Puesto que el Calculo Vectorial tiene gran aplicación en las áreas de la Ingeniería Mecánica, Industrial, Eléctrica, Electrónica y en la ciencias como la Física, Química , etc.

El presente trabajo es una serie de ejercicios resueltos de un selecto grupo de temas de calculo vectorial que tiene como propósito primordial contribuir a la mejora de la enseñanza de Calculo Vectorial

Al igual que otras asignaturas de Matemáticas, el Calculo Vectorial se aprende resolviendo ejercicios, por lo que se ha tenido el cuidado de seleccionar un gran numero de ellos.

Esta serie puede servir como practica adicional para aquellos quienes estén interesados en las áreas antes mencionadas, ya que se pretende que la misma sea un auxiliar didáctico y convertirse en una colaborador mas de la tarea docente y del aprendizaje

1- f (t) = 2i + 4 j – 2 k y que pasa por ( -2, 0. 4)

para

(-2, 0, 4) x = xo + tA

Ai + Bj + Ck = 2i + 4 j – 2 k y= yo + tB

Z= zo + tC

x =-2 + 2 t

y = 4t

z = 4 – 2 t

2- f (t) = 4i -3j +7k y que pasa por ( -3, 3 , -3)

x= -3 + 4t

y = 2 – 3t

z = -3 + 7 t

3- f (t) = 4i -3j +7k y que pasa por ( 1,-1, 4)

x= 1 + 4t

y = -1 – 3t

z = 4 + 7t

4- Ai+ Bj + Ck

P ( -3, 3, -3) Q (1 , -1, 4)

x= -3 + 4t

y = 2– 3t

z = -3+7t

5- f (t) = a sent i + b cos t j

x= a sen t

y= b cost

z= 0

L(p)= (4 (sen2t)² + 4 (cos2t)² + 5 )^1/2 = 9^1/2 = 3

L (p) = 

2 – f (t) = (1- cost, sent)

f (t) =

3- La longitud del arco de la curva y=f(x) entre a y b es: 

4 - y=ln(1-x²) en [1/3, 2/3].

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5- ( t, t -1/2, 0) en (-1,1) en R³

-1= t_o < t₁ < ½= t2 < 1 = t₃

(-1,0), (0,1/2) y (1/2,1)

(-1,0)

x(t) = -t

y (t)= -t

z (t) = 0

ds = 2^1/2

dt

(0,1/2)

x(t) = t

y (t)= -t + 1/2

z (t) = 0

ds = ( 2^1/2) 1/2

dt

(1/2, 1)

x(t) = t

y (t)= t – 1/2

z (t) = 0

= 2 (2) ^1/2

1. Si Ø : t ( cost, sent, t)

v (t)= Ø’(t)=

v=(-sent, cost, 1)

S (t) = v(t) = ( sen²t+cos²t + 1)^1/2 = 2 ^½

rapidez =

2. Considerer la patricula movimeindose donde t= Π y hallar la ubicacion en 2 Π

Si Ø : t ( cost, sent, t)

v(Π) = Ψ’(Π)

c(Π)= Ψ= (-1, 0, Π)

w+tv(Π) = w + t ( 0, -1, 1)

c(Π)= w + Π ( 0, -1, 1) = Ψ= (-1, 0, Π)

w= (-1, 0, Π)- ( 0- Π- Π) = (-1, Π,0) + t ( 0, -1, 1)

c (2Π)= ( -1, Π,0)+2 Π (0, -1, 1)= ( -1, - Π , 2 Π)

