04 em 07 emcved uda1 s2 material de estudio

Page 1

A ⊂ ℝn k>0

x0 x0 Vk (x0

𝐴

Vx0

x0 ∈ ℝn

x ∈ ℝn A ⊂ ℝn

x

A

A

C

ℝ2 ℝ3

ℝn

⊆ ℝ2

{f(x, y) = z│(x, y) ∈ D ⊆ ℝ2 , z ∈ ℝ} : D ⊆ ℝ2 → ℝ

ℝ2 = {(x, y)│x ∈ ℝ, y ∈ ℝ}

f = {f(x1 , x2 , x3 , … , xn ) = z│(x1 , x2 , x3 , … , xn ) ∈ D ⊆ ℝn , z ∈ ℝ} n

f(x, y) =

𝑥 2 +1 𝑥+𝑦

𝑥2 + 1 ∀x ∈ ℝ

xy

𝑥+𝑦 =0

x≠y

x+y≠0

Df = {(x, y) ∈ ℝ2 │x ≠ y,

x ∈ ℝ,

Df = {(x, y) ∈ ℝ2 \x ≠ y }

y∈ℝ}

f(x, y)

xy

f

f(x, y) xy x=y

ℝ𝟑

ℝ3

{f(x, y) = z│(x, y) ∈ D ⊆ ℝ2 , z ∈ ℝ}

Gf = {(x, y, f(x, y)) ∈ ℝ3 │(x, y) ∈ D} f(x, y) = x 2 + y 2

xy ℝ2

f(x, y)

𝐷𝑓 = (−100, 0) × (0, 100)

{(x, y) ∈ D│(x, y) = c } ⊆ ℝ2

Pxy ∩ Gf = {(x, y, z)│y = 0, z = f(x, 0)} Pyz ∩ Gf = {(x, y, z)│x = 0, z = f(0, y)} Pxy

xy Pyz

yz

Gf

f

f(x, y) = z = x 2 − y 2 z=0

z=4

x=0

y=0

z = 0,4

z=0

𝑥2 − 𝑦2 = 0

𝑥2 = 𝑦2

x=y

z=0

z=4

x2 − y2 = 4 y = ±√ x 2 − 4 x = ±2

x = ±√ y 2 + 4

x≥2 y

y = ±2

z=4

x2 − y2 = c

xy

f(x, y) = x 2 − y 2 = c

y=0 xz

y=0

x=0 yz

x=0

ℝ3

f(x, y) = z ℝ3

ℝ2

(a, b) f(x, y)

L=

lim

(x,y)→(a,b)

f(x, y)

(a, b) δ>0 √(x − a)2 + (y − b)2 < δ

D ∈ ℝ2 ε>0

(x, y) ∈ D |f(x, y) − L| < ε

|f(x, y) − L| < ε z xy

𝐷𝛿 (a, b)

y2

f(x, y) = x2 +y2

y2

lim

(x,0)→(0,0)

lim

(0,y)→(0,0)

y2

x2 +y2

(0)2

= x2 +(0)2 = 0

y2

x2 +y2

= (0)2 +y2 = 1

f(x, y)

1

f(x, y) = y 2 sen (y2 +|x|)

f(x, y) (0,0) [−1,1]

−1 ≤ sen (

y2

1 )≤1 + |x|

y2

y2 ≥ 0

y −y 2 ≤ sen (

y2

1 ) ≤ y2 + |x|

(x, y) → (0,0)

lim

(x,y)→(0,0)

− y2 = 0

lim

(x,y)→(0,0)

