1. 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 = x 2 is γ(t) = (t, t 2 ), t ∈ R. The curve γ 1 (t) =
(t 2 , t 4 ), t ∈ R is not a parametrization of y = x 2 . The curve γ(t) = (t 3 , t 6 ), t ∈ R is
a parametrization of y = x 2 .
2
y2
(ii) The curve γ(t) = (a cos t, b sin t), t ∈ R is a parametrization of the ellipse x 2 + b2 =
a
1.
(iii) The curve γ(t) = (a sec t, b tan t), t ∈ (− π , π ) is a parametrization of the hyper2 2
2
y2
bola x 2 − b2 = 1.
a
(iv) The curve γ(t) = (a cos3 t, a sin3 t), t ∈ (0, 2π] or t ∈ R is a parametrization of the
2
2
2
astroid x 3 + y 3 = a 3
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}. Then
2
y2
the image of γ is an ellipse in R2 . Its Cartesian equation is x 2 + b2 = 1. In
a
particular when a = b, then the image of γ is a circle in R2 .
(ii) Let γ : R → R2 be defined by γ(t) = (t, t 2 ). Then the image of γ is a parabola in
R2 . Its Cartesian equation is y = x 2 .
(iii) Let γ : R → R2 be defined by γ(t) = (a cosh t, b sinh t), where a, b ∈ R{0}. Then
2
y2
the image of γ is a part of the hyperbola x2 − b2 = 1.
a
(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 = 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 = y−b = z−c .
l
m
n

2. 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}.
2
y2
Then the image of γ is a helix in R3 . It is a helix with the base ellipse x 2 + b2 = 1.
a
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 in
some plane of R3 . It is called non-planar or twisted if it is not planar.
Let γ : (a, b) → Rn be a parametrized curve. Then γ = (γ1 , γ2 , . . . , γn ), where each
˙
γi is a mapping from (a, b) to R. The symbol γ is the derivative of γ and it means
˙
γ = (γ1 , γ2 , . . . , γn ).
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, t 2 , t 3 ) is not planar.
Suppose that γ is planar. Then the exist a, b, c, d ∈ R, a2 + b2 + c2 = 0, such that
at + bt 2 + ct 3 = d for all t. But then a = b = c = 0 (how???). This contradicts
the fact a2 + b2 + c2 = 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.
3
(iii) The curve γ 1 (t) = 4 cos t, 1 − sin t, − 5 cos t) is a plane curve.
5
˙
Definition 1.1.6. Let γ be a parametrized curve. Then the vector γ (t) is called the
tangent vector to γ at the point γ(t).
˙
If γ (t) = 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)
(ii)
(iii)
(iv)
γ(t) = (a cos t, a sin t, bt), (a, 0, 2πb).
γ(t) = (t, t 2 , t 3 ), (1, 1, 1).
1 1
γ(t) = (cos2 t, sin2 t), ( 2 , 2 ).
γ(t) = (et , t 2 ), (1, 0).
Example 1.1.8. For a logarithmic spiral γ(t) = (et cos t, et sin t), show that the angle
˙
between 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). Let
˙
θ(t) be the angle between γ(t) and γ (t). Then
θ(t) = cos−1
˙
γ(t)γ (t)
˙
γ(t) γ (t)

3. 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 = 0, then
γ 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 curves
Definition 1.2.1. The arc-length of a curve γ starting at the point γ(t0 ) is the function
s(t) given by
t
s(t) =
˙
γ (u) du.
t0
Thus, s(t0 ) = 0 and s(t) is positive or negative according to whether t is larger or
smaller than t0 .
Example 1.2.2. Compute the arc-length 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 arc-length of γ starting at the point (1, 0) is
t
√
√
1 + k2 kt
(e − 1).
s(t) = eku 1 + k2 du =
k
0
Definition 1.2.3. Let γ : (a, b) → Rn be a parametrized curve. Then γ is called regular
˙
˙
at γ(t) if γ (t) = 0 (or γ (t) > 0). A point γ(t) of the curve γ is called a singular point
if γ is not regular at that point. A curve γ is said to be regular if all its points are
regular.
˙
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 unit-speed curve if
˙
γ (t) = 1 for all t.
It follows from the definitions that every unit-speed curve is regular. The converse
is not true. For example γ(t) = (cos 2t, sin 2t) is regular but not unit-speed.

