products:
λ(s) = u(s) · n′(s), µ(s) = n(s) · n′(s), ν(s) = b(s) · n′(s).
• Now u(s) · n(s) = 0 for all s, which upon differentiating with respect to s yields
0 = u′(s) · n(s) + u(s) · n′(s) = κ(s)n(s) · n(s) + u(s) · n′(s) = κ(s) + λ(s),
using equation (1). Hence λ(s) = −κ(s).
• Similarly, n(s) · n(s) = 1 for all s, and differentiating yields
0 = n′(s) · n(s) + n(s) · n′(s) = 2µ(s).
Hence µ(s) = 0.
• Finally, b(s) · n(s) = 0 implies
0 = b′(s) · n(s) + b(s) · n′(s) = −τ(s)n(s) · n(s) + b(s) · n′(s) = −τ(s) + ν(s),
using equation (2). Hence ν(s) = τ(s).
Assembling the pieces, we have the formula
n′(s) = −κ(s)u(s) + τ(s)b(s) — (3)
Taking (1),(2),(3) together, we have just proved:
Theorem 61 (The Frenet formulae) Let γ : I → R3 be a unit speed curve ofnonvanishing curvature. Then its Frenet frame satisfies the formulae
u′(s) = κ(s)n(s)n′(s) = −κ(s)u(s) +τ(s)b(s)b′(s) = −τ(s)n(s) �
We will use the Frenet formulae to prove two fundamental results:
• That a USCNVC in R3 is planar if and only if its torsion is zero.
51
• That a USCNVC in R3 is uniquely determined by its scalar curvature and torsion(almost).
Definition 62 A plane P ⊂ R3 is the set of points x ∈ R3 satisfying the equation
B · x = ν,
where B ∈ R3 and ν ∈ R are constants, and B is a unit vector (|B| = 1). Geometri-cally, B determines the orientation of the plane P , while |ν| is its distance from theorigin, (0, 0, 0). [Note that the pairs (B, ν) and (−B,−ν) describe the same plane.]
A curve γ : I → R3 is planar if its image is contained in some plane P , that is, ifthere exist constants B ∈ R3, |B| = 1 and ν ∈ R, such that
B · γ(s) = ν
for all s ∈ I
Example 63 The USCNVC γ(s) = 12(1+2 sin s,
√3+cos s, 1−
√3 cos s) is a planar
curve. �
How on earth do I know this? One way to settle the issue is to use:
Theorem 64 Let γ : I → R3 be a USCNVC. Then γ is planar if and only if τ(s) = 0for all s ∈ I.
Proof: We must prove: (A) γ planar ⇒ τ = 0(B) τ = 0 ⇒ γ planar
(A) Assume γ is planar. Then there exist constants B ∈ R3 and ν ∈ R, as indefinition 62 such that, for all s ∈ I,
γ(s) · B = ν
Diff. w.r.t. s: u(s) · B = 0
Diff. w.r.t. s again: k(s) · B = 0
⇒ κ(s)n(s) · B = 0
⇒ n(s) · B = 0
since γ has nonvanishing curvature. So B is a unit vector, and is orthogonal to bothu(s) and n(s) for all s. Compare this with b(s) = u(s)× n(s). This has precisely thesame properties. Furthermore, in R3, the unit vector orthogonal to both u(s) andn(s) is unique up to sign. Hence
b(s) = ±B,
a constant vector. Hence b′(s) = 0 and so τ(s) = 0 for all s ∈ I by the last Frenetformula.
(B) Assume τ ≡ 0. Then b′ ≡ 0 and hence b(s) = b0, a constant unit vector (by theMean Value Theorem). But then
d
ds(γ(s) · b0) = u(s) · b0 = u(s) · b(s) = 0.
52
Hence γ(s) · b0 = a, some constant (by the Mean Value Theorem again). So γ lies inthe plane determined by the unit vector B = b0 and the constant ν = a. �
[Note: we only used the first and last Frenet formulae in this proof.]
Example 63 (revisited) Let’s use theorem 64 to show that
γ(s) =1
2(1 + 2 sin s,
√3 + cos s, 1−
√3 cos s)
is planar. First, check it’s a USCNVC:
γ′(s) =1
2(2 cos s,− sin s,
√3 sin s)
|γ′(s)| =1
2(4 cos2 s+ sin2 s+ 3 sin2 s)
1
2 = 1
k(s) = γ′′(s) =1
2(−2 sin s,− cos s,
√3 cos s)
κ(s) = |k(s)| =1
2(4 sin2 s+ cos2 s+ 3 cos2 s)
1
2 = 1
Now construct its Frenet frame, compute b′(s) and extract τ :
u(s) = γ′(s) =1
2(2 cos s,− sin s,
√3 sin s)
n(s) =k(s)
|k(s)| =1
2(−2 sin s,− cos s,
√3 cos s)
b(s) = u(s)× n(s) =1
4(0,−2
√3,−2)
=1
2(0,−
√3,−1)
b′(s) = (0, 0, 0)
τ(s) = 0
and hence γ is planar.
So what kind of curve is this?
• It’s planar.
• It has constant curvature.
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Is it, perhaps, a circle? The answer is yes, but we will need to develop another pieceof theory to demonstrate this.
Recall that a USC in R2 is uniquely determined by its signed curvature κ and itsinitial position γ0 and initial velocity (cos θ0, sin θ0) (Theorem 33). It turns out thata similar result holds for a USCNVC in R3, except now we have to specify the scalarcurvature κ, the torsion τ , the initial position γ0, the initial velocity u0 and the initialnormal vector n0.
Theorem 65 Given smooth functions κ : I → (0,∞), τ : I → R (0 ∈ I) andconstants γ0, u0, n0 ∈ R3, |u0| = |n0| = 1, u0 · n0 = 0, there exists a unique unit speedcurve γ : I → R3 with
(a) γ(0) = γ0 (b) γ′(0) = u0 (c) n(0) = n0 (d) curvature κ (e) torsion τ .
Partial proof: Existence – too hard. Uniqueness: Let γ, γ be any pair of curvessatisfying (a)–(e). We will show that γ(s) = γ(s) for all s ∈ I. Since both curveshave identical torsion and (nonvanishing) curvature, their Frenet frames, call them
[u, n, b] and [u, n, b] respectively, both satisfy the Frenet formulae for κ and τ . Itfollows that
1
2
d
ds
{|u− u|2 + |n− n|2 + |b− b|2
}
= (u− u) · (u′ − u′) + (n− n) · (n′ − n′) + (b− b) · (b′ − b′)
= −[u · u′ + u′ · u+ n · n′ + n′ · n + b · b′ + b′ · b]= −[u · κn+ κn · u+ n · (−κu + τ b) + (−κu+ τb) · n+ b · (−τn)− τn · b]= 0
by the Frenet formulae. Hence
|u− u|2 + |n− n|2 + |b− b|2 = C (1)
some constant. Substituting s = 0 in (1) and using properties (b) and (c), one seesthat C = 0. Since all terms on the left hand side of (1) are non-negative, it followsthat each is identically zero. Hence |u(s)− u(s)| = 0 for all s. It follows that for alls,
d
ds(γ(s)− γ(s)) = 0
⇒ γ(s)− γ(s) = v
a constant vector, which by property (a) must be 0. Hence γ ≡ γ. �
A less formal way of stating this result is that, up to rigid motions (translations androtations of R3), a USCNVC is uniquely determined by κ(s) and τ(s).
So, since a circle has constant scalar curvature (Example 34) and zero torsion (it’splanar), Example 63 must be a circle, by Theorem 65.
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Summary
• Given a RPC γ : I → R3 of nonvanishing curvature we define its unit tangentvector u(t), principal unit normal n(t) and binormal b(t) by
u(t) =γ′(t)
|γ′(t)| , n(t) =k(t)
|k(t)| , b(t) = u(t)× n(t).
This triple of vectors forms an orthonormal basis for R3 called the Frenet frame.
• The curvature |k(t)| is usually denoted κ(t). Note κ ≥ 0.
• For a unit speed curve γ : I → R3 of nonvanishing curvature we define thetorsion τ : I → R by
b′(s) = −τ(s)n(s).
• A USCNVC is planar if and only if τ ≡ 0.
• The rate of change of the Frenet frame as one travels along a USCNVC is deter-mined by the torsion and curvature according to the Frenet formulae
u′(s) = κ(s)n(s)n′(s) = −κ(s)u(s) +τ(s)b(s)b′(s) = −τ(s)n(s)
• A USCNVC is uniquely determined (up to rigid motions) by its curvature andtorsion.
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6 Regularly parametrized surfaces
6.1 The basic definition
Our approach to studying the geometry of curves in Rn was to think of them assmooth maps γ : I → Rn. Recall that not every such map gives a nice, smooth,regular curve. We needed to impose a constraint on the derivative of γ, namely γ′(t)should never vanish, in order to guarantee the curve was well behaved.
