”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

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

∙ 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.

3

https://en.wikipedia.org/wiki/Implicit_surfacehttps://academo.org/demos/3d-surface-plotter/https://en.wikipedia.org/wiki/Rotation_matrix

∙ 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

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

5

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

∙ 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)

7

https://en.wikipedia.org/wiki/Affine_transformation

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.

8

https://en.wikipedia.org/wiki/Surface_integral

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

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)

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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∘!!!

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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.

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8 Suprafete. Sfera