3. r (t)= 6t² i – t³j + t² k

x(t)i + y (t)j + z(t)k

v= x’(t) i + y’ (t) j + z’ (t) k

a= x’’(t) i + y’’ (t) j + z’’ (t) k

v= 12 t i – 3 t² j + 2t k

a= 12 i – 6t j + 2 k

4. r (t) = (3cost)i + (3sent)j + t² k

v= x’(t) i + y’ (t) j + z’ (t) k

a= x’’(t) i + y’’ (t) j + z’’ (t) k

v= -(3sent)i + (3cost) j + (2t)k = -3sent i + 3cost j + 2t k

a = -(3cost)i -3(sent)j + (2) k = -3 cost i - 3 sent j + 2 k

5- r (t) = (sent) i+ (cost) j + 3^1/2 k

v= x’(t) i + y’ (t) j + z’ (t) k

a= x’’(t) i + y’’ (t) j + z’’ (t) k

v= (cos t)i - (sen t) j + (o)k = (cos t)i - (sen t) j

a= -(sent) i - ( cost) j + o = -sen t i - cost j

1- g(x,y)= xy, 5x, y³ f(x,y)= 3x²+y²+z², 5xyz

J(fog)= d (fog)₁ d (fog)₁ = Jf (g(x,y)) Jg(x,y) =

dx dy

d (fog)₂ d (fog)₂

dx dy

f₁= 3x²+y²+z² , f₂= 5xyz , g₁= xy, g₂= 5x , g₃= y³

6x 2y 2z y x

5xy 5xz 5xy 5 0

0 3y²

sustituyendo las variables y multiplicando por la matriz obtengo:

= 6xy² + 50x 6x²y+ 6y⁵

50xy⁴ 100x²y³

2- f(x,y)= x²+3y² f₂ (x,y)= 5x³+2y⁶

Jf= df₁ df₁ = 2x 6y

dx dy 15x² 12y⁵

df₂ df₂

dx dy

3- f(x,y,z)= x²+2, x+y²+z³ g(x,y,z)= x+y+z, xyz, x²+y³ en P(1,1,1)

J(fog) (1,1,1) = Jf (g(1,1,1)) Jg(1,1,1) = Jf(3,1,2) Jg(1,1,1)

Jf(3,1,2)= 2x 0 0 = 6 0 0

1 2y 3x² 1 2 12

Jg(1,1,1) = 1 1 1 = 1 1 1

yz xz xy 1 1 1

2x 3y² 0 2 3 0

= 6 0 0 1 1 1 = 6 6 6

1 2 12 1 1 1 27 39 3

2 3 0

4- f(x,y)= sen (x+y) f₂(x,y)= xe^x+y f₃(x,y)= x+y en P(0,0)

Jf(0,0)= df₁ df₁ = cos(x+y) cos(x+y) ₌ 1 1

dx dy e^x+y(x+1) xe^x+y 1 0

df₂ df₂ 1 1 1 1

dx dy

df₃ df₃

x dy

5- f(x) = sen (x²) f(g) = sen(g)

D_x[sen(x²)] = D_g[sen(g)]·D_x[g(x)] =

cos(g)·2x =cos(x²)·2x =

2xcos(x²)

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1- f(x,y,z) = 2x³+7y²+9z² v = (a,b,c)

x = x_o+ as

y = y_o+bs

z = z_o+cs

gs = 2 (x_o+ as)³+7(y_o+bs)²+9(z_o+cs)²

= 2 (x_o³+3x²as+3x_oa²s²+a³s³)+7(y_o²+2 y_obs+b²s²)+9(z_o²+2 z_ocs+c²s²)

= 2 x_o³+6 x²as+6 x_oa²s²+2 a³s³+7 y_o²+14y_obs+7 b²s²+9 z_o²+18 z_ocs+9 c²s²

g’(s) = 6ax²+12a²xs+6a³s²+14by_o+14b²s+18cz_o+18c²s

= 6ax²+14by_o+18cz_o

2- f(x,y) = 3x-2y v = (1/, 1/)

x = x_o+ 1/s

y = y_o+ 1/s

gs = 3(x_o+ 1/s)-2(y_o+ 1/s)

= 3x_o+3/ s- 2 y_o-2/ s)

g’(s) = 3/ - 2/

3- f(x,y) = x²+y² v = (a,b) p= (0,0)

x = 0+as

y = 0+bs

gs = (0+as)²+( 0+bs)²

= 0²+2 0as+a²s²+ 0²+2 0bs+b²s²

g’(s) = 2 0a+2 a²s+2 0b+2 b²s

4- f(x,y) = x y²+ x²y v=(1,0)

x = x_o+ s

y = y_o+0

gs = x_o+ s (y_o+0)²+( x_o+ s)² y_o+0

g’(s) = (y_o+0)² + y_o+0 .2(x_o+ s)