We now want to develop a similar method for studying surfaces, roughly speaking,nice smooth two-dimensional sets in R3. Actually, the techniques we will studygeneralize quite easily to work for (n − 1)-dimensional surfaces in Rn for any n, butwe will stick to two-dimensional surfaces in R3 because these are easiest to visualize.Again, we will think of surfaces as smooth maps φ : U → R3, where U is (a subset of)R2, so a surface will be a R3-valued function of two variables, φ(u, v). Once again,not every such map gives a nice smooth surface, and we need to impose constraints onthe behaviour of φ(u, v) to guarantee its image set is well behaved. Just as for curves,the constraints amount to the requirement that the map φ : U → R3 be regular, butto make this notion precise requires a little more work.
First, we identify the class of domains U we will be using.
Definition 66 For each x ∈ R2 and r > 0 let Br(x) = {y ∈ R2 : |y − x| < r} bethe disk of radius r centred at x. Then a subset U ⊆ R2 is open if for all x ∈ Uthere exists δ > 0 such that Bδ(x) ⊆ U . �
The idea is that no point in U lies at the edge of U , because every point can besurrounded by some disk in U . This is a generalization of the idea of an open intervalin R (one which has no endpoints).
Example 67 The sets
(−1, 1)× R, (−π, π)× (−1, 1), {x ∈ R2 : 1 < |x| < 2}, {x ∈ R
2 : x1 > −1}
are all open sets, while
[−1, 1)× R, (−π, π)× (−1, 1], {x ∈ R2 : 1 ≤ |x| ≤ 2}, {x ∈ R
2 : x1 = −1}
56
are not. �
Definition 68 Given a smooth map φ : U → R3, where U is an open subset of R2,the coordinate basis vectors φu, φv : U → R3 are
φu(u, v) =∂φ
∂u(u, v), φv(u, v) =
∂φ
∂v(u, v).
A point (u, v) ∈ U is a regular point of φ if the vectors φu(u, v) and φv(u, v) arelinearly independent. The map φ is regular if every (u, v) ∈ U is a regular point. �
Since φu(u, v), φv(u, v) are just two vectors, they are linearly independent if andonly if they are not parallel. Two vectors in R3 are parallel if and only if their vectorproduct vanishes, so another (more convenient) way to state Definition 68 is
Definition 68 (*) φ : U → R3 is regular if for all (u, v) ∈ U ,
φu(u, v)× φv(u, v) 6= (0, 0, 0).
Definition 69 A regularly parametrized surface (RPS) is an injective (i.e. one-to-one), regular map φ : U → R3. �
We will often denote the image set of the map by M :
M := φ(U) = {φ(u, v) ∈ R3 : (u, v) ∈ U}.
Example 70 (The graph of a function) Let f : R2 → R be any smooth function.Then
φ : R2 → R3, φ(u, v) = (u, v, f(u, v))
is a RPS. Let’s check it:
φ(u, v) = φ(u′, v′) =⇒ (u, v, f(u, v)) = (u′, v′, f(u′, v′))
=⇒ u = u′ (from first component) and
v = v′ (from second component).
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So φ is injective. Furthermore
φu =
(1, 0,
∂f
∂u
)
φv =
(0, 1,
∂f
∂v
)
⇒ φu × φv =
(−∂f
∂u,−∂f
∂v, 1
).
The third component never vanishes, so φu × φv never vanishes and φ is regular.Hence φ is a RPS. It is known as the graph of the function f .
x1 x2
x3
(u, v, 0)
(u, v, f(u, v))
Example 71 (A sphere (almost)) Let U = (0, π)× (0, 2π). Note this is an opensubset of R2 (it is a rectangle without boundary). Consider the mapping
φ : U → R3, φ(u, v) = (sin u cos v, sin u sin v, cosu)
I claim this also is a RPS. Let’s check:
(A) Is φ injective?Suppose φ(u, v) = φ(u′, v′). Then
(1) sin u cos v = sin u′ cos v′ (2) sin u sin v = sin u′ sin v′ (3) cosu = cosu′
(3) implies that u = u′, because 0 < u, u′ < π and cos is strictly decreasing on thisinterval. Then sin u = sin u′ 6= 0, so we can divide through in (1) and (2) to obtain
(1’) cos v = cos v′ (2’) sin v = sin v′
These imply that v′ = v + 2nπ for some n ∈ Z. However, 0 < v, v′ < 2π so in factv = v′.
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(B) Is φ regular?
φu(u, v) = (cosu cos v, cosu sin v,− sin u)
φv(u, v) = (− sin u sin v, sin u cos v, 0)
φu × φv = (sin2 u cos v, sin2 u sin v, sin u cosu)
= sin u(sin u cos v, sin u sin v, cosu)
|φu × φv|2 = sin2 u(sin2 u cos2 v + sin2 u sin2 v + cos2 u)
= sin2 u(sin2 v + cos2 v)
= sin2 u
Since sin2 u 6= 0 for 0 < u < π, φu × φv never vanishes.What does the image M = φ(U) of this RPS look like? Notice that
|φ(u, v)|2 = sin2 u cos2 v + sin2 u sin2 v + cos2 u = sin2 u+ cos2 u = 1.
So the image of φ is a subset of the sphere of radius 1. Is the image the whole of thesphere?
Consider fixing u = u0 ∈ (0, π) and varying v. Doing so traces out a curve in R3.The third component of φ(u, v) doesn’t depend on v, so this curve is contained in thehorizontal plane x3 = cosu0. The first two components are (sin u0 cos v, sin u0 sin v),and these trace out a circle of radius sin u0 as v varies in the interval (0, 2π). However,v is not allowed to equal 0 or 2π, so the point
(sin u0 cos 0, sin u0 sin 0, cosu0) = (sin u0, 0, cosu0)
is missing from the circle.Now consider fixing v = v0 ∈ (0, 2π) and varying u. This traces out a vertical
semicircular arc on the surface of the sphere as u varies in the interval (0, π). Howeverthe points with u = 0 and u = π are missing, and these are
(0, 0, 1) and (0, 0,−1).
Putting things together, the image of φ is the whole of the two-sphere except for thesemi-circular arc described by
{(sin u, 0, cosu) : 0 ≤ u ≤ π}.Consider again the curve γ(v) = φ(u0, v) for fixed u0. The tangent vector to this
curve is just
γ′(v) =∂φ
∂v(u0, v) = φv(u0, v).
So φv can be interpreted as the velocity of a curve obtained by varying v with uheld constant. This observation should help you to visualise φv. Similarly, φu is thevelocity of a curve obtained by varying u with v constant.
59
This picture of the sphere illustratesthe preceding discussion. The curvesdrawn on the surface are the curves ofconstant u and of constant v. The dashedline is the semi-circular arc that is miss-ing from our parametrization. The twoarrows are the vectors φu and φv at a par-ticular point φ(u, v).
Remark 72 A similar idea applies to any RPS M : U → R3. If we fix v = b, someconstant, we get a curve αb(u) = φ(u, b) along which u varies. Changing the fixedvalue b generates a family of curves which sweep out the surface M = φ(U). Similarlyif we fix u = a, some constant, we get a curve βa(v) = φ(a, v) along which v varies.Changing the fixed value a generates a family of curves which sweep out the surfaceM . Furthermore
α′b(u) =∂
∂uφ(u, b) = φu(u, b), β ′a(v) =
∂
∂vφ(a, v) = φv(a, v)
which gives a geometric interpretation of the coordinate basis vectors.
Remark 73 Since a RPS φ : U → R3 is one-to-one, given a point p = (p1, p2, p3) onM = φ(U), there is one and only one point (u, v) ∈ U such that φ(u, v) = (p1, p2, p3).We call the numbers (u, v) the local coordinates of the point p. In this way, we canspecify a point on M by giving two numbers (its local coordinates) rather than three(its coordinates in R3), in much the same way that we can specify a point on a curveγ(t) by giving the single number t.
Example 70 (revisited) For the sphere
φ : (0, π)× (0, 2π) → R3, φ(u, v) = (sin u cos v, sinu sin v, cosu)
what are the local coordinates of the point p = (0, 1, 0)? We seek u ∈ (0, π) andv ∈ (0, 2π) such that
(0, 1, 0) = φ(u, v) = (sin u cos v, sin u sin v, cosu)
3rd component =⇒ cosu = 0 =⇒ u = π2.
2nd component =⇒ sin v = 1 =⇒ v = π2.
1st component: check sin u cos v = sin π2cos π
2= 0 as required.
So (u, v) =(π2, π2
).
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6.2 The tangent and normal spaces
Recall that every point on a regularly parametrized curve has a well defined tangentline. The generalization of this to regularly parametrized surfaces is called the tangentspace:
Definition 74 Let p ∈ M be a point on a RPS M = φ(U) where φ : U → R3. Thena curve in M through p is a smooth map α : I → M (0 ∈ I) with α(0) = p. Thetangent space to M at p ∈ M is
TpM = {x ∈ R3 : there exists a curve α in M through p with α′(0) = x}.
