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”You don’t see something until you have the right metaphor to
let youperceive it .”
James Gleick
8Suprafete. Sfera
Design cu ajutorul suprafetelor Bézier
Suprafetele Bézier reprezinta ometoda eleganta de a construi
osuprafata, fiind initial folosite indesign-ul automobilelor.
Astazi suntfolosite pe scara larga in grafica com-puterizata si
computer-aided design(CAD). Ideea principala este sa folosimpuncte
de control pentru a genera oriceforma sau imagine. O ecuatie
para-metrica generala, in cazul a (𝑛 + 1) ×(𝑚 + 1) puncte de
control, este dataprin formula
𝑆 : 𝑟(𝑢, 𝑣) =
𝑛∑︁𝑖=0
𝑚∑︁𝑗=0
𝐵𝑖,𝑛(𝑢)𝐵𝑗,𝑚(𝑣)𝑃𝑖𝑗
unde 𝐵𝑖,𝑛(𝑢) = 𝐶𝑖𝑛𝑢
𝑖(1 − 𝑢)𝑛−𝑖 sunt polinoamele Bernstein, 𝑃𝑖𝑗 punctele decontrol,
iar 𝑢, 𝑣 ∈ [0, 1]. In grafica computerizata cele mai utilizate
suprafeteBézier sunt suprafetele Bézier bicubice (m=n=3), in
principal deoarece sunt opunte intre simplitate si complexitate,
asigurand libertatea dorita artistului cuun nivel minimal de
complexitate pentru programator si renderer.
1
https://en.wikipedia.org/wiki/Computer-aided_designhttps://en.wikipedia.org/wiki/Computer-aided_design
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Suprafete. Exemple
Suprafete regulate:A set 𝑆 ⊂ R3 is a regular surface if each of
its points 𝑝 has a neighborhood 𝑉and there exists an open set 𝐷 ⊂
R2 and a mapping 𝑟 : 𝐷 → 𝑉 ⊂ 𝑆:
𝑟(𝑢, 𝑣) = 𝑥(𝑢, 𝑣)𝑖 + 𝑦(𝑢, 𝑣)𝑗 + 𝑧(𝑢, 𝑣)𝑘, (𝑢, 𝑣) ∈ 𝐷
with the following properties:i) 𝑟 is bijective and continuous
with a continuous inverse
ii) every component 𝑥(𝑢, 𝑣), 𝑦(𝑢, 𝑣), 𝑧(𝑢, 𝑣) is continuously
differentiable
iii) in every point of 𝐷:𝑟𝑢 × 𝑟𝑣 ̸= (0, 0, 0)
The mapping 𝑟 will be called a parametrization (surface patch)
of 𝑉.
∙ the above definition means that locally a surface can be
deformed into aplane
∙ from now on by a surface we will understand a regular
surface.∙ a parametric surface is a surface that can be described
by a single parametriza-
tion 𝑟 : 𝐷 → 𝑆
If 𝑓 : 𝐷 ⊂ R2 → R is continuously differentiable then:
𝑟(𝑥, 𝑦) = 𝑥 · 𝑖 + 𝑦 · 𝑗 + 𝑓(𝑥, 𝑦) · 𝑘
is a global surface patch (parametrization) of the graph of 𝑓 ,
because:
𝑟𝑥 × 𝑟𝑦 =(︂−𝜕𝑓𝜕𝑥
,−𝜕𝑓𝜕𝑦
, 1
)︂̸= (0, 0, 0).
Grafice de functii
2
https://proofwiki.org/wiki/Definition:Continuously_Differentiable
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∙ the graph is a surface givenin the explicit equation 𝑧 = 𝑓(𝑥,
𝑦)and every point on the graph hascoordinates (𝑥, 𝑦, 𝑓(𝑥, 𝑦))
∙ in particular for
𝑓(𝑥, 𝑦) = 𝑥2 − 𝑦2
one gets the surface:
∙ you can use this link to generate surfaces as graphs of
functions.