4. 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 unit-speed
¨
˙
curve, 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,
˙ for all t.
i.e., α(t) ⊥ α(t)
˙
¨
˙
It follows that if γ is a unit-speed curve, then γ (t) ⊥ γ (t) for all t as γ (t) = 1 for
all t.
Lemma 1.2.6. Let γ : (a, b) → Rn be a regular curve. Then the arc-length of γ,
starting at any point of the curve, is a smooth map.
t
˙
˙
PROOF. We have s(t) = t0 γ (u) du. Then ds = γ . Let γ = (γ1 , γ2 , . . . , γn ). Since γ is
dt
regular, it follows that γ12 + γ22 + · · · + γn2 > 0. Define f : (0, ∞) → R and g : (a, b) → R
√
by f(t) = t and g(t) = γ12 (t) + · · · + γn2 (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 a
smooth map. But f ◦ g = ds . Therefore ds is a smooth map and hence s is a smooth
dt
dt
map.
Definition 1.2.7. A parametrized curve γ : (˜ , b) → Rn is called a reparametrization
a ˜
a ˜
of γ : (a, b) → Rn if there exits a bijective smooth map φ : (˜ , b) → (a, b), whose inverse
is also smooth, such that γ = γ ◦ φ. The map φ is called the reparametrization map
for 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 the
same 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 an
equivalence relation.
Proposition 1.2.9. Any reparametrization of a regular curve is regular.
PROOF. Let γ : (˜ , b) → Rn be a reparametrization of a regular curve γ : (a, b) → Rn .
a ˜
Then there is a bijective smooth map φ : (˜ , b) → (a, b), whose inverse is also smooth,
a ˜
such that γ = γ ◦ φ. Let ψ : (a, b) → (˜ , b) be the inverse of φ. Then φ ◦ ψ(t) = t for all
a ˜
dφ dψ
t ∈ (a, b). Therefore d˜ dt = 1 and hence dφ (˜ = 0 for any ˜ ∈ (˜ , b). Since γ = γ ◦ φ,
t)
t
a ˜
t
d˜
t
dγ
dγ dφ
dγ
dφ
we have d˜ = dt d˜ . Since γ is regular dt = 0. Since d˜ is never vanishing, it follows
t
t
t
that dγ is never zero (vector), i.e., γ is regular.
d˜
t

5. iv
1. CURVE THEORY
Proposition 1.2.10. A parametrized curve has a unit-speed reparametrization iff it
is regular.
PROOF. Let γ be a unit-speed 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 unit-speed) and any
reparametrization 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 arc-length of
t
˙
γ, starting at the point γ(t0 ), i.e., s(t) = t0 γ (u) du. We note that s is a smooth
˙
map. Since γ is regular, ds (t) = γ (t) > 0 for all t, and hence s is strictly increasing
dt
ds
and so it is one-one. Since dt (t) = 0 for all t, it follows from the inverse function
theorem that s−1 : (˜ , b) → (a, b) is smooth, where (˜ , b) is the range of s. Consider
a ˜
a ˜
n
the reparametrization γ : (˜ , b) → R given by γ = γ ◦ s−1 , i.e., γ ◦ s = γ. Now
a ˜
dγ ds
γ
dγ ds
= dt . Then d˜ dt = dγ = ds , i.e., dγ = 1. Hence γ has a unit-speed
dt
dt
d˜ dt
t
t
d˜
t
reparametrization.
Corollary 1.2.11. Let γ be a regular curve, and let γ be a unit-speed reparametrization of γ given by γ ◦ u = γ, where u is a smooth map. Then u = ±s + c, where
s is the arc-length of γ, starting at any point, and c is a constant. Conversely, if
u = ±s + c, then γ is a unit-speed reparametrization of γ.
PROOF. Since γ ◦ u = γ, we have dγ du = dγ . This implies dγ du = dγ . Therefore
dt
dt
dt
d˜ dt
t
d˜
t
du
= ds , as γ is unit-speed. This proves that u = ±s + c.
dt
dt
Conversely, assume that u = ±s + c. Clearly u is bijective, smooth and its inverse
is also smooth. We have γ(±s + c) = γ. Hence ± dγ ds = dγ . This implies dγ ds =
dt
d˜ dt
t
d˜ dt
t
dγ
= ds . Therefore dγ = 1, i.e., γ is unit-speed reparametrization of γ.
dt
dt
d˜
t
1.3. Curvature and Torsion
Definition 1.3.1. Let γ : (a, b) → R3 be a unit-speed 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 unit-speed curve
γ(t) = ta + b is zero everywhere.
(ii) Let (x0 , y0 ) ∈ R2 , and let R > 0. Then the curvature of the unit-speed curve
t
t
γ(t) = (x0 + R cos( R ), y0 + R sin( R )), which is a circle with center (x0 , y0 ) and
1
radius R, is R everywhere.