−y 2 ≤

lim

(x,y)→(0,0)

lim

(x,y)→(0,0)

sen (

y2

y2 = 0

1 ) ≤ lim y 2 (x,y)→(0,0) + |x|

lim

1

(x,y)→(0,0)

sen (y2 +|x|) = 0 1

f(x, y) = y 2 sen (y2 +|x|)

f(x, y): → ℝ

∃

lim

(x,y)→(a,b)

g(x, y): D → ℝ (x, y) ∈ D

a, b, d, c ∈ ℝ

f(x, y) = b

lim

(x,y)→(a,b)

f(x, y) = b

c

lim

g(x, y) = d

lim

g(x, y) = b

(x,y)→(a,b)

lim

(x,y)→(a,b)

f(x, y) = cb

lim

(x,y)→(a,b)

f(x, y) +

g(x, y) = b + d

∃

lim

(x,y)→(a,b)

f(x, y) = a

(x,y)→(a,b)

lim

(x,y)→(a,b)

f(x, y)g(x, y) = ab

∃

lim

(x,y)→(a,b) 1

lim

(x,y)→(a,b) f(x,y)

f(x, y) = b

b, f(x, y) ≠ 0

(x, y)

1

=b

1

f(x, y) = 2 x 2 + 2y 2 + 1 1 2 x 2

+ 2y 2 + 1

f(x, y)

x(x) = x 2

lim

(x,y)→(a,b)

lim

x2 = 1 2

(x,y)→(1,0) 2

lim

(x,y)→(a,b)

x(

lim

(x,y)→(a,b)

x) = a2

y2

1

x =2 1 2 1 2 1 3 x + lim y + lim 1 = + 1 = (x,y)→(1,0) 2 (x,y)→(1,0) 2 (x,y)→(1,0) 2 2 lim

1 2 1 3 x + 2y 2 + 1 = lim (1)2 + 2(0)2 + 1 = (x,y)→(1,0) 2 2 2 lim

(x,y)→(1,0)

D (a, b) ∈ D

L = f(a, b) (a, b)

lim

(x,y)→(a,b)

f(x, y) = f(a, b)

đ?‘“(đ?‘Ľ, đ?‘Ś) =

x3 2x 2 − xy − y 2

1

y = −2x Âą 3 2

x

y = x + x2

x3 lim f(x, y) = lim (x,y)→(0,0) (x,x+x2 )→(0,0) 2x 2 − x(x + đ?‘Ľ 2 ) − (x + đ?‘Ľ 2 )2

x3 1 1 = lim − = 2 2 2 2 2 )→(0,0) 2x − x(x + đ?‘Ľ ) − (x + đ?‘Ľ ) (x,x+x )→(0,0) đ?‘Ľ+3 3

lim 2

(x,x+x

y = 3x

x3 x lim = lim − =0 (x,2x)→(0,0) 2x 2 − x(3x) − (3x)2 (x,3x)→(0,0) 9đ?‘Ľ 4 + 1

lim

(x,y)→(0,0)

f(x, y) = f(0,0)

(a, b)

đ?œ€âˆ’đ?›ż

f(x, y): → �

g(x, y): D → � (x, y) ∈ D

(x, y)

f

f

(cf)(x, y) = c[f(x)]

(x, y)

g

f

c∈�

(f + g)(x, y) = f(x, y) + g(x, y)

(x, y)

g

(fg)(x, y) = f(x, y)g(x, y)

f(x, y): D ⊆ �2 → �

1

( f ) (x, y) = f(x)

g(x0 , y0 ): B ⊆ �2 → � (f ∘ g)(x, y) = f(g(x0 , y0 ))

g(B) ⊂ D g

1

(x, y)

f≠0

g

(x0 , y0 ) ∈ đ??ľ

(f ∘ g)(x, y)

f

(x, y) = g(x0 , y0 )

(x0 , y0 )

f(x, y) = (x 4 + y 4 )8 + cos(x)

f f â„?2

z = x4 + y4

� 8 = (x 4 + y 4 )8

z(x, y) = x 4 + y 4

w = cos(x)

â„?2

f

(x0 , y0 ) ∈ �2

lim

(x,y)→(x0 ,y0 )

lim

(x,y)→(a,b)

f(x, y) = f(a, b)

(x 4 + y 4 )8 â„?

x 4 + y 4 = x0 4 + y0 4

cos(x) f(x, y)

𝛾(t): T → ℝ3 (𝑥(t), y(t), z(t)) ∈ ℝ3

t∈T⊆ℝ 𝑥(t), y(t)

z(t)