∙ if we rotate a regular 3𝐷 curve 𝑐(𝑢) =
⎛⎜⎜⎜⎝𝑐1(𝑢)
𝑐2(𝑢)
𝑐3(𝑢)
⎞⎟⎟⎟⎠, 𝑢 ∈ (𝑎, 𝑏), whose imagedoes not intersect the Oz-axis,
around the 𝑂𝑧-axis we get a parametricsurface with the following
parametrization:
𝑟(𝑢, 𝑣) =
⎛⎜⎜⎜⎝cos 𝑣 − sin 𝑣 0
sin 𝑣 cos 𝑣 0
0 0 1
⎞⎟⎟⎟⎠⎛⎜⎜⎜⎝𝑐1(𝑢)
𝑐2(𝑢)
𝑐3(𝑢)
⎞⎟⎟⎟⎠, 𝑢 ∈ (𝑎, 𝑏), 𝑣 ∈ (0, 2𝜋).
∙ the matrix
⎛⎜⎜⎜⎝cos 𝑣 − sin 𝑣 0
sin 𝑣 cos 𝑣 0
0 0 1
⎞⎟⎟⎟⎠ acting, in the above formula, on thecurve coordinates is
called the rotation matrix around the 𝑂𝑧-axis.
∙ with the other rotation matrices we can rotate around any axis
of anOxyz frame.
Suprafete de revolutie
Torus:∙ a torus can be obtained rotating a circle, lying in the
𝑂𝑥𝑧-plane with originat (𝑎, 0, 0) and radius 0 < 𝑅 < 𝑎,
around the 𝑂𝑧-axis.
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https://en.wikipedia.org/wiki/Implicit_surfacehttps://academo.org/demos/3d-surface-plotter/https://en.wikipedia.org/wiki/Rotation_matrix
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∙ such a surface will be parametrized as 𝑟 : [0, 2𝜋] × [0, 2𝜋] →
R3 :
𝑟(𝑢, 𝑣) =
⎛⎜⎜⎜⎝cos 𝑣 − sin 𝑣 0
sin 𝑣 cos 𝑣 0
0 0 1
⎞⎟⎟⎟⎠⎛⎜⎜⎜⎝𝑎 + 𝑅 cos𝑢
0
𝑎 + 𝑅 sin𝑢
⎞⎟⎟⎟⎠= (𝑎 + 𝑅 cos𝑢)cos 𝑣 · 𝑖 + (𝑎 + 𝑅 cos𝑢)sin 𝑣 · 𝑗 + (𝑎 + 𝑅
sin𝑢) · 𝑘
Sfera:∙ the sphere with center 𝑂(𝑥0, 𝑦0, 𝑧0) and radius 𝑅 is a
surface but it doesn’t
admit a single global parametrization, since it can be covered
only by at leasttwo surface patches: the northern and the southern
hemisphere.
∙ the northern hemisphere is a parametric surface with the
parametrization𝑟 :[︀0, 𝜋2
]︀× [0, 2𝜋] → R3 :
𝑟(𝑢, 𝑣) = (𝑥0 + 𝑅sin𝑢 cos 𝑣) · 𝑖 + (𝑦0 + 𝑅sin𝑢 sin 𝑣) · 𝑗 + (𝑧0
+ 𝑅cos𝑢) · 𝑘.
∙ the sphere can be seen as a surface of revolution.∙ daca sfera
este data prin ecuatia carteziana
𝑆 : 𝑥2 + 𝑦2 + 𝑧2 + 𝑚𝑥 + 𝑛𝑦 + 𝑝𝑧 + 𝑞 = 0
4
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atunci centrul sau are coordonatele 𝐶(︀−𝑚2 ,−
𝑛2 ,−
𝑝2
)︀iar raza este 𝑅 = 12
√︀𝑚2 + 𝑛2 + 𝑝2 − 4𝑞
Straight Circular Cylinder:
∙ a parametrization of a cylinderwith radius 𝑅 is:
𝑟(𝑢, 𝑣) = 𝑅 cos𝑢 · 𝑖 + 𝑅 sin𝑢 · 𝑗 + 𝑣 · 𝑘
where 0 ≤ 𝑢 ≤ 2𝜋, 𝑎 ≤ 𝑣 ≤ 𝑏
Flat Approximations of Surfaces:
∙ in the sequel we’ll consider only parametric surfaces in order
to simplify sometedious computations
Coordinate curves: If 𝑟 is a parametrization of a parametric
surface 𝑆, thenfor fixed values 𝑢0, 𝑣0 the mappings:
𝑢 → 𝑟(𝑢, 𝑣0), 𝑣 → 𝑟(𝑢0, 𝑣)
are called coordinate curves on 𝑆.