Any x ∈ TpM is called a tangent vector to M at p. �
So the tangent space at p is the set of all possible velocity vectors α′(0) ∈ R3 of curvesα passing through p. Note that the curve α is not assumed to be regular.
p
αα′(0)
Example 75 (paraboloid) Let φ : R2 → R3, φ(u, v) = (u, v, u2 + v2) and p =(2, 1, 5) = φ(2, 1). This the graph of the smooth function f : R2 → R, f(u, v) =u2 + v2, and hence is a RPS (it’s a particular case of Example 70). The easiest wayto define a curve through p in M is to write it as
α(t) = φ(u(t), v(t)).
The function α : I → R2 α(t) = (u(t), v(t)) is known as the coordinate expression forα. In order to be a curve through p, this must satisfy
α(0) = (2, 1)
where (2, 1) are the coordinates of p. For example,
α(t) = (2, 1 + t) ⇒ α(t) = φ(α(t)) = (2, 1 + t, 5 + 2t+ t2)
β(t) = (2 + t, 1− t) ⇒ β(t) = φ(β(t)) = (2 + t, 1− t, 5 + 2t+ 2t2)
γ(t) = (2 + t2, 1 + cos t) ⇒ γ(t) = φ(α(t)) = (2 + t2, 1 + cos t, (2 + t2)2 + (1 + cos t)2)
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all give curves through p in M . The corresponding tangent vectors are:
α′(0) = (0, 1, 2)
β ′(0) = (1,−1, 2)
γ′(0) = (0, 0, 0).
[Note that γ is not a RPC]. �
A handy way to calculate with tangent vectors is to express them in terms of thecoordinate basis vectors φu, φv. If α is a curve through p = φ(u, v), then
α(t) = φ(u(t), v(t))
⇒ α′(0) = u′(0)∂φ
∂u
∣∣∣∣(u(0),v(0))
+ v′(0)∂φ
∂v
∣∣∣∣(u(0),v(0))
= u′(0)φu(u, v) + v′(0)φv(u, v).
So every tangent vector can be expressed as a linear combination of φu(u, v), φv(u, v).Furthermore, every linear combination of φu(u, v), φv(xu, v) is a tangent vector, as weshall now prove. Let x = aφu(u, v) + bφv(u, v) where a, b ∈ R. I claim that x ∈ TpM ,that is, that there exists a curve through p whose velocity vector at time t = 0 is x.One such curve is
α(t) = φ(u+ at, v + bt).
Check:
α′(0) = a∂φ
∂u
∣∣∣∣(u(0),v(0))
+ b∂φ
∂v
∣∣∣∣(u(0),v(0))
= aφu(u, v) + bφv(u, v) = x
HenceTpM = {aφu(u, v) + bφv(u, v) | a, b ∈ R}.
It follows that the subset TpM ⊂ R3 is closed under vector addition and scalarmultiplication: TpM is a subspace of R3. Further, {φu, φv} is a spanning set for thissubspace. Since M is regular, this spanning set is linearly independent, hence a basisfor TpM . This explains why we call φu, φv the coordinate basis vectors of the surface.
To summarize, we have proved the following:
Theorem 76 TpM is a vector space of dimension 2 spanned by {φu(u, v), φv(u, v)}.�
Definition 77 The normal space at p ∈ M is
NpM = {y ∈ R3 : y · x = 0 for all x ∈ TpM}.
Any y ∈ NpM is said to be normal to M at p. �
62
Remark 78 By its definition, NpM is also a vector space, that is, it is closed undervector addition and scalar multiplication: let y, z ∈ NpM and a, b ∈ R. Then for allx ∈ TpM ,
x · (ay + bz) = a(x · y) + b(x · z) = 0 + 0
so ay + bz ∈ NpM . Clearly NpM is one-dimensional, so any non-zero normal vector,for example φu × φv, is a basis for NpM .
Furthermore, the tangent space TpM is precisely the two dimensional space ofvectors orthogonal to NpM , or equivalently, to any nonzero vector in NpM . Hence
x ∈ TpM ⇔ x · (φu × φv) = 0.
φu
φv
φu × φv
TpM
NpM
This gives us a sneaky way of checking whether a given vector is a tangent vector.
Example 79 (saddle surface) Let φ : R2 → R3 such that
φ(u, v) = (u, v, uv).
Again, this is certainly an RPS since it’s a particular case of Example 70 (for f(u, v) =uv).
The coordinate basis vectors are
φu = (1, 0, v) φv = (0, 1, u).
The point p = (3,−2,−6) ∈ M has local coordinates (3,−2), so the coordinate basisfor TpM is
φu(3,−2) = (1, 0,−2) φv(3,−2) = (0, 1, 3)
and NpM is spanned by
φu × φv = (1, 0,−2)× (0, 1, 3) = (2,−3, 1)
Let’s determine whether the following vectors are in TpM , NpM or neither:
x = (1, 2, 4), y = (2, 1, 1), z = (−4, 6,−2).
63
x: x · (φu × φv) = 2− 6 + 4 = 0
So x ∈ TpM . Hence (1, 2, 4) = aφu + bφv for some a, b ∈ R. In fact
(1, 2, 4) = (1, 0− 2) + 2(0, 1, 3)
so a = 1 and b = 2 .
y: y · (φu × φv) = 4− 3 + 1 = 2
So y /∈ TpM . What about NpM?
y · φu = 2 + 0− 2 = 0 y · φv = 0 + 1 + 3 = 4
So y /∈ NpM either.
z: z · (φu × φv) = −8− 18− 2 = −28
So z /∈ TpM . What about NpM?
z · φu = −4 + 0 + 4 = 0 z · φv = 0 + 6− 6
So z ∈ NpM . Hence z = c φu × φv for some c ∈ R. In fact
(−4, 6,−2) = −2(2,−3, 1)
so c = −2 . �
Summary
• Given a smooth mapping φ : U → R3, where U ⊆ R2 is an open set, itscoordinate basis vectors are
φu =∂φ
∂u, φv =
∂φ
∂v.
φ is regular if for all (u, v) ∈ U , φu(u, v)× φv(u, v) 6= 0.
• A regularly parametrized surface is a regular, one-to-one map φ : U → R3.
• The tangent space TpM at p = φ(u, v) is the set of all velocity vectors of curvesin the surface M = φ(U) passing through the point p. It is a two-dimensionalvector space spanned by {φu(u, v), φv(u, v)}.
• The normal space NpM at p = M(u, v) is the set of vectors in R3 orthogonalto TpM . It is a one-dimensional vector space spanned by φu(u, v)× φv(u, v).
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7 Calculus on surfaces
7.1 Directional derivatives
A mapping f : R3 → R is smooth if all its partial derivatives (of all orders) existeverywhere. Given a regularly parametrized surface M ⊂ R3 we can define functionsf : M → R: they are mappings which assign a real number f(p) to each point p onthe surface M . What does it mean to say that such a function is smooth?
Definition 80 Let φ : U → R3 be a regularly parametrized surface with M = φ(U)and let f : M → R be a function on M . We say that f is smooth if f ◦ φ : U → R
is smooth in the usual sense. We refer to f = f ◦ φ
f(u, v) = f(φ(u, v)).
as the coordinate expression of f .
Example 81 Consider the RPS defined in Example 71, whose image is (almost allof) the unit sphere:
φ : (0, π)× (0, 2π) → R3, φ(u, v) = (sin u cos v, sin u sin v, cosu).
The following maps M → R
f : (y1, y2, y3) 7→ y1 + y2 + y3, g : (y1, y2, y3) 7→ y21 + y22 − y23
have coordinate expressions
f(u, v) = f(φ(u, v)) = sin u cos v + sin u sin v + cosu
g(u, v) = g(φ(u, v)) = sin2 u cos2 v + sin2 u sin2 v − cos2 u = sin2 u− cos2 u = − cos(2u).
These are both smooth functions of u and v, so f, g are smooth functions M → R.
Given a smooth function f : M → R on a regularly parametrized surface M , anda tangent vector x ∈ TpM , we can ask “what is the rate of change of f at p in thedirection of x?” The answer is given (almost) by the following definition:
Definition 82 Let f : M → R be a smooth function and x ∈ TpM . The directionalderivative of f along x is
∇xf = (f ◦ α)′(0) = d
dtf(α(t))
∣∣∣∣t=0
where α is a generating curve1 for x. [Such a curve exists by the definition of TpM ,Definition 74.] Note that ∇xf is a single number, associated with the point p, not afunction on M . �
1Recall this is a curve α(t) in M with α(0) = p and α′(0) = x
65
Example 83 Let M be the unit sphere (parametrized as in Example 71 say), letp = (0, 1, 0) and x = (0, 0, 1). Then x ∈ TpM (check it!). Consider the functionf : M → R given by
f(y1, y2, y3) = y3
which assigns to each point on the sphere its “height” above the (y1, y2) plane. Whatis ∇xf?