6. 3. Curvature and Torsion
v
Proposition 1.3.3. If γ is a regular curve in R3 , then its curvature κ is given by the
formula
¨ ˙
γ×γ
κ=
.
˙ 3
γ
PROOF. Let γ be a unit-speed reparametrization of γ given by γ ◦ s = γ. We note
that the curvature of γ(s(t)) is same as the curvature of γ at the point γ(t). We have
2
ds 2
˙ 2
˙¨
˙¨
= γ = γ γ , and hence ds d s = γ γ . Let γ denote the derivative of γ with
dt
dt dt 2
˙
γ
ds
˙
respect to s. Since γ ◦ s = γ, γ
= γ , i.e., γ = ds . Differentiating it again, we have
dt
γ
=
=
Hence κ = γ
=
ds ¨
γ
dt
dt
d2 s ˙
γ
dt 2
ds 2
dt
−
dt
=
ds
ds 2
dt
ds d2 s ˙
γ
dt dt 2
ds 4
dt
¨
γ−
˙¨ ˙
˙
¨ ˙
˙˙ ¨
γ × (γ × γ )
(γ γ )γ − (γ γ )γ
=
.
˙ 4
˙ 4
γ
γ
˙
¨ ˙
γ × (γ × γ )
4
˙
γ
=
˙
γ
¨ ˙
(γ × γ )
4
˙
γ
=
¨ ˙
γ×γ
. The last step follows
3
˙
γ
¨ ˙
˙
because γ × γ ⊥ γ .
Example 1.3.4. Let γ(t) = (a cos t, a sin t, bt), where a, b > 0. Then computations
a
shows that κ(t) = 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, t 2 , t 3 ).
Now we define a special type of curvature for plane curves only.
Definition 1.3.6. Let γ : (a, b) → R2 be a unit-speed curve. Let t be the unit tangent
˙
of γ, i.e., t = γ . The unit vector ns obtained by rotating the tangent vector t anticlockwise by an angle π is called the signed unit normal of γ. Since γ is unit-speed,
2
¨
γ is perpendicular to t and hence it is parallel to ns . Therefore there exists a map
¨
κs : (a, b) → R such that γ = κs ns . 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 anti-clockwise by
an angle θ). We note that Rθ is linear and Rθ (1, 0) = (cos θ, sin θ) and Rθ (0, 1) =

7. 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).
Hence
Rθ (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 = π , R π (x, y) = (−y, x).
2
2
Fix a vector a ∈ R2 . Define Ta : R2 → R2 by Ta (x) = x + a. Then Ta is called a
translation 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 unit-speed. The unit tangent will be the unit vector in the direc˙
˙
tion of the tangent vector, i.e., t(t) = γ (t) . Now γ (t) = (−ekt sin t+kekt cos t, ekt cos t+
˙
γ (t)
√
1
˙
kekt sin t) and γ (t) = ekt 1 + k2 . Hence t(t) = √1+k2 (− sin t + k cos t, cos t + k sin t).
Since ns (t) = R π (t(t)), we have
2
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 unit-speed curve, then show that both signed
unit normal and signed curvature are smooth maps.
Proposition 1.3.10. Let γ : (a, b) → R2 be a unit-speed curve, and let s0 ∈ (a, b). Let
˙
φ0 ∈ R such that γ (s0 ) = (cos φ0 , sin φ0 ). Then there exists a unique smooth map
˙
φ : (a, b) → R such that γ (s) = (cos φ(s), sin φ(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 smooth
˙
maps. Since γ is unit-speed, f 2 + g 2 = 1 and hence f f + g g = 0. Define φ : (a, b) → R
˙
by
s
φ(s) = φ0 +
(f g − fg).
˙ ˙
s0

8. 3. Curvature and Torsion
vii
Since f and g are smooth, φ is a smooth map and φ(s0 ) = φ0 .
Let F = f cos φ + g sin φ and G = f sin φ − g cos φ. Then
˙
˙
F = f cos φ − f sin φφ + g sin φ + g cos φφ
˙ ˙
˙
˙
= (f + g φ) cos φ + (g − f φ) sin φ.
˙
˙
˙
Now,
˙
˙
˙
f + g φ = f + g(f g − fg) = f(1 − g 2 ) + fg g = ff 2 + fg g = f(f f + g g) = 0.
˙ ˙
˙ ˙
˙ ˙
˙
˙
˙
By similar arguments it follows that g − f φ = 0. Hence F = 0, i.e., F is a constant map.
˙
˙
Therefore F(s) = F(s0 ) for all s. Now
F(s0 ) = f(s0 ) cos φ(s0 ) + g(s0 ) sin φ(s0 ) = cos2 φ0 + sin2 φ0 = 1.
This shows that F = 1. Similarly, we get G = 0. Hence f cos φ + g sin φ = 1 and
f sin φ − g cos φ − 0. Solving these equations we get f = cos φ and g = sin φ. Therefore
˙
γ (s) = (cos φ(s), sin φ(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 ) = φ0
˙
and γ (s) = (cos ψ(s), sin ψ(s)) for all s. Then there is a map n : (a, b) → Z such that
φ(s)−ψ(s) = 2πn(s) for all s. Since both φ and ψ are smooth, n is smooth and hence it
is continuous. Since (a, b) is connected, n is continuous and the non-empty connected
subsets of Z are singletons only, it follows that n is a constant map, say, c. Therefore
φ(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 unit-speed curve. Then there is a smooth
˙
map φ : (a, b) → R such that γ (s) = (cos φ(s), sin φ(s)) for all s, which is determined
uniquely 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 unit-speed curve, and let φ be the
turning angle of γ. Then the signed curvature of γ can be determined by the
formula κs = φ.
˙
˙
PROOF. Since φ is the turning angle of γ, we have γ (s) = (cos φ(s), sin φ(s)) for all s.
¨
˙
¨
Hence γ = (− sin φ, cos φ)φ. Now ns = R π (γ ) = (− sin φ, cos φ). Therefore γ = φns
˙
˙
2
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 which
the tangent vector of the curve rotates.
Example 1.3.13. If γ : (a, b) → R2 is a unit-speed curve, then its signed curvature and
signed unit normal are smooth maps.