∙ meridians and parallels are coordinate curves on a sphere
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Tangent plane :
The tangent plane at a point 𝑀(𝑢0, 𝑣0)to a surface 𝑆 is the
plane that containsthe point 𝑀 and the tangent vectors𝑟𝑢(𝑢0, 𝑣0)
and 𝑟𝑣(𝑢0, 𝑣0) to the coor-dinate curves. It has the normal
direc-tion given by 𝑟𝑢(𝑢0, 𝑣0) × 𝑟𝑣(𝑢0, 𝑣0).
𝛼 :
⃒⃒⃒⃒⃒⃒⃒⃒⃒𝑥− 𝑥𝑀 𝑦 − 𝑦𝑀 𝑧 − 𝑧𝑀
𝑥′𝑢(𝑢0, 𝑣0) 𝑦′𝑢(𝑢0, 𝑣0) 𝑧
′𝑢(𝑢0, 𝑣0)
𝑥′𝑣(𝑢0, 𝑣0) 𝑦′𝑣(𝑢0, 𝑣0) 𝑧
′𝑣(𝑢0, 𝑣0)
⃒⃒⃒⃒⃒⃒⃒⃒⃒ = 0
∙ ecuatia planului tangent la sfera:
𝑆 : 𝑥2 + 𝑦2 + 𝑧2 + 𝑚𝑥 + 𝑛𝑦 + 𝑝𝑧 + 𝑞 = 0
in punctul 𝑀(𝑥0, 𝑦0, 𝑧0) se poate obtine prin dedublare:
𝛼 : 𝑥𝑥0 + 𝑦𝑦0 + 𝑧𝑧0 + 𝑚𝑥 + 𝑥0
2+ 𝑛
𝑦 + 𝑦02
+ 𝑝𝑧 + 𝑧0
2+ 𝑞 = 0
∙ the graph of a function 𝑓 : 𝐷 ⊂ R2 → R has the
parametrization:
𝑟(𝑥, 𝑦) = 𝑥 · 𝑖 + 𝑦 · 𝑗 + 𝑓(𝑥, 𝑦) · 𝑘
thus the tangent plane at 𝑀(𝑎, 𝑏, 𝑓(𝑎, 𝑏)) is:
𝛼 :
⃒⃒⃒⃒⃒⃒⃒⃒⃒𝑥− 𝑎 𝑦 − 𝑏 𝑧 − 𝑓(𝑎, 𝑏)
1 0 𝜕𝑓𝜕𝑥 (𝑎, 𝑏)
0 1 𝜕𝑓𝜕𝑦 (𝑎, 𝑏)
⃒⃒⃒⃒⃒⃒⃒⃒⃒ = 0
which implies:
𝛼 : 𝑧 = 𝑓(𝑎, 𝑏) +𝜕𝑓
𝜕𝑥(𝑎, 𝑏)(𝑥− 𝑎) + 𝜕𝑓
𝜕𝑦(𝑎, 𝑏)(𝑦 − 𝑏)⏟ ⏞
𝑔(𝑥,𝑦)
∙ the tangent plane is the graph of the function:
𝑔(𝑥, 𝑦) = 𝑓(𝑎, 𝑏) +𝜕𝑓
𝜕𝑥(𝑎, 𝑏)(𝑥− 𝑎) + 𝜕𝑓
𝜕𝑦(𝑎, 𝑏)(𝑦 − 𝑏)
Affine Approximation of a Function
6
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∙ 𝑔 contains the first terms of the Taylor expansion of 𝑓 in (𝑎,
𝑏), thus ina neighborhood of (𝑎, 𝑏) we have:
𝑓(𝑥, 𝑦) ≈ 𝑓(𝑎, 𝑏) + 𝜕𝑓𝜕𝑥
(𝑎, 𝑏)(𝑥− 𝑎) + 𝜕𝑓𝜕𝑦
(𝑎, 𝑏)(𝑦 − 𝑏)
∙ the function 𝑔 is not linear, because of the term 𝑓(𝑎, 𝑏), but
it is affineand we call it the affine approximation of 𝑓 around (𝑎,
𝑏).
