Mathématiques : applications 2005 Université de Technologie de Belfort Montbéliard

Description

Examen du Supérieur Université de Technologie de Belfort Montbéliard. Sujet de Mathématiques : applications 2005. Retrouvez le corrigé Mathématiques : applications 2005 sur Bankexam.fr.

Sujets

• 2005

Informations

• Publié par: bankexam
• Publié le: 18 août 2008
• Nombre de lectures: 53
• Langue: Français

Extrait

1
xe +1
f(x) =
1
xe −1
n
g(x) = x cosx
(S) A
  
ax+y +z = 1 a 1 1  (S) x+ay +z = 1 A = 1 a 1 a∈R
x+y +az = 1 1 1 a
a = 1
A
(S)
a = 0
A
(S)
(S) a
V
 3 xz e −z y sinx
2f :R → R 
V(x, y, z) = f(x,z) 
(x,z) → f(x,z)
y2 x3z e +ycosx+ +2z y
z
f
V
f V ∇∧V = 0
∂f ∂f
(x,z) (x,z)
∂x ∂z
f (z)1
f(x,z) = zcosx+f (z)1
c1
2f (z) = lnz +z +c1 1
f ∇∧V = 0


f c = 01
∇∧V = 0
φ
∂φ
(x,y,z) = V (x,y,z),1
∂x
φ (y,z)1
3 xφ(x,y,z) = z e +z y cosx+φ (y,z)1
∂φ
(x,y,z) = V (x,y,z),2
∂y
φ (z)2
2φ (y,z) = y lnz +z y +φ (z).1 2
∂φ
(x,y,z) = V (x,y,z),3
∂z
c2
φ (z) = c .2 2
φ(x,y,z)









1
xe +1f(x) = 1
ex−1
a = 1
A  
1 1 1 A = 1 1 1
1 1 1
A L = L1 2
(S)
a = 0
A  
0 1 1 1 0 1A =
1 1 0
A detA = 2 = 0
(S)
 
ax+y +z = 1 ax+y +z = 1 
2(S) x+ay +z = 1 ⇔ (a −1)y +(a−1)z = a−1 
x+y +az = 1 (1−a)y +(a−1)z = 0
 ax+y +z = 1
⇔ (a−1)(a+1)y +(a−1)z = a−1
(1−a)(y−z) = 0
a = 1

x+y +z = 1
(S) 0 = 0
0 = 0
a = 1
 
ax+y +z = 1 ax+y +z = 1 
(S) (a+1)y +z = 1 ⇔ (a+1)y +z = 1 
y−z = 0 (a+2)y = 1
a =−2






a =−2 
1x = a+2
1y =
a+2 1z =
a+2
a = 1 (S)
a =−2 (S)
(S)
V
 
3 xz e −z y sinx
2f :R → R 
V(x, y, z) = f(x,z) 
(x,z) → f(x,z)
2 x y3z e +ycosx+ +2zy
z
∇∧V  
∂V ∂V3 2
(x,y,z)− (x,y,z))   ∂y ∂z  ∂f
1  cosx+ +2z− (x,z)
z   ∂z     ∂V ∂V 2 x 2 x1 3  ∇∧V (x,y,z) = =  3z e −y sinx−(3z e −ysinx)(x,y,z)− (x,y,z)   ∂z ∂x    ∂f  (x,z)+z sinx ∂V ∂V ∂x2 1
(x,y,z)− (x,y,z)
∂x ∂y
 
∂f
1cosx+ +2z− (x,z) z ∂z  ∇∧V (x,y,z) = 0  ∂f
(x,z)+z sinx
∂x
f V ∇∧V = 0
∇∧V = 0
∂f
1(x,z) = cosx+ +2z
z∂z
∂f
(x,z) = −zsinx
∂x
∂f
(x,z) =−zsinx⇒ f(x,z) = zcosx+f (z)1
∂x
∂f 1
(x,z) = cosx+ +2z
∂z z
f(x,z) = zcosx+f (z)1
1
′cosx+ +2z = cosx+f (z)1
z
1
′f (z) = +2z1
z
2f (z) = lnz +z +c1 1

f
2f(x,z) = zcosx+lnz +z +c1
2f(x,z) = zcosx+lnz +z
∇∧V = 0
∇φ =V
V1
∂φ
V (x,y,z) = (x,y,z)1
∂x
∂φ
3 x(x,y,z) = z e −z y sinx
∂x
x
3 x
φ(x,y,z) = z e +z y cosx+φ (y,z)1
∂φ 2(x,y,z) = V (x,y,z) = zcosx+lnz +z2
∂y
∂φ ∂φ1
(x,y,z) = zcosx+ (y,z)
∂y ∂y
∂φ ∂φ1 12 2zcosx+ (y,z) = zcosx+lnz +z ⇒ (y,z) = lnz +z
∂y ∂y
2φ (y,z) = ylnz +yz +φ (z)1 2
∂φ y
2 x(x,y,z) = V (x,y,z) = 3z e +ycosx+ +2z y3
∂z z
∂φ y y
2 ′ 2 x(x,y,z) = 3z +ycosx+ +2zy +φ (z) = 3z e +ycosx+ +2z y
2∂z z z
′φ (z) = 02
φ (z) = c2 2
φ
3 x 2φ(x,y,z) = z e +z y cosx+ylnz +yz +c2