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Chapter 1. Curves in Rn 3
1. Definitions and basic properties 3
2. Moving frames. The Frenet frame 5
Chapter 2. Plane curves (local theory) 11
1. Basic definitions 11
2. Tangent an normal in a point of a plane curve 11
3. The length of a plane curve. Natural parametrizations 12
4. The Frenet (Serret-Frenet) frame of a plane curve 13
5. The equations of the curvature 15
6. The fundamental theorem of plane curves 16
7. The envelope of a family of plane curves 19
8. The evolute of a plane curve 20
Chapter 3. Space curves 23
1. Basic definitions 23
2. The tangent and the normal 24
3. The length of a arc. Natural parametrization 25
4. The Frenet frame 26
Chapter 4. Plane curves 33
1. The equation of a plane curve 33
2. Asymptotes. The folium of Descartes 34
3. The cycloid. The astroid 361
2 CONTENTS
4. The tangent to a plane curve 37
5. Asymptotes in polar coordinates. The hyperbolic spiral. 39
6. The cardioid. The lemniscate of Bernoulli 40
Chapter 5. 43
1. Ruled surfaces 43
2. Surfaces of revolution 51
CHAPTER 1
Curves in Rn
Vom studia prop. de baza al curbelor in Rn, vom introduce curburile si reperul Frent.
1. Definitions and basic properties
Definition 1.1. Fie I ⊆ R Un interval . O curba (parametrizata) in Rn este o aplc.
C∞ c : I −→ Rn. Curba c s.n. regulara daca t ∈ I c′(t) 6= 0.
Observations
a) If I is not as open interval by we should explain what it means for c to be C∞.
That is there exists an open interval I1 containing I and a C∞ curve c1 : I1 −→ Rn
such that c1|I = c
b) The variable t will be called the parameter of the curve
c) The tangent space Rt0 = Tt0R of R at t0 ∈ R has a distinguished basis 1 = (t0, 1).
As an alternate notation the basis (t0, 1) is denoted by ddt .
d) For a curve c : I −→ Rn, the vector dct0(1) ∈ Tc(t0)Rn is well defined and given
by
dct0(1) = limc(t)− c(t0)
t− t0= c(t0)
Definition 1.2. a) A vector field along c : I −→ Rn is a differentiable mapping
X : I −→ Rn. The vector X(t) is understand to lie in a copy of Rn identified
with Tc(t)Rn.
b) The tangent vector field of c : I −→ Rn is the vector field along c : I −→ Rn
given by t −→ C ′(t).3
4 1. CURVES IN Rn
Let c : I −→ Rn be a curve, t0 ∈ I fixed and t ∈ I and M0,M the corresponding points of
the values t0, t of the parameter on the curve c.
The line M0M will be said to be a secant line through M0. What does it happen when
t −→ t0, that is M goes to M0 on the curve c.
The limit position of the secant line is a line through M0 parallel with c(t0) called the
tangent line through M0.
At every point of the curve the tangent vector c′(t), t ∈ I is defined. The vector
n = 1|c′(t)|c
′(t) is called the unit tangent vector at c corresponding to the value t of the
parameter.
Definition 1.3. Let I1, I be intervals and α : I1 −→ I a diffeomorphism. We obtain
a new curve c1 = c ◦ α : I1 −→ Rn. The map α is called parameter transformation or a
change of variables relating c to c1.
The map α is called orientation preserving if α′ > 0 and orientation reversing if α′ > 0.
Definition 1.4. a) A vector field along c : I −→ Rn is a differentiable mapping
X : I −→ Rn which assign to every t ∈ I a vector X(t) in Rn with the origin in
c(t) (the vector X(t) is understand to be in the copy of Rn identified with Tc(t)Rn).
b) The tangent vector field of c : I −→ Rn is the vector field along c : I −→ Rn
given by t −→ c(t).
Example. On the set of curve a relation is induces by the parameter transformation.
Show that this relation is an equivalence relation.
Definition 1.5. a) The curve c(t), t ∈ I is said to be parametrized by arc length
if |c′(t)| = 1. Such kind of curve is sometimes called a unit speed curve and the
parametrization, natural parametrization.
b) The 1- form ds = |c′(t)|dt is called the arc-form of c.
Let c : I −→ Rn be a regular curve and t0, t ∈ I and M0,M be the corresponding
points of the curve. It is prove that the length of the curve segment joining M0 and M is
2. MOVING FRAMES. THE FRENET FRAME 5
given by
M0M =∫ t
t0
|c′(t)|dt
The length of c is given by L(c) =∫I |c′(t)|dt.
Proposition 1.1. Every regular curve c : I −→ Rn can be parametrized by arc length.
That is, given a regular curve c : I −→ Rn there is a charge of variables α : I1 −→ I such
that |(c ◦ α)′(s)| = 1.
Proof. If we have such kind of change of variables we have that∣∣∣dc
ds
∣∣∣ =∣∣∣dc
dt
∣∣∣ ·∣∣∣dα
ds
∣∣∣ = 1
We define s(t) =∫ tt0|s′(t)|dt, to ∈ I and set s(t) = α−1(t). Since c is regular, α exists and
satisfy the desired equation. Clearly c ◦ α is parametrized by arc-length. ¤
Remark 1.2. Practically one cannot find a arc-length parametrization.
2. Moving frames. The Frenet frame
We are study the local differential geometry of a curve it is useful to consider a frame
in every point, which varies smooth on the curve. Special such frame is the Frenet frame
which allows us to define some basic objects associated to a curve, the curvatures.
Definition 2.1. Let c : I −→ Rn be a curve.
a) A moving n-frame along c is a collection of n differentiable mapping ei : I −→Rn, i = 1, n such that < ei(t), ej(t) >= δij, for all t ∈ I, where δij is the
Kronecker symbol. Each ei(t) is a vector field along c, and ei(t) is considered as
a vector of Tc(t)Rn.
b) A moving n-frame is called a Frenet frame, if for all k, 1 ≤ k ≤ n, the k-th
derivative of c(k)(t) lies in the span of the vectors e1(t)0 . . . , ek(t).
In the next proposition we shall prove the existence and uniqueness of a distinguished
Frenet under suitable conditions.
6 1. CURVES IN Rn
Proposition 2.1. Let c : I −→ Rn be a curve such that for all t ∈ I, the vectors
c′(t), c(2)(t), . . . , c(n−1)(t) are linearly independent. Then there exists a unique Frenet frame
with the following properties:
a) fn 1 ≤ k ≤ n − 1, the families of vectors {c(i)(t)}i=1,k and {ei(t)}i=1,k have the
same orientation.
b) The frame {ei(t)}i=1,n has a positive orientation.
Proof. see yourself please z We shall use the Gram-Schimidt orthogeneralization
process. c′(t) 6= 0 by the hypothesis and we set e1(t) = c′(t)|c′(t)| .
Suppose that e1(t), . . . , ej−1(t), j < n are defined.
Let
ej(t) = −γ−1∑
k=1
< c(j)(t), ek(t) > ek(t) + cj(t)
and ej(t) = ej(t)|ej(t)| Clearly (ej(t)), j < n are well defined and verify the conditions in the
theorem. We define en(t) such that e1(t) . . . , en(t) has positive orientation. The diferen-
tiability of ej(t) is clear from the definition.
The components (ein(t)) of en(t) are expressed as minors of rank (n−1) in the n×(n−1)
matrix (ej(t)) 1 ≤ i ≤ n, 1 ≤ j ≤ n− 1, so en(t) is a differentiable .
¤
Now we are ready to prove the Frenet equations and to introduce the curvatures. These
will be used later for plane and space curves.
Proposition 2.2. Let c : I −→ Rn be a curve and consider a moving frame (ei(t)), i =
1, n on the curve. The derivatives c′(t) and e′i(t) satisfy the following equations
c′(t) =n∑
i=1
αi(t)ei(t)
e′i(t) =n∑
j=1
ωij(t)ej(t)
2. MOVING FRAMES. THE FRENET FRAME 7
where
ωij(t) = −ωji(t) =< e′i(t), ej(t) >
In the particular case when (ei(t)) is the distinguished Frenet frame α1(t) =
|c′(t)|, αi(t) > 0 for i > 1 and ωij(t) = 0 for j > i + 1
Proof. By differentiating 〈ei(t), ej(t)〉 = δij we obtain the form of ωij .
The relations ωij(t) = 0 for j > j + 1 follows from the fact that ei(t) is a linear
combination of c′(t), . . . , c(i)(t) implies that e′i(t) is a linear combination of c′(t), . . . , ci+1(t)
and hence of e1(t), . . . , ei+1(t). ¤
Remark 2.3. Denote by ω = (ωij(t))1≤i,j≤n we can express c′(t) = ω(t)e(t).