From the definition, we must choose a curve through (0, 1, 0) in M whose initialvelocity is (0, 0, 1). One obvious choice is
α(t) = (0, cos t, sin t).
Check: |α(t)| =√
cos2+ sin2 t = 1
α(0) = (0, 1, 0)
α′(0) = (0,− sin 0, cos 0) = (0, 0, 1)
Therefore α is a generating curve for x. Now
(f ◦ α)(t) = sin t
(f ◦ α)′(t) = cos t
so ∇xf = (f ◦ α)′(0) = 1
Note that ∇xf isn’t quite just the “rate of change of f in the direction of x,”because it depends on the length of x as well as its direction. For example, lety = 2x = (0, 0, 2) ∈ TpM . Then this points in the same direction as x, but is theinitial velocity vector of the curve
β(t) = (0, cos 2t, sin 2t)
through p.
Now: (f ◦ β)(t) = sin 2t
(f ◦ β)′(t) = 2 cos 2t
So: ∇yf = 2
which is different from ∇xf . In fact ∇yf = 2∇xf . Coincidence? �
One nice thing about Definition 82 is that it doesn’t use the specific parametriza-tion of the surface involved (we never actually had to worry about the defining map
66
φ(u, v) for the unit sphere in the above example). There is something a bit worryingabout it, however. To compute ∇xf we have to choose a generating curve α for x.There are infinitely many such curves. In the example above, we could equally wellhave chosen
α(t) = (0,√1− t2, t), or α(t) = (t2,
√1− t4 − sin2 t, sin t)
for example. How do we know that the answer doesn’t depend on our choice?
Lemma 84 ∇xf is independent of the choice of generating curve α for the tangentvector x.
Proof: Let α1, α2 : I → M be two curves through p = φ(u, v) with α′1(0) = α′2(0) = x.We can write
αi(t) = φ(ui(t), vi(t)), for i = 1, 2.
Then
x = α′1(0) = u′1(0)∂φ
∂u(u1(0), v1(u)) + v′1(0)
∂φ
∂v(u1(0), v1(0))
= u′1(0)φu(u, v) + v′1(0)φv(u, v)
Similarly,x = α′2(0) = u′2(0)φu(u, v) + v′2(0)φv(u, v)
Since φu(u, v) and φv(u, v) are linearly independent, u′1(0) = u′2(0) and v′1(0) = v′2(0).For i = 1 or 2,
f ◦ αi(t) = f(φ(ui(t), vi(t))) = f ◦ φ(ui(t), vi(t)) = f(ui(t), vi(t)).
So
(f ◦ αi)′(0) = u′i(0)
∂f
∂u(ui(0), vi(0)) + v′i(0)
∂f
∂v(ui(0), vi(0))
= u′i(0)∂f
∂u(u, v) + v′i(0)
∂f
∂v(u, v).
Since u′1(0) = u′2(0) and v′1(0) = v′2(0),
(f ◦ α1)′(0) = (f ◦ α2)
′(0).
�
This is reassuring. However, Definition 82 is rather cumbersome to use in practicebecause it’s tiresome to have to keep inventing generating curves for tangent vectors.Recall that we may also think of TpM as
TpM = {a φu + b φv : a, b,∈ R}.
The next Lemma will allow us to reduce the calculation of directional derivatives tostraightforward partial differentiation with respect to the local coordinates (x1, x2).
Lemma 85 ∇xf is linear with respect to both x and f . That is, for all x, y ∈ TpM ,a, b ∈ R and f, g : M → R,
67
(A) ∇ax+byf = a∇xf + b∇yf
(B) ∇x(af + bg) = a∇xf + b∇xg
Proof:(A) Let α1, α2 : I → M be curves through p = φ(u, v) with α′1(0) = x, α′2(0) = y. Asbefore we can write these as
αi(t) = φ(ui(t), vi(t)).
Consider the curve
α3(t) = φ(u3(t), v3(t)),
(u3(t), v3(t)) = a(u1(t), v1(t)) + b(u2(t), v2(t))− (u, v).
Then α3(0) = p and α′3(0) = ax+ by. [Check it!]Now we try to calculate ∇ax+byf = (f ◦ α3)
′(0). We have that, for i = 1, 2, 3,
(f ◦ αi)(t) = f(φ(ui(t), vi(t))) = f(ui(t), vi(t)),
so
(f ◦ αi)′(0) = u′i(0)
∂f
∂u(ui(0), vi(0)) + v′i(0)
∂f
∂v(ui(0), vi(0))
= u′i(0)∂f
∂u(u, v) + v′i(0)
∂f
∂v(u, v).
Therefore
∇ax+byf = (f ◦ α3)′(0)
= (au′1(0) + bu′2(0))∂f
∂u(u, v) + (av′1(0) + bv′2(0))
∂f
∂v(u, v)
= a
(u′1(0)
∂f
∂u+ v′1(0)
∂f
∂v
)+ b
(u′2(0)
∂f
∂u+ v′2(0)
∂f
∂v
)
= a(f ◦ α1)′(0) + b(f ◦ α2)
′(0)
= a∇xf + b∇yf.
(B) Follows immediately from the definition: if α is a generating curve for x then
∇x(af + bg) = ((af + bg) ◦ α)′(0) = (af ◦ α+ bg ◦ α)′(0)= a(f ◦ α)′(0) + b(g ◦ α)′(0) = a∇xf + b∇xg. �
The upshot of Lemma 85 is that if we know the directional derivatives ∇φuf and
∇φvf with respect to the coordinate basis vectors, then we know ∇xf for all tangent
vectors x ∈ TpM . For any x ∈ TpM may be written as
x = aφu + bφv
so applying Lemma 85 part (A) gives
∇xf = a∇φuf + b∇φv
f. (♠)
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But φu(u, v) and φv(u, v) may be represented by the curves
φ(u+ t, v) and φ(u, v + t)
respectively. Hence
∇φuf =
∂f
∂u, ∇φv
=∂f
∂v.
Example 83 (revisited) Let’s re-do Example 83 using the trick of reducing direc-tional derivatives to partial differentiation. Recall M is the unit sphere parametrizedby
φ(u, v) = (sin u cos v, sin u sin v, cosu).
The function is f(p1, p2, p3) = p3 and we wish to compute ∇xf where p = (0, 1, 0)and x = (0, 0, 1) ∈ TpM .
• The local coordinates of p are (u, v) = ( π/2 , π/2 )[Check: φ( π/2 , π/2 ) = (0, 1, 0).
• The local coordinate expression for f is
f(u, v) = cos(u)
• The coordinate basis vectors are
φu(u, v) = (cosu cos v, cosu sin v,− sinu)
φv(u, v) = (− sin u sin v, sin u cos v, 0).
At p they are
φu( π/2 , π/2 ) = (0, 0,−1)
φv( π/2 , π/2 ) = (−1, 0, 0).
• Hence, x = aφu + bφv wherea = −1, b = 0
• Finally,
∇xf = a∇φuf + b∇φv
f = a∂f
∂u( π/2 , π/2 ) + b
∂f
∂v( π/2 , π/2 )
= (−1)(− sin(π/2)) + 0 = 1
We got the same answer! Which calculation was easier, Example 83 or Example 83(revisited)?
69
The fact that we can reduce the calculation of ∇xf to partial differentiation istheoretically convenient, too. For example, we know that partial derivatives obey theproduct (or Leibniz) rule
∂
∂u(f g) = f
∂g
∂u+ g
∂f
∂uand similarly for ∂/∂v. We leave it as an exercise for you to use this to prove that
Corollary 86 For all f, g : M → R and x ∈ TpM ,
∇x(fg) = f(p)∇xg + g(p)∇xf
�
7.2 Vector fields
As usual, let U be an open subset of R2, let φ : U → R3 be a RPS and let M = φ(U).
Definition 87 A vector field on a RPS M is a smooth map X : M → R3 (wheresmooth means that each of its component functions X1, X2, X3 : M → R is smooth).IfX(p) ∈ TpM for all p ∈ M thenX is called a tangent vector field. IfX(p) ∈ NpMfor all p ∈ M then X is a normal vector field.
Remark 88 Just as for functions, we define the coordinate expression of a vectorfield X : M → R3 to be
X : U → R3, X(u, v) = V (φ(u, v)).
