Ps02 cmth03 unit 1

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CHAPTER 1
Curve Theory
1.1. What is a curve?Definition 1.1.1. A parametrized curve in Rn is a continuous function γ : I → Rn,where I is an interval in R.
Examples 1.1.2. Parametrization of Cartesian curves.
(i) A parametrization of a parabola y = x2 is γ(t) = (t, t2), t ∈ R. The curve γ1(t) =(t2, t4), t ∈ R is not a parametrization of y = x2. The curve γ(t) = (t3, t6), t ∈ R isa parametrization of y = x2.
(ii) The curve γ(t) = (a cos t, b sin t), t ∈ R is a parametrization of the ellipse x2
a2 + y2
b2 =1.
(iii) The curve γ(t) = (a sec t, b tan t), t ∈ (−π2 ,
π2 ) is a parametrization of the hyper
bola x2
a2 − y2
b2 = 1.(iv) The curve γ(t) = (a cos3 t, a sin3 t), t ∈ (0, 2π] or t ∈ R is a parametrization of the
astroid x 23 + y 2
3 = a 23
Examples 1.1.3.
(i) Let γ : R → R2 be defined by γ(t) = (a cos t, b sin t), where a, b ∈ R\{0}. Thenthe image of γ is an ellipse in R2. Its Cartesian equation is x2
a2 + y2
b2 = 1. Inparticular when a = b, then the image of γ is a circle in R2.
(ii) Let γ : R→ R2 be defined by γ(t) = (t, t2). Then the image of γ is a parabola inR2. Its Cartesian equation is y = x2.
(iii) Let γ : R→ R2 be defined by γ(t) = (a cosh t, b sinh t), where a, b ∈ R\{0}. Thenthe image of γ is a part of the hyperbola x2
a2 − y2
b2 = 1.(iv) Let γ : R → R2 be defined by γ(t) = (et cos t, et sin t). Then the image of γ is a
logarithmic spiral in R2.(v) Let γ : R→ R3 be defined by γ(t) = (a+lt, b+mt, c+nt), where a, b, c, l,m, n ∈ R
and l2 +m2 +n2 6= 0. Then the image of γ is a line in R3 passing through (a, b, c)having direction (l,m, n). Its Cartesian equation is x−a
l = y−bm = z−c
n .
1. What is a curve? i
(vi) Let γ : R → R3 be defined by γ(t) = (a cos t, b sin t, ct), where a, b, c ∈ R\{0}.Then the image of γ is a helix in R3. It is a helix with the base ellipse x2
a2 + y2
b2 = 1.When a = b, it is called circular helix.
Definition 1.1.4. A parametrized curve γ in R3 is called planar if it is contained insome plane of R3. It is called nonplanar or twisted if it is not planar.
Let γ : (a, b)→ Rn be a parametrized curve. Then γ = (γ1, γ2, . . . , γn), where eachγi is a mapping from (a, b) to R. The symbol γ is the derivative of γ and it means
γ = (γ ′1, γ ′2, . . . , γ ′n).Throughout it is assumed that all curves are smooth, i.e., all the derivatives of γ exist.
Examples 1.1.5.(i) The curve γ(t) = (t, t2, t3) is not planar.
Suppose that γ is planar. Then the exist a, b, c, d ∈ R, a2 +b2 + c2 6= 0, such thatat + bt2 + ct3 = d for all t. But then a = b = c = 0 (how???). This contradictsthe fact a2 + b2 + c2 6= 0.
(ii) The curve γ(t) = (cos t, sin t, 3 sin t + 4 cos t) is planar.One can see that the coordinates of γ satisfy the equation of the plane z = 4x+3y.Hence γ is planar.
(iii) The curve γ1(t) = 45 cos t, 1− sin t,−3
5 cos t) is a plane curve.
Definition 1.1.6. Let γ be a parametrized curve. Then the vector γ(t) is called thetangent vector to γ at the point γ(t).
If γ(t) 6= 0, then the equation of the tangent to γ at the point γ(t) is R − γ(t) =uγ(t), u ∈ R.
Exercise 1.1.7. Find the equation of the tangent to the following curves.(i) γ(t) = (a cos t, a sin t, bt), (a, 0, 2πb).(ii) γ(t) = (t, t2, t3), (1, 1, 1).(iii) γ(t) = (cos2 t, sin2 t), (1
2 ,12 ).
(iv) γ(t) = (et, t2), (1, 0).
Example 1.1.8. For a logarithmic spiral γ(t) = (et cos t, et sin t), show that the anglebetween γ(t) and γ(t) is independent of t.
Here γ(t) = (et cos t, et sin t) and γ(t) = (et cos t − et sin t, et sin t + et cos t). Letθ(t) be the angle between γ(t) and γ(t). Then
θ(t) = cos−1(
γ(t)γ(t)‖γ(t)‖‖γ(t)‖
)
ii 1. CURVE THEORY
= cos−1(
e2t√
2e2t
)= cos−1
(1√2
).
Therefore the angle between γ(t) and γ(t) is independent of t.
Proposition 1.1.9. Let γ be a parametrized curve in Rn. If γ = a, where a 6= 0, thenγ is a part of a line.
PROOF. Since γ = a, we have γ(t) = at + b. Hence γ is part of a line. �
1.2. Regular curvesDefinition 1.2.1. The arclength of a curve γ starting at the point γ(t0) is the functions(t) given by
s(t) =t∫
t0
‖γ(u)‖du.
Thus, s(t0) = 0 and s(t) is positive or negative according to whether t is larger orsmaller than t0.
Example 1.2.2. Compute the arclength of the logarithmic spiral γ(t) = (ekt cos t, ekt sin t)starting at the point (1, 0).
We have γ(t) = (−ekt sin t + kekt cos t, ekt cos t + kekt sin t). Therefore
‖γ(t)‖ = ekt√
(− sin t + k cos t)2 + (cos t + k sin t)2 = ekt√
1 + k2.
Since γ(0) = (1, 0), the arclength of γ starting at the point (1, 0) is
s(t) =t∫
0
eku√
1 + k2du =√
1 + k2
k (ekt − 1).
Definition 1.2.3. Let γ : (a, b)→ Rn be a parametrized curve. Then γ is called regularat γ(t) if γ(t) 6= 0 (or ‖γ(t)‖ > 0). A point γ(t) of the curve γ is called a singular pointif γ is not regular at that point. A curve γ is said to be regular if all its points areregular.
Definition 1.2.4. Let γ : (a, b)→ Rn be a parametrized curve. Then the scalar ‖γ(t)‖is called the speed of γ at the point γ(t). The curve γ is called a unitspeed curve if‖γ(t)‖ = 1 for all t.
It follows from the definitions that every unitspeed curve is regular. The converseis not true. For example γ(t) = (cos 2t, sin 2t) is regular but not unitspeed.