ω is a shew symmetric matrice.
Moreover, the relation ωij(t) = 0 for j > i + 1 implies that ω is of the form
ω =
0 ω12 0 . . . 0 0
−ω12 0 ω23 . . . 0 0
0 −ω23 0 . . . 0 0
. . . . . . . . . . . . . . .
0 0 0 . . . 0 ωn−1,n
0 0 0 . . . −ωn−1,n 0
The matrix ω is sometimes called the curvature matrix of the curve c. We prove the
form of the curvature matrix for a parametrization of the curve.
What does it happen if we change the variable?
If we consider an isometry of Rn, how the curvature matrix change?
To be really invariants of the curve the quantities defined have not to change under
such kind of transformations.
Proposition 2.4. a) Let c : I −→ Rn be a curve and A : Rn −→ Rn an isometry
of Rn which has the orthogonal component is R.
8 1. CURVES IN Rn
Let c = A◦c : I −→ Rn, and let ei(t))i=1,n a moving frame on c then (ei(t))i=1,n =
(Rei(t))i=1,n is a moving frame on C and if ωij(t) and ω(t) are the curvature
matrixes of associated Frenet frames of c and c then
|c′(t)| = |c′(t)| and ω = ω
b) Let c : I −→ Rn and c : J −→ Rn related by an orientation preserving change of
variables α, that is c = c ◦ α, α : J −→ I.
Let (ei(t)i=1,n be a moving frame on c. Then ei(s) = (ei ◦α)(s) is a moving frame
on C, and, if |c′(s)| 6= 0 then
ωij(s)|c′(s)| =
ωij(α(s))|c′(α(s))|
Proof. ωij =< e′i(t), ej(t) >=< Oe′i(t), Oej(t) >=< e′i(t), ej(t) >= ωij(t) because A
is an isometry
(2.1)ωij(s)|c′(s)| =
⟨e′(s),
ej(s)|c′(s)|
⟩=
⟨e′i(α(s))α′(s),
ej(α(s))|c′(α(s))|α(s)
⟩=
ωij(α(s))|c′(α(s))|
¤
Definition 2.2. Let c : I −→ Rn be a curve satisfying the existence of a distinguished
Frenet frame. The i-th curvature of c, i = 1, n− 1 is the function
ki(s) =ωi,i+1(t)|c′(t)|
The definition is motivated by the second part of the above theorems.
The matrix ω can be written
ω = |c′|
0 k1 0 . . . 0 0
−k1 0 k2 . . . 0 0
0 −k2 0 . . . 0 0
. . . . . . . . . . . . . . . . . .
0 0 0 . . . −kn−1 0
2. MOVING FRAMES. THE FRENET FRAME 9
Proposition 2.5. Let ki(t) i = 1, n− 1 be the curvature functions defined above. Then
ki(t) > 0 for 0 ≤ i ≤ n− 2
Proof.
c(k)(t) =k∑
i=1
akiei ek =k∑
i=1
bkec(e)
where akk > 0, bkk = a−1kk > 0 for 0 ≤ k ≤ n− 1. There for
|c|ki = 〈e′i, ei+1〉 = bii〈c(i+1), ei+1〉 = biiai+1,i+1 > 0
¤
We now prove that curvatures determines the curve in the following sense.
Theorem 2.6. Let c, c : I −→ Rn be two curves satisfying the conditions of existence
of a distinguished Frenet frame . Denote these Frenet frames by (ei(t)) and (ei(t)) resp,
i = 1, n. If ki(t) = ki(t), i = 1, n− 1 and c′(t)| = |c′(t)| for these frames, then exists a
unique isometry A : Rn −→ Rn such that
c = A ◦ c
Furthermore A is a congruence; its orthogonal component has determinant +1.
Proof. Let t0 ∈ I. There exists exactly one isometry with the properties
Ac(t0) = c(t0)
Rei(t0) = ei(t0) i ≤ i ≤ n
where R is the orthogonal component of A. detR = 1 because both Frenet frames are
positively oriented.
By the hypothesis ωij = ωij(t), fact which implies
ei(t) =∑
j
ωij(t)ej(t)
10 1. CURVES IN Rn
On the other hand
Re′i(t) =∑
j
ωij(t)Rej(t)
ei(t) and Rei(t) satisfy the some system of linear differential equation and ei(t) = Oei(t)
since they are equal at t = t0.
Rc′(t) = |c′(t)|Rei(t) = |c′(t)| · |c1(t) = c′(t)
It follows that
Ac(t)−Ac(t0) =∫ t
t0
Rc′(t)dt = c(t)− c(t0)
which proves that Bc(t) = c(t).
The next theorem of curves whiter prescribed curvature functions. Let
k1(s), . . . , kn−1(s) be differentiable functions defined in a neighborhood O ∈ R with
ki(s) ≥ 0 1 ≤ i ≤ n − 2. Then there exists an interval I containing O and a unit speed
curve c : I −→ Rn which satisfy the conditions of the existence of the distinguished Frenet
frame and where curvature functions an ki(s), 1 ≤ i ≤ n− 1.
¤
CHAPTER 2
Plane curves (local theory)
In this chapter we study the simplest curves, namely plane ones. We restrict our study
to local theory of plane curves, the next chapter being devoted to some global properties.
1. Basic definitions
Throughout the chapter we consider curves c : I −→ R2. We consider the euclidian
space R2, with the origin O, basis i, j and coordinates x, y.
We denote C = imc = {c(t)|t ∈ I}.For t ∈ I let A(x(t), y(t); OA = r(t) = x(t)i + y(t)j , |r′(t)| = |dr
dt | 6= 0.
First we give without details the three representations of a plane curve.
Definition 1.1. a) The parametric representation of the plane curve is x =
x(t); y = y(t), t ∈ I, x′(t)2 + y′(t)2 > 0∀t ∈ I
b) r(t) = x(t)i + y(t)j, t ∈ I, |r′(t)| > 0 for all t ∈ I
c) The implicit representation is F (x, y) = 0, F : D −→ R, D ⊆ R2, F has F ′2x +
F ′2y > 0∀(x, y) ∈ D.
2. Tangent an normal in a point of a plane curve
Using the notions from the previous chapter for the special case n = 2 have that
Definition 2.1. Let c : I −→ R2 be a plane curve and t0 ∈ I. The tangent line through
A(x(t0), y(t0)) is
X − x(t0)x′(t0)
=Y − y(t)
y′(t0)11
12 2. PLANE CURVES (LOCAL THEORY)
For a curve given in the vector representation is
r = r(t0) + λr′(t0), λ ∈ R
For a curve given in the implicit form, the tangent through a point (x0, y0) is
(x− x0)F ′x(x0, y0) + (y − y0)F ′
y(x0, y0) = 0
Definition 2.2. Let c : I −→ R2 be a curve t0 ∈ I, and P the corresponding point on
the curve, namely P (x(t0), y(t0)).
The normal line through P is the orthogonal line to the tangent line in P .
The equations of the normal line are
x− x(t0)y′(t0) + (y − y(t0))y′(t0) = 0
for the curve given in explicit form and
(x− x0)F ′y(x0, y0)− (y − y0)F ′
y(x0, y0) = 0
for the curve given by implicit equation.
The vectors T = r′(t0) = x′(t0)i + y′(t)j and N = −y′(t)i + x′(t)j the tangent and the
normal vector fields related to the curves and t = 1|T | T and n = 1
|N |N are the unit tangent
and the unit normal vector fields to the curve c.
3. The length of a plane curve. Natural parametrizations
Let c : I −→ R2 be a plane curve. a, b ∈ I and A, B the corresponding points on the
curve.
Using again the notions for the previous chapter we have the following:
Definition 3.1. The arc form (or arc element) of the curve c is ds =√
x′2(t) + y′2(t)dt and the length of the arc of the curve joining A and B is
l(AB) =∫ b
a
√x′2(t) + y′2(t)dt
4. THE FRENET (SERRET-FRENET) FRAME OF A PLANE CURVE 13
The natural parametrization of c is c ◦ α : J −→ R2 where α is the inverse function of
s : I −→ J, s(t) =∫ tt0
√x′2(t) + y′2(t)dt.
4. The Frenet (Serret-Frenet) frame of a plane curve
Let c : I −→ R2 be a plane curve. The condition c′(t) 6= 0 is equivalent to the existence
of the existence of the distinguished Frenet frame . We shall always choose this frame as
the moving 2- frame on the curve c.