ℝ

𝛾(t): t → (x(𝑡), y(t), z(t))

t∈ℝ

(x(𝑡), y(t), z(t)) ∈ ℝ3

𝛾(t): x(t)𝐢 − y(t)𝐣 + z(t)𝐤

𝛾(t): ℝn → ℝm

𝛾: 𝐒 ∈ ℝn → (x1 (𝑡), x2 (t), x3 (t), … , xm (t)) ∈ ℝm

𝛾(t) = (

x(t) =

t+1

t=0

t

t+1 t

, t, t 2 − 1)

Dom𝛾 =

{t ∈ ℝ│t ≠ 0}

𝛾(t) = (t + 1, 5t + 1, 3t − 1)

x=t+1 y= 5t + 1

z = 3t − 1

𝑡=0

(1, 1, −1)

(1, 5, 3)

𝐢 𝐣 𝐤 |1 1 −1| = (3 + 5)𝐢 − (3 + 1)𝐣 + (5 − 1)𝐤 = 0 1 5 3

𝛾(t) = (1, 1, −1) + t(1, 5, 3)

𝛾(t)

𝑥2 + 𝑦2 − 1 = 0

𝑧 = 2−𝑦

xy

x y 2đ?œ‹

z

x = cost y = sent

z = 2 − sent x = cost y = sent

0 ≤ t ≤ 2π

z

đ?›ž(t) = costđ??˘ + sentđ??Ł + (2 − sent)đ??¤

0 ≤ t ≤ 2π

0≤�≤

x

y

�(t): t → (x(�), y(t)) �(t): t → (x(�), y(t), z(t))

Îł(t) = (3cost, 3sent)

Îł x(đ?‘Ą) = 3đ?‘?đ?‘œđ?‘ đ?‘Ą

y(t) = 3sent

x 2 (đ?‘Ą) = 9cos2 đ?‘Ą y 2 (đ?‘Ą) = 9sen2 đ?‘Ą

x 2 (đ?‘Ą) + y 2 (đ?‘Ą) = 9cos 2 đ?‘Ą + 9sen2 đ?‘Ą

Îł(t) = (3cost, 3sent)

Îł(t)

lim �(t) t→a

1

Es una plataforma computacional online dedicado a la bĂşsqueda de respuestas, entre las que se encuentran los cĂĄlculos matemĂĄticos.

lim đ?›ž(x(t), y(t), z(t)) = (lim x(t) đ??˘ + lim y(t) đ??Ł + lim z(t) đ??¤) = (L1 , L2 , L3 ) t→a

t→a

t→a

t→a

đ?‘’ đ?‘Ą +1

1

đ?›ž(t) = (đ?‘’ đ?‘Ą −1 , 1+đ?‘Ą , √1 + đ?‘Ą)

đ?‘’ đ?‘Ą +1

1

x(t) = đ?‘’ đ?‘Ą −1 y(t) = 1+đ?‘Ą z(t) = √1 + đ?‘Ą

lim

1

�→0 1+�

lim

=1

lim √1 + đ?‘Ą = 1

�→0

đ?‘’ đ?‘Ą +1

đ?‘Ąâ†’0 đ?‘’ đ?‘Ą −1

�(t)

lim đ?›ž(t) = lim x(a) đ??˘ + lim y(a) đ??Ł + lim z(a) đ??¤ = (L1 , L2 , L3 ) t→a

t→a

t→a

t→a

�(t) = ( 3

đ?‘Ą2

√đ?‘Ą 2 +1

x(t) =

đ?‘Ą2 3

√đ?‘Ą 2 +1

, eđ?‘Ą ,

t−1 đ?‘Ą4

)

t−1

y(t) = eđ?‘Ą z(t) = đ?‘Ą 4 +2

02 0−1 1 lim đ?›ž(t) = lim 3 đ??˘ + lim e0 đ??Ł + lim 4 đ??¤ = (0,1, − ) t→0 t→0 √02 + 1 t→0 t→a 0 + 2 2