9. 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 map
and hence the map Rπ/2 ◦ t = ns is a smooth map, i.e., the signed unit normal is a
smooth 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 unit-speed
curve γ : (a, b) → R2 whose signed curvature is k. Moreover, if γ : (a, b) → R2 is any
other unit-speed curve whose signed curvature is k, then there is a direct isometry
M of R2 such that γ = M ◦ γ.
s
PROOF. Let s0 ∈ (a, b). Define φ : (a, b) → R by φ(s) = s0 k(u)du. Since k is smooth,
φ is a smooth map. Note that φ(s0 ) = 0 Define γ : (a, b) → R2 by
 s

s
γ(s) = 
cos φ(u)du,
s0
sin φ(u)du .
s0
˙
Then γ (s) = (cos φ(s), sin φ(s)) for all s. Therefore γ is unit-speed curve. It follows
form the Proposition 1.3.10 that φ is the turning angle of γ satisfying φ(s0 ) = 0 and
hence the signed curvature of γ is φ. But φ = k. Hence the signed curvature of γ is
˙
˙
k.
Let γ : (a, b) → R2 be a unit-speed curve with signed curvature k. Let ψ be the
s
˙
turning angle of γ. Then ψ = k. But then ψ(s) = s0 k(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) = (cos ψ(s), sin ψ(s)) for all s. Let a = γ(s0 ). Integrating the above equation, we
have
 s

s
γ(s) = 
cos ψdu,
s0

s0
s
s0
sin ψdu
s0
s
= Ta 

s
cos(φ + θ)du,
s0


s
cos ψdu,
= Ta 

sin ψdu + a
sin(φ + θ)du
s0
s
= Ta 
s0
s
(cos φ cos θ − sin φ sin θ)du,

(sin φ sin θ + cos φ cos θ)du
s0

10. 3. Curvature and Torsion
ix

s
= Ta cos θ

s
cos φdu − sin θ
s0
= Ta ◦ Rθ 
sin φdu, cos θ
s0
s
s
cos φdu,
s0
s
s
cos φdu + sin θ
s0


sin φdu
s0
sin φdu
s0
= Ta ◦ Rθ (γ(s)).
Therefore γ(s) = M(γ(s)) for all s, where M = Ta ◦ Rθ is a direct isometry. Hence
γ = 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 unit-speed. Let k be the curvature of γ. Let k < 0. Then
k = − for some > 0. Let γ : (a, b) → R2 be defined by γ(t) = ( 1 cos( t), − 1 sin( t)).
¨
˙
Then γ is a unit-speed 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 the
signed curvature of γ is − = k. Since γ and γ have the same signed curvature, there
exists a direct isometry M of R2 such that γ = M ◦ γ. Since N ◦ γ is (part of) a circle
for 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 x-axis and γ (t).
Then
(1, 0)(1, sinh t)
1
cos φ(t) =
=
.
2
cosh t
1 + sinh t
Therefore sin φ(t) = tanh t and tan φ(t) = sinh t. Let s be the arc-length of γ,
t
˙
i.e., s(t) = 0 γ (u) du = sinh t. Differentiating tan φ(t) = sinh t = s with respect
1
1
to s we have sec2 φφ = 1, i.e, (1 + tan2 φ)φ = 1. Therefore φ = 1+tan2 φ = 1+s2 =
˙
˙
˙
1
1
= cosh2 t . Hence κs (t) = cosh2 t .
1
(ii) Give an example of a plane curve having signed curvature 1+s2 .
It is
 s

s
1
1+sinh2 t
cos(tan−1 u)du,
γ(s) = 
0
See the proof of Theorem 1.3.14.
sin(tan−1 u)du .
0