= y_o²+2 x_o y_o+2s y_o

= y_o²+2 x_o y_o

5- f(x,y,z) = xyz v= (1/3, -2/3,- 2/3)

x = x_o+ 1/3s

y = y_o-2/3s

z = z_o-2/3s

gs = (x_o+ 1/3s)( y_o-2/3s)( z_o-2/3s)

= x_o y_o- x_o 2/3s+1/3s y_o-2/9s²(z_o-2/3s)

= z_o x_o y_o- z_o x_o 2/3s+ z_o1/3s y_o- z_o2/9s² -2/3s x_o y_o+4/9s²-2/9 yos³+4/27 s³

g’(s) = - z_o x_o 2/3+ z_o1/3 y_o-4/9s z_o-8/9s-6/9syo+12/27s²

= -2 z_o x_o + y_o z_o- 2 x_o y_o

3

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1- f(x,y) = x²y³

x²(0)+ y³2x

x² 3 y²+ y³(0)

2 - f(x,y) = Sen ( )

Cos ( ) . 1/2 ( )^-1/2 . 6x²

( )

Cos ().1/2 ( )^-1/2 . 2y

3- f(x,y) = x^y+y^x

4- f(x,y) = (2y)^x + 2^y

si

yx^y-1 + y^xlny

si

x^ylnx + xy^x-1

5- f(x,y) = x ln y – y ln x

x(0)+lny(1)

-y(1/x)+lnx(0)

x(1/y)+(-y)(0)+lnx(-1)

1- f(x, y, z)= x²y³z⁴

df = 2xy³z⁴ df = 3x²y²z⁴ df = 4x²y³z³

dx dy dz

En el punto (1,1,1) son:

df (1,1,1)= 2 df (1,1,1)= 3 df (1,1,1)= 4

dx dy dz

grad f (1,1,1)= (2,3,4)

2- f(x,y)= 3x²y+cos(xy) en p=(1,1)

df = 6xy -sen (xy) y df = 3x² –sen (xy) x

dx dy

df (1,1) = 5.982 df (1,1)= 2.98

dx dy

grad f (1,1)= (5.982 , 2.98 )

3- f(x,y)= x^y en p= (2,2)

df = yx^y-1 df = x^y ln x

dx dy

df (2,2) = 4 df (2,2)= 4 ln 2

dx dy

grad f (1,1)= (4, 4 ln 2 )

4- f(x,y)=

df = -x df = -x dx

En el punto (1,1)

df = -1 df = - 1

1 1

df (1,1) = -1 df (1,1)= -1

dx dy

grad f (1,1)= (-1, -1 )

5- f(x, y, z)= ln (x, y, z)

En el punto (1,1,1)

df = 1 df = 1 df = 1

dx x dy y dz z

df = 1 df = 1 df = 1

dx 1 dy 1 dz 1

df = 1 df = 1 df = 1

dx dy dz

1- f(x,y)= 3x+ 8y – 2xy + 4

df = 3 – 2y= 0 -2y =-3 y= 3

dx 2

df = 8 – 2x = 0 -2x= -8 x= 4

dy

pc= ( 4, 3 )

2

2- f(x,y)= x²+x+ y²+1

df = 2x+1 = 0 2x = -1 x= -1

dx 2

df = 2y = 0 2y= 0 y= 0

dy

pc= ( -1 , 0 )

2

3- f(x,y)= x² + 2x + y² – 4y + 10

df = 2x+2 = 0 2x = -2 x= -1

dx

df = 2y- 4 = 0 2y= -4 y= -2

dy

pc= ( -1 , -2 )