Taking t(t) = e1(t) and n(t) = e2(t), the Frenet equations become
r′(t) = |r′(t)|t(t)
t′(t) = ω12(t)n(t)
n′(t) = −ω12(t)t(t)
In the matrix form r′(t) = |r′(t)|t(t)
e′(t) =
0 ω12(t)
−ω12(t) 0
e(t) ,
where e(t) =( t(t)n(t)
)
We have only one curvature
k(t) =ω12(t)|r′(t)| dt
For a curve with natural parametrization r′(t) = 1, r′(t) = n′(t) and r′′(t) = t′(t) =
ω12(t)n(t) = ki(t)n(t) so k(t) = r′′(t). Graphically k(t) > 0 (k(t) < 0) mean that n(t)
points outward (invard) the convex (concave) side of the curve c at c(t).
Example:Graph of the sine
c(t) = (t, sin t) t ∈ R
k(t) < 0 t ∈ (0, π)
k(t) > 0 t ∈ (π, 2π)
14 2. PLANE CURVES (LOCAL THEORY)
There is the posibility that k(t) = 0. if in addition k′(t) 6= 0 (and hence the zero of
k is isolated) c(t) is called an inflexion point of the curve. In the above example c(0) and
c(π) are inflexion points.
desen
We give now a geometrical interpretation of the curvature function.
Fix a vector v of unit length. Define θ(t) by
cos θ(t) = 〈t(t); v〉
sin θ(t) = −〈n(t); v〉
Up to a multiple of 2π, θ(t) is the angle between v and t(t), measured in the positive
direction. In a sufficiently small neighborhood of t0 ∈ I, θ(t) may be considered to be
differentiable. Clearly θ(t) is a well defined function.
Proposition 4.1. Let θ(t) locally defined a as above. Then
θ′(t) = ω12(t) = k(t)|c′(t)|
If |c′(t)| = 1, k(t) = θ′(t).
Proof. We differentiate the expression of θ and use the Frenet equations
− sin θ(t)θ′(t) = ω12(t)〈n(t), v〉 = − sin θ(t)ω12(t)
− cos θ(t)θ′(t) = ω12(t)〈t(t), v〉 = cos θ(t)ω12(t)
¤
From these relations easily follows the proposition
Proposition 4.2. (Characterizations of straight lines) Let c : I −→ R2 be a plane
curve. The following conditions are equivalent.
a) k(t) = 0 for all t ∈ I
b) c is a straight line
5. THE EQUATIONS OF THE CURVATURE 15
Proof. Assume |r′(t)| = 1 k(t) = 0 =⇒ r′′(t) = 0 =⇒ r(t) = (t− t0)r′(t0)+ r(t0) ∀t ∈I. The converse is obvious. ¤
Proposition 4.3. (Characterization of the circle) Let c : I −→ R be a plane curve.
The following conditions are equivalent.
a) k(t) = 1/r = const > 0
b) c is a piece of a circular arc, that is, there exists an x0 ∈ R2 with |c(t)−x0| = r > 0
for all t ∈ I
Proof. Assume |r′(t)| = 1
r′(t) = t(t)
t′(t) =ε
rn(t) ε = ±1
n′(t) = −ε
rt(t)
(c(t)+εre2(t))′ = c′(t)−e1(t) = 0 and c(t)+εre2(t) = x0 =⇒ c(t)−x0 = εre2(t) implying
|c(t)− x0|2 = r2
b) =⇒ a) By direct computation we obtain
|k(t)| = 1r
¤
5. The equations of the curvature
In this chapter we derive explicit formulas for curvature of a plane curve.
Let c : I −→ R2 be a plane curve and r(t) a position vector of a point on the curve
From the Frenet equations we have
r′(t) = |r′(t)|t(t)
16 2. PLANE CURVES (LOCAL THEORY)
Differentiating this equation we have
r′′(t) =|r′(t)|′t(t) + |r′(t)|t′(t) =
= |r′(t)|′t(t) + |r′(t)|ω12n(t)
= |r′(t)|′t(t) + |r′(t)|k(t)|r′(t)|n(t)
It follows that
r′′(t)× r′(t) =|r′(t)|2k(t)n(t)× r′(t) =
= |r(t)|3k(t)n(t)× t(t)
It follows that:
k(t) =|r′′(t)× r′(x)|
|r′(t)|3 =det(r′(t), r′′(t))
|v′(t)|3In parametric form the equation of the curvature is
k(t) =x′(t)y′′(t)− y′(t)x′′(t)(√
x′2(t) + y′2(t))3
If the curve is parametrized by arc-length the curvature is
k(s) = x′(s)y′′(s)− y′(s)x′′(s)
6. The fundamental theorem of plane curves
Let c : I −→ R2 be a plane curve. We can associate to the curve the curvature function.
k : I −→ R k(t) =x′(t)y′′(t)− x′′(t)y′(t)
(x′2(t) + y′2(t))3/2
This function determine uniquely the curve
Theorem 6.1. Let k : [0, L] −→ R, k −→ k(s) of class Cr, r ≥ 1 there exists a unique
up to a isometry such that s is the arc-length and k is the curvature function of the curve.
6. THE FUNDAMENTAL THEOREM OF PLANE CURVES 17
The curvature k(s) can be interpreted as the instantaneous rotation velocity of the
unit tangent vector t(s). The curve is ”more curved” when this velocity is great.
Example: The ellipse
Considering two fixed points F and F ′ in plane, with FF ′ = c and a ∈ R, a > c, the set
of the points M in plane such that
MF + MF ′ = 2a
is called an ellipse.
Choosing the coordinate axes such that Ox is FF ′ and O is the middle of the segment
[FF ′], the implicit ecuation of the ellipse with the axes a and b (b2 = a2 − c2) is
x2
a2+
y2
b2= 1.
From this, the explicit representation can also be deduced:
y =b
a
√a2 − x2, x ∈ [−a, a] the arch of the ellipse above Ox
y = − b
a
√a2 − x2, x ∈ [−a, a] the arch of the ellipse below Ox
Very useful in applications are the parametric equations
x(t) = a cos t
y(t) = b sin t, t ∈ [0, 2π).
The curvature at an arbitrary point A(x(t), y(t)), t ∈ [0, 2π), is
k(t) =x′(t)y′′(t)− y′(t)x′′(t)(√
x′2(t) + y′2(t))3=
ab
(√
a2 sin2 t + b2 cos2 t)3
At two of the intersection points with the axes, we have, for t = 0, k(0) =a
b2and for
t = π2 , k(π
2 ) =b
a2.
The tangent at an arbitrary point has the equation
x− a cos t
−a sin t=
y − b sin t
b cos t
or xb cos t + ya sin t− ab = 0. (for instance, at t = 0 we get x = a, a vertical tangent).
18 2. PLANE CURVES (LOCAL THEORY)
The equation of the normal at an arbitrary point is
(x− a cos t)(−a sin t) = (y − b sin t)b cos t
or xa sin t + yb cos t− (a2 + b2) sin t cos t = 0.
It is interesting that, while the area of an ellipse has a simple expresion (abπ), there is
no exact formula for the length of an ellipse in elementary functions. It can be expresed
using infinite series expansions.
7. THE ENVELOPE OF A FAMILY OF PLANE CURVES 19
7. The envelope of a family of plane curves
Consider a family of curves (given by the implicit equations) depending on a parameter
λ:
(Cλ) F (x, y; λ) = 0
Let Cλ and Cλ+h be two curves of the family, close to each other, and M = Cλ ∩ Cλ+h
their intersection point. Denote by Nλ the limit position of the point M , when h → 0.
The geometric place of the points Nλ is called the envelope of the family (Cλ).
The coordinates of Nλ satisfy the system
F (x, y; λ) = 0
F ′λ(x, y; λ) = 0
Eliminating the parameter λ from the system the equation of the envelope is obtained.
The envelope Γ of the family (Cλ) has the following properties:
- each curve Cλ has exactly one common point with Γ,
- through each point of Γ passes exactly one curve from the family,
- at the common points, Γ and Cλ have common tangents.
We will prove the last statement. Let P the common point of Γ and Cλ.We can consider
the coordinates of a point on Γ as functions of the parameter λ: x = x(λ), y = y(λ). We
have that
F (x(λ), y(λ);λ) = 0
Differentiating with respect to λ, we have F ′x · x′(λ) + F ′
y · y′(λ) + F ′λ = 0. Taking account
that F ′λ = 0 it follows
−F ′x(x(λ), y(λ);λ)
F ′y(x(λ), y(λ);λ)
=y′(λ)x′(λ)
The left-hand term is the slope of the tangent at P of Cλ and the right-hand term is the
slope of the tangent of Γ.
20 2. PLANE CURVES (LOCAL THEORY)
Example Determine the envelope for a family of straight lines for which the length
of the segment cut by the axes is equal to a constant k.