11. x
1. CURVE THEORY
Definition 1.3.17. Let γ : (a, b) → R3 be a unit-speed curve with nowhere vanishing
¨
curvature. Then the function n = γ is called the unit normal or principal normal of
κ
γ.
¨
t
It is clear that n(t) = 1 for all t. Since t(t) = 1 for all 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 unit-speed curve γ, we have a right handed system of
orthonormal 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. Therefore
˙
˙
˙
˙
˙
˙
b = ˙ × n + t × n = κ(n) × n + t × n = t × n. This implies that b ⊥ t. Hence b n.
t
˙
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 × ˙ = −τ(n × t) + κ(b × n) = τb − κt.
t
Proposition 1.3.18. Let γ : (a, b) → R3 be a regular curve with nowhere vanishing
curvature. Then the torsion τ of γ is given by
...
˙ ¨
(γ × γ ) γ
.
τ=
˙ ¨ 2
γ×γ
˙
¨
PROOF. First assume that γ is unit-speed. Then γ = t, γ = κn and
...
˙ ˙
γ = κn + κn = κn + κ(τb − κt). Now
˙
...
˙ ¨
(γ × γ )γ = (t × κn)(˙ n + κ(τb − κt)) = κb(˙ n + κ(τb − κt)) = κ2 τ,
κ
κ
...
2
2
˙ ¨
˙ ¨ 2
and γ × γ = κb = κ2 . Hence (γ ×γ ) γ 2 = κκ2τ = τ.
˙ ¨
γ×γ
Let γ be a regular curve with nowhere vanishing curvature (not necessarily unitspeed). Let γ be its unit-speed reparametrization given by γ ◦ s = γ, where s is the
arc-length of γ starting at any point of γ. Note that the torsion of γ at γ(s(t)) is same
˙
γ
˙
as the torsion of γ at the point γ(t). Since γ ◦ s = γ, we have γ ds = γ , i.e,. γ = ds .
dt
dt
...
2
2
2
3
¨
Therefore γ ( ds )2 + γ d s = γ and γ ( ds )3 + γ 2 ds d s + γ ds d s + γ d s = γ. It follows that
dt
dt dt 2
dt dt 2
dt 2
dt 3
... dt
˙ × γ = (γ × γ )( ds )3 and (γ × γ )γ = (γ × γ )γ ( ds )6 . Also γ × γ 2 = γ × γ 2 ( ds )6 .
¨
˙ ¨
˙ ¨
γ
dt
dt
dt
...
˙ ¨
Hence (γ ×γ ) γ 2 = (γ ×γ )γ 2 = τ, as γ is a unit-speed curve.
˙ ¨
γ ×γ
γ×γ
Exercises 1.3.19. Compute curvature and torsion of the following curves.
(i) γ(t) = (t, t 2 , t 3 ).
(ii) γ(t) = (a cos t, a sin t, bt)
(iii) γ(t) = (et cos t, et sin t, et )
4
(iv) γ(t) = ( 5 cos t, 1 − sin t, 3 cos t)
5
1
1
t
(v) γ(t) = ( 3 (1 + t)3/2 , 3 (1 − t)3/2 , √2 )

12. 3. Curvature and Torsion
xi
Example 1.3.20.
(i) If γ : (a, b) → R3 is a unit-speed curve with nowhere vanishing curvature, then
the 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) .
2
2
2
Let γ = (γ1 , γ2 , γ3 ). Since κ is nowhere vanishing, γ1 (t) + γ2 (t) + γ3 (t) > 0
√
for all t. Define f : (0, ∞) → R by f(t) = t and g : (a, b) → R by g(t) =
γ1 2 (t) + γ2 2 (t) + γ3 2 (t). Then both f and g are smooth maps. Since g(t) > 0 for
all t, f ◦ g is a smooth map. But f ◦ g = κ and hence κ is a smooth map.
1
Since κ is a smooth map and κ(t) > 0 for all t, κ is a smooth map. Therefore
¨
γ
= n is a smooth map.
κ
By definition b = t × n and both t and n are smooth maps Therefore b is a
smooth 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 is
planar iff the torsion of γ is identically zero.
We may assume that γ is unit-speed. Suppose that γ is planar. Then there exist
a unit vector a ∈ R3 and a constant d ∈ R such that γa = d. Differentiating, we
get ta = 0. Differentiating it again, we have ˙ = 0, i.e., κ(n a) = 0. This gives
ta
n a = 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 all
˙
˙
t. 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 constant
d
˙
unit vector. Now 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 γ
is (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.
d
1
1˙
1
Let κ > 0. Note that τ = 0. We see that dt (γ + κ n) = t + κ n = t + κ (τb − κt) = 0.
1
1
Therefore γ + κ n = a for some constant vector a ∈ R3 . But then γ − a = k ,
1
i.e., γ is lying on the sphere R − a = 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) a
circle.
(iv) Let γ be a regular curve in R3 with nowhere vanishing curvature. Show that γ
d
˙
is spherical iff τ = dt κκτ .
2
κ

13. xii
1. CURVE THEORY
We may assume that γ is unit-speed. Assume that γ is spherical. Then there
exist 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 we
1
κ
˙
have (γ − a)n = − κ . Again differentiation gives (γ − a)b = τκ2 . 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
1
κ
˙
γ − a = − n + 2 b.
κ
τκ
2
κ2
˙
1
2
But then R = γ − a = κ2 + τ 2 κ4 . Differentiating it we obtain
2¨
2˙˙
τ 2 κ4 2˙ κ−˙ 2 (2τ τ κ4 +τ 2 4κ2 κ)
κ¨ κ
˙
˙
κ
κ2
d
˙
− 2˙ = 0, i.e., κ κτ−2κ˙τ 2τ−κ κτ − τ = 0, i.e., τ = dt κκτ .
2
κ
κ
τ 4 κ8
κ3
κ4
d
˙
1
κ
˙
Conversely, assume that τ = dt κκτ . Let a = γ + κ n − τκ2 b. Then
2
κ
κ
˙
˙
a = t − 2n +
κ
κ
˙
= t − 2n +
κ
τ
d
=
−
κ dt
1˙
d
κ
˙
κ ˙
˙
n−
b − 2b
2τ
κ
dt κ
τκ
1
d
κ
˙
κ
˙
(τb − κt) −
b − 2 (−τ)n
2τ
κ
dt κ
τκ
κ
˙
b = 0.
κ2 τ
˙
Hence a is a constant vector. Now γ − a 2 = κ12 + τκκ4 . Again using τ =
2
κ
κ2
˙
1
2
it follows that κ2 + τ 2 κ4 is a positive constant, say, R . Therefore γ − a
and hence γ is spherical.
2
d
dt
2
κ
˙
κ2 τ
,
=R ,
2
OR
1
1
Let ρ = κ and σ = τ . 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 = −σ ρ. Now
˙
γ − a = ((γ − a)t)t + ((γ − a)n)n + ((γ − a)b)b
= −ρn − σ ρb.
˙
2
Therefore R2 = γ − a 2 = − ρn − σ ρb = ρ2 + (σ ρ)2 . Differentiation of the
˙
˙
ρ
•
last equation gives σ + (σ ρ) = 0, which is required.
˙
ρ
Conversely, assume that σ + (σ ρ)• = 0. Then ρ2 + (σ ρ)2 = d for some constant
˙
˙
d > 0(???). Let a = γ + ρn + σ ρb. Then a = t + ρn + ρ(τb − κt) + (σ ρ)• b −
˙
˙
˙
˙
ρ
•
τσ ρn = ( σ + (σ ρ) )b = 0. Therefore a is a constant vector. Now γ − a 2 =
˙
˙
2
− ρn − σ ρb = ρ2 + (σ ρ)2 = d. Hence γ is spherical.
˙
˙
(v) Let γ be a unit-speed curve in R3 with nowhere vanishing curvature. Show that
the curve α = t is a regular curve and find the curvature and torsion of α.