=⇒ the tangent plane at the graph of 𝑓 (surface) can be seen as
ageometric visualization of the Taylor expansion of order 1 !
Analysis on a Surface:
∙ we call the expressions:
𝐸(𝑢, 𝑣) = ⟨𝑟𝑢(𝑢, 𝑣), 𝑟𝑢(𝑢, 𝑣)⟩
𝐹 (𝑢, 𝑣) = ⟨𝑟𝑢(𝑢, 𝑣), 𝑟𝑣(𝑢, 𝑣)⟩
𝐺(𝑢, 𝑣) = ⟨𝑟𝑣(𝑢, 𝑣), 𝑟𝑣(𝑢, 𝑣)⟩
the coefficients of the first fundamental form of 𝑆.
∙ a regular plane curve
𝑐 :
{︃𝑢 = 𝑢(𝑡)
𝑣 = 𝑣(𝑡)
in 𝐷 generates a regular space curve:
𝑟 ∘ 𝑐 : 𝑟(𝑢(𝑡), 𝑣(𝑡)) = 𝑥(𝑢(𝑡), 𝑣(𝑡)) · 𝑖 + 𝑦(𝑢(𝑡), 𝑣(𝑡)) · 𝑗 +
𝑧(𝑢(𝑡), 𝑣(𝑡)) · 𝑘
on a parametric surface with parametrization 𝑟 : 𝐷 → 𝑆.
Length of a curve: The length of the arc between 𝐴(𝑡 = 𝑎) and
𝐵(𝑡 = 𝑏),determined on the curve 𝑟 ∘ 𝑐, is given by:
ℓ𝐴𝐵 =
∫︁ 𝑏𝑎
√︀[�̇�(𝑡)]2𝐸(𝑢(𝑡), 𝑣(𝑡)) + 2�̇�(𝑡)�̇�(𝑡)𝐹 (𝑢(𝑡), 𝑣(𝑡)) +
[�̇�(𝑡)]2𝐺(𝑢(𝑡), 𝑣(𝑡)) 𝑑𝑡
∙ for a closed curve like in the last figure one has to specify
which of thetwo possible arcs between 𝐴 and 𝐵 are considered. (or
the curve needs anorientation)
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https://en.wikipedia.org/wiki/Affine_transformation
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Integration on parametric surfaces:For a parametric surface 𝑟 :
𝐷 → 𝑆 we define the scalar surface integral of afunction 𝑓 : 𝑆 → R
as:∫︁∫︁𝑆
𝑓(𝑥, 𝑦, 𝑧)𝑑𝑆 =
∫︁∫︁𝐷
𝑓(𝑟(𝑢, 𝑣))·‖𝑟𝑢×𝑟𝑣‖ 𝑑𝑢𝑑𝑣 =∫︁∫︁
𝐷
𝑓(𝑟(𝑢, 𝑣))·√︀
𝐸𝐺− 𝐹 2 𝑑𝑢𝑑𝑣
and we call the formal expression 𝑑𝑆 = ‖𝑟𝑢×𝑟𝑣‖ 𝑑𝑢𝑑𝑣 the surface
area element.
∙ a scalar surface integral reduces to a double integral over 𝐷
!
∙ the area of a surface patch 𝑟 : 𝐷 → 𝑉 can be obtained via the
formula:
Area(𝑉 ) =
∫︁∫︁𝑉
1 𝑑𝑆 =
∫︁∫︁𝐷
‖𝑟𝑢 × 𝑟𝑣‖ 𝑑𝑢𝑑𝑣
Solved Problems
Problema 1. Compute the mass of a cylindrical surface
parametrizedby:
𝑟(𝑢, 𝑣) = 𝑎 cos𝑢𝑖 + 𝑎 sin𝑢𝑗 + 𝑣𝑘,
where 0 ≤ 𝑢 ≤ 2𝜋, 0 ≤ 𝑣 ≤ 𝐻.The density in every point (𝑥, 𝑦, 𝑧)
is given by 𝜇(𝑥, 𝑦, 𝑧) = 𝑧2(𝑥2 + 𝑦2).