We write the equations ”by cuts” for an arbitrary line from the family:x
a+
y
b= 1.
We know that a2 + b2 = k2 so taking a = k cosλ and b = k sinλ the family of lines can be
expresed as depending of the parameter λ:
x
k cosλ+
y
k sinλ= 1.
The system that gives the envelope is:
x
k cosλ+
y
k sinλ= 1
x sinλ
k cos2 λ− y cosλ
k sin2 λ= 0
From here we get x = k cos3 λ and y = k sin3 λ and by eliminating λ, x2/3 + y2/3 = k2/3,
the equation of an astroid.
Remark 7.1. There exist families of curves that have no envelope (for instance a
family of concentric circles or a family of parallel lines).
Problems
Find the envelope of a family of straight lines that determine on the axes 0x and 0y
segments with the product equal to a constant k. (Answer: 4xy=k).
8. The evolute of a plane curve
Consider a plane curve C and the family of normals to the curve. The envelope of this
family is called the evolute of the curve C.
If the curve is given in parametric form: x = x(t), y = y(t), then the equation of the
normal at an arbitrary point Pt is
y − y(t) = −x′(t)y′(t)
(x− x(t)).
8. THE EVOLUTE OF A PLANE CURVE 21
Using the method described in the previous paragraph, we can determine the parametric
equations of the evolute:
x = x(t)− y′(t)(x′2(t) + y′2(t))x′(t)y′′(t)− x′′(t)y′(t)
y = y(t) +x′(t)(x′2(t) + y′2(t))
x′(t)y′′(t)− x′′(t)y′(t)
These equations have also another meaning: the point Nt(x, y) (with the coordinates x, y
given by the formulas above) is called the curvature center of the curve C, corresponding
to the point Pt. The distsnce NtPt = R(t) = 1k(t) is the curvature radius of the curve at
Pt. The circle with the center at Nt and of radius R(t) is called the osculating circle of C,
at the point Pt.
Example
CHAPTER 3
Space curves
1. Basic definitions
In this chapter we continue our study with space curves c : I −→ R3. We consider the
euclidian space R3 with the origin O and the orthonormal basis {i, j, k}
Definition 1.1. The vectorial equation of the curve c is r : I −→ R3 r(t) = x(t)i +
y(t)j + z(t)k, t ∈ I
(x(t), y(t), z(t)) are the coordinates of the vector Oc(x) in the basis {i, j, k} that is
Oc(t) = x(t)i + y(t)j + z(t)k
Throughout the chapter we suppose that the map c is differentiable of a class Cs(s ≥ 0),
that is the functions t −→ x(t), t −→ y(t), t −→ z(t) are differentiable of class Cs(s ≥ 0)
We shall give now the representations of a space curve: explicit parametric and implicit.
Definition 1.2. Let I ⊆ R and α, β : I −→ R be a function of class Cs(s ≥ 1). The
set {(x, y, z)|x ∈ I y = α(x), y = β(x)} is a curve of class Cs in R3.
This is the explicit form of the curve.
Definition 1.3. Let D ⊆ R3 and F, G : D −→ R be functions of class Cs(s ≥ 1 with
rang
F ′
x F ′y F ′
z
G′x G′
y G′z
= 2 on D.
The set {(x, y, z) ∈ R3|F (x, y, z) = 0, G(x, y, z) = 0} is a Cs curve in R3.
The above representations of a space curve are equivalent.
The proof is analytical and will be omitted here.23
24 3. SPACE CURVES
2. The tangent and the normal
Let r : I −→ R3 be a Cs curve
Using the results of the Ch... the equation of the tangent line through P0, corresponding
to the value t0 of the parameter is
x− x(t0)x′(t0)
=y − y(t0)
y′(t0)=
z − z(t0)z′(t0)
t ∈ I
or in the vector form
r(t) = r′(t0) + λr′(t0) λ ∈ I
If the curve C is given explicitly, (x ∈ I, y = α(x), z = β(x)) with α, β : I −→ R3 of class
Cs and we consider a point P0(x0, y0 = α(x0), z0 = β(x0)) on the curve the equation of
the tangent line isx− x0
1=
y − α(x0)α′(x0)
=z − β(x0)
β′(x0)
In the case of a curve which is represented by the implicit equation F,G : D ⊆ R3 −→
R F (x, y, z) = 0, G(x, y, z) = 0, rang
F ′
x F ′y F ′
z
G′x G′
y G′z
= 2. the equation of the tangent
line through a point (x0, y0, z0) to the curve (F (x0, y0, z0) = 0, G(x0, y0, z0) = 0 is
x− x0∣∣∣∣∣∣F ′
y F ′z
G′y G′
z
∣∣∣∣∣∣
=y − y0∣∣∣∣∣∣F ′
z F ′x
G′z G′
x
∣∣∣∣∣∣
=z − z0∣∣∣∣∣∣F ′
x F ′y
G′x G′
y
∣∣∣∣∣∣
the paralel derivatives being calculated in (x0, y0, z0).
Definition 2.1. The normal plane at t0 the curve C through a point P0 is the plane
through P0 and orthogonal the tangent line through P0.
In the case of parametric equation of the curve, the equation of the normal plane is
〈r − r(t0), r′(t0)〉 = 0 or
(x− x(t0))x′(t0) + (y − y(t0))y′(t0) + (z − z(t0))z′(t0) = 0
3. THE LENGTH OF A ARC. NATURAL PARAMETRIZATION 25
For the explicit equations we have the equation of the normal plane.
x− x0 + (y − α(x0))α′(x0) + (z − β(x0))β′(x0) = 0
For a curbe given by implicit equations, the equation of the normal plane becomes
(x− x0)
∣∣∣∣∣∣F ′
y F ′z
G′y G′
z
∣∣∣∣∣∣+ (y − y0)
∣∣∣∣∣∣F ′
z F ′x
G′z G′
x
∣∣∣∣∣∣+ (z − z0)
∣∣∣∣∣∣F ′
x F ′y
G′x G′
y
∣∣∣∣∣∣= 0
or ∣∣∣∣∣∣∣∣∣
x− x0 y − y0 z − z0
F ′x F ′
y F ′z
G′x G′
y G′z
∣∣∣∣∣∣∣∣∣= 0
the partial derivatives being again calculated in (x0, y0, z0).
3. The length of a arc. Natural parametrization
Let r = r(t), t ∈ I be a curve, r(t) being differentiable of class Cs(s ≥ 1). Let a, b ∈ I
and A,B the corresponding points on the curve.
According to the results in chapter .......... the arc-form of the curve is given by
ds =√
x′2(t) + y′2(t) + z′2(t)dt
and the length of the arc of the curve joining A and B is given by
l(AB) =∫ b
a
√x′2(t) + y′2(t) + z′2(t)dt =
∫ b
a||F ′′(t)||dt
we obtain the function s : I −→ [0, L].
s(t) =∫ tt0
√x′2(t) + y′2(t) + z′2(t)dt the length function and the natural parametrization
of the curve is
r(α(s))
where α = s−1. If the curve is given with the natural parametrization the length of the
tangent vector is 1 |r′(s)| = 1.
26 3. SPACE CURVES
4. The Frenet frame
Let r : I −→ R3 be a curve. To consider the Frenet frame consider that the vectors
r′(t), r′′(t) are linearly independent (sec....). By the study in ch () the distinguished Frenet
frame exists in these conditions.
Remark 4.1. Note the straight lines are excluded from our consideration.
Definition 4.1. Let r : I −→ R3 be a curve. The curvatures k1(t), k2(t) defined in
Theorem() will be denoted by k(t) and τ(t) and called the curvature and the ”torsion” of
c, respectively.
We have the formulas
k(t) =〈e′1(t), e2(t)〉
|r′(t)| > 0
T (t) =〈e′2(t), e3(t)〉
|r′(t)|The Frenet equations in matrix form are
e′(t) = |r′(t)|
0 k(t) 0
−k(t) 0 τ(t)
0 −τ(t) 0
e(t)
Proposition 4.2. Let r : I −→ R3 be a curve parametrized by arc-length. We have
k(t) = |r′′(t)|, τ(t) =det(r′(t), r′′(t), r′′′(t))
k2(t)
Proof. We have that r′(t) = e1(t),
e2(t) = r′′(t)/|r′′(t)|, e3(t) = e1(t)× e2(t) = r′(t)×r′′(t)|r′′(t)|
It follows that k(t) = |r′′(t)| and r′′(t) = k(t)e2(t).