14. 4. Fundamental Theorem of Space Curves
xiii
˙
¨
¨
Here α = t. Therefore α = ˙ = γ . Since the curvature κ = γ > 0, the
t
˙ = κn, α = κn + κn = κn + κ(τb − κt) and
˙
˙
¨
curve α is regular. Now α = t
˙
˙
...
2
3
˙ ¨
α = κn ... κ(τb − κt) + κτb + κ˙ b − κτ n − 2κ˙ t − κ n. √
¨ +˙
˙
τ
κ
Now α × α = κ2 τt + κ3 b,
3
4
2
˙
¨
˙ ¨
(α × α)α = −κ κτ + κ τ , α = κ and α × α = κ κ2 + τ 2 . Let κ1 be the
˙
˙ ˙
curvature of α and τ1 be the torsion of α. Then
√
√
˙ ¨
α×α
κ2 κ2 + τ 2
κ2 + τ 2
κ1 =
=
=
3
κ3
κ
α
˙
and
...
˙ ¨
(α × α)α
κ(κ˙ − κτ)
τ ˙
τ1 =
= √
.
˙ ¨ 2
κ2 + τ 2
α×α
Theorem 1.3.21. Let γ be a unit-speed curve in R3 . Let t, n and b be its unit
tangent, unit normal and unit binormal respectively. Let κ and τ be the curvature
and torsion of γ respectively. Then
˙ =
t
κn
˙ = −κt
n
τb
˙ =
b
−τn
The equations in Theorem 1.3.21 are called Frenet-Serret equations or SerretFrenet equations.
Example 1.3.22. Verify the Serret-Frenet 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 Curves
Theorem 1.4.1. (Existence) Let k, t : (α, β) → R be smooth maps with κ(s) > 0 for
all s. Then there exists a unit-speed curve γ : (α, β) → R3 whose curvature is k and
torsion is t.
PROOF. Fix s0 ∈ (α, β). It follows from the theory of Differential Equations that the
˙
˙
˙
equations T = kN, N = −kT + tB and B = −tN have a unique solution T, N and 
B

0
k 0
such that {T(s0 ), N(s0 ), B(s0 )} is a standard basis of R3 . Since the matrix −k 0 t 
0 −t 0
˙
˙
˙ N and B in terms of T, N and B is skew-symmetric, it follows that T,
expressing T,
N and B are orthogonal unit vectors. Since B is a smooth map orthogonal to both T

15. 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. Since
B(s0 ) = T(s0 ) × N(s0 ), we have λ(s) = 1 for all s, i.e., B = T × N. Define γ : (α, β) → R3
by
s
γ(s) =
T(u)du.
s0
˙
Then γ = T. Therefore γ is a unit-speed curve and T is the unit tangent of γ. Now
˙ = kN. Therefore γ = k, i.e., k is the curvature of γ. It follows from
¨
¨
γ = T
˙
¨
γ = T = kN that N is the unit normal of γ. Since B = T × N, B is the unit binormal
˙
of 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 unit-speed curves with same
curvature 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 let
t, 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 that
maps t(s0 ), n(s0 ) and b(s0 ) to t(s0 ), n(s0 ) and b(s0 ) respectively. Further, there is a
translation which takes γ(s0 ) to γ(s0 ) (and this has no effect on t, n and b). By applying
rotation 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 prove
γ = 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 observe
˙
that A(s) = 0 for all s (you need to verify this). Therefore A is a constant map. Since
A(s0 ) = 3, we have A(s) = 3 for all s. Hence it follows that t = t, n = n and b = b. The
equation 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 and
constant torsion τ.
Let γ : (a, b) → R3 be a unit-speed curve with constant curvature κ > 0 and
κ
τ
constant torsion τ. Let a = κ2 +τ 2 and b = κ2 +τ 2 . Define γ : (a, b) → R3 by
γ(t) =
a cos
√
t
b2
, a sin
√
t
, b√
t
.
+
+
+ b2
a
Then γ is a unit-speed curve. The curvature of γ is a2 +b2 = κ and its torsion is
b
= τ. Therefore γ, γ : (a, b) → R3 are unit-speed curves with same curvature
a2 +b2
a2
a2
b2
a2