�→0

𝑡→0

𝑡→

0

𝛾(t): ℝ → ℝ3 𝛾(t): ℝ → ℝm

𝛾(𝐭): ℝn → ℝm

f(x, y): ℝ2 → ℝ (x0 , y0 ) ∈ ℝ2

y

(x0 , y0 ) ∈ ℝ2 f(x0 + h, y0 ) − f(x0 , y0 ) h→0 h

fx (x0 , y0 ) = lim

f(x0 , y0 + h) − f(x0 , y0 ) h→0 h

fy (x0 , y0 ) = lim

f(x, y) = 2x 2 − y + 4

fx (0,1)

fy (0,1)

f(0 + h, 1) − f(0,1) f(h, 1) − f(0,1) = lim = lim 2h = 0 h→0 h→0 h→0 h h f(0,1 + h) − f(0,1) f(0,1 + h) − f(0,1) fy (0,1) = lim = lim = lim −1 = −1 h→0 h→0 h→0 h h fx (0,1) = lim

fx (0,1) = 0

fy (0,1) = −1

x

∂f ∂x

(x0 , y0 , z0 )

f(x, y) = z ∂f

∂f

∂y

∂x

𝜕𝑓

𝜕

𝑓𝑥 (𝑥, 𝑦) = 𝜕𝑥 = 𝜕𝑥 𝑓(𝑥, 𝑦) 𝜕𝑓

𝜕

𝑓𝑦 (𝑥, 𝑦) = 𝜕𝑦 = 𝜕𝑦 𝑓(𝑥, 𝑦)

z = f(x, y) ∂z ∂x

∂

= ∂x f(x, y)

∂f

∂

∂x

∂y

∂z ∂y

∂

= ∂y f(x, y)

xy2

f(x, y) = x+y

x+y>0 u = xy 2 v = x + y ≠ 0 y

x

∂f u´v − uv y 2 (x + y) − xy 2 y3 = = = (x + y)2 (x + y)2 ∂x v2

y

x

∂f u´v − uv x2y(x + y) − xy 2 (2x − y)(xy) = = = (x + y)2 (x + y)2 ∂y v2

f(x, y)

∂f

∂f

∂x

∂y

(x0 , y0 , z0 )

𝑧 = f(x, y)

z − z0 = fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 )

𝑧 = f(x, y) = 3𝑥 2 + 6𝑦 2

(2,2,6) f(x, y)

fx (x, y) = 6𝑥 fy (2,2) = 12

fy (x, y) = 12𝑦 fy (2,2) = 24

z − 6 = 12(x − 2) + 24(y − 2) z = 12x + 24y − 66

f(x, y): ℝ2 → ℝ

∇f(x, y) =

∂f ∂f 𝐢 + 𝐣 = 〈fx (x, y), fy (x, y)〉 ∂x ∂y

f(x, y) (x0 , y0 , z0 ) c

��

f

f(x, y) = √x 2 + y 2

f ∂f

= ∂x

x √x2 +y2

∂f

= ∂y

y

∇f(x, y) =

√x2 +y2

x √x2 +y2

đ??˘+

y √x2 +y2

f(x, y): â„?2 → â„? đ??Ž = ⌊a, bâŒŞ

f(x0 + ha, y0 + hb) − f(x0 , y0 ) h→0 h

Du f(x0 , y0 ) = lim

đ??Ł

(x0 , y0 )

Du f(x0 , y0 ) = ∇f ∙ 𝐮 = fx (x, y)x0 + fy (x, y)y0

𝐮

C

𝑧 = f(x, y) 𝑃 = (x0 , y0 , z0 𝐮

T

C

x0 , y0 , z0 f

𝐮

𝑃′ , 𝑄′

𝑄∈𝐶

𝑃, 𝑄 𝑃′𝑄 ′ 𝐮

f(x, y) Du f(x0 , y0 ) = fx (x, y)x0 + fy (x, y)y0

𝐮 = (x0 , y0 )

f(x, y) = x 2 y 4 + 12x (2,2)

v = (3,2)

∇f(x, y) = (fx , fy ) = (2xy 4 + 12, 4x 2 y 3 ) ∇f(1,2) = (fx , fy ) = (2(1)(2)4 + 12, 4(1)2 (2)3 ) = (48,32)

v = (3,2) v

𝐮 = ‖v‖ =

1 √13

3

(3,2) = (

,

2

)