2. Regular curves iii
Lemma 1.2.5. Let α : (a, b) → Rn be a parametrized curve. If ‖α(t)‖ = 1 for all t ,then α(t) and α(t) are perpendicular for all t. In particular, if γ is a unitspeedcurve, then γ(t) ⊥ γ(t) for all t.
PROOF. Since ‖α(t)‖ = 1 for all t , we have α(t)α(t) = 1. Hence α(t)α(t) = 0 for all t ,i.e., α(t) ⊥ α(t) for all t.
It follows that if γ is a unitspeed curve, then γ(t) ⊥ γ(t) for all t as ‖γ(t)‖ = 1 forall t. �
Lemma 1.2.6. Let γ : (a, b) → Rn be a regular curve. Then the arclength of γ,starting at any point of the curve, is a smooth map.
PROOF. We have s(t) =∫ tt0 ‖γ(u)‖du. Then ds
dt = ‖γ‖. Let γ = (γ1, γ2, . . . , γn). Since γ isregular, it follows that γ ′21 + γ ′22 + · · ·+ γ ′2n > 0. Define f : (0,∞)→ R and g : (a, b)→ Rby f (t) =
√t and g(t) = γ ′21 (t) + · · ·+γ ′2n (t). Then both f and g are smooth maps. Since
the range of g is contained in (0,∞), the domain of f , it follows that the map f ◦ g is asmooth map. But f ◦ g = ds
dt . Therefore dsdt is a smooth map and hence s is a smooth
map. �
Definition 1.2.7. A parametrized curve γ : (a, b)→ Rn is called a reparametrizationof γ : (a, b)→ Rn if there exits a bijective smooth map φ : (a, b)→ (a, b), whose inverseis also smooth, such that γ = γ ◦ φ. The map φ is called the reparametrization mapfor the above reparametrization.
If γ is a reparametrization of γ with the reparametrization map φ, then γ = γ◦φ−1.Hence γ is a reparametrization of γ.
We note that two curves which are reparametrizations of each other have thesame image. Hence they have the same geometric properties.
Exercise 1.2.8. Let C be the collection of all parametrized curve in Rn. Let γ, β ∈ C.We say γ ∼ β if γ is a reparametrization of β. Show that the above relation on C is anequivalence relation.
Proposition 1.2.9. Any reparametrization of a regular curve is regular.
PROOF. Let γ : (a, b) → Rn be a reparametrization of a regular curve γ : (a, b) → Rn.Then there is a bijective smooth map φ : (a, b)→ (a, b), whose inverse is also smooth,such that γ = γ ◦φ. Let ψ : (a, b)→ (a, b) be the inverse of φ. Then φ ◦ψ(t) = t for allt ∈ (a, b). Therefore dφ
dtdψdt = 1 and hence dφ
dt (t) 6= 0 for any t ∈ (a, b). Since γ = γ ◦ φ,we have dγ
dt = dγdt
dφdt . Since γ is regular dγ
dt 6= 0. Since dφdt is never vanishing, it follows
that dγdt is never zero (vector), i.e., γ is regular. �
iv 1. CURVE THEORY
Proposition 1.2.10. A parametrized curve has a unitspeed reparametrization iff itis regular.
PROOF. Let γ be a unitspeed reparametrization of the curve γ. Since γ is a reparametrization of γ, γ is a reparametrization of γ. Since γ is regular (as it is unitspeed) and anyreparametrization of a regular curve is regular, it follows that γ is regular.
Conversely, assume that γ : (a, b) → Rn be regular. Let s be the arclength ofγ, starting at the point γ(t0), i.e., s(t) =
∫ tt0 ‖γ(u)‖du. We note that s is a smooth
map. Since γ is regular, dsdt (t) = ‖γ(t)‖ > 0 for all t , and hence s is strictly increasing
and so it is oneone. Since dsdt (t) 6= 0 for all t , it follows from the inverse function
theorem that s−1 : (a, b) → (a, b) is smooth, where (a, b) is the range of s. Considerthe reparametrization γ : (a, b) → Rn given by γ = γ ◦ s−1, i.e., γ ◦ s = γ. Nowdγdt
dsdt = γ
dt . Then ‖dγdt ‖dsdt = ‖dγdt ‖ = ds
dt , i.e., ‖dγdt ‖ = 1. Hence γ has a unitspeedreparametrization. �
Corollary 1.2.11. Let γ be a regular curve, and let γ be a unitspeed reparametrization of γ given by γ ◦ u = γ, where u is a smooth map. Then u = ±s + c, wheres is the arclength of γ, starting at any point, and c is a constant. Conversely, ifu = ±s + c, then γ is a unitspeed reparametrization of γ.
PROOF. Since γ ◦ u = γ, we have dγdt
dudt = dγ
dt . This implies ‖dγdt ‖∣∣dudt
∣∣ = ‖dγdt ‖. Therefore∣∣dudt
∣∣ = dsdt , as γ is unitspeed. This proves that u = ±s + c.
Conversely, assume that u = ±s+c. Clearly u is bijective, smooth and its inverseis also smooth. We have γ(±s + c) = γ. Hence ±dγ
dtdsdt = dγ
dt . This implies ‖dγdt ‖dsdt =
‖dγdt ‖ = dsdt . Therefore ‖dγdt ‖ = 1, i.e., γ is unitspeed reparametrization of γ. �
1.3. Curvature and TorsionDefinition 1.3.1. Let γ : (a, b)→ R3 be a unitspeed curve. Then the curvature of κ(t)of γ at the point γ(t) is defined by κ(t) = ‖γ(t)‖.
Examples 1.3.2.(i) Let a, b ∈ R3, and let ‖a‖ = 1. Then the curvature of the unitspeed curve
γ(t) = ta + b is zero everywhere.(ii) Let (x0, y0) ∈ R2, and let R > 0. Then the curvature of the unitspeed curve
γ(t) = (x0 + R cos( tR ), y0 + R sin( tR )), which is a circle with center (x0, y0) andradius R, is 1
R everywhere.
3. Curvature and Torsion v
Proposition 1.3.3. If γ is a regular curve in R3, then its curvature κ is given by theformula
κ = ‖γ × γ‖‖γ‖3 .
PROOF. Let γ be a unitspeed reparametrization of γ given by γ ◦ s = γ. We notethat the curvature of γ(s(t)) is same as the curvature of γ at the point γ(t). We have(dsdt)2 = ‖γ‖2 = γ γ, and hence ds
dtd2sdt2 = γ γ. Let γ ′ denote the derivative of γ with
respect to s. Since γ ◦ s = γ, γ ′ dsdt = γ, i.e., γ ′ = γdsdt
. Differentiating it again, we have
γ ′′ =dsdt γ −
d2sdt2 γ(ds
dt)2
dtds =
(dsdt)2 γ − ds
dtd2sdt2 γ(ds
dt)4
= (γ γ)γ − (γ γ)γ‖γ‖4 = γ × (γ × γ)
‖γ‖4 .