The Frenet equations imply
r′′′(t) =k′(t)e2(t) + k(t)e′2(t)
k′(t)e2(t) + k(t)(−k(t)e1(t) + T (t)e3(t))
k′(t)e2(t)− k2(t)e1(t) + k(t)T (t)e3(t)
4. THE FRENET FRAME 27
and it follows the equation for T (t).
From the th. in the Chapter () K(t) and T (t) are geometric objects, that is are invariant
under isometries and orientation preserving changes of variables. ¤
Proposition 4.3. (The local equivalence problem for space curve) Let k, T : [0, L] −→R, k(s) > 0 There exists a unique curve parametrized by arc length defined on [0, L] which
has the curvature and the torsion functions respe. k and r.
Proof. If is exactly the reformulation of the local equivalence of curves in Rn for
n = 3. ¤
28 3. SPACE CURVES
We have seen that, for curves in R, the Frenet moving frame becomes a moving
triedhron (which defines, at each point of the curve a system of axes of coordinates).
For a curve given in parametric form r = r(t), we construct the Frenet triedhron at
an arbitrary point, according to the proof of Proposition 2.1.
We have e1(t) = t(t) =r′
‖r′‖ , the tangent unit vector.
Then, following the Gramm-Schmidt method, e2(t) will have the form:
e2(t) = r′′ − 〈r′′, r′〉‖r′‖2
· r′
=r′′〈r′, r′〉 − 〈r′′, r′〉r′
‖r′‖2=
r′ × (r′′ × r′)‖r′‖2
=(r′ × r′′)× r′
‖r′‖2
and
e2(t) = n(t) =(r′ × r′′)× r′
‖(r′ × r′′)× r′‖
called the principal normal vector.
For the third vector of the triedhron, we simply notice that e3(t) = r′ × r′′ is orthogonal
to both t and n, so be take
e3(t) = b(t) =r′ × r′′
‖r′ × r′′‖ ,
called the binormal vector.
The three axes detemine three planes:
(1) the normal plane, determined by the primcipal normal and the binormal, or-
thogonal to the tangent. the equation of this plane is,
x′(t)(x− x(t)) + y′(t)(y − y(t)) + z′(t)(z − z(t)) = 0
(for r(t) = x(t)i + y(t)j + z(t)k)
4. THE FRENET FRAME 29
(2) the osculating plane, determined by the tangent and the principal normal,
with the equation∣∣∣∣∣∣∣∣∣
x− x(t) y − y(t) z − z(t)
x′(t) y′(t) z′(t)
x′′(t) y′′(t) z′′(t)
∣∣∣∣∣∣∣∣∣= 0,
or (r − r(t), r′(t), r′′(t)) = 0.
(3) the rectifying plane, determined by the tangent and the binormal, with the
equation:
(r − r(t), r′(t), r′(t)× r′′(t)) = 0.
Also, simple formulas for the curvature and the torsion can be deduced.
k(t) =〈e′1(t), e2(t)〉‖r′(t)‖ =
〈r′′(t), (r′(t)× r′′(t))× r′(t)〉‖r′(t)‖2‖(r′(t)× r′′(t))× r′(t)‖
‖r′(t)× r′′(t)‖2
‖r′(t)‖2 · ‖r′(t)× r′′(t)‖ · ‖r′(t)‖ =‖r′(t)× r′′(t)‖
‖r′(t)‖3
The torsion is:
(4.1) τ(t) =(r′(t)× r′′(t)) · r′′′(t)‖r′(t)× r′′(t)‖2
.
For curves in R3, the Frenet equations can be written explicitly:
t′ = kn
n′ = −kt + τ b
b′ = −τ n.
Remark 4.4. The torsion represents the rotation velocity of the binormal vector. A
curve has τ ≡ 0 if and only if it is a plane curve.
A special example is the cylindric helix, obtained by the movement of a point P on
the surface of a circular cylinderin such a way that the coordinate z of P is proportional
to the length of the circle arch AM , M beeing the projection of P on the circle passing
through A.
30 3. SPACE CURVES
The coordinate system is chosen such that Oz is the axis of the cylinder and Ox is the
perpendicular from A on Oz. The curve has the parametric equations:
x = R cos θ
y = R sin θ
z = hθ
with the parameter θ ∈ [0, 2π], R the radius of the cylinder.
We compute the curvature
k =‖r′ × r′′‖‖r′‖3
We have:
r′ × r′′ =
∣∣∣∣∣∣∣∣∣
i j k
−R sin θ R cos θ h
−R cos θ −R sin θ 0
∣∣∣∣∣∣∣∣∣= iRh sin θ − jRh cos θ + R2k,
‖r′ × r′′‖ = R√
h2 + R2, ‖r′‖ =√
h2 + R2.
It follows k =R
R2 + h2.
In the same way, (see formula 4.1) we get the torsion τ =h
R2 + h2.
One can notice that the curvature and the torsion do not depend on the parameter θ. It
can be proved that, conversly, every curve whose curvature and torsion are constant is a
circular helix.
The unit vectors of the Frenet frame are:
• the tangent: t =r′
‖r′‖ =1√
R2 + h2
(−R sin θi + R cos θj + hk)
• the binormal: b =r′ × r′′
‖r′ × r′′‖ =1√
R2 + h2
(h sin θi− h cos θj + Rk
)
• the principal normal: n = b× t = − cos θi− sin θj.
It is a simple exercise to see that the Frenet equations are satisfied.
Also, the equations of the planes in the Frenet triedhron are following imediatly:
• the normal plane: R sin θx−R cos θy − hz + h2θ = 0,
• the osculating plane: h sin θx− h cos θy + Rz − hθR = 0,
4. THE FRENET FRAME 31
• the rectifying plane: cos θx + sin θy −R = 0.
Computing the inner product between the principal normal n and the unit vector k we
get 〈n, k〉 = 0, which means that, for the circular helix, the principal normal at any point
is orthogonal to the axis of the cylinder.
We have 〈b, k〉 =R√
R2 + h2and 〈t, k〉 =
h√R2 + h2
, which means that the angle between
the binormal and the cylinder axis (or the tnagent and the axis) is constant.
CHAPTER 4
Plane curves
1. The equation of a plane curve
1. The explicit equation of a plane curve is of the form y = y(x), x ∈ I, I ⊂ R.
The reader is assumed to have some knowledge concerning the representation of such a
curve.
2. The implicit equation: F (x, y) = 0
3. The parametric equations of a plane curve are x = x(t), y = y(t), t ∈ I, I ⊂ R.
Usually we shall assume that the functions x(t) and y(t) are continuous and have piecewise
continuous derivatives x(t), y(t) in the interval I.
As is customary in differential geometry, we shall suppose that the functions x(t), y(t)
also possess continuous derivatives of an order r, higher than one (according as the problem
under investigation may require) without stating explicitly this condition.
4. The vector equation: r = r(t), t ∈ I, I ⊂ R. Here r(t) = x(t)i+ y(t)j is the so-called
radius vector, or position vector of the point (x(t), y(t)) on the curve.
As t runs through the interval I, the end point of the radius vector describes the given
curve.
5. The equation in polar coordinates. The length ρ = ||r|| and the polar angle θ of the
polar coordinates of P . The connection between polar coordinates (ρ, θ) and Cartesian
coordinates (x, y) is given by the formulas
x = ρ cos θ, y = ρ sin θ
ρ =√
x2 + y2, cos θ =x
ρ, sin θ =
y
ρ
These can be read off Fig.133
34 4. PLANE CURVES
desen
The equation of a plane curve in polar coordinates is of the form ρ = ρ(θ), θ ∈ I, I ⊂ R.
From it we deduce easily the parametric equations: x = ρ(θ) cos θ, y = ρ(θ) sin θ.
Example.The circle, shown in Fig.2, has the equation x2 + y2 = a2. The arc ABC can
be represented by the explicit equation y =√
a2 − x2, x ∈ [−a, a].
The arc CDA has equation y = −√a2 − x2, x ∈ [−a, a].
The parametric equations of the circle are x = a cos t, y = a sin t, t ∈ [0, 2π].
The vector equation is r = a cos ti + a sin tj.
The equation in polar coordinates reads ρ = a.
2. Asymptotes. The folium of Descartes
Let us consider the curve (c) : x = x(t), y = y(t). The straight line x = x0 is a vertical
asymptote of (c) if there exists t0 ∈ R = R ∪ {±∞} such that
(2.1) limt↗t0
x(t) = x0 and limt↗t0
y(t) = ±∞
or
(2.2) limt↘t0
x(t) = x0 and limt↘t0
y(t) = ±∞
We can characterize similarly the horizontal asymptotes, having equation of the form
y = y0.