It follows from Theorem 76 that the coordinate expression of any tangent vectorfield X takes the form
X(u, v) = f(u, v)φu(u, v) + g(u, v)φv(u, v)
where f, g are smooth real-valued functions on U . Similarly, it follows from Remark78 that any normal vector field X : M → R3 takes the form
X(u, v) = f(u, v)φu(u, v)× φv(u, v)
where f is a smooth real-valued function on U .We will often refer to φu(u, v), φv(u, v) as tangent vector fields, even though they
are really coordinate expressions for tangent vector fields. Similarly, we will normallyrefer to φu × φv as a normal vector field (even though it’s actually a coordinateexpression for a normal vector field).
We may define directional derivatives of vector fields along tangent vectors exactlyas we defined directional derivatives of functions:
Definition 89 Let X : M → R3 be a vector field on M and y ∈ TpM . Then thedirectional derivative of X with respect to y is
∇yX = (X ◦ α)′(0)where α : I → M is a curve through p ∈ M with α′(0) = x. �
70
Notes:
• X is a vector field, but y and ∇yX are just vectors.
• X can be any vector field, but y must be a tangent vector.
• Even if X is a tangent vector field, there is no reason to expect ∇yX to be atangent vector.
• We can write X = (X1, X2, X3) where X1, X2, X3 : M → R are ordinary func-tions. Then
∇yX = ((X1 ◦ α)′(0), (X2 ◦ α)′(0), (X3 ◦ α)′(0)) = (∇yX1,∇yX2,∇yX3). (♦)
So we can write ∇yX in terms of the directional derivatives of the functionsX1, X2, X3. Since ∇yX1,∇yX2,∇yX3 are independent of the choice of curve α(lemma 85), ∇yX is also independent of the choice of curve.
This observation allows us to transfer our knowledge about directional derivativesof functions to directional derivatives of vector fields:
Lemma 90 Let X, Y be vector fields on M , z, w ∈ TpM , f be a smooth function onM and a, b ∈ R. Then
(a) ∇az+bwX = a∇zX + b∇wX.
(b) ∇z(aX + bY ) = a∇zX + b∇zY.
(c) ∇z(fX) = (∇zf)X(p) + f(p)∇zX.
(d) ∇z(X · Y ) = (∇zX) · Y (p) +X(p) · (∇zY ).
Proof: We prove (d) as follows. Write X = (X1, X2, X3) and Y = (Y1, Y2, Y3) forfunctions Xi, Yi. Then
∇z(X · Y ) = ∇z(X1Y1 +X2Y2 +X3Y3)
= ∇z(X1Y1) +∇z(X2Y2) +∇z(X3Y3) (Lemma 85)
= (∇zX1)Y1(p) +X1(p)(∇zY1) + (∇zX2)Y2(p)
+X2(p)(∇zY2) + (∇zX3)Y3(p) +X3(p)(∇zY3) (Corollary 86)
= (∇zX1,∇zX2,∇zX3) · (Y1(p), Y2(p), Y3(p))
+ (X1(p), X2(p), X3(p)) · (∇zY1,∇zY2,∇zY3)
= (∇zX) · Y (p) +X(p) · (∇zY ).
We leave it as an exercise to prove parts (a), (b) and (c) using Lemma 85, Corollary86 and equation (♦). �.As with directional derivatives of functions, we can use Lemma 90 to reduce calcula-tion of ∇yX to partial differentiation with respect to (u, v). We just need to expressy in terms of the coordinate basis and find the coordinate expression for X .
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Example 91 Let φ(u, v) = (sin u cos v, sin u sin v, cosu) be our standard parametri-sation for the unit sphere M . Let p = (0, 1, 0) ∈ M , y = (1, 0, 0) ∈ TpM , and let Xbe the vector field X(q) = (q2,−q1, 0). Calculate ∇yX .
Rather than working directly from definition (which involves finding a generatingcurve for y), we calculate∇yX by expressing y in terms of φu and φv. The coordinatesof p are (π/2, π/2). We calculate:
φu(u, v) = (cosu cos v,− cosu sin v, sinu)
φv(u, v) = (− sin u sin v, sin u cos v, 0)
φu(π/2, π/2) = (0, 0, 1)
φv(π/2, π/2) = (−1, 0, 0).
So y = aφu + bφv witha = 0 , b = −1 .
Now
X(u, v) = (sin u sin v,− sin u cos v, 0)
∂X
∂u(u, v) = (cosu sin v,− cosu cos v, 0)
∂X
∂v(u, v) = (sin u cos v, sin u sin v, 0)
∂X
∂u(π/2, π/2) = (0, 0, 0)
∂X
∂v(π/2, π/2) = (0, 1, 0) .
So
∇yX = 0∂X
∂u+ (−1)
∂X
∂v= (0,−1, 0) .
�
Given a tangent vector field Y : M → R3 and another vector field X : M → R3
(tangent, normal or neither), we can define a 3rd vector field Z : M → R3 as follows:at each p ∈ M ,
Z(p) = ∇Y (p)X.
We shall denote this vector field Z = ∇YX .
Example 92 Consider again the unit sphere M with the parametrization from ex-ample 91. Let X : M → R3 be the tangent vector field X(q) = (q3q1, q3q2, q
23 − 1).
We will calculate the vector field ∇XX .
72
We calculate
φu(u, v) = (cos u cos v, cosu sin v,− sin u)
φv(u, v) = (− sin u sin v, sin u cos v, 0)
X(u, v) = (sin u cosu cos v, sin u cosu cos v,− sin2 u) .
Since we were told that X is a tangent vector field, we should be able to find functions
f(u, v), g(u, v) such that X = fφu + gφv. We find
f(u, v) = sin u , g(u, v) = 0 .
So at the point p with coordinates (u, v),
∇X(p)X = sin u∂X
∂u+ 0× ∂X
∂v
= sin u((cos2 u− sin2 u) cos v, (cos2 u− sin2 u) sin v,−2 sinu cosu)
= sin u(cos 2u cos v, cos 2u sin v, sin 2u) .
This formula gives us the coordinate expression for ∇XX .
In the particular case that Y is the coordinate vector fields φu, ∇YX is easy toevaluate: since Y = 1× φu + 0× φv the coordinate expression for ∇φu
X is
∇φuX = 1× ∂X
∂u(u, v) + 0× ∂X
∂v(u, v) =
∂X
∂u(u, v).
Similarly ∇φvcorresponds to taking the partial derivative in v.
Lemma 93∇φu
φv = ∇φvφu.
Proof: The coordinate expression for φv is ∂φ/∂v, so the coordinate expression for∇φu
φv is
∇φuφv =
∂2φ
∂u∂v
Similarly,
∇φvφu =
∂2φ
∂v∂u.
Since partial derivative commute, these two expressions must be equal. �
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Summary
• A function f : M → R is smooth if its coordinate expression f = f ◦φ : U → R
is smooth.
• Given a smooth function f : M → R and a tangent vector y ∈ TpM , thedirectional derivative of f with respect to (or along) y is
∇yf = (f ◦ α)′(0)
where α(t) is any generating curve for y.
• The directional derivative is linear, that is
∇ax+byf = a∇xf + b∇yf, ∇y(af + bg) = a∇yf + b∇yg.
• Directional derivatives along coordinate basis vectors reduce to partial deriva-tives
∇φuf =
∂f
∂u, ∇φv
f =∂f
∂v.
• Vector fields are smooth maps X : M → R3.
• We can extend the definition of directional derivative to vector fields. If X is avector field and Y is a tangent vector field then the directional derivative ∇YXis another vector field.
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8 Curvature of an oriented surface
8.1 The shape operator on an oriented surface
Definition 94 An orientation on a RPS φ : U → R3 with image M = φ(U) isa choice of unit normal vector field N . In other words, N : M → R3 must satisfy|N(p)| = 1 and N(p) ∈ NpM for all p ∈ M . �
For example, if M is the plane z = 0 then N(p) = (0, 0, 1) is a unit normal vector field,so is an orientation for M . Another choice of orientation could be N(p) = (0, 0,−1).
If M is the unit sphere (i.e. the set of vectors p of unit length) then we couldchoose N(p) = p. This is a vector field that points radially outwards. Then for anyp ∈ M , |N(p)| = |p| = 1 and N(p) is normal to M at p, so N is an orientation. Canyou think of another orientation of the sphere?
In general, we can choose an orientation for an RPS as follows: recall that φu×φv
is a normal vector field which never vanishes. It follows that
N =φu × φv
|φu × φv|is a unit normal vector field. This N is known as the canonical orientation of φ.The only other possible choice of orientation is
N = − φu × φv
|φu × φv|.
The key point about oriented surfaces is that, as we noted in Remark 78, the unitnormal at p, N(p), determines the tangent space at p:
TpM = {x ∈ R3 : x ·N(p) = 0}.
So we can glean information about how the tangent space varies with p by computingdirectional derivatives ∇xN where x ∈ TpM . Just as with the curvature of a regularlyparametrized curve, it’s essential that N is a unit normal vector field – if it weren’tthen ∇xN would contain information about the rate of change of the length of N inthe direction u, as well as the rate of change of the direction of N .