√13 √13

Du f(x0 , y0 ) = fx (x, y)x0 + fy (x, y)y0

D(3,2) f(48,32) = ∇f ∙ v = (76,128) ∙ ( D(3,2) f(48,32) =

3

,

2

√13 √13 400 √13

)=

144 √13

+

256 √13

𝑧 = f(x, y): D ⊆ ℝ2 → ℝ

lim

(x0 , y0 )

f(x, y) − f(x0 , y0 ) − (fx (x0 , y0 ) ∗ (x − x0 )) − (fy (x0 , y0 ) ∗ (y − y0 ))

(x,y)→(x0 ,y0 )

√(x − x0 )2 + (y − y0 )2

dz =

=0

∂z ∂z dx + dy = fx (x, y)dx + fy (x, y)dy ∂x ∂y dx = ∆x

𝑧 = f(x, y) = 𝑥 4 + 𝑥𝑦 − 2𝑦 4

x

y=2

fx (x, y)

fy (x, y)

fx (x, y) = 4𝑥 3 + 𝑦 fy (x, y) = 𝑥 − 8𝑦 3 dz = (4𝑥 3 + 𝑦)𝑑𝑥 + (𝑥 − 8𝑦 3 )𝑑𝑦 x=2

y = 2 dx = 1

dy = 0.5

dz

dz = (4(2)3 + 2)(1) + (2 − 8(2)3 )(0.5) = 34 − 31 = 3 dz = 3

f(x, y): ℝ2 → ℝ ∂f

∂f

∂x

∂y

(x0 , y0 ) (x0 , y0 )

f(x, y): ℝ2 → ℝ

(x0 , y0 )

f(x, y) = exy + xy

(0,1)

(0,1) ∂f = yexy + y = y(exy + 1) ∂x ∂f = xexy + x = x(exy + 1) ∂y

∂f ∂x ∂f ∂y

= g(x, y) = y(exy + 1)

= h(x, y) = x(exy + 1) (0,1) lim

g(x, y) =

lim

h(x, y) =

(x,y)→(0,1) (x,y)→(0,1)

lim

y(exy + 1) = 2 = g(0,1)

(0,1)

lim

x(exy + 1) = 0 = h(0,1)

(0,1)

(x,y)→(0,1) (x,y)→(0,1)

f(x, y) (0,1)

f(x, y): ℝ2 → ℝ2

∂f1 ∂f1 (x0 , y0 ) (x , y ) ∂x ∂y 0 0 Df(x0 , y0 ) = ∂f2 ∂f2 (x0 , y0 ) (x0 , y0 ) ∂x ∂y ( )

(x0 , y0 ) ∈ ℝ2

f(x, y): ℝ2 → ℝ ∂f

∂f

Df(x0 , y0 ) = ∂x + ∂y

f(x, y) = 5ex y

(1,1)

g(x, y) = (5ex y, xy)

Df(1,1) =

∂f ∂f (1,1) + (1,1) = 5e + 5e = 10e ∂x ∂y

∂f1 ∂f1 (1,1) (1,1) ∂x ∂y 5e Dg(1,1) = =( ∂f2 ∂f2 1 (1,1) (1,1) ∂y ( ∂x )

f: ℝ2 → ℝ2

5e ) 1

f: ℝn → ℝm

f(x, y): ℝ2 → ℝ2 ℝ2

f: (x, y) → (g(x, y), h(x, y))

Df

𝑓1 (𝑥, 𝑦) = 𝑒 𝑥 y 𝑓2 (𝑥, 𝑦) = 2xcosy Df(x, y)

f(x, y) = (𝑒 𝑥 y, 2xcosy)