Hence κ = ‖γ ′′‖ = ‖γ × (γ × γ)‖‖γ‖4 = ‖γ‖‖(γ × γ)‖
‖γ‖4 = ‖γ × γ‖‖γ‖3 . The last step follows
because γ × γ ⊥ γ. �
Example 1.3.4. Let γ(t) = (a cos t, a sin t, bt), where a, b > 0. Then computationsshows that κ(t) = a
a2+b2 .
Exercises 1.3.5. Compute the curvature of the following curves.(i) γ(t) = (ekt cos t, ekt sin t).(ii) γ(t) = (t, cosh t). Also determine a point having maximum curvature.(iii) γ(t) = (t, t2, t3).
Now we define a special type of curvature for plane curves only.
Definition 1.3.6. Let γ : (a, b) → R2 be a unitspeed curve. Let t be the unit tangentof γ, i.e., t = γ. The unit vector ns obtained by rotating the tangent vector t anticlockwise by an angle π
2 is called the signed unit normal of γ. Since γ is unitspeed,γ is perpendicular to t and hence it is parallel to ns. Therefore there exists a mapκs : (a, b)→ R such that γ = κsns. The map κs is called the signed curvature of γ.
Since κ = ‖γ‖, it follows that κ = κs, i.e., the absolute value of the signed curvature is the curvature
Let Rθ : R2 → R2 be a rotation operator (it rotates a vector anticlockwise byan angle θ). We note that Rθ is linear and Rθ(1, 0) = (cos θ, sin θ) and Rθ(0, 1) =
vi 1. CURVE THEORY
(− sin θ, cos θ). Let (x, y) ∈ R2. Then
Rθ(x, y) = Rθ(x(1, 0) + y(0, 1))= xRθ(1, 0) + yRθ(0, 1)= x(cos θ, sin θ) + y(− sin θ, cos θ).
HenceRθ(x, y) = (x cos θ − y sin θ, x sin θ + y cos θ).
In fact, Rθ is an isometry of R2, i.e., it satisfies
‖Rθ(x1, y1)− Rθ(x2, y2)‖ = ‖(x1, y1)− (x2, y2)‖ ((x1, y1), (x2, y2) ∈ R2).
In particular, when θ = π2 , R π
2(x, y) = (−y, x).
Fix a vector a ∈ R2. Define Ta : R2 → R2 by Ta(x) = x + a. Then Ta is called atranslation by a vector a (it translates given vector by a) and it is an isometry.
Definition 1.3.7. An isometry of R2 of the form Ta ◦ Rθ is called a direct isometry,where a ∈ R2 and θ ∈ [0, 2π).
Example 1.3.8. Compute the signed unit normal of the curve γ(t) = (ekt cos t, ekt sin t).The curve is not unitspeed. The unit tangent will be the unit vector in the direc
tion of the tangent vector, i.e., t(t) = γ(t)‖γ(t)‖ . Now γ(t) = (−ekt sin t+kekt cos t, ekt cos t+
kekt sin t) and ‖γ(t)‖ = ekt√
1 + k2. Hence t(t) = 1√1+k2 (− sin t + k cos t, cos t + k sin t).
Since ns(t) = R π2(t(t)), we have
ns(t) = 1√1 + k2
(− cos t − k sin t,− sin t + k cos t)
Exercise 1.3.9. If γ : (a, b) → R2 is a unitspeed curve, then show that both signedunit normal and signed curvature are smooth maps.
Proposition 1.3.10. Let γ : (a, b)→ R2 be a unitspeed curve, and let s0 ∈ (a, b). Letφ0 ∈ R such that γ(s0) = (cosφ0, sinφ0). Then there exists a unique smooth mapφ : (a, b)→ R such that γ(s) = (cosφ(s), sinφ(s)) for all s ∈ (a, b) and φ(s0) = φ0.
PROOF. Let γ(s) = (f (s), g(s)) (s ∈ (a, b)). Since γ is smooth, both f and g are smoothmaps. Since γ is unitspeed, f2 + g2 = 1 and hence f f + gg = 0. Define φ : (a, b)→ Rby
φ(s) = φ0 +s∫
s0
(fg − fg).
3. Curvature and Torsion vii
Since f and g are smooth, φ is a smooth map and φ(s0) = φ0.Let F = f cosφ + g sinφ and G = f sinφ − g cosφ. Then
F = f cosφ − f sinφφ + g sinφ + g cosφφ= (f + gφ) cosφ + (g − fφ) sinφ.
Now,
f + gφ = f + g(fg − fg) = f (1− g2) + fgg = f f2 + fgg = f (f f + gg) = 0.
By similar arguments it follows that g− fφ = 0. Hence F = 0, i.e., F is a constant map.Therefore F (s) = F (s0) for all s. Now
F (s0) = f (s0) cosφ(s0) + g(s0) sinφ(s0) = cos2 φ0 + sin2 φ0 = 1.
This shows that F = 1. Similarly, we get G = 0. Hence f cosφ + g sinφ = 1 andf sinφ−g cosφ−0. Solving these equations we get f = cosφ and g = sinφ. Thereforeγ(s) = (cosφ(s), sinφ(s)) for all s ∈ (a, b) and φ(s0) = φ0.
For the uniqueness, let ψ : (a, b) → R be a smooth map satisfying ψ(s0) = φ0and γ(s) = (cosψ(s), sinψ(s)) for all s. Then there is a map n : (a, b) → Z such thatφ(s)−ψ(s) = 2πn(s) for all s. Since both φ and ψ are smooth, n is smooth and hence itis continuous. Since (a, b) is connected, n is continuous and the nonempty connectedsubsets of Z are singletons only, it follows that n is a constant map, say, c. Thereforeφ(s)− ψ(s) = 2πc for all s. Since φ(s0) = ψ(s0) = φ0, we get c = 0. Hence ψ(s) = φ(s)for all s. �
Definition 1.3.11. Let γ : (a, b) → R2 be a unitspeed curve. Then there is a smoothmap φ : (a, b) → R such that γ(s) = (cosφ(s), sinφ(s)) for all s, which is determineduniquely by the condition φ(s0) = φ0. The map φ is called the turning angle of γ.
Proposition 1.3.12. Let γ : (a, b) → R2 be a unitspeed curve, and let φ be theturning angle of γ. Then the signed curvature of γ can be determined by theformula κs = φ.
PROOF. Since φ is the turning angle of γ, we have γ(s) = (cosφ(s), sinφ(s)) for all s.Hence γ = (− sinφ, cosφ)φ. Now ns = R π
2(γ) = (− sinφ, cosφ). Therefore γ = φns
and hence the signed curvature κs of γ is φ, i.e., κs = φ. �
It follows from the Proposition 1.3.12 that the signed curvature is the rate at whichthe tangent vector of the curve rotates.