4- f(x,y)= 2x³ + 3x² + 6x +y³ + 3y + 12

df = 6x²+6x +6 = 0 6x²= -6x – 6 x² = -x -1 raiz negativa

dx

df = 3y²+3y = 0 3y( y + 1) = 0 y = -1

dy y= 0

No hay puntos criticos

5- f(x,y)= x²y – x² – 3xy + 3x + 2y -2

df = 2xy – 2x – 3y +3= 0 2x (y -1 ) -3 (y -1) = 0 y= 1

dx x= 1

df = x² – 3x + 2 = 0 x(x-3) +2 = 0

dy

pc= ( 1 , 1 )

1- f(x, y, z)= x²y³z⁴

df = 2xy³z⁴ df = 3x²y²z⁴ df = 4x²y³z³

dx dy dz

En el punto (1,1,1) son:

df (1,1,1)= 2 df (1,1,1)= 3 df (1,1,1)= 4

dx dy dz

grad f (1,1,1)= (2,3,4)

2- f(x,y)= 3x²y+cos(xy) en p=(1,1)

df = 6xy -sen (xy) y df = 3x² –sen (xy) x

dx dy

df (1,1) = 5.982 df (1,1)= 2.98

dx dy

grad f (1,1)= (5.982 , 2.98 )

3- f(x,y)= x^y en p= (2,2)

df = yx^y-1 df = x^y ln x

dx dy

df (2,2) = 4 df (2,2)= 4 ln 2

dx dy

grad f (1,1)= (4, 4 ln 2 )

1- f(x,y) = x²+y² si df = 2x , df = 2y

dx dy

d²f = d ( df ) = d (2x) = 2 d²f = d ( df ) = d (2y) = 0

dx² dx dx dx dxdy dx dy dx

d²f = d ( df ) = d (2x) = 0 d²f = d ( df ) = d (2y) = 2

dydx dy dx dy dy² dy dy dy

2- f(x,y) = x²e ^x2+y2 si df = x²e ^x2+y2 (2x) + 2xe ^x2+y2 = 2xe ^x2+y2(x³+x)

dx

df = x²e ^x2+y2 (2y) = 2x²ye ^x2+y2

dy

d²f = d ( df ) = d 2x (x³+x) = 2xe ^x2+y2(3x²+1)+4 x²e ^x2+y2 (x³+x)=

dx² dx dx dx

2 e ^x2+y2(2x⁴+5x²+)

d²f = d ( df ) = d (2x e ^x2+y2(x³+x)= 4y e ^x2+y2(x³+x)

dydx dy dx dy

d²f = d ( df ) = d (2x² ye ^x2+y2) = 2x² ye ^x2+y2(2x)+ 4x ye ^x2+y2= 4y e ^x2+y2(x³+x)

dxdy dx dy dx

d²f = d ( df ) = d (2x² ye ^x2+y2) = 2x² ye ^x2+y2(2y)+ 2x² e ^x2+y2 = 2x² e ^x2+y2(2y²+1)

dy² dy dy dy

3- f(x,y) = x³+6x²y⁴+7xy⁵+10x³y

si df = 3x²+12xy4+7y5+30x²y

dx

d²f = d ( df ) = d (3x²+12xy4+7y5+30x²y ) = 6x+12y⁴+60xy

dx² dx dx dx

d²f = d ( df ) = d (24y³x²+35xy⁴+10x³) = 48xy³+35y⁴+30x

dxdy dx dy dx

d²f = d ( df ) = d (3x²+12xy4+7y5+30x²y ) = 6x+12y⁴+60xy

dydx dy dx dy

d²f = d ( df ) = d (24y³x²+35xy⁴+10x³) = 72x²y²+140xy³

dy² dy dy dy

4- f(x,y) = x+y Verificar que satisfaga lo siguiente: d²f + d²f = 0

x²+y² dx² dy²

df = y²-2xy-x²

dx (x²+y²)²

d²f = d ( df ) = 2x³-2y³+6x²y-6xy²

dx² dx dx (x²+y²)³

df = x²-2xy-y²

dy (x²+y²)²

d²f = 2y³-2x³+6y²x-6xy²

dy² (x²+y²)³

5- f(x,y) = x^y

df = yx^y-1 df = x^y lnx

dy dx

d²f = x^y-2(y²-y) d²f = x^yln²x

dy² dx²

d²f = x^y-1(ylnx+1) d²f = x^y-1(ylnx+1)