Solution: The mass of this cylindrical surface is given by the
scalar surfaceintegral:
𝑚 =
∫︁∫︁𝑆
𝜇(𝑥, 𝑦, 𝑧)𝑑𝑆,
where 𝑆 is the surface (side surface of the cylinder in this
figure) and 𝑑𝑆 is thesurface area element.
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The surface area element has the formula 𝑑𝑆 = ‖𝑟𝑢 × 𝑟𝑣‖𝑑𝑢𝑑𝑣 and
one canreduce the surface integral to the double integral:
𝑚 =
∫︁∫︁𝑆
𝜇(𝑥, 𝑦, 𝑧)𝑑𝑆 =
∫︁∫︁𝐷
𝜇(𝑟(𝑢, 𝑣)) · ‖𝑟𝑢 × 𝑟𝑣‖𝑑𝑢𝑑𝑣
In our case the set 𝐷 is the set of (𝑢, 𝑣), thus 𝐷 = [0, 2𝜋] ×
[0, 𝐻] (have a lookat the cylinder parametrization). First of
all:
𝑟𝑢 = −𝑎 sin𝑢𝑖 + 𝑎 cos𝑢𝑗 + 0𝐾.
and:𝑟𝑣 = 0𝑖 + 0𝑗 + 1𝑘
implies:
𝑟𝑢 × 𝑟𝑣 =
⃒⃒⃒⃒⃒⃒⃒⃒⃒
𝑖 𝑗 𝑘
−𝑎 sin𝑢 𝑎 cos𝑢 0
0 0 1
⃒⃒⃒⃒⃒⃒⃒⃒⃒ = 𝑎 cos𝑢𝑖 + 𝑎 sin𝑢𝑗
=⇒ ‖𝑟𝑢 × 𝑟𝑣‖ =√︀
(𝑎 cos𝑢)2 + (𝑎 sin𝑢)2 = 𝑎.
Hence 𝑑𝑆 = 𝑎 𝑑𝑢𝑑𝑣 and by its very definition:
𝑚 =
∫︁∫︁𝑆
𝜇(𝑥, 𝑦, 𝑧)𝑑𝑆 =
∫︁∫︁[0,2𝜋]×[0,𝐻]
𝜇(𝑟(𝑢, 𝑣))𝑎 𝑑𝑢𝑑𝑣
Formulas of 𝜇 and r=
∫︁∫︁[0,2𝜋]×[0,𝐻]
𝑣2((𝑎 cos𝑢)2 + (𝑎 sin𝑢)2)𝑎𝑑𝑢𝑑𝑣
𝐹𝑢𝑏𝑖𝑛𝑖=
∫︁ 2𝜋0
(︃∫︁ 𝐻0
𝑣2((𝑎 cos𝑢)2 + (𝑎 sin𝑢)2)𝑎𝑑𝑣
)︃𝑑𝑢
=
∫︁ 2𝜋0
(︃∫︁ 𝐻0
𝑣2𝑎2𝑎 𝑑𝑣
)︃𝑑𝑢 =
∫︁ 2𝜋0
𝑎3𝐻3
3𝑑𝑢 = 2𝜋𝑎3
𝐻3
3
9
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Bonus: Let us compute the area of the cylindrical surface. By
its verydefinition:
𝐴𝑟𝑒𝑎(𝑆) =
∫︁∫︁𝑆
1 𝑑𝑆 =
∫︁∫︁𝐷
1 · ‖𝑟𝑢 × 𝑟𝑣‖𝑑𝑢𝑑𝑣 =∫︁∫︁
[0,2𝜋]×[0,𝐻]
1 · 𝑎 𝑑𝑢𝑑𝑣
𝐹𝑢𝑏𝑖𝑛𝑖=
∫︁ 2𝜋0
(︃∫︁ 𝐻0
𝑎𝑑𝑣
)︃𝑑𝑢 =
∫︁ 2𝜋0
𝐻𝑎 𝑑𝑢 = 2𝜋𝑎𝐻
This formula is the famous 2𝜋𝑅𝐺. Here R=a is the base circle
radius and 𝐺 = 𝐻the length of the cylinder’s generatrix.