∂f1

Df(x, y) =

∂x (∂f 2 ∂x

∂f1 ∂x ∂f2 ∂x

= đ?‘’đ?‘Ľy

∂f2 ∂y

(x, y) = đ?‘’ đ?‘Ľ y

(x, y) = 2cosy

∂f1

(x, y) = đ?‘’ đ?‘Ľ

đ?‘’đ?‘Ľy ) Df(x, y) = ( ∂f2 2cosy (x, y) = −2xseny ∂y ∂y

= đ?‘’đ?‘Ľ ) −2xseny

∂f1

â„?2

= −2xseny

= 2cosy

∂y

â„?

f(x, y): �2 → �2

g(x, y): �2 → �2

f(x, y): �2 → � g(x, y): �2 → �

h(x, y): �2 → � ≠0

f

f

D(f + g)(x0 , y0 ) = Df(x0 , y0 ) + Dg(x0 , y0 )

D(hf)(x0 , y0 ) = h(x0 , y0 )Df(x0 , y0 ) + f(x0 , y0 )Dh(x0 , y0 )

f

D (h) (x0 , y0 ) =

h(x0 ,y0 )Df(x0 ,y0 )−f(x0 ,y0 )Dh(x0 ,y0 ) ℎ2 (x0 ,y0 )

z = f(x, y) = x(g(t)) + y(h(t))

g(t) h(t)

t

z

t

= đ?‘’đ?‘Ľ

∂z ∂f dx ∂f dy = + ∂t ∂x dt ∂y dt

z = f(x, y) = đ?‘Ľđ?‘Ś 2 + 4đ?‘Ľ 4 đ?‘Ś

g(t) = 2cost

h(t) = t +

1 z = f(x, y)

g(t)

h(t) t0 ∈ �

lim 2đ?‘?đ?‘œđ?‘ đ?‘Ą = 2cos t 0 lim đ?‘Ą + 1 = t 0 + 1 t0

t0

z = f(g(t), h(t)) = (2đ?‘?đ?‘œđ?‘ đ?‘Ą)đ?‘Ś(đ?‘Ą + 1)2 + 4(2đ?‘?đ?‘œđ?‘ đ?‘Ą)4 (đ?‘Ą + 1)

∂z ∂t

= (đ?‘Ś 2 + 12đ?‘Ľ 3 đ?‘Ś)(−2sent) + (2xy) = −2sent(đ?‘Ą + 1)[(đ?‘Ą + 1) + (2đ?‘?đ?‘œđ?‘ đ?‘Ą)3 ] + 4(đ?‘?đ?‘œđ?‘ đ?‘Ą)(đ?‘Ą + 1)

z = f(x, y) = đ?‘Ľ(g(s, t)) + y(h(s, t))

g(s, t) h(s, t)

z

t ∂z ∂s

∂z ∂x

∂z ∂y

= ∂x ∂s + ∂y ∂s

∂z ∂t

∂z ∂x

∂z ∂y

= ∂x ∂t + ∂y ∂t

z ∂z

∂z

∂x

∂y

x

y

∂x

∂x

∂y

∂y

∂s

∂t

∂s

∂t

s

t

s

t

f: (x, y) = 5𝑥 2 + 2𝑦

f: (g(t), h(t))

g(t) =

𝑐𝑜𝑠𝑡 h(t) = 2𝑒 𝑡 + 1

∂z ∂f dx ∂f dy = + = (10𝑥)(−𝑠𝑒𝑛𝑡) + (2)(2𝑒 𝑡 ) ∂t ∂x dt ∂y dt ∂z = −10𝑐𝑜𝑠𝑡(𝑠𝑒𝑛𝑡) + 4𝑒 𝑡 ∂t g(x, y): D ⊆ ℝ2 → ℝ2

(f ∘ g) (𝑥0 , 𝑦0 ) ∈ 𝐴 𝑔(𝑥0 , 𝑦0 )

(f ∘ g)

f(x, y): A ⊆ ℝ2 → ℝ2

(𝑠0 , 𝑡0 ) =

(𝑥0 , 𝑦0 )

𝐷(f ∘ g)(𝑥0 , 𝑦0 ) = 𝐷𝑓(𝑔(𝑥0 , 𝑦0 ))𝐷𝑔(𝑥0 , 𝑦0 )

f∘g f: (x, y) = (senx + 𝑦 2 , 2xy)

g: (x, y) = (t𝑒 𝑠 , 𝑠 − 𝑐𝑜𝑠𝑡)