16. 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 a
helix and hence M ◦ γ is a helix. This implies that γ is a helix. Hence any curve with
constant (non-zero) curvature and constant torsion is a helix.
1.5. Isoperimetric Inequality and Four Vertex Theorem
Definition 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 not
before 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 the
disjoint union of two subsets of R2 , denoted by int(γ) and ext(γ), with the following
properties.
(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 unit-speed
reparametrization is (γ)- periodic, where (γ) is the length of γ.
PROOF. Since γ is a- periodic, the length, (γ), of γ is
a
˙
γ (u) du.
(γ) =
0
Let γ be a unit-speed 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 is
k ∈ N ∪ {0} such that t − t = ka + ε, where 0 ≤ ε < a. Now
t
s(t ) − s(t) =
t
˙
γ (u) du −
0
˙
γ (u) du =
˙
γ (u) du
t
0
t+a
t+2a
˙
γ (u) du +
=
t
t
t+ka
˙
γ (u) du + · · · +
t+a
t+ε
= k (γ) +
˙
γ (u) du.
t
t+(k−1)a
t+ka+ε
˙
γ (u) du +
˙
γ (u) du
t+ka

17. xvi
1. CURVE THEORY
t+ε
˙
Hence s(t ) − s(t) = k (γ) iff t
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). Therefore
γ is (γ)- periodic.
Let γ be a simple closed curve in R2 . Let A(int(γ)) be the area of the interior of
γ. Then A(int(γ)) = int(γ) dxdy.
Definition 1.5.3. A simple closed curve in R2 is called positively oriented if its unit
signed 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 let
γ be positively oriented. Let f and g be continuous on the closure of int(γ) and
smooth on int(γ). Then
∂g
∂f
−
∂x ∂y
[f(x, y)dx + g(x, y)dy].
dxdy =
γ
int(γ)
Proposition 1.5.5. Let γ(t) = (x(t), y(t)) be a simple closed curve in R2 with period
a. Let γ be positively oriented. Then
a
1
A(int(γ)) =
2
(x y − y x )dt.
˙
˙
0
1
1
PROOF. Take f(x, y) = − 2 y and g(x, y) = 2 x. Then both f and g are smooth functions
2
on R . It follows from the Green’s theorem that
1
1
1
( 1 − (− 2 )dxdy = γ [ 2 xdy − 2 ydx]. This implies
int(γ) 2
1
1
1 a
˙
˙
A(int(γ)) = γ [ 2 xdy − 2 ydx] = 2 0 (x y − y x )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
π
˙
F 2 dt ≥
0
F 2 dt,
0
with equality holding iff F(t) = A sin t for some constant A.
F(t)
.
sin t
PROOF. Define G : [0, π] → R by G(t) =
π
Note that G is a smooth map. Then
π
˙
F 2 dt =
0
˙
(G sin t + G cos t)2 dt
0
π
π
˙2
2
G sin tdt + 2
=
0
π
G 2 cos2 tdt
˙
G G sin t cos tdt +
0
0

18. 5. Isoperimetric Inequality and Four Vertex Theorem
π
π
˙
G 2 sin2 tdt −
=
0
π
G 2 (cos2 t − sin2 t)dt +
0
π
2
π
2
˙2
G sin tdt +
0
π
0
0
π
˙
G 2 sin2 tdt ≥ 0.
2
F dt −
˙
G 2 sin2 tdt.
F dt +
0
π
˙2
π
2
G sin tdt =
0
Hence
G 2 cos2 tdt
0
π
2
=
xvii
F dt =
0
(1.5.6.1)
0
π
0
π
π ˙
˙
˙
By equation (1.5.6.1) it follows that
F 2 dt = 0 F 2 dt iff 0 G 2 sin2 tdt = 0 iff G 2 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 area
of 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(u) du =
(γ) =
0
DM(γ(u)) du
0
˙
DM(γ(u)) ◦ γ (u) du
=
0
˙
M(γ (u)) du =
=
0
˙
γ (u) du = (γ).
0
1
Since M is an isometry , there are a, b, c, d, e, f ∈ R2 such that a2 +b2 = c2 +d2 =
a + 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
2
A(int(γ)) =
1
2
((ax + by + e)(cx + dy + f)• − (ax + by + e)• (cx + dy + f))dt
0
1
We 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 an
isometry of R2 if and only if there is a 2 × 2 orthogonal matrix A and a ∈ R2 such that N(x) = Ax + a

19. xviii
1. CURVE THEORY
=
1
2
((ax + by + e)(cx + dy ) − (ax + b˙ )(cx + dy + f))dt
˙
˙
˙
y
0