Lemma 95 Let φ : U → R3 be an RPS and let N be an orientation for M = φ(U).Then for any p ∈ M and x ∈ TpM , ∇xN ∈ TpM also.
Proof: It suffices to show that N(p) · (∇xN) = 0. Now N is a unit vector field soN ·N = 1. Taking the directional derivative of this (constant) function with respectto x gives
0 = ∇x(N ·N) = (∇xN) ·N(p) +N(p) · (∇xN) = 2N(p) · (∇xN),
where we have used Lemma 90 part (d). The result immediately follows. �
Reminder 96 A map L : V → V , where V is a vector space, is said to be linear iffor all a, b ∈ R and x, y ∈ V ,
L(ax + by) = aL(x) + bL(y).
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Definition 97 Let φ : U → R3 and let N be an orientation of M = φ(U). Theshape operator at p ∈ M is the map
Sp : TpM → TpM, Sp(x) = −∇xN.
Sp really does map TpM to itself, by Lemma 95, and is a linear map by Lemma 90.It is often called the Weingarten map in honour of its discoverer. �
Example 98 (a cylinder) Let R > 0and let φ : (0, 2π)× R → R3 be the map
φ(u, v) = (R cosu,R sin u, v).
This is a cylinder of radius R > 0. Let’scalculate the shape operator of φ. Firstwe find the canonical orientation:
φu = (−R sin u,R cos u, 0)
φv = (0, 0, 1)
φu × φv = (R cosu,R sin u, 0)
|φu × φv|2 = R2 cos2 u+R2 sin2 u = R2
N =φu × φv
|φu × φv|= (cosu, sinu, 0)
Now we find the shape operator Sp. We start by calculating Sp(φu) and Sp(φv):
Sp(φu) = −∇φuN = (sin u,− cosu, 0)
= − 1
Rφu + 0φv
Sp(φv) = −∇φvN = (0, 0, 0)
= 0φu + 0φv.
If we wanted to calculate Sp(x) for any other x ∈ TpM we could start by finding a, bsuch that x = aφu + bφv. Then we would know that
Sp(x) = Sp(aφu + bφv) = aSp(φu) + bSp(φv) = − a
Rφu
76
because Sp is linear. So Sp(x) is easily worked out from Sp(φu) and Sp(φv). As isusually done with linear maps, we can completely describe Sp by writing down its
matrix Sp with respect to the basis φu, φv for TpM :
Sp =
↑ ↑Sp(φu) Sp(φv)
↓ ↓
=
(− 1
R0
0 0
).
Then applying Sp to x = aφu + bφv corresponds to multiplying the vector
(ab
)with
Sp. �
Example 99 (a plane) Let φ : R2 → R3 such that φ(u, v) = (u, v, 0). The image isthe plane y3 = 0. As discussed above, N(u, v) = (0, 0, 1) is a unit normal. Then N isconstant, so ∇xN = 0 for any x ∈ TpM . Therefore Sp sends every tangent vector xto zero.Question: what is the matrix of Sp in this case? �
Example 100 (unit sphere) Let’s calculate the shape operator for the unit sphereM , parametrized as usual by φ(u, v) = (sin u cos v, sin u sin v, cosu). Then
φu = (cosu cos v, cosu sin v,− sin u)
φv = (− sin u sin v, sin u cos v, 0)
φu × φv = (sin2 u cos v, sin2 sin v, sin u cosu)
= sin u(sinu cos v, sin u sin v, cosu)
|φu × φv| = sin u
N(u, v) =φu × φv
|φu × φv|= (sin u cos v, sin u sin v, cosu).
So
Sp(φu) = −∇φuN = −(cos u cos v, cosu sin v,− sin u) = −φu + 0φv
Sp(φv) = −∇φvN = −(− sin u sin v, sin u cos v, 0)) = 0φu − φv.
So the matrix of Sp is
Sp =
(−1 00 −1
).
For any x ∈ TpM , Sp(x) = −x, so Sp = −Id, where Id : TpM → TpM is the identitymap.
Note that we could have taken a shortcut to get this answer: as stated earlier,we can choose the orientation to be N(p) = p, and this has coordinate expression
N(u, v) = φ(u, v). Then ∇φuN = −∂N/∂u = −∂φ/∂u = −φu, and similarly for φv.
Our measure(s) of the curvature of a surface will be defined in terms of the eigenvaluesof the linear map Sp : TpM → TpM . It turns out to be crucial that Sp has a propertycalled self-adjointness.
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Definition 101 A linear map L : TpM → TpM is self adjoint if for all x, y ∈ TpM ,x · L(y) = y · L(x). �
Theorem 102 Let M be an oriented surface and Sp : TpM → TpM be its shapeoperator at the point p. Then Sp is self adjoint.
Proof: Let x = aφu + bφv and y = cφu + dφv. Then
x.Sp(y)− y.Sp(x) = acφu · Sp(φu) + adφu · Sp(φv) + bcφv · Sp(φu) + bdφv · Sp(φv)
−acφu · Sp(φu)− adφv · Sp(φu)− bcφu · Sp(φv)− bdφv · Sp(φv)
= (ad− bc)(φu · Sp(φv)− φv · Sp(φu)).
So we just need to show that φu ·Sp(φv) = φv ·Sp(φu), i.e. that φu ·∇φvN = φv ·∇φu
N .Since φu is a tangent vector field and N is a normal vector field, φu ·N = 0, so
0 = ∇φv(φu ·N) = (∇φu
φv) ·N + φv · (∇φuN)
by lemma 90(d). Similarly, (∇φvφu) ·N + φu · (∇φv
N) = 0. So
φu ·Sp(φv)−φv ·Sp(φu) = −φu · (∇φvN)+φv · (∇φu
N) = (∇φvφu) ·N−(∇φu
φv) ·N = 0
by lemma 93. �
Example 103 (a saddle surface) Let φ : R2 → R3 be the map φ(u, v) = (u, v, uv).This is a RPS, as we have observed previously. Let’s construct its shape operator
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Sp : TpM → TpM at the point p = φ(1, 0) = (1, 0, 0).
φu = (1, 0, v)
φv = (0, 1, u)
φu × φv = (−v,−u, 1)
|φu × φv| =√1 + u2 + v2
N =1√
1 + u2 + v2(−v,−u, 1)
∂
∂u
1√1 + u2 + v2
=−u
(1 + u2 + v2)3/2
∂
∂v
1√1 + u2 + v2
=−v
(1 + u2 + v2)3/2
Sp(φu) = −∇φuN =
u
(1 + u2 + v2)3/2(−v,−u, 1) +
1√1 + u2 + v2
(0, 1, 0)
=1
23/2(0,−1, 1) +
1
21/2(0, 1, 0) at (u, v) = (1, 0)
= 2−3/2(0, 1, 1)
Sp(φv) = −∇φvN =
v
(1 + u2 + v2)3/2(−v,−u, 1) +
1√1 + u2 + v2
(1, 0, 0)
= 2−1/2(1, 0, 0) at (u, v) = (1, 0)
Now we express these in terms of the basis φu, φv at p:
φu(1, 0) = (1, 0, 0)
φv(1, 0) = (0, 1, 1)
Sp(φu) = 0φu + 2−3/2φv
Sp(φv) = 2−1/2φu + 0φv.
So the matrix of the shape operator is
Sp =
↑ ↑Sp(φu) Sp(φv)
↓ ↓
=
(0 2−1/2
2−3/2 0
).
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Let’s check that Sp is self-adjoint at p = φ(1, 0):
φu · φu = 12 + 02 + 02 = 1
φv · φv = 02 + 12 + 12 = 2
φu · Sp(φv) = 2−1/2φu · φu = 2−1/2
φv · Sp(φu) = 2−3/2φv · φv = 2−3/2 × 2 = 2−1/2
8.2 The principal curvatures of an oriented surface
Let V be a vector space (e.g. V = TpM) and L : V → V be a linear map (e.g.Sp : TpM → TpM). An eigenvalue of L is a number λ such that
L(x) = λx (♦)
for some x ∈ V , x 6= 0. The vector x is called an eigenvector corresponding to theeigenvalue λ. Clearly, given one such x any multiple ax (a 6= 0) is also an eigenvector,so we are free to choose our eigenvectors to have unit length. Let Id : V → V be theidentity map on V (Id(x) = x). Then we may re-arrange (♦) to read
(L− λ Id)(x) = 0, x 6= 0,
so λ is an eigenvalue if and only if the linear map L− λ Id fails to be injective (bothu 6= 0 and 0 get mapped to 0 by L−λ Id), that is, if and only if L−λ Id is not invertible.