(f ∘ g)(s, t) = f(e2s , s − cost) = (sen(es ) + (s − cost)2 , 2es (s − cost)) 𝐷(𝑓 ∘ 𝑔)(0,0) = 𝐷𝑓(𝑔(0,0))𝐷𝑔(0,0)

Dg(0,0) = (

tes 1

g(0,0) = (0, −1) Df(0, −1) = (

0 1 es ) =( ) 1 0 sent (s,t)=(0,0)

cosx 2y 1 −2 ) =( ) 2y 2x (x,y)=(0,0) −2 0

1 −2 0 1 −2 1 )( )=( ) −2 0 1 0 0 −2

đ??ˇ(f ∘ g)(0,0) = (

đ??š=0

F(x, y): D ⊆ �2 → � F(x, y) = 0 x

y = f(x) F(x, f(x)) = 0

x ∈ Domf

F

F(x, y) = 0 ∂F ∂y Fx = − ∂x = − ∂F ∂x Fy ∂y ∂F ∂y

≠0

f(x): � → �

F(x, y): D ⊆ �3 → � F(x, y, z) = 0

z

� = f(x, y)

F(x, y, f(x, y)) = 0

(x, y) ∈ Domf

F

f

F(x, y, z) = 0 ∂z ∂x

∂F ∂z

≠0

=−

f(x): : �2 → �

∂F ∂x ∂F ∂z

∂z ∂y

=−

∂F ∂y ∂F ∂z

x 2 + y 2 + z 2 + kxyz = 5

𝐹(𝑥, 𝑦, 𝑧) = x 2 + y 2 + z 2 + xyz − 5

z ∂z ∂x

=−

∂F ∂x ∂F ∂z

2𝑥+𝑦𝑧

= 2𝑧+𝑥𝑦

x ∂z ∂x

=−

∂F ∂y ∂F ∂z

2𝑦+𝑥𝑧

= 2𝑧+𝑥𝑦

f(𝐱): ℝn → ℝm

№ О(t): т т т 3

t0

№ О(№ Ё + т ) т № О(№ Ё) =0 hт 0 т

№ О т В (t) = lim

№ О т В (t) =

d№ О dt

№ О(t) = (x(t), y(t), z(t))

№ О(t) = (x(t), y(t), z(t)) = x(t)№ Ђ + y(t)№ Ѓ + z(t)№ Є

x(t) y(t)

z(t)

№ О т В (t) = (x т В (t), y т В (t), z т В (t)) = xт В(t)№ Ђ + yт В(t)№ Ѓ + zт В(t)№ Є

№ О(t) = (sen2t, № Ё 2 , t + 6) № О т В (t) = (cost, 2t, 1)

№ О(t)

№ (t)

(№ О(t) + № (t) )т В = № От В(t) + № т В(t) (k№ О(t) )т В = k№ От В(t) (f(t)№ (t) )т В = f т В (t)№ (t) + f(t)№ т В(t) (№ О(t) т № (t) )т В = № От В(t) т № (t) + № О(t) т № т В(t) (№ О(t) У № (t) )т В = № От В(t) У № (t) + № О(t) У № т В(t) (№ О(f(t) )т В = f т В (t)№ т В(f(t))

kт т

f(t): т т т

đ?›ž(đ?‘Ą): [đ?‘Ž, đ?‘?] → â„?2

đ?›ž(đ?‘Ą): [đ?‘Ž, đ?‘?] → â„?3

C1

b

âˆŤ √(x′(t))2 + (y′(t))2 dt,

b

si su rango esta en â„?2

a

â„’ = âˆŤâ€–Îłâ€˛(t)‖ dt =

b

a

{

si su rango esta en â„?3

âˆŤ √(x′(t))2 + (y′(t))2 + (z′(t))2 dt, a

C1

Cn

Dâ„’ = √(x′(t))2 + (y′(t))2 + (z′(t))2 dt

Dâ„’