=

1
(ad − bc)
2
(x y − x y)dt + (ec − af)(x( ) − x(0)) + (ed − bf)(y( ) − y(0))
˙ ˙
0
=
1
2
(x y − x y)dt = A(int(γ)),
˙ ˙
0
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 equality
holding iff γ is a circle.
PROOF. We may assume that γ is unit-speed. 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 tπ , i.e.,
γ = γ ◦ s or γ ◦ s−1 = γ, where s(t) = tπ . [Note that γ is a unit-speed reparametrization
of γ with the parametrization map s−1 (t) = tπ , i.e., s(t) = tπ .] Then γ is a simple closed
curve with the same length and period π. Also the area of interior of γ is same as
the area of interior of γ. Since the length and the area of a curve are invariant under
isometry, 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). This
˙
˙
gives r(0) = r(π) as γ(0) = (0, 0). Now x 2 + y 2 = r 2 + r 2 θ 2 and x y − y x = r 2 θ. Since
˙
˙
˙
˙
˙
2
2 ds 2
2
˙
˙
˙
˙
˙
γ = γ ◦ s, we have γ = γ ds . But then r 2 + r 2 θ 2 = x 2 + y 2 = γ = γ ( dt ) = π 2 . We
˙
˙
˙
dt
π
π 2˙
also note that A = A(int(γ)) = 1 (x y − y x )dt = 1
˙
˙
r θ. Now,
0
2
0
2
π
2
1
−A =
4π
4
π
1
˙
(˙ 2 + r 2 θ 2 )dt −
r
2
0
˙
r 2 θdt
0
π
=
1
4
˙
˙
(˙ 2 + r 2 θ 2 − 2r 2 θ)dt
r
0
π
1
=
4
π
1
˙
r (θ − 1) dt +
4
2
2
0
(˙ 2 − r 2 )dt ≥ 0.
r
0
π
1 π
˙
r
It follows from the above equation that 4π −A = 0 iff 4 0 r 2 (θ −1)2 dt + 0 (˙ 2 −r 2 )dt = 0
π 2 ˙
π 2
2
2
˙
iff 0 r (θ − 1) dt = 0 (˙ − r )dt = 0 iff θ = 1 and r = A sin t for some A iff θ = t + α
r
2

20. 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
(where p and q are positive constants), prove that
2π
x2
p2
+
y2
q2
= 1
x2
p2
+
y2
q2
= 1.
√
p2 sin2 t + q 2 cos2 t dt ≥ 2π pq,
0
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
Note the γ is 2π-periodic. Therefore length of γ is
2π
2π
p2 sin2 t + q 2 cos2 t dt.
˙
γ (t) dt =
=
0
0
Also, the area A of the interior of γ is
2π
1
A=
2
2π
1
(x y − y x )dt =
˙
˙
2
0
pq(cos t cos t + sin t sin t)dt =
pq
2π = pqπ.
2
0
It follows from the Isoperimetric inequality that
2
 2π
p2 sin2 t + q 2 cos2 t dt  ≥ 4πpqπ = 4π 2 pq,

0
i.e.,
2π
√
p2 sin2 t + q 2 cos2 t dt ≥ 2π pq.
0
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 a
convex subset of R2 .
Definition 1.5.11. The vertex of a curve γ is a point of the curve where its signed
curvature 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.

21. 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 > 0
and a = b.
Let s be the arc-length of γ starting at some point of γ. Denoting the derivative
dt
of γ with respect to s by γ , we have γ (t) = (−a sin t, b cos t) ds . Since γ = 1,
dt
= √ 2 2 1 2 2 . Therefore
ds
a sin t+b cos t
1
t(t) = γ (t) =
a2 sin2 t + b2 cos2 t
Differentiating γ with respect to s, we get
γ (t) =
Now ns (t) = R π (t(t)) =
2
(a2
ab
(−b cos t, −a sin t).
sin t + b2 cos2 t)2
2
1
1
(a2 sin2 t+b2 cos2 t) 2
γ =
This implies that κs (t) =
(−a sin t, b cos t).
(−b cos t, −a sin t). Hence
ab
3
(a2 sin2 t + b2 cos2 t) 2
ab
3
(a2 sin2 t+b2 cos2 t) 2
. Now κs (t) =
˙
ns .
−3ab(a2 −b2 ) sin t cos t
5
(a2 sin2 t+b2 cos2 t) 2
sin 2t = 0 iff t = nπ
2
. Since a, b > 0
and a = b, we have κs (t) = 0 iff sin t cos t = 0 iff
˙
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.
y
Example 1.5.14. The ellipse x 2 + b2 = 1 (or the curve γ(t) = (a cos t, b sin t)) is a
a
convex curve.
2
2
y2
y
z2
Let (x, y), (z, w) ∈ int(γ), i.e., x 2 + b2 < 1 and a2 + w2 < 1. Let X = x , b and
a
a
b
z
Y = a , w . Then X < 1 and Y < 1. Let t ∈ [0, 1]. Then tX + (1 − t)Y ≤
b
t X + (1 − t) Y < t + 1 − t = 1. Now
2
2
(tx + (1 − t)z)2 (ty + (1 − t)w)2
+
= tX + (1 − t)Y
a2
b2
Therefore t(x, y) + (1 − t)(z, w) ∈ int(γ) and hence γ is convex.
2
< 1.