Choose a basis e1, . . . , em for V and let L be the m×m matrix representing L. NowId is represented by the m × m identity matrix Im, so the linear map L − λId is
represented by the square matrix L− λIm. Hence, it fails to be invertible if and onlyif
det(L− λIm) = 0. (♣)
Equation (♣) is a degree m polynomial equation in λ, called the characteristicequation of the linear map L. Although it looks like it depends on the matrix
L chosen to represent L (i.e. the choice of basis for V ), in fact it doesn’t. Theequation has exactly m solutions, counted with multiplicity. However, even thoughthe coefficients of the polynomial are all real, these solutions may, in general, becomplex. In this case, the solution λ is still called an eigenvalue of L, but itsinterpretation is rather subtle.
Example 104 On R2 consider the linear map L : (a, b) 7→ (−b, a). Relative to thestandard basis e1 = (1, 0), e2 = (0, 1) this has matrix representation
L =
[0 −11 0
]
so the characteristic equation is∣∣∣∣−λ −11 −λ
∣∣∣∣ = λ2 + 1 = 0.
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So L has only complex eigenvalues, namely ±i. �
Luckily, this can never happen for the shape operator Sp : TpM → TpM because it isself-adjoint.
Theorem 105 Let L : TpM → TpM be a self-adjoint linear map. Then
(A) its eigenvalues are all real, and
(B) its eigenvectors form an orthonormal basis for TpM .
Proof: (A) Let e1, e2 be an orthonormal basis for TpM and
L =
[L11 L12
L21 L22
]
be the matrix representing L relative to this basis. Then, by definition,
e1 · L(e2) = e1 · (L12e1 + L22e2) = L12,
ande2 · L(e1) = e2 · (L11e1 + L21e2) = L21.
But L is self adjoint, so L12 = L21. Hence, the characteristic equation of L is
0 = det(L− λI2) =
∣∣∣∣L11 − λ L12
L12 L22 − λ
∣∣∣∣ = λ2 − (L11 +L22)λ+L11L22 −L212. (♠)
The discriminant of this quadratic polynomial is
“b2 − 4ac” = (L11 + L22)2 − 4L11L22 + 4L2
12 = (L11 − L22)2 + 4L2
12 ≥ 0.
Hence (♠) has two real solutions.(B) Let these real eigenvalues be λ1, λ2 and denote their corresponding eigenvectorsu1, u2. If λ1 = λ2, equation (♠) has a repeated root, hence its discriminant vanishes,
so L11 = L22 = λ and L12 = 0. But then L = λI2 ⇒ L = λ Id, so every vectoru ∈ TpM is an eigenvector corresponding to eigenvalue λ. So we may choose u1 = e1and u2 = e2, which are orthonormal.
If λ1 6= λ2 then
0 = u1 · L(u2)− u2 · L(u1) = (λ2 − λ1)u1 · u2
so u1 · u2 = 0. �
This allows us to make the following definition:
Definition 106 Let M be an oriented surface, Sp : TpM → TpM be its shape opera-tor at p ∈ M . Then the principal curvatures of M at p are κ1, κ2, the eigenvaluesof Sp. The principal curvature directions of M at p are the corresponding eigen-vectors e1, e2 (normalized to have unit length). By Theorem 105, κ1, κ2 are real andthe eigenvectors e1, e2 form an orthonormal basis for TpM . �
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Example 107 (saddle surface) Recall (Example 103) that the saddle surface
φ(u, v) = (u, v, uv)
at p = φ(1, 0) = (1, 0, 0) has shape operator
Sp =
(0 2−
1
2
2−3/2 0
)
relative to the coordinate basis φu = (1, 0, 0), φv = (0, 1, 1) for T(1,0,0)M . Note thismatrix is not symmetric, because the coordinate basis is not orthonormal. Nonethe-less, its eigenvalues must still be real, and its eigenvectors orthonormal. Let’s check.The characteristic equation is
0 =
∣∣∣∣−λ 2−
1
2
2−3
2 −λ
∣∣∣∣ = λ2 − 1
4
so the principal curvatures are:
κ1 = −1
2, κ2 =
1
2.
To find e1, solve the linear system (Sp − κ1I2)e1 = 0:
(2−1 2−
1
2
2−3
2 2−1
)(ab
)=
(00
)⇐⇒ a + b
√2 = 0 so choose e1 =
(√2
−1
)
We won’t worry about the length of e1 yet. Similarly for e2:
(−2−1 2−
1
2
2−3
2 −2−1
)(ab
)=
(00
)⇐⇒ a− b
√2 = 0 so choose e2 =
(√21
)
Are these vectors orthogonal? They certainly don’t look orthogonal. But you mustremember that these 2×1 column vectors represent e1 and e2 with respect to the basisφu, φv. The vectors e1, e2 themselves lie in TpM ⊂ R3, that is, they are 3-dimensionalvectors. In fact,
e1 =√2(1, 0, 0)− (0, 1, 1) = (
√2,−1,−1)
e2 =√2(1, 0, 0) + (0, 1, 1) = (
√2, 1, 1)
whence we see that e1 · e2 = 0 as expected. Note that when we normalize them tohave unit length, we must again think of them as vectors in R3, not 2 × 1 columnmatrices:
|(√2,−1,−1)| = |(
√2, 1, 1)| = 2 =⇒ e1 =
1
2(√2,−1,−1), e2 =
1
2(√2, 1, 1)
�
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Example 108 (a cylinder) Recall (Example 98) that the shape operator Sp cylin-der of radius R has matrix
Sp =
(− 1
R0
0 0
)
This matrix obviously has eigenvalues κ1 = − 1Rand eigenvectors
(10
)and
(01
). The
principle curvature directions at p = φ(u, v) are therefore
1× φu + 0× φv = (−R sin u,R cosu, 0), 0× φu + 1× φv = (0, 0, 1).
We normalise these to get
e1 = (− sin u, cosu, 0), e2 = (0, 0, 1).
Notice that as R grows large, κ1 = − 1Rapproaches zero. In this sense wide cylinders
are “more curved” than narrow cylinders. �
Example 109 (unit sphere) Recall (Example 100) that in the case of the unitsphere, we found that Sp(x) = −x for any x ∈ TpM . Thus any non-zero vectorx ∈ TpM is an eigenvector of Sp with eigenvalue -1. So Sp has a single eigenvalue-1, and κ1 = κ2 = −1. In this situation there are no preferred principal curvaturedirections. �
8.3 Normal curvature
Definition 110 Let M be a regularly parametrized surface. The unit tangentspace at p ∈ M is
UpM = {x ∈ TpM : |x| = 1},the set of unit vectors in TpM . Geometrically, we can think of UpM as the unit circlein TpM . �
Note that UpM is not a vector space.
Definition 111 Let M be an oriented surface. The normal curvature functionof M at p ∈ M is
kp : UpM → R, kp(x) = x · Sp(x)
where Sp denotes the shape operator as usual. �
Why call this “normal curvature”?
Lemma 112 Let M be a surface, let N be an orientation for M , and x ∈ UpM . Letα : I → M be any unit speed curve in M through p generating x. Then the componentof the curvature vector of α in the direction of N(p) is kp(x).
Proof: We have a curve α in M with α(0) = p, α′(0) = x. Since α is a unit speedcurve, its curvature vector at p is k(0) = α′′(0). Now α stays in M , so its velocity isalways tangent to M , and hence
α′(t) ·N(α(t)) = 0 . ((♣))
83
for all t ∈ I. Differentiate (♣) with respect to t and set t = 0:
α′′(0) ·N(α(0)) + α′(0) · (N ◦ α)′(0) = 0
=⇒ k(0) ·N(p) + x · (N ◦ α)′(0) = 0.
But (N ◦ α)′(0) = ∇xN by the definition of directional derivatives (Definition 89).Hence
k(0) ·N(p) = −u · ∇xN = u · Sp(x) = kp(x)
as was to be proved. �
Example 113 (hyperboloid of one sheet)The hyperboloid
M = {y ∈ R3 : y21 + y22 − y23 = 1}
may be parametrized by
φ : (−π, π)× R → R3,
φ(u, v) = (cosh v cosu, cosh v sin u, sinh v).
Then
φu = (− cosh v sin u, cosh v cosu, 0)
φv = (sinh v cos u, sinh v sin u, cosh v)
φu × φv = (cosh2 v cosu, cosh2 v sin u,− sinh v cosh v)
|φu × φv| = cosh v√cosh2 v + sinh2 v
= cosh v√2 cosh2 v − 1
N =1√
2 cosh2 v − 1(cosh v cos u, cosh v sin u,− sinh v)
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Let’s calculate the shape operator at p = φ(0, 0) = (1, 0, 0):
φu(0, 0) = (0, 1, 0)
φv(0, 0) = (0, 0, 1)
Sp(φu) = −∇φuN(0, 0) =
1√2− 1
(− cosh 0 sin 0, cosh 0 cos 0,− sinh 0)
= (0, 1, 0)
= φu
Sp(φv) = −∇φuN(0, 0) =
1√2− 1
(sinh 0 cos 0, sinh 0 sin 0,− cosh 0) + sinh 0× . . .