đ?‘Ľđ?‘–

đ?‘Ľđ?‘—

đ?‘–<đ?‘—

𝜋 𝛾(t) = (sen2t, cos2t + 1 )

𝛾 ′ (t)

𝛾 ′ (t) = (2cos2t, −2sen2t) 𝜋

‖γ′(t)‖ = ∫

𝜋

√(2cos2t)2

0

+

(−2sen2t)2 dt

= ∫ √4dt = 2𝜋 0

𝐹: 𝐴 ⊆ ℝ3 → ℝ3 (x, y, z ) ∈ 𝐴

𝐹(𝑥, 𝑦, 𝑧) = (𝐹1 , 𝐹2 , 𝐹3 ) ℝ3 → ℝ

𝐹: 𝐴 ⊆ ℝ2 → ℝ2

(x, y ) ∈ 𝐴 𝐹(𝑥, 𝑦) =

(𝐹1 , 𝐹2 )

ℝ2 → ℝ

∇F(x, y, z) =

∂F ∂F ∂F (x, y, z)𝐢 + (x, y, z)𝐣 + (x, y, z)𝐤 ∂x ∂y ∂z

∇F(x, y) =

∂F ∂F (x, y, z)𝐢 + (x, y, z)𝐣 ∂x ∂y

𝑧= F(x, y)

𝑛𝑖 ∈ ℝ

(xi , yi )

F

𝛾(t)

𝛾 ′ (t) = 𝐹(𝛾(t)) 𝛾(t) = (𝑠𝑒𝑛𝑡, 𝑐𝑜𝑠𝑡) F(x, y) = (y, −x) 𝛾 ′ (t)

𝛾 ′ (t) = (𝑐𝑜𝑠𝑡, −𝑠𝑒𝑛𝑡) 𝐹(𝛾(t)) = (𝑐𝑜𝑠𝑡, −𝑠𝑒𝑛𝑡)

𝐹(𝛾(t))

𝛾 ′ (t) = 𝐹(𝛾(t))

𝛾(t)

divF(x, y, z) = ∇ ∙ F =

F(x, y)

∂𝐹1 ∂𝐹2 ∂𝐹3 (x, y, z) + (x, y, z) + (x, y, z) ∂x ∂y ∂z

divF(x, y) = ∇ ∙ F =

∂𝐹1 ∂𝐹2 (x, y, z) + (x, y, z) ∂x ∂y

F(x, y, z) = (x, y, z)𝐢 + 𝐹2 (x, y, z)𝐣 + 𝐹3 (x, y, z)𝐤

𝐢 𝐣 𝐤 ∂ ∂ ∂ ∇×F =| | ∂x ∂y ∂z F1 F2 F3 𝐢 ∂

F(x, y) = (x, y)𝐢 + F2 (x, y)𝐣

𝐣

𝐤

Rot F = |∂x

∂

∂

∂y

∂z

P

Q

R

|

∂F

∇ × F = ( ∂x2 − 𝐤

F(x, y, z) = (2xz, y 2 z, yxz)

∂F1 ∂y

)𝐤

F(x, y, z) ∇F(x, y, z) =

∂F ∂F ∂F (x, y, z)𝐢 + (x, y, z)𝐣 + (x, y, z)𝐤 ∂x ∂y ∂z

∇F(x, y, z) = (2z + yz)𝐢 + (2yz + yx)𝐣 + (2x + y 2 + xy)𝐤

div F(x, y, z) =

∂𝐹1 ∂𝐹2 ∂𝐹3 (x, y, z) + (x, y, z) + (x, y, z) ∂x ∂y ∂z

div F(x, y, z) = 2z + 2yz + yx

𝐢 ∂ ∇ × F = || ∂x 2xz

𝐣 𝐤 ∂ ∂ || = (xz − y 2 )𝐢 + (2x − yz)𝐣 + (0)𝐤 ∂y ∂z y 2 z xy

Turn static files into dynamic content formats.

Create a flipbook
Issuu converts static files into: digital portfolios, online yearbooks, online catalogs, digital photo albums and more. Sign up and create your flipbook.