More generally, the line y = mx + n is an asymptote of the given curve if there exists
t0 ∈ R such that
(2.3) limt↗t0
x(t) = ±∞,m = limt↗t0
y(t)x(t)
, n = limt↗t0
(y(t)−mx(t))
or
(2.4) limt↘t0
x(t) = ±∞,m = limt↘t0
y(t)x(t)
, n = limt↘t0
(y(t)−mx(t))
2. ASYMPTOTES. THE FOLIUM OF DESCARTES 35
These facts are similar to those (well-known to the reader) which hold for curves having
explicit equation y = y(x).
Now let us consider the curve
(c) : x3 + y3 − xy = 0
It is called the folium of Descartes. We easily see that it admits the parametric equations:
(c) : x =t
1 + t3, y =
t2
1 + t3, t 6= −1.
I) For y = 0 we have t = 0 and hence x = 0. Similarly, x = 0 implies t = 0 and
hence y = 0.
We conclude that the curve intersects the axes Ox and Oy only in the origin.
II) The functions x(t), y(t) have infinite one-sided limits only at t0 = −1. In fact,
limt↗−1
x(t) = +∞, limt↗−1
y(t) = −∞; limt↘−1
x(t) = −∞, limt↘−1
x(t) = +∞ Hence the
curve has neither vertical, nor horizontal asymptotes. But we have
m = limt→−1
y(t)x(t)
= −1, n = limt→−1
(y(t) + x(t)) = −13.
Therefore, the line y = −x− 13 is asymptote of the curve.
III) x′(t) =1− 2t3
(1 + t3)2, y′(t) =
2t− t4
(1 + t3)2,dy
dx=
dydtdxdt
=2t− t4
1− 2t3.
The curve is shown on the next page. It is symmetrical about the straight line
y = x. At the point O, it has a node with the x-axis and y-axis as tangents; at
the point A(12 , 1
2), it has a vertex. The straight line x+y+ 13 = 0 is its asymptote.
Example. A periodic curve. Consider the curve
(c)
x = 3a cos t− 2a cos3 t
y = 3a sin2 t
where a > 0 is a given constant.
The functions x(t) and y(t) are periodic with period 2π, so it suffices to study the
curve for t ∈ [−π, π]. Moreover, x(t) and y(t) are even functions, i.e.,
(x(−t), y(−t)) = (x(t), y(t)).
36 4. PLANE CURVES
Consequently we shall consider t ∈ [0, π].
Since x(t) and y(t) are bounded functions, the curve has no asymptote.
The derivatives are
x′(t) = −3a sin t(1− 2 cos2 t)
y′(t) = 6a sin t cos t
and therefore
dy
dx= − 2 cos t
1− 2 cos2 t.
3. The cycloid. The astroid
By rolling a circle C along a straight line l without slipping, a given point of the circle
describes a cycloid. Let l be the Ox axis. Suppose that the generating point M is initially
in O. Let a be the radius of the circle C.
Due to these assumptions, the length of the segment OA is equal to the length of the arc
AM .
It follows that OA = at. Then x = OA−AM ′ = at−PI = at−a cos(t− π2 ) = at−a sin t.
y = M ′P + PM = a + a sin(t− π2 ) = a− a cos t.
Hence the parametric equations of the cycloid are
x = a(t− sin t)
y = a(1− cos t)t ∈ R
The arc OT of the cycloid corresponds to t ∈ [0, 2π].
desen
Points like O and T are called cuspidal points of the cycloid.
If a circle C of radius a4 rolls along the interior circumference of a fixed circle of radius a,
then a given point of C describes a curve called astroid.
desen
4. THE TANGENT TO A PLANE CURVE 37
It can be shown that the parametric equations of the astroid are
x = a cos3 t
y = a sin3 t
By eliminating t we obtain the implicit equation of the astroid:
x23 + y
23 = a
23
4. The tangent to a plane curve
a) Consider the curve (C) : y = y(x). The tangent to this curve at its point P (x0, y0)
has the equation
(4.1) y − y0 = y′(x0)(x− x0)
We know also that tanα = y′(x0)
b) Let now (C) : f(x, y) = 0. Suppose that ∂f∂x and ∂f
∂y are continuous in a neigh-
borhood of the point P (x0, y0) of the curve; moreover, suppose that ∂f∂y (x0, y0) 6=
0.Consider the implicit function y(x) defined by the equation f(x, y) = 0. It
is possible to compute y′. Indeed, let us differentiate the equation f(x, y) = 0
partially with respect to x, considering y as a function of x:
f ′x(x)(x, y) + f ′y(x, y)y′ = 0.
We obtain immediately
y′ = −f ′x(x)(x, y)f ′y(x, y)y′
or
(4.2) f ′x(x0, y0) + f ′y(x0, y0)(y − y0) = 0
We obtain the same equation if ∂f∂y (x0, y0) = 0 but ∂f
∂y (x0, y0) 6= 0. Generally
speaking, a point (x0, y0) on a plane curve f(x, y) = 0 is called a regular (ordinary)
point if at least one of the numbers
∂f
∂x(x0, y0),
∂f
∂y(x0, y0)
38 4. PLANE CURVES
is non-zero. Every other point on the curve is said to be singular.Thus, equation
(4.2) is written under the hypothesis that (x0, y0) is a regular point.
c) Consider (C) : x = x(t), y = y(t). Then dydx =
dydtdxdt
= y′(t)x′(t) . The equation of the
tangent at the point (x(t0), y(t0)) is
y − y(t0) =y′(t0)x′(t0)
(x− x(t0)),
or
(4.3)x− x9t0)
x′(t0)=
y − y(t0)y′(t0)
Here we also suppose that (x(t0), y(t0)) is a regular point, that is, at least one of
the numbers x′(t0), y′(t0) is non-zero.
d) The line through P , which is perpendicular to the tangent , will be called the
normal at P . If the equation of the tangent is y − y0 = m(x − x0), then the
equation of the normal will be
(4.4) y − y0 = − 1m
(x− x− 0).
e) If the curve is defined by (C) : ρ = ρ(θ), it admits the parametric representation
(C) :
x = ρ(θ) cos θ
y = ρ(θ) sin θ
The slope of the tangent is
tanα =dy
dx=
dydθdxdθ
=ρ′ sin θ + ρ cos θ
ρ′ cos θ − ρ sin θ
Let us compute the angle V between the position vector OP and the tangent
to the curve through P . We have V = α− θ, hence
tanV =tanα− tan θ
1 + tanα tan θ=
=ρ′ sin θ+ρ cos θρ′ cos θ−ρ sin θ − sin θ
cos θ
1 + ρ′ sin θ+ρ cos θρ′ cos θ−ρ sin θ · sin θ
cos θ
5. ASYMPTOTES IN POLAR COORDINATES. THE HYPERBOLIC SPIRAL. 39
Finally we obtain tanV = ρρ′ .
5. Asymptotes in polar coordinates. The hyperbolic spiral.
Consider the curve (C) : ρ = ρθ. Suppose that the straight line l is an asymptote of
this curve.
The position of l is determined by the angle θ0 and the distance d = ON . Clearly
(5.1) limθ→θ0
ρ(θ) = +∞
Using this equality we determine θ0. Then d = limθ→θ0
OM = limθ→θ0
ρ(θ) cos[π2 − (θ − θ0)] =
limθ→θ0
ρ(θ) sin(θ − θ0)
Hence
(5.2) d = limθ→θ0
ρ(θ) sin(θ − θ0)
According to these results, we find an asymptote (if it exists) of a curve as follows:
First we determine its direction defined by the angle θ0 (given by (5.1) and then its distance
d from the origin (given by (5.2).
Besides the asymptotes, some plane curves may admit so-called asymptotic points.
An asymptotic point of a plane curve is a point which does not lie on the curve but to
which a point moving along the curve approaches to within an arbitrary small distance.
This occurs in the following example.
The hyperbolic spiral is the curve
(C) : ρ =k
θ, k = const. > 0.
Since ρ ≥ 0, we must have θ > 0. Clearly limθ↘0
ρ(θ) sin θ = limθ↘0
k sin θθ = k. Thus the
asymptote is a line parallel to, and at a distance k from, the polar axis. Now limθ→+∞
ρ(θ) = 0;
therefore the origin is an asymptotic point of the curve.
Since ρ′(θ) = − kθ2 , we have the following table:
The graph is the following:
40 4. PLANE CURVES
6. The cardioid. The lemniscate of Bernoulli
If a circle C of radius b rolls along the exterior circumference of a fixed circle of the
same radius b, then a given point of the circle C describes a curve called cardioid.
Let A be the centre of the fixed circle and M the given point of C. Initially, M is in
a position O on the fixed circle. Let O be the pole and AO the polar semi-axis.