Example 1.3.13. If γ : (a, b)→ R2 is a unitspeed curve, then its signed curvature andsigned unit normal are smooth maps.
viii 1. CURVE THEORY
The map ns : (a, b) → R2 is given by ns(t) = Rπ/2(t(t)). The map Rπ/2 : R2 → R2
defined by Rπ/2(x, y) = (−y, x) is a smooth map and the map t = γ is a smooth mapand hence the map Rπ/2 ◦ t = ns is a smooth map, i.e., the signed unit normal is asmooth map.
Now the map κs : (a, b) → R is given by γ(t) = κs(t)ns(t). Therefore κs(t) =γ(t)ns(t). Since both γ and ns are smooth maps, it follows that κs is a smooth map.
Theorem 1.3.14. Let k : (a, b) → R be a smooth map. Then there is a unitspeedcurve γ : (a, b)→ R2 whose signed curvature is k. Moreover, if γ : (a, b)→ R2 is anyother unitspeed curve whose signed curvature is k, then there is a direct isometryM of R2 such that γ = M ◦ γ.
PROOF. Let s0 ∈ (a, b). Define φ : (a, b) → R by φ(s) =∫ ss0k(u)du. Since k is smooth,
φ is a smooth map. Note that φ(s0) = 0 Define γ : (a, b)→ R2 by
γ(s) =
s∫
s0
cosφ(u)du,s∫
s0
sinφ(u)du
.
Then γ(s) = (cosφ(s), sinφ(s)) for all s. Therefore γ is unitspeed curve. It followsform the Proposition 1.3.10 that φ is the turning angle of γ satisfying φ(s0) = 0 andhence the signed curvature of γ is φ. But φ = k. Hence the signed curvature of γ isk.
Let γ : (a, b) → R2 be a unitspeed curve with signed curvature k. Let ψ be theturning angle of γ. Then ψ = k. But then ψ(s) =
∫ ss0k(u)du + ψ(s0) = φ(s) + θ, where
θ = ψ(s0) is a constant. So we have ψ = φ + θ. Since ψ is the turning angle of γ,˙γ(s) = (cosψ(s), sinψ(s)) for all s. Let a = γ(s0). Integrating the above equation, wehave
γ(s) =
s∫
s0
cosψdu,s∫
s0
sinψdu
+ a
= Ta
s∫
s0
cosψdu,s∫
s0
sinψdu
= Ta
s∫
s0
cos(φ + θ)du,s∫
s0
sin(φ + θ)du
= Ta
s∫
s0
(cosφ cos θ − sinφ sin θ)du,s∫
s0
(sinφ sin θ + cosφ cos θ)du
3. Curvature and Torsion ix
= Ta
cos θs∫
s0
cosφdu − sin θs∫
s0
sinφdu, cos θs∫
s0
cosφdu + sin θs∫
s0
sinφdu
= Ta ◦ Rθ
s∫
s0
cosφdu,s∫
s0
sinφdu
= Ta ◦ Rθ(γ(s)).
Therefore γ(s) = M(γ(s)) for all s, where M = Ta ◦ Rθ is a direct isometry. Henceγ = M ◦ γ. �
Example 1.3.15. Let γ : (a, b)→ R2 be a regular curve with constant signed curvature.Show that γ is (part of) a circle.
We may assume that γ is unitspeed. Let k be the curvature of γ. Let k < 0. Thenk = −` for some ` > 0. Let γ : (a, b) → R2 be defined by γ(t) = (1
` cos(`t),−1` sin(`t)).
Then γ is a unitspeed and t(t) = ˙γ(t) = (− sin(`t),− cos(`t)), ¨γ(t) = (−` cos(`t), sin(`t)).Hence ns(t) = (cos(`t),− sin(`t)). Therefore ¨γ(t) = −`ns(t). This implies that thesigned curvature of γ is −` = k. Since γ and γ have the same signed curvature, thereexists a direct isometry M of R2 such that γ = M ◦ γ. Since N ◦ γ is (part of) a circlefor any direct isometry N of R2, it follows that γ is also a circle.
Similarly one can show that γ is a circle when k > 0.
Examples 1.3.16.(i) Compute the signed curvature of γ(t) = (t, cosh t).
We have γ(t) = (1, sinh t). Let φ(t) be the angle between positive xaxis and γ(t).Then
cosφ(t) = (1, 0)(1, sinh t)√1 + sinh2 t
= 1cosh t .
Therefore sinφ(t) = tanh t and tanφ(t) = sinh t. Let s be the arclength of γ,i.e., s(t) =
∫ t0 ‖γ(u)‖du = sinh t. Differentiating tanφ(t) = sinh t = s with respect
to s we have sec2 φφ = 1, i.e, (1 + tan2 φ)φ = 1. Therefore φ = 11+tan2 φ = 1
1+s2 =1
1+sinh2 t = 1cosh2 t . Hence κs(t) = 1
cosh2 t .(ii) Give an example of a plane curve having signed curvature 1
1+s2 .It is
γ(s) =
s∫
0
cos(tan−1 u)du,s∫
0
sin(tan−1 u)du
.
See the proof of Theorem 1.3.14.
x 1. CURVE THEORY
Definition 1.3.17. Let γ : (a, b) → R3 be a unitspeed curve with nowhere vanishingcurvature. Then the function n = γ
κ is called the unit normal or principal normal ofγ.
It is clear that ‖n(t)‖ = 1 for all t. Since ‖t(t)‖ = 1 for all t , t t = 0, i.e., γ t = 0.This means t ⊥ n. The map b defined by b = t× n is called the unit binormal of γ.
Thus at each point γ(t) of a unitspeed curve γ, we have a right handed system oforthonormal basis {t(t),n(t),b(t)}, i.e, t(t)×n(t) = b(t), n(t)×b(t) = t(t), b(t)×t(t) = n(t),t(t) t(t) = n(t) n(t) = b(t) b(t) = 1.
Since b(t) b(t) = 1 for all t , b b = 0, i.e, b ⊥ b. Also, b = t × n. Thereforeb = t × n + t × n = κ(n) × n + t × n = t × n. This implies that b ⊥ t. Hence b ‖ n.Therefore there is a map τ such that b = −τn. The map τ is called the torsion of γ.
Since n = b× t, we have n = b× t + b× t = −τ(n× t) + κ(b× n) = τb− κt.
Proposition 1.3.18. Let γ : (a, b) → R3 be a regular curve with nowhere vanishingcurvature. Then the torsion τ of γ is given by
τ = (γ × γ)...γ
‖γ × γ‖2 .