Problema 2. Compute the length of the curve obtained for:{︃𝑢 =
𝑡
𝑣 = 𝑡, 𝑡 ∈
[︁0,
𝜋
2
]︁on the surface 𝑥2 + 𝑦2 + 𝑧2 = 𝑅2.
Solution: The given equation is the implicit equation of a
sphere with center𝑂(0, 0, 0) and radius 𝑅. The general implicit
equation of a sphere is:
(𝑥− 𝑥𝑂)2 + (𝑦 − 𝑦𝑂)2 + (𝑧 − 𝑧𝑂)2 = 𝑅2
We consider now the spherical coordinates on the sphere. The
given curve islocated in the northern hemisphere because 𝑢 = 𝑡
∈
[︀0, 𝜋2
]︀. The northern hemi-
sphere is a parametric surface with the parametrization given by
the sphericalcoordinates:
𝑟(𝑢, 𝑣) = 𝑅sin𝑢 cos 𝑣 · 𝑖 + 𝑅sin𝑢 sin 𝑣 · 𝑗 + 𝑅cos𝑢 · 𝑘.
Remember: the whole sphere is not a parametric surface !The 3𝐷
curve obtained for 𝑢 = 𝑣 = 𝑡 is:
𝑐 :
⎧⎪⎨⎪⎩𝑥 = 𝑅 sin 𝑡 cos 𝑡
𝑦 = 𝑅 sin 𝑡 sin 𝑡
𝑧 = 𝑅 cos 𝑡
, 𝑡 ∈[︁0,
𝜋
2
]︁The curve starts at 𝐴(0, 0, 𝑅) = 𝐴(𝑡 = 0) and ends at 𝐵(0, 𝑅,
0) = 𝐵(𝑡 = 𝜋2 ).
In order to find its length one needs to compute the
coefficients of the firstfundamental form:
𝐸(𝑢, 𝑣) = 𝑟𝑢 · 𝑟𝑢 = 𝑅2(cos2 𝑢 cos2 𝑣 + sin2 𝑣 sin2 𝑢 + sin2 𝑢) =
𝑅2
𝐹 (𝑢, 𝑣) = 𝑟𝑢 · 𝑟𝑣 = 𝑅2 sin𝑢(− cos𝑢 cos 𝑣 sin 𝑣 + cos𝑢 sin 𝑣 cos
𝑣) = 0𝐺(𝑢, 𝑣) = 𝑟𝑣 · 𝑟𝑣 = 𝑅2 sin2 𝑢(sin2 𝑣 + cos2 𝑣) = 𝑅2 sin2
𝑢
Now the length of c is:
ℓ𝐴𝐵 =
∫︁ 𝜋2
0
√︀[�̇�(𝑡)]2𝐸(𝑢(𝑡), 𝑣(𝑡)) + 2�̇�(𝑡)�̇�(𝑡)𝐹 (𝑢(𝑡), 𝑣(𝑡)) +
[�̇�(𝑡)]2𝐺(𝑢(𝑡), 𝑣(𝑡)) 𝑑𝑡
=
∫︁ 𝜋2
0
√︁[𝑡′]2𝑅2 + 2(𝑡′)(𝑡′)0 + [𝑡′]2𝑅2 sin2 𝑡 𝑑𝑡
= 𝑅
∫︁ 𝜋2
0
√︀1 + sin2 𝑡 𝑑𝑡
10
-
and so forth... :))
Problema 3. Consider the triangle whose sides are the
curves:
𝑢 = 0, 𝑣 = 0, 𝑢 + 𝑣 = 1
lying on the surface:
𝑆 :
⎧⎪⎨⎪⎩𝑥 = 𝑢 cos 𝑣
𝑦 = 𝑢 sin 𝑣
𝑧 = 2 sin 2𝑣
, (𝑢, 𝑣) ∈ R× (0, 2𝜋)
Computer the perimeter and the measures of the angles
corresponding tothis triangle.