Let us remark that the length of the arc OT is equal to the length of the arc TM .
It follows that OM is parallel to AB, hence OAB = MBA = θ. We obtain immediately
OM = AB − 2b cos θ, i.e., ρ = 2b(1− cos θ). Denoting 2b = a, we have the equation of the
cardioid:
ρ = a(1− cos θ)
I) Due to the periodicity of the function cosine, we may consider θ ∈ [−π, π]. Since
the same function is even, we may restrict to the case θ ∈ [0, π]; indeed, we have
ρ(−θ) = ρ(θ) and hence the curve is symmetric with respect to the polar axis.
II) ρ′(0) = 0, ρ(π) = 2a. The function ρ(θ) has no infinite limits, hence there are no
asymptotes.
III) ρ′ = a sin θ
IV) tanV = tan θ2 . For θ = 0 we have V = 0, and for θ = π, V = π
2 . This means that
for θ = 0, the tangent of the curve is the polar axis, while for θ = π the tangent
is perpendicular to the polar axis.
V)
Now let us consider the points A(a, 0), A′(−a, 0). The locus of a point M(x, y) which moves
so that MA′ ·MA = a2 is a curve called the lemniscate of Bernoulli.
We have [(x + a)2 + y2] · [(x− a)2 + y2] = a4. This implies
(x2 + y2)2 = 2a2(x2 − y2)
6. THE CARDIOID. THE LEMNISCATE OF BERNOULLI 41
This is the implicit equation of the lemniscate. Set x = ρ cos θ, y = ρ sin θ and simplify.
This yields the equation in polar coordinates:
ρ2 = 2a2 cos 2θ
or
ρ = a√
2√
cos 2θ
I) We have ρ(θ + π) = ρ(θ), hence it suffices to consider θ ∈ [−π2 , π
2 ]. Since cos 2θ
must be positive, we have 2θ ∈ [−π2 , π
2 ], hence θ ∈ [−π4 , π
4 ]. Finally, ρ(−θ) = ρ(θ);
this means that the curve is symmetric with respect to the polar axis and we
shall consider θ ∈ [0, π4 ].
II) ρ(0) = a√
2, ρ(π4 ) = 0. There are no asymptotes.
III) ρ′ = −2a sin 2θ√2 cos 2θ
≤ 0 ∀θ ∈ [0, π4 )
IV) tanV = ρρ′ = − cot 2θ. For θ = π
4 we have V = 0 and for θ = 0, V = π2 .
V)
CHAPTER 5
1. Ruled surfaces
If a surface can be generated by the movement of a straight line which satisfies certain
conditions, it is called a ruled surface. The straight lines that generate the surface are
called the rectilinear generators of the surface.
The conditions satisfied by the generators determine several types of ruled surfaces.
We call cylindrical surface (or cylinder) a ruled surface generated by a straight line
which keeps a fixed direction and intersects a fixed curve-called directrix of the cylinder.
It is clear that the rectilinear generators are parallel to one another.
The directrix of a cylinder is not unique. In fact, every curve located on the cylindrical
surface and having the property that each generator intersects it exactly in one point, is
a directrix.
Any plane which is not parallel to the generators intersects the cylinder by a plane
directrix.
Theorem 1.1. A cylindrical surface with the generators parallel to the straight line
P (x, y, z) = 0
Q(x, y, z) = 0has the equation of the form
(1.1) f(P, Q) = 0
Proof. The straight line that gives the fixed direction is given as an intersection of
two planes: P = a1x+ b1y + c1z +d1 = 0 and Q = a2x+ b2y + c2z +d2 = 0. Any generator43
44 5
d beeing parallel to this line, has the equation of the form
P (x, y, z) = λ
Q(x, y, z) = µ, with λ, µ
real parameters (P = λ and Q = µ are planes parallel to P = 0, respective Q = 0).
Let the curve (C) :
F (x, y, z) = 0
G(x, y, z) = 0be a directrix. The condition that the generator
d intersects the curve (C) is that the system:
P (x, y, z) = λ
Q(x, y, z) = µ
F (x, y, z) = 0
G(x, y, z) = 0
is compatible. Eliminating the variables x, y, z between the four equations we get a com-
patibility condition of the form f(λ, µ) = 0. Then, substituting λ and µ we get the equation
of the cylinder: f(P,Q) = 0. ¤
Example 1. Find the equation of the cylindrical surface with the generators parallel to
the direction ~v(1,−1, 1) and the directrix (C)
x2 + (y − 1)2 = 4
z = 0(a circle in the plane
x0y).
The equations of the fixed direction are x1 = y
−1 = z1 , or written as intersection of two
planes (P ) and (Q):
x− z = 0
y + z = 0.
In order that a generator intersects the directrix, the system
x− z = λ
y + z = µ
x2 + (y − 1)2 = 4
z = 0
must have a solution. Eliminating x, y and z we get the compatibility relationship λ2 +
1. RULED SURFACES 45
(µ− 1)2 = 4 which gives the equation of the cylinder:(x− z)2 +(y + z− 1)2 = 4, obviously
of type (1.1). Another method to obtain this equation is the following:
Let M(a, b, c) be a point on the directrix, which means that
(1.2) a2 + (b− 1)2 = 4; c = 0.
The generator through M has the equations
(1.3)x− a
1=
y − b
−1=
z − c
1
Eliminating a, b, c between (1.2) and (1.3) we get
(x− z)2 + (y + z − 1)2 = 4
The converse of Theorem 1.1 is also true:
Theorem 1.2. An equation of the form f(P,Q) = 0, where P (x, y, z) and Q(x, y, z)
are polynomials of degree one with respect to x, y, z (P = ax + by + cz + d), represents a
cylindrical surface with the generators parallel to
P (x, y, z) = 0
Q(x, y, z) = 0.
Proof. Let λ and µ be real parameters such that f(λ, µ) = 0. Then, the straight lines
given by
P = λ
Q = µare generating the surface and are parallel to the direction
P = 0
Q = 0,
which means the surface is a cylindrical one. ¤
Remark 1.3. If there are not such real numbers λ, µ for which f(λ, µ) = 0, the cylinder
is an imaginary one.
In particular, an equation of the form f(x, y) = 0 represents a cylindrical surface with
the generators parallel to the axis oz.
We present some examples of equations representing cylindrical surfaces.
1) x2 + y2 = R2 has the generators parallel to Oz, and the circle of center O, radius
R in the plane xOy is the directrix.
46 5
2) x2
a2 + z2
b2= 1 the generators are parallel to Oy, and an ellipse is a directrix.
3) y2
a2 − z2
c2= 1 the generators are parallel to Ox, a directrix is a hyperbola
4) y2 = 2px the generators are parallel to Oz, a directrix is a parabola.
Application. The projection of a curve on a plane. Considering a curve
(C)
F (x, y, z) = 0
G(x, y, z) = 0and a plane P : Ax + By + Cz + D = 0, we want to determine
the projection of this curve on the plane, by a given direction d.
The projection is obtained by intersecting the plane P with the cylindrical surface that
has the generators parallel with d and the curve (C) as a directrix.
Most frequentely met are the orthogonal projections, when the direction d is the normal
to the plane P : ~N(A,B,C).
Example 2. The circle (C)
x2 + y2 + z2 = 1
x− y + z = 0is projected orthogonally on the coordi-
nate planes. Find the equations of the three curves that are obtained.
To find the projection on the plane xOy, we write the equation of the cylinder that
has (C) as a directrix and the generators parallel to Oz. The system
x = λ
y = µ
x2 + y2 + z2 = 1
x− y + z = 0
gives the compatibility condition λ2 + µ2 + (µ − λ)2 = 1, and the
cylindrical surface is x2 + y2 + (y − x)2 = 1. The projection is the intersection between
this surface and the plane xOy :
2x2 + 2y2 − 2xy = 1
z = 0.
In the same way, one can obtain the projections on xOz and yOz :
2x2 + 2z2 + 2xz = 1
y = 0.respectively
2y2 + 2z2 − 2yz = 1
x = 0.
1. RULED SURFACES 47
We call conic surface (or cone) a surface generated by a straight line that passes
through a fixed point called vertex and intersects a fixed curve, called directrix.
As for the cylinder, the directrix of a cone is not unique.
Theorem 1.4. A conic surface with the vertex V (x0, y0, z0) has the equation of the
form
(1.4) f(x− x0
z − z0,y − y0
z − z0) = 0,
or in a similar form, where the variables x, y, z are permuted.