= (0, 0,−1)
= −φv
Now φu and φv are both unit vectors, so both belong to U0M . Let’s calculate theirnormal curvature:
kp(φu) = φu · Sp(φu) = φu · φu = (0, 1, 0) · (0, 1, 0) = 1
kp(φv) = φv · Sp(φv) = −φv · φv = −(0, 0, 1) · (0, 0, 1) = −1
In general, if x = cos θφu + sin θφv = (0, cos θ, sin θ),
kp(x) = x · (0, cos θ,− sin θ) = cos2 θ − sin2 θ = cos 2θ
Notice that the maximum and minimum values of kp are ±1 = κ1, κ2. . . �
85
Normal curvature allows us to give a new interpretation of the principal curvaturesof a surface.
Theorem 114 Let M be an oriented surface and κ1, κ2 be its principal curvatures atp ∈ M , ordered so that κ1 ≤ κ2. Then
κ1 = min{kp(x) : x ∈ UpM}κ2 = max{kp(x) : x ∈ UpM}.
Proof: Let e1, e2 be the orthonormal pair of eigenvectors corresponding to κ1, κ2 (seeTheorem 105). These span TpM so any unit vector x ∈ UpM may be written
x = cos θ e1 + sin θ e2
for some angle θ. The normal curvature of x is
kp(x) = (cos θ e1 + sin θ e2) · Sp(cos θ e1 + sin θ e2)
= κ1 cos2 θ + κ2 sin
2 θ (♠)
= κ1 + (κ2 − κ1) sin2 θ.
Now κ2 − κ1 ≥ 0, and the maximum and minimum values of sin2 θ are 1 and 0, sothe maximum and minimum values of kp(x) are κ2 and κ1 respectively. �
Example 115 (hyperboloid of one sheet) For M and p as in Example 113, con-struct vectors x ∈ UpM with the following properties:
(a) kp(x) = 0, (b) kp(x) = 2, (c) Sp(x) = 0.
(a) Recall that we calculated kp(0, cos θ, sin θ) = cos 2θ. So kp = 0 is solved by
2θ =π
2+ nπ ⇐⇒ θ =
π
4+
nπ
2= . . . ,
π
4,3π
4,5π
4,7π
4, . . . .
So we obtain four solutions:
x =1√2(0, 1, 1),
1√2(0,−1, 1),
1√2(0,−1,−1),
1√2(0, 1,−1).
(b) No such u can exist by Theorem 114. For all x ∈ UpM , −1 ≤ kp(x) ≤ 1.
(c) Again, no such x can exist. If it did, then since x 6= 0, it follows that 0 is aneigenvalue of Sp, and hence a principle curvature. But the principal curvaturesare −1 and 1, not 0. �
8.4 Mean and Gauss curvatures
Definition 116 The mean curvature of M at p ∈ M is
H(p) =1
2(κ1 + κ2)
where κ1, κ2 are the principal curvatures of M at p. The Gauss curvature of M atp is
K(p) = κ1κ2 �
86
One need not solve the eigenvalue problem for Sp to compute H(p) and K(p):
Proposition 117 For all p ∈ M ,
H(p) =1
2tr Sp, K(p) = det Sp,
where Sp is the matrix representing Sp : TpM → TpM relative to some choice of basisfor TpM .
[Recall that the trace of a square matrix L is the sum of its diagonal elements.]
Proof: Let Sp =
(S11 S12
S21 S22
)relative to our chosen basis. Then κ1, κ2 are roots of
the polynomial
p(λ) =
∣∣∣∣S11 − λ S12
S21 S22 − λ
∣∣∣∣ = λ2−(S11+S22)λ+(S11S22−S12S21) = λ2−tr Sy λ+det Sy. (♣)
But p(κ1) = p(κ2) = 0 and the coefficient of λ2 is one, so
p(λ) = (λ− κ1)(λ− κ2) = λ2 − (κ1 + κ2)λ+ κ1κ2. (♠)
Comparing coefficients of λ and unity in (♠) and (♣), we see that κ1 + κ2 = tr Sp
while κ1κ2 = det Sp. �
Example 118 (a) The unit sphere has shape operator Sp : v 7→ −v at every point p(see Example 100). Hence, relative to any basis for TpM ,
Sp =
(−1 00 −1
).
It follows that
H(p) =1
2(−1 + −1) = −1
K(p) = (−1)(−1) = 1
for all p ∈ M .(b) For the cylinder of radius R (Example 98), the shape operator has matrix
Sp =
(− 1
R0
0 0
). So at any point p,
H(p) =1
2
(− 1
R+ 0
)= − 1
2R, K(p) = − 1
R× 0 = 0
The cylinder has the interesting property that it can be made by rolling up a sheetof paper. It is a deep and amazing fact that a surface can be made by bending aflat sheet of paper paper if and only if its Gauss curvature vanishes everywhere.This is a consequence of Gauss’ Theorema Egregrium. You can learn more about this
87
remarkable theorem by studying MATH3113 Differential Geometry or MATH5113Advanced Differential Geometry.
(c) For the hyperboloid (Example 113), the principle curvatures at p = (1, 0, 0)where κ1 = −1 and κ2 = 1. So
H(1, 0, 0) =1
2(−1 + 1) = 0, K(1, 0, 0) = −1× 1 = −1
�
Example 119 Let M = a cone, outwardly oriented. What can we deduce about Kand H just by looking at the surface?
N
u
kv
N
v
Pick orthonormal pair u, v at any pointp ∈ M . Then N is constant along thestraight line in M generating u ⇒∇uN = 0 ⇒ Sp(u) = 0. Hence, uis an eigenvector of Sp, with eigenvalue 0,so at least one principal curvature κ2 = 0,and u is the corresponding principal cur-vature direction. HenceK(p) = κ1κ2 = 0.
The other principal curvature direction isorthogonal to u, so must be v (or −v).Hence, v is an eigenvector of Sp witheigenvalue κ2, the other principal curva-ture. So
kp(v) = v · Sp(v) = v · (κ1v) = κ1
But kp(v) = k · N < 0 since the circlegenerating v curves away from the out-ward unit normal. Hence κ2 < 0, soH(p) = (κ1 + κ2)/2 = κ2/2 < 0 also. �
Fact 120 If we change orientation N 7→ N = −N then Sp 7→ Sp = −Sp ⇒ κi 7→κi = −κi ⇒ H 7→ H = −H but K 7→ K = K. The only measure of curvaturewe’ve introduced so far which is independent of the choice of orientation on M is theGauss curvature. For all other types of curvature (including normal curvature), thesign of the curvature has no intrinsic meaning (i.e. no meaning independent of ourmore or less arbitrary choice of orientation).
The sign of K does have an intrinsic geometric meaning!
K(p) > 0 ⇒ κ1, κ2 have same sign⇒ kp(u) strictly positive (if κ1, κ2 > 0) or
kp(u) strictly negative (if κ1, κ2 < 0)⇒ all curves through p curve towards N(p) or
all curves through p curve away from N(p)⇒ all curves curve in the same “sense”.
88
N all away from N
k < 0y
all towards N
k > 0y
N
K(p) < 0 ⇒ κ1, κ2 differ in sign⇒ kp(u) takes both positive and negative values⇒ ∃ curves through p which curve in opposite senses.
N
Ntowards
awaytowards
away
Example 121 Let’s test our intuition on the following surfaces:
89
(a) (b)
(c) (d)
(e) (f)
90
Summary
• An oriented surface is a RPS together with a choice of unit normal vector fieldN . Usually we choose
N =φu × φv
|φu × φv|.
• The shape operator of an oriented surface is
Sp : TpM → TpM, Sp(x) = −∇xN.
The shape operator is linear,
Sp(ax+ by) = aSp(x) + bSp(y) ∀a, b ∈ R, x, y ∈ TpM,
and self adjoint,x · Sp(y) = y · Sp(x) ∀x, y ∈ TpM.
• The principal curvatures of M at p are the eigenvalues κ1, κ2 of Sp. The prin-cipal curvature directions are the corresponding (normalized) eigenvectors.
• The normal curvature of a unit vector x ∈ TpM is
kp(x) = x · Sp(x).
This coincides with k(0) · N(p) where k(t) is the curvature vector of any gen-erating curve for x. The principal curvatures are the maximum and minimumvalues of kp(x) as x takes all values in the unit tangent space at p.
• The mean curvature at p is
H =1
2(κ1 + κ2).
• The Gauss curvature at p is
K = κ1κ2.
The sign of K has intrinsic meaning, independent of the choice of N : if K(p) > 0then either all curves in M through p curve towards N(p), or they all curve awayfrom N(p); if K(p) < 0 then some curves curve towards N(p) and some curveaway.
• Both H and K can be computed directly from any matrix Sp representing Sp:
H =1
2tr Sp, K = det Sp.
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