PROOF. First assume that γ is unitspeed. Then γ = t, γ = κn and...γ = κn + κn = κn + κ(τb− κt). Now
(γ × γ)...γ = (t× κn)(κn + κ(τb− κt)) = κb(κn + κ(τb− κt)) = κ2τ,
and ‖γ × γ‖2 = ‖κb‖2 = κ2. Hence (γ×γ)...γ
‖γ × γ‖2 = κ2τκ2 = τ.
Let γ be a regular curve with nowhere vanishing curvature (not necessarily unitspeed). Let γ be its unitspeed reparametrization given by γ ◦ s = γ, where s is thearclength of γ starting at any point of γ. Note that the torsion of γ at γ(s(t)) is sameas the torsion of γ at the point γ(t). Since γ ◦ s = γ, we have γ ′ dsdt = γ, i.e,. γ ′ = γ
dsdt
.Therefore γ ′′(dsdt )2+γ ′ d2s
dt2 = γ and γ ′′′(dsdt )3+γ ′′2dsdt
d2sdt2 +γ ′′ dsdt
d2sdt2 +γ ′ d3s
dt3 =...γ. It follows that
γ × γ = (γ ′ × γ ′′)(dsdt )3 and (γ × γ)...γ = (γ ′ × γ ′′)γ ′′′(dsdt )6. Also ‖γ × γ‖2 = ‖γ ′ × γ ′′‖2(dsdt )
6.Hence (γ×γ)
...γ
‖γ × γ‖2 = (γ′×γ′′)γ′′′
‖γ ′ × γ ′′‖2 = τ , as γ is a unitspeed curve. �
Exercises 1.3.19. Compute curvature and torsion of the following curves.
(i) γ(t) = (t, t2, t3).(ii) γ(t) = (a cos t, a sin t, bt)(iii) γ(t) = (et cos t, et sin t, et)
(iv) γ(t) = (45 cos t, 1− sin t, 3
5 cos t)(v) γ(t) = (1
3 (1 + t)3/2, 13 (1− t)3/2, t√
2 )
3. Curvature and Torsion xi
Example 1.3.20.
(i) If γ : (a, b) → R3 is a unitspeed curve with nowhere vanishing curvature, thenthe unit normal, unit binormal, curvature and torsion are smooth maps.We know that n,b : (a, b)→ R3 and κ, τ : (a, b)→ R. By definition κ(t) = ‖γ(t)‖.Let γ = (γ1, γ2, γ3). Since κ is nowhere vanishing, γ ′′21 (t) + γ ′′22 (t) + γ ′′23 (t) > 0for all t. Define f : (0,∞) → R by f (t) =
√t and g : (a, b) → R by g(t) =
γ ′′21 (t) + γ ′′22 (t) + γ ′′23 (t). Then both f and g are smooth maps. Since g(t) > 0 forall t , f ◦ g is a smooth map. But f ◦ g = κ and hence κ is a smooth map.Since κ is a smooth map and κ(t) > 0 for all t , 1
κ is a smooth map. Thereforeγκ = n is a smooth map.By definition b = t × n and both t and n are smooth maps Therefore b is asmooth map.Now b = −τn. It means τ = −bn. Since both b (so is b) and n are smooth maps,τ is a smooth map.
(ii) Let γ be a regular curve in R3 with nowhere vanishing curvature. Then γ isplanar iff the torsion of γ is identically zero.We may assume that γ is unitspeed. Suppose that γ is planar. Then there exista unit vector a ∈ R3 and a constant d ∈ R such that γa = d. Differentiating, weget ta = 0. Differentiating it again, we have ta = 0, i.e., κ(na) = 0. This givesna = 0, as κ is nowhere vanishing. Therefore a ⊥ t and a ⊥ n. It means b = ±a.Since b is continuous, it follows that either b(t) = a for all t or b(t) = −a for allt. So, in either case, b = 0. Since b = −τn, it follows that τ = 0.Conversely, assume that τ = 0. Since b = −τn, we have b = 0, i.e, b is a constantunit vector. Now d
dt (γ b) = γ b = t b = 0. This means γ b is a constant function,say, d, i.e., γ b = d. Hence γ is lying on the plane R b = d, i.e., γ is planar.
(iii) Let γ be a regular curve in R3 with constant curvature and zero torsion. Then γis (part of) a line or (part of) a circle.Assume that the curvature κ = 0. Since γ = κn, γ = 0. Therefore γ is (part of)a line.Let κ > 0. Note that τ = 0. We see that d
dt (γ + 1κn) = t + 1
κ n = t + 1κ (τb− κt) = 0.
Therefore γ + 1κn = a for some constant vector a ∈ R3. But then ‖γ − a‖ = 1
k ,i.e., γ is lying on the sphere ‖R − a‖ = 1
κ . Since τ = 0, γ is lying on some plane.Therefore γ is lying on a sphere as well as a plane and hence it is (part of) acircle.
(iv) Let γ be a regular curve in R3 with nowhere vanishing curvature. Show that γis spherical iff τ
κ = ddt( κκ2τ).
xii 1. CURVE THEORY
We may assume that γ is unitspeed. Assume that γ is spherical. Then thereexist a constant vector a ∈ R3 and a constant R ∈ R such that (γ − a)(γ − a) =‖γ − a‖2 = R2. Differentiating it we get (γ − a)t = 0. Differentiating it again wehave (γ − a)n = − 1
κ . Again differentiation gives (γ − a)b = κτκ2 . Since {t,n,b}
is an orthonormal basis of R3, γ − a = ((γ − a)t)t + ((γ − a)n)n + ((γ − a)b)b.Therefore
γ − a = −1κn + κ
τκ2 b.
But then R2 = ‖γ − a‖2 = 1κ2 + κ2
τ2κ4 . Differentiating it we obtainτ2κ42κκ−κ2(2ττκ4+τ24κ2κ)
τ4κ8 − 2κκ3 = 0, i.e., κ2κτ−2κκ2τ−κ2κτ
κ4τ2 − τκ = 0, i.e., τ
κ = ddt( κκ2τ).
Conversely, assume that τκ = d
dt( κκ2τ). Let a = γ + 1
κn−κτκ2 b. Then
a = t− κκ2 n + 1
κ n− ddt
(κκ2τ
)b− κ
τκ2 b
= t− κκ2 n + 1
κ (τb− κt)− ddt
(κκ2τ
)b− κ
τκ2 (−τ)n
=(τκ −
ddt
(κκ2τ
))b = 0.
Hence a is a constant vector. Now ‖γ − a‖2 = 1κ2 + κ2
τ2κ4 . Again using τκ = d
dt( κκ2τ),
it follows that 1κ2 + κ2
τ2κ4 is a positive constant, say, R2. Therefore ‖γ − a‖2 = R2,and hence γ is spherical.