Proof. Any generator that passes through the point V has the equation x−x0l =
y−y0
m = z−z0n , with l2 + m2 + n2 6= 0. Supposing that n 6= 0, we denote λ = l
n , µ =
mn and the family of generators is given by
x− x0 = λ(z − z0)
y − y0 = µ(z − z0).If the directrix is
(C)
F (x, y, z) = 0
G(x, y, z) = 0,the condition that the generator intersects the directrix is that the
system
x− x0 = λ(z − z0)
y − y0 = µ(z − z0)
F (x, y, z) = 0
G(x, y, z) = 0
is compatible. Eliminating x, y, z we get a relationship of
the form f(λ, µ) = 0 and then the equation of the cone,
f(x− x0
z − z0,y − y0
z − z0) = 0.
¤
In the same way, one can prove:
48 5
Theorem 1.5. A conic surface with the vertex V given as intersection of three planes
P = 0
Q = 0
R = 0
has the equation of the form f(PR , Q
R ) = 0.
Example 3. Find the equation of the conic surface with the vertex V (1, 0, 2) and
(C)
x2 + y2 − 2x− 2y + 1 = 0
z = 0as directrix.
A straight line passing through V has the equation d
x− 1 = λ(z − 2)
y = µ(z − 2)
If the line intersects the curve (C)
x2 + y2 − 2x− 2y + 1 = 0
z = 0, then the system con-
sisting of the four equations above is compatible.
Because z = 0, we get x = 1−2λ and y = −2µ. Replacing these in the first equation of
(C), we have the compatibility condition (1− 2λ)2 +(−2µ)2− 2(1− 2λ)− 2(−2µ)+1 = 0,
or λ2 + µ2 + µ = 0.
But λ = x−1z−2 , µ = y
z−2 , so the cone has the equation:(x− 1
z − 2
)2+
( y
z − 2
)2+
y
z − 2= 0, or (x− 1)2 + y2 + y(z − 2) = 0.
The converse of Theorem 1.4 is true.
Theorem 1.6. If P, Q and R are polynomials of degree one with respect to x, y, z such
that the system
P = 0
Q = 0
R = 0
has a unique solution (x0, y0, z0), then an equation of the form
f(
PR , Q
R
)= 0 represents a conic surface, with the vertex V (x0, y0, z0) the intersection of
the three planes P, Q, R.
1. RULED SURFACES 49
Remark 1.7. An equation of the form:
a11x2 + a22y
2 + a33z2 + 2a12xy + 2a13xz + 2a23yz = 0, aij ∈ R
(homogeneous of order II) represents a conic surface with the vertex at the origin.
Indeed, dividing the equation by z2, we get an equation of the form(
xz , y
z
), which
means that the vertex is the intersection point of the planes x = 0, y = 0, z = 0, that is
the origin.
Application. The cone of tangents to a sphere. Consider a point V (x0, y0, z0)
and a sphere (S) of equation
(1.5) x2 + y2 + z2 + 2ax + 2by + 2cz + d = 0.
Which is the geometric locus of the tangents from V to the sphere?
To find it, we write the equations of a straight line that passes through V , as before:
(1.6) dλ,µ :
x− x0 = λ(z − z0)
y − y0 = µ(z − z0).
Then we select from all these lines (depending on the real parameters (λ, µ) those that
intersect the sphere at only one point.
Replacing x and y with their expressions from (1.6), in (1.5), we get a degree two equation,
with the unknown z. The condition to be satisfied is ∆(λ, µ) = 0. But λ = x−x0z−z0
, µ = y−y0
z−z0,
so we get the equation of a cone: ∆(
x−x0z−z0
, y−y0
z−z0
)= 0. As an example, the cone of tangents
from the origin O to the sphere
(x + 2)2 + (y − 1)2 + (z − 3)2 = 9
has the equation
x2 + 4y2 − 4z2 + 4xy + 12xz − 6yz = 0.
50 5
We call conoid surface (or conoid) a ruled surface generated by a straight line that
is parallel to a fixed plane (the director plane), intersects a fixed line (axis) and a fixed
curve (directrix).
Theorem 1.8. A conoid surface with the director plane given by P (x, y, z) = 0, the
axis given as intersection of two planes d0
Q(x, y, z) = 0
R(x, y, z) = 0has the equation of the form
f(P, Q
R
)= 0.
Proof. The straight lines that are parallel to the plane P and intersect the line d0
have the equations
(1.7)
P = λ
Q− µR = 0,
with λ and µ real parameters. To select from these the ones that intersect a directrix given
as
(1.8)
F (x, y, z) = 0
G(x, y, z) = 0
we put the condition that the system (1.7)+(1.8) has a solution. Eliminating the variables
x, y, z between the four equations, we get a relationship f(λ, µ) = 0.
But λ = P, µ = QR so the equation of the conoid is f
(P, Q
R
)= 0. ¤
Theorem 1.9. Conversely, any equation of the form f(P, Q
R
)= 0, where
P (x, y, z), Q(x, y, z) and R(x, y, z) are polynomials of degree one in x, y, z; represents a
conoid surface.
Example 4. Write the equation for a conoid surface that has the director plane P : y = 0;
the axis x = y−1 = z and the curve
x2 − z2 = 1
x + y + z = 0as a directrix.
2. SURFACES OF REVOLUTION 51
The parameters λ and µ must be such that the system:
y = λ
x + y − µ(x− z) = 0
x2 − z2 = 1
x + y + z = 0
is com-
patible. We eliminate x, y and z : x + z = −λ; (x − z)(x + z) = 1 implies x − z = − 1λ
and then, x = − 12λ − λ
2 . Substituting now x, y and x− z in the second equation; we haveλ2 − 1
2λ + µλ = 0, finally y2
2 − 12 + x+y
x−z = 0.
2. Surfaces of revolution
We call surface of revolution a surface generated by the rotation of a plane curve
around a fixed straight line-called the revolution axis.
Each point of the curve describes a circle with the center on the revolution axis, in a
plane that is perpendicular on the axis.
Each plane that contains the axis, intersects the surface by two curves, equal to the given
curve.
Some of the well-known surfaces, studied before, are in fact surfaces of revolution:
• a sphere can be obtained by rotating a circle around a straight line that passes
through its center.
• by the rotation of a straight line around another straight line, we obtain: a circular
cylinder if the two lines are parallel and a circular cone if the two lines have a
common point.
Theorem 2.1. A surface of revolution having the straight line d : x−x0l = y−y0
m = z−z0n
as axis, has the equation of the form
(2.1) f(lx + my + nz, (x− x0)2 + (y − y0)2 + (z − z0)2) = 0.
Proof. we consider the surface generated by circle (Γ) that have the centers on the
axis (d), belong to perpendicular planes on (d) and intersect the rotating curve (C).
52 5
The family of circles depending on the parameters λ and µ is described by the equations
(2.2) (Γ)
lx + my + nz = λ
(x− x0)2 + (y − y0)2 + (z − z0)2 = µ
(intersection between a sphere with the center at (x0, y0, z0) and a plane)
If the curve is given by the equations,
(2.3) (C)
F (x, y, z) = 0
G(x, y, z) = 0
the condition that the circle (Γ) intersects the curve (C) is that the system (2.2)+(2.3) is
compatible. Eliminating x, y, z we get an equation f(λ, µ) = 0, and replacing λ, µ by their
expressions in (2.2); an equation of the form (2.1). ¤
Theorem 2.2. Conversely, every equation of the form (2.1) is a surface of revolution
with the axis d : x−x0l = y−y0
m = z−z0n .
Example 5.Find the equation that represents the surface of revolution obtained when the
circle (C)
(y − a)2 + z2 = R2
x = 0, with a > R, moves around the axis Oz (this surface is
called a torus).
The axis Oz has the direction ~v(0, 0, 1) and the origin O(0, 0, 0) is a point on the axis.
Then a circle in a plane perpendicular to Oz, with the center on Oz has the equations:
z = λ
x2 + y2 + z2 = µ.
2. SURFACES OF REVOLUTION 53
In order that this circle intersects the rotating circle (C), the system
z = λ
x2 + y2 + z2 = µ
(y − a)2 + z2 = R2
x = 0
must have a solution. We eliminate x, y, z : y =√
µ− λ2
(we take only the positive solution y, because for a point on (C), clearly y > 0).
Substituting in the third equation we get (√
µ− λ2 − a)2 + λ2 = R2, and finally the
equation of the torus:(√
x2 + y2 − a)2 + z2 = R2.
Example 6.Find the equation for the surface of revolution obtained by rotating the
straight line
x = a
y = 0around the axis Oz.
In the same way as for the previous example, we get that the system
z = λ
x2 + y2 + z2 = µ
x = a
y = 0
must be compatible. Eliminating x, y and z we have a2 + λ2 = µ,
and the equation x2 + y2 = a2 which represents, as expected, a cylinder.