Solution: First of all we have to observe the following
parametrizations forthese three curves:
𝑐1 :
{︃𝑢 = 0
𝑣 = 𝑠, 𝑠 ∈ R
𝑐2 :
{︃𝑢 = 𝑡
𝑣 = 0, 𝑡 ∈ R
𝑐3 :
{︃𝑢 = 𝜏
𝑣 = 1 − 𝜏, 𝜏 ∈ R
In the parameter domain R× (0, 2𝜋) these curves will intersect
at the points𝐴(0, 1) = 𝑐1 ∩ 𝑐3, 𝐵(0, 0) = 𝑐1 ∩ 𝑐2 and 𝐶(1, 0) = 𝑐2
∩ 𝑐3.
The parametrization 𝑟 will attach to these points the points
𝐴′(0, 0, 2 sin 2),𝐵′(0, 0, 0) and 𝐶′(1, 0, 0) lying on the surface.
The triangle whose perimeterand angles we have to compute is ∆𝐴′𝐵′𝐶
′.
In order to apply the length formula we have to compute the
coefficients ofthe first fundamental form:
𝐸(𝑢, 𝑣) = 𝑟𝑢 · 𝑟𝑢 = (cos 𝑣, sin 𝑣, 0) · (cos 𝑣, sin 𝑣, 0) = cos2
𝑣 + sin2 𝑣 = 1
𝐹 (𝑢, 𝑣) = 𝑟𝑢 · 𝑟𝑣 = (cos 𝑣, sin 𝑣, 0) · (−𝑢 sin 𝑣, 𝑢 cos 𝑣, 4
cos(2𝑣)) = 0
𝐺(𝑢, 𝑣) = 𝑟𝑣·𝑟𝑣 = (−𝑢 sin 𝑣, 𝑢 cos 𝑣, 4 cos(2𝑣))·(−𝑢 sin 𝑣, 𝑢
cos 𝑣, 4 cos(2𝑣)) = 𝑢2 + 16 cos2(2𝑣)
11
-
For the length of 𝐴′𝐵′ we need the corresponding parameters,
with respectto the curve:
𝑟 ∘ 𝑐1 :
⎧⎪⎨⎪⎩𝑥 = 0
𝑦 = 0
𝑧 = 2 sin(2𝑠)
, 𝑠 ∈ R
We get easily 𝐴′(𝑠 = 1) and 𝐵′(𝑠 = 0), thus:
ℓ𝐴′𝐵′ =
∫︁ 10
√︀[�̇�(𝑠)]2𝐸(𝑢(𝑠), 𝑣(𝑠)) + 2�̇�(𝑠)�̇�(𝑠)𝐹 (𝑢(𝑠), 𝑣(𝑠)) +
[�̇�(𝑠)]2𝐺(𝑢(𝑠), 𝑣(𝑠)) 𝑑𝑠
𝑢(𝑠) = 0, 𝑣(𝑠) = 𝑠=
∫︁ 10
√︀0 · 𝐸(0, 𝑠) + 0 · 𝐹 (0, 𝑠) + 1 ·𝐺(0, 𝑠) 𝑑𝑠
=
∫︁ 10
√︀16 cos2(2𝑠) 𝑑𝑠
and so forth....For the length of 𝐵′𝐶 ′ we need the
corresponding parameters, with respect
to the curve:
𝑟 ∘ 𝑐2 :
⎧⎪⎨⎪⎩𝑥 = 𝑡
𝑦 = 0
𝑧 = 0
, 𝑡 ∈ R
We get easily 𝐵′(𝑡 = 1) and 𝐶 ′(𝑡 = 0), thus ....For the length
of 𝐴′𝐶 ′ we need the corresponding parameters, with respect
to the curve:
𝑟 ∘ 𝑐3 :
⎧⎪⎨⎪⎩𝑥 = 𝑝 cos(1 − 𝑝)𝑦 = 𝑝 sin(1 − 𝑝)𝑧 = 2 sin(2(1 − 𝑝))
, 𝑝 ∈ R
We get easily 𝐴′(𝑝 = 0) and 𝐶 ′(𝑝 = 1), thus ....To compute the
angle between the arcs 𝐵′𝐴′ and 𝐴′𝐶 ′ we need to find for
both curves the tangent vector at 𝐴′.On 𝑟 ∘ 𝑐1 the point 𝐴′
corresponds to the parameter 𝑠 = 1 thus the tangent
vector at 𝐴′ on 𝑟 ∘ 𝑐1 is:
(𝑟 ∘ 𝑐1)′(1) = (0, 0, 4 cos 2)
On 𝑟 ∘ 𝑐3 the point 𝐴′ corresponds to the parameter 𝑝 = 0 thus
the tangentvector at 𝐴′ on 𝑟 ∘ 𝑐3 is:
(𝑟 ∘ 𝑐3)′(1) = (cos 1, sin 1,−4 cos 2)
By its very definition the angle between the arcs is the angle
between thesetangent vectors:
cos 𝜃 =(0, 0, 4 cos 2) · (cos 1, sin 1,−4 cos 2)
‖(0, 0, 4 cos 2)‖ · ‖(cos 1, sin 1,−4 cos 2)‖≈ 0.85
Hence the angle 𝜃 ≈ arccos 0.85 ≈ 0.5∘!!!
12
-
Probleme propuse
Problema 1. Consider the surface:
𝑆 :
⎧⎪⎨⎪⎩𝑥 = sin𝑢 cos 𝑣
𝑦 = sin𝑢 sin 𝑣
𝑧 = cos𝑢
, (𝑢, 𝑣) ∈ (0, 2𝜋) × (0, 2𝜋)
i) Find the coefficients of the first fundamental form
ii) Compute the curvature and the torsion of the coordinate
curve 𝑐 : 𝑣 = 𝜋4
iii) Compute the measure of the angle between the tangent line
at 𝐴(𝑢0 = 0)to the curve c and the line:
𝑥− 21
=𝑦 + 1
0=
𝑧√2
Problema 2. Consider the surface:
𝑆 : 𝑟(𝑢, 𝑣) = 𝑒𝑢 cos 𝑣𝑖 + 𝑒−𝑢 sin 𝑣𝑗 + 𝑒2𝑢𝑘, (𝑢, 𝑣) ∈ R× (0,
2𝜋)
Find:
i) The equation of the tangent plane and tangent line at 𝑀(𝑢0 =
0, 𝑣0 =𝜋2 )
ii) The equation of the binormal line and the osculating plane
at 𝐴(𝑢0 = 0)belonging to the curve 𝑐 : 𝑣 = 𝜋4 lying on the surface
𝑆.
iii) The distance from the abovementioned binormal line to the
line:
𝑥− 21
=𝑦 + 1
0=
𝑧
2
Problema 3. Consider the surface:
𝑆 : 𝑟(𝑢, 𝑣) = 𝑢𝑖 + 𝑣𝑗 + sin(𝑢 + 𝑣)𝑘
i) Find the equation of the normal line and the tangent plane at
𝑂(0, 0, 0)and compute the angle between the normal line and the
unit vector 𝑘.
ii) Compute the area element of this surface and the area of the
surface patchobtained for (𝑢, 𝑣) ∈
[︀0, 𝜋2
]︀×[︀0, 𝜋2
]︀.
Problema 4. Sa se determine ecuatia sferei de raza 𝑅 = 2 cu
centrul pe dreapta
𝑑 :𝑥− 1
1=
𝑦 + 2
1=
𝑧 + 1
2
si care este tangenta la planul
𝛼 : 2𝑥− 2𝑦 + 𝑧 − 1 = 0.
Problema 5. Sa se determine ecuatia sferei care contine punctele
𝐴(1, 0, 0), 𝐵(0, 2, 0), 𝐶(0, 0,−1)si 𝐷(−1, 2, 1) si sa se determine
centrul si raza sa.
13
8 Suprafete. Sfera