ORLet ρ = 1
κ and σ = 1τ . Assume that γ is spherical. Then there exist a constant
vector a ∈ R3 and a constant R > 0 such that (γ − a)(γ − a) = ‖γ − a‖2 = R2.Differentiating it we get (γ − a)t = 0. Differentiating it again we have (γ − a)n =− 1
κ = −ρ. One more differentiation gives (γ − a)b = −σρ. Now
γ − a = ((γ − a)t)t + ((γ − a)n)n + ((γ − a)b)b= −ρn− σρb.
Therefore R2 = ‖γ − a‖2 = ‖ − ρn− σρb‖2 = ρ2 + (σρ)2. Differentiation of thelast equation gives ρ
σ + (σρ)• = 0, which is required.Conversely, assume that ρ
σ + (σρ)• = 0. Then ρ2 + (σρ)2 = d for some constantd > 0(???). Let a = γ + ρn + σρb. Then a = t + ρn + ρ(τb − κt) + (σρ)•b −τσρn = ( ρσ + (σρ)•)b = 0. Therefore a is a constant vector. Now ‖γ − a‖2 =‖ − ρn− σρb‖2 = ρ2 + (σρ)2 = d. Hence γ is spherical.
(v) Let γ be a unitspeed curve in R3 with nowhere vanishing curvature. Show thatthe curve α = t is a regular curve and find the curvature and torsion of α.
4. Fundamental Theorem of Space Curves xiii
Here α = t. Therefore α = t = γ. Since the curvature κ = ‖γ‖ > 0, thecurve α is regular. Now α = t = κn, α = κn + κn = κn + κ(τb − κt) and...α = κn + κ(τb − κt) + κτb + κτb − κτ2n − 2κκt − κ3n. Now α × α = κ2τt + κ3b,(α × α)
...α = −κ3κτ + κ4τ , ‖α‖ = κ and ‖α× α‖ = κ2
√κ2 + τ2. Let κ1 be the
curvature of α and τ1 be the torsion of α. Then
κ1 = ‖α× α‖‖α‖3 = κ2
√κ2 + τ2
κ3 =√κ2 + τ2
κand
τ1 = (α× α)...α
‖α× α‖2 = κ(κτ − κτ)√κ2 + τ2
.
Theorem 1.3.21. Let γ be a unitspeed curve in R3. Let t, n and b be its unittangent, unit normal and unit binormal respectively. Let κ and τ be the curvatureand torsion of γ respectively. Then
t = κnn = −κt τbb = −τn
The equations in Theorem 1.3.21 are called FrenetSerret equations or SerretFrenet equations.
Example 1.3.22. Verify the SerretFrenet equations for the curve
γ(a) =(a cos
(a√
a2 + b2
), a sin
(a√
a2 + b2
), b√
a2 + b2
).
1.4. Fundamental Theorem of Space CurvesTheorem 1.4.1. (Existence) Let k, t : (α, β) → R be smooth maps with κ(s) > 0 forall s. Then there exists a unitspeed curve γ : (α, β)→ R3 whose curvature is k andtorsion is t.
PROOF. Fix s0 ∈ (α, β). It follows from the theory of Differential Equations that theequations T = kN, N = −kT + tB and B = −tN have a unique solution T, N and B
such that {T(s0),N(s0),B(s0)} is a standard basis of R3. Since the matrix
0 k 0−k 0 t0 −t 0
expressing T, N and B in terms of T, N and B is skewsymmetric, it follows that T,N and B are orthogonal unit vectors. Since B is a smooth map orthogonal to both T
xiv 1. CURVE THEORY
and N, there exists a smooth map λ : (α, β)→ R such that B = λT×N. Note that λ(s)is 1 or −1. Since λ is continuous, either λ(s) = 1 for all s or λ(s) = −1 for all s. SinceB(s0) = T(s0)×N(s0), we have λ(s) = 1 for all s, i.e., B = T×N. Define γ : (α, β)→ R3
by
γ(s) =s∫
s0
T(u)du.
Then γ = T. Therefore γ is a unitspeed curve and T is the unit tangent of γ. Nowγ = T = kN. Therefore ‖γ‖ = k, i.e., k is the curvature of γ. It follows fromγ = T = kN that N is the unit normal of γ. Since B = T × N, B is the unit binormalof γ. It follows from the equation B = −tN that t is the torsion of γ. �
Theorem 1.4.2. (Uniqueness) If γ, γ : (α, β) → R3 are unitspeed curves with samecurvature and same torsion, then there is an isometry M of R3 such that γ = M ◦γ.
PROOF. Let t, n and b be the unit tangent, unit normal and unit binormal of γ, and lett, n and b be those of γ. Fix s0 ∈ (α, β). Since {t(s0),n(s0),b(s0)} and {t(s0), n(s0), b(s0)}are both right handed orthonormal basis of R3, there is a rotation about origin thatmaps t(s0), n(s0) and b(s0) to t(s0), n(s0) and b(s0) respectively. Further, there is atranslation which takes γ(s0) to γ(s0) (and this has no effect on t, n and b). By applyingrotation followed by translation, we can therefore assume that γ(s0) = γ(s0), t(s0) =t(s0), n(s0) = n(s0) and b(s0) = b(s0). Hence to prove the theorem we need to proveγ = γ. Define A : (α, β)→ R by
A(s) = t(s)t(s) + n(s)n(s) + b(s)b(s).Then A is a smooth map. Note that A(s) ≤ 3 for all s and A(s0) = 3. Also observethat A(s) = 0 for all s (you need to verify this). Therefore A is a constant map. SinceA(s0) = 3, we have A(s) = 3 for all s. Hence it follows that t = t, n = n and b = b. Theequation t = t implies that γ = γ + c. Since γ(s0) = γ(s0), we get c = 0. Hence γ = γ.This proves the theorem. �
Example 1.4.3. Describe all curves in R3 which have constant curvature κ > 0 andconstant torsion τ.
Let γ : (a, b) → R3 be a unitspeed curve with constant curvature κ > 0 andconstant torsion τ. Let a = κ
κ2+τ2 and b = τκ2+τ2 . Define γ : (a, b)→ R3 by
γ(t) =(a cos
(t√
a2 + b2
), a sin
(t√
a2 + b2
), b t√
a2 + b2
).
Then γ is a unitspeed curve. The curvature of γ is aa2+b2 = κ and its torsion is
ba2+b2 = τ. Therefore γ, γ : (a, b) → R3 are unitspeed curves with same curvature
5. Isoperimetric Inequality and Four Vertex Theorem xv
and same torsion. So, there is an isometry M of R3 such that γ = M ◦ γ. But γ is ahelix and hence M ◦ γ is a helix. This implies that γ is a helix. Hence any curve withconstant (nonzero) curvature and constant torsion is a helix.
1.5. Isoperimetric Inequality and Four Vertex TheoremDefinition 1.5.1. Let a ∈ R be a positive constant. A simple closed curve in R2 is a(regular) curve γ : R→ R2 such that γ(t) = γ(t ′) iff t ′ − t = ka for some k ∈ Z.
Thus, the point γ(t) returns to its starting point when t is increased by a, but notbefore that.
By Jordan Curve Theorem any simple closed curve in R2 has an ‘interior’ and‘exterior’. More precisely, the set of points in R2 that are not on the curve γ is thedisjoint union of two subsets of R2, denoted by int(γ) and ext(γ), with the followingproperties.
(i) int(γ) is bounded,(ii) ext(γ) is unbounded,(iii) both int(γ) and ext(γ) are connected, but any curve joining a point of int(γ) to a
point of ext(γ) crosses γ.
Proposition 1.5.2. Let γ be a regular curve with period a. Then its unitspeedreparametrization is `(γ) periodic, where `(γ) is the length of γ.
PROOF. Since γ is a periodic, the length, `(γ), of γ is
`(γ) =a∫
0
‖γ(u)‖du.
Let γ be a unitspeed reparametrization of γ given by γ ◦ s = γ, where s is the arclength of γ starting at the point γ(0). Let t, t ′ ∈ R, and let t ≤ t ′. Then there isk ∈ N ∪ {0} such that t ′ − t = ka + ε , where 0 ≤ ε < a. Now
s(t ′)− s(t) =t ′∫
0
‖γ(u)‖du −t∫
0
‖γ(u)‖du =t ′∫
t
‖γ(u)‖du
=t+a∫
t
‖γ(u)‖du +t+2a∫
t+a
‖γ(u)‖du + · · ·+t+ka∫
t+(k−1)a
‖γ(u)‖du +t+ka+ε∫
t+ka
‖γ(u)‖du
= k`(γ) +t+ε∫
t
‖γ(u)‖du.
xvi 1. CURVE THEORY
Hence s(t ′) − s(t) = k`(γ) iff∫ t+εt ‖γ(u)‖du = 0 iff ε = 0 iff t ′ − t = ka. Now γ(s(t)) =
γ(s(t ′)) iff γ(t) = γ(t ′) iff t ′ − t = ka for some k ∈ Z iff s(t ′) − s(t) = k`(γ). Thereforeγ is `(γ) periodic. �
Let γ be a simple closed curve in R2. Let A(int(γ)) be the area of the interior ofγ. Then A(int(γ)) =
∫∫int(γ) dxdy.
Definition 1.5.3. A simple closed curve in R2 is called positively oriented if its unitsigned normal points towards the interior of the curve at ach point of the curve.
Theorem 1.5.4. (Green’s Theorem) Let γ be a simple closed curve in R2, and letγ be positively oriented. Let f and g be continuous on the closure of int(γ) andsmooth on int(γ). Then
∫∫
int(γ)
(∂g∂x −
∂f∂y
)dxdy =
∫
γ
[f (x, y)dx + g(x, y)dy].
Proposition 1.5.5. Let γ(t) = (x(t), y(t)) be a simple closed curve in R2 with perioda. Let γ be positively oriented. Then
A(int(γ)) = 12
a∫
0
(xy − yx)dt.
PROOF. Take f (x, y) = −12y and g(x, y) = 1
2x. Then both f and g are smooth functionson R2. It follows from the Green’s theorem that∫∫
int(γ)(12 − (−1
2 )dxdy =∫γ[ 1
2xdy −12ydx]. This implies
A(int(γ)) =∫γ[ 1
2xdy −12ydx] = 1
2∫ a
0 (xy − yx)dt. �
Proposition 1.5.6 (Wirtinger’s Inequality). Let F : [0, π]→ R be a smooth map satisfying F (0) = F (π) = 0. Then
π∫
0
F2dt ≥π∫
0
F2dt,
with equality holding iff F (t) = A sin t for some constant A.
PROOF. Define G : [0, π]→ R by G(t) = F (t)sin t . Note that G is a smooth map. Then
π∫
0
F2dt =π∫
0
(G sin t +G cos t)2dt
=π∫
0
G2 sin2 tdt + 2π∫
0
GG sin t cos tdt +π∫
0
G2 cos2 tdt
5. Isoperimetric Inequality and Four Vertex Theorem xvii
=π∫
0
G2 sin2 tdt −π∫
0
G2(cos2 t − sin2 t)dt +π∫
0
G2 cos2 tdt
=π∫
0
G2 sin2 tdt +π∫
0
G2 sin2 tdt =π∫
0
F2dt +π∫
0
G2 sin2 tdt.
Henceπ∫
0
F2dt −π∫
0
F2dt =π∫
0
G2 sin2 tdt ≥ 0. (1.5.6.1)
By equation (1.5.6.1) it follows that∫ π
0 F2dt =∫ π
0 F2dt iff∫ π
0 G2 sin2 tdt = 0 iff G2 sin2 t =0 iff G sin t iff G = 0. But G = 0 implies G(t) = A for all t , for some constant A, i.e.,F (t) = A sin t for all t. �
Example 1.5.7. The length and the are of a simple closed are invariant under isometry.Let γ : R→ R2 be a simple closed curve with period ` , and let M : R2 → R2 be an
isometry. Let γ : R→ R2 be γ = M ◦γ. We need to prove that the lengths and the areaof interiors of γ and γ are same. Let t, t ′ ∈ R. Then γ(t ′) = γ(t) iff M(γ(t ′)) = M(γ(t))iff γ(t ′) = γ(t) iff t ′ − t = k` for some k ∈ Z. Therefore γ is `periodic. Now
`(γ) =`∫
0
‖ ˙γ(u)‖du =`∫
0
‖DM(γ(u))‖du
=`∫
0
‖DM(γ(u)) ◦ γ(u)‖du
=`∫
0
‖M(γ(u))‖du =`∫
0
‖γ(u)‖du = `(γ).
Since M is an isometry 1, there are a, b, c, d, e, f ∈ R2 such that a2 +b2 = c2 +d2 =a2 + c2 = b2 +d2 = 1, ac+bd = ab+ cd = 0 and M(x, y) = (ax+by + e, cx+dy + f ).Let γ(t) = (x(t), y(t)). Then γ(t) = (ax(t) + by(t) + e, cx(t) + dy(t) + f ). Now
A(int(γ)) = 12
`∫
0
((ax + by + e)(cx + dy + f )• − (ax + by + e)•(cx + dy + f ))dt
1We know that N : R2 → R2 is an isometry iff there are a, b, c, d, e, f ∈ R2 such that a2 +b2 = c2 +d2 =a2 + c2 = b2 + d2 = 1, ac + bd = ab + cd = 0 and N(x, y) = (ax + by, cx + dy) + (e, f ) OR N is anisometry of R2 if and only if there is a 2× 2 orthogonal matrix A and a ∈ R2 such that N(x) = Ax + a
xviii 1. CURVE THEORY
= 12
`∫
0
((ax + by + e)(cx + dy)− (ax + by)(cx + dy + f ))dt
= 12
(ad − bc)`∫
0
(xy − xy)dt + (ec − af )(x(`)− x(0)) + (ed − bf )(y(`)− y(0))
= 12
`∫
0
(xy − xy)dt =A(int(γ)),
as ad − bc = ±1, x(0) = x(`) and y(0) = y(`) (???).
Theorem 1.5.8 (The Isoperimetric Inequality). Let γ be a simple closed curve, let `be its length, and let A be the area of interior of γ. Then `2 ≥ 4πA, with equalityholding iff γ is a circle.
PROOF. We may assume that γ is unitspeed. Then the length of γ is its period, i.e.,γ is ` periodic. Consider the curve γ(t) = (x(t), y(t)) defined by γ(t) = γ
( t`π), i.e.,
γ = γ ◦ s or γ ◦ s−1 = γ, where s(t) = t`π . [Note that γ is a unitspeed reparametrization
of γ with the parametrization map s−1(t) = tπ` , i.e., s(t) = t`
π .] Then γ is a simple closedcurve with the same length ` and period π. Also the area of interior of γ is same asthe area of interior of γ. Since the length and the area of a curve are invariant underisometry, we may assume that γ(0) = γ(0) = (0, 0). Let x = r cos θ and y = r sin θ.Note that r, θ : [0, π] → R are smooth maps. Since γ is π periodic, γ(0) = γ(π). Thisgives r(0) = r(π) as γ(0) = (0, 0). Now x2 + y2 = r2 + r2θ2 and xy − yx = r2θ. Sinceγ = γ ◦ s, we have ˙γ = γ ds
dt . But then r2 + r2θ2 = x2 + y2 = ‖ ˙γ‖2
= ‖γ‖2(dsdt )2 = `2
π2 . Wealso note that A =A(int(γ)) = 1
2∫ π
0 (xy − yx)dt = 12∫ π
0 r2θ. Now,
`2
4π −A = 14
π∫
0
(r2 + r2θ2)dt − 12
π∫
0
r2θdt
= 14
π∫
0
(r2 + r2θ2 − 2r2θ)dt
= 14
π∫
0
r2(θ − 1)2dt + 14
π∫
0
(r2 − r2)dt ≥ 0.
It follows from the above equation that `2
4π−A = 0 iff 14∫ π
0 r2(θ−1)2dt+∫ π
0 (r2−r2)dt = 0iff∫ π
0 r2(θ − 1)2dt =∫ π
0 (r2 − r2)dt = 0 iff θ = 1 and r = A sin t for some A iff θ = t + α
5. Isoperimetric Inequality and Four Vertex Theorem xix
for some constant α and r = A sin t for some A iff r = A sin(θ−α). But r = A sin(θ−α)is a circle. Hence `2 − 4πA = 0 iff γ is a circle. �
Example 1.5.9. By applying the isoperimetric inequality to the ellipse x2
p2 + y2
q2 = 1(where p and q are positive constants), prove that
2π∫
0
√p2 sin2 t + q2 cos2 t dt ≥ 2π√pq,
with equality holding if and only if p = q.Consider γ(t) = (p cos t, q sin t). Then the trace of γ is the ellipse x2
p2 + y2
q2 = 1.Note the γ is 2πperiodic. Therefore length ` of γ is
` =2π∫
0
‖γ(t)‖dt =2π∫
0
√p2 sin2 t + q2 cos2 t dt.
Also, the area A of the interior of γ is
A = 12
2π∫
0
(xy − yx)dt = 12
2π∫
0
pq(cos t cos t + sin t sin t)dt = pq2 2π = pqπ.
It follows from the Isoperimetric inequality that
2π∫
0
√p2 sin2 t + q2 cos2 t dt
2
≥ 4πpqπ = 4π2pq,
i.e.,2π∫
0
√p2 sin2 t + q2 cos2 t dt ≥ 2π√pq.
Equality holds in above inequality iff γ is a circle iff p = q. [Because γ(t) =(p cos t, q sin t) is a circle iff p = q.]
Definition 1.5.10. A simple closed curve in R2 is called convex if its interior is aconvex subset of R2.
Definition 1.5.11. The vertex of a curve γ is a point of the curve where its signedcurvature κs has a stationary point, i.e, κs = 0 at that point.
Theorem 1.5.12 (Four Vertex Theorem). Every convex, simple closed curve in R2
has at least four vertices.
xx 1. CURVE THEORY
Example 1.5.13. Find the vertices of the curve γ(t) = (a cos t, b sin t), where a, b > 0and a 6= b.
Let s be the arclength of γ starting at some point of γ. Denoting the derivativeof γ with respect to s by γ ′, we have γ ′(t) = (−a sin t, b cos t) dtds . Since ‖γ ′‖ = 1,dtds = 1√
a2 sin2 t+b2 cos2 t. Therefore
t(t) = γ ′(t) = 1√a2 sin2 t + b2 cos2 t
(−a sin t, b cos t).
Differentiating γ ′ with respect to s, we get
γ ′′(t) = ab(a2 sin2 t + b2 cos2 t)2
(−b cos t,−a sin t).
Now ns(t) = R π2(t(t)) = 1
(a2 sin2 t+b2 cos2 t)12(−b cos t,−a sin t). Hence
γ ′′ = ab(a2 sin2 t + b2 cos2 t) 3
2ns.
This implies that κs(t) = ab(a2 sin2 t+b2 cos2 t)
32. Now κs(t) = −3ab(a2−b2) sin t cos t
(a2 sin2 t+b2 cos2 t)52
. Since a, b > 0and a 6= b, we have κs(t) = 0 iff sin t cos t = 0 iff sin 2t = 0 iff t = nπ
2 for some n ∈ Z.Since γ is 2π periodic, it follows that the vertices of γ are γ(0) = (a, 0), γ(π/2) = (0, b),γ(π) = (0,−a) and γ(3π/2) = (0,−b) only.
Example 1.5.14. The ellipse x2
a2 + y2
b2 = 1 (or the curve γ(t) = (a cos t, b sin t)) is aconvex curve.
Let (x, y), (z,w) ∈ int(γ), i.e., x2
a2 + y2
b2 < 1 and z2
a2 + w2
b2 < 1. Let X =(xa ,
yb)
andY =
( za ,
wb). Then ‖X‖ < 1 and ‖Y‖ < 1. Let t ∈ [0, 1]. Then ‖tX + (1− t)Y‖ ≤
t‖X‖+ (1− t)‖Y‖ < t + 1− t = 1. Now(tx + (1− t)z)2
a2 + (ty + (1− t)w)2
b2 = ‖tX + (1− t)Y‖2 < 1.
Therefore t(x, y) + (1− t)(z,w) ∈ int(γ) and hence γ is convex.