dx dy dydx

1- f(x,y) = xy²

Si es diferenciable

2-f(x,y) = x²+y²

Si es diferenciable

3- f(x,y) = e^-(x2+y2)

df = -2xe^-(x2+y2)

dx

df = -2y e^-(x2+y2)

dy

Si es diferenciable

4- f(x,y,z) = cos (x+y²+z³)

df = -sen (x+y²+z³)

dx

df = -2ysen (x+y²+z³)

dy

df = -3z² sen (x+y²+z³)

dz

Si es diferenciable

5- f(x,y) = 3x

Δf= 3(x+Δx)-3x =

3x+3 Δx-3x = 3 Δx

Si es diferenciable

1. f (xyz) = x i + xy j + k

∇ x F = i j k

d d d =(0-0)i- (0-0)j + (y-0)k

x xy 1

∇ x F = yk

2. f (xyz) = -wy i + wx j

rot = i j k

d d d = 2wk

-wy wx 0

rot = 2w

3. F = x ²y i + z j + xyz k

div F = d (x ²y) + d (z) + d(xyz) = 2xy + 0 + xy = 3xy

dx dy dz

4. F = F₁i + F₂ j + F₃ k

∇² f = ∇. (∇f) = d²f + d²f + d²f

d x ² dy ² dz ²

∇² f = ∇² F₁i +∇² F₂ j + ∇²F₃ k

5. F = 3 x ²y i + 5xz³j – y²k

div F = d (x ²y) + d (z) + d(xyz) = 5xy + 0 + xy = 6xy

dx dy dz

5xy + xy = 6xy

T (s) = f’’(s) = -1 sen s , 1 cos s , s

f’’ (s) = -1 cos s , -1 sen s , 0

2 2

k ( s) = ½

N (s) = 1 f’’ (s) = 1 -1 cos s , -1 sen s , 0

k ( s) ½ 2 2

= - cos s , - sen s , 0

B (s) = T (s) x N (s) = det i j k

-1 sen s 1 cos s s

- cos s - sen s 0

= 1 sen s , -1 cos s , 1

1 ( x-0) + 0 ( y-1) + 0 ( z- П) = 0 x = 0 Rectificante

2- F (t) = ( t , t² , t³ ) p= f (2) = ( 2, 4, 8 )

u = f’(t) = (1, 2t, 3t² ) , f’’’(t) =( 0, 2, 6t)

v= f’(t) x f´´(t) = det i j k

1 2t 3t² = ( 6t² – 6t, 2)

w = v x u = det i j k

1 2t 3t²

12x – 6y + z = 8

1 ( x-2) + 4 ( y – 4) + 12 ( z – 8 ) = 0 Normal

x + 4y + 12 z = 114

-152 ( x-2) – 286 ( y – 4)+ 108 ( z – 8 ) = 0 Rectificante

76x + 143 y – 54z= 292

3- f ( t) = ( cost, sent , 2 ) p= (1 , 1, 2 )

f’(t) x f´´(t) = (- sent, cost , 0 ) x (- cost, -sent , 0)

= det i j k

- sent cost 0 = ( 0, 0, 2)

0 ( x-1) + 0 ( y-1) + 2 ( z-2) = 0 Osculador

4 - si T ( - 3/5, 0, 4/5) ; p ( 0, 3 , 2 П)

5 5

5 5 5

- 3x + 4z – 8 П = 0

3x – 4z + 8 П = 0 Normal

4x + 3z – 6 П =0 Osculador

2

x’ = 1 + sen t = 2

y’= 3 + sen2t = -2

z’= 1 + cos 3t = 3

2 2 2

y= 3 + sen 2t = 3

z= 1 + cos 3t = 1

2x- П – 2y + 6 + 3z – 3 = 0

2x- 2y + 3z + 3 – П = 0 Normal

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1. f(x,y)= 2 (x-1)² + 3(y-2)²

df = 4 (x-1) = 0 df = 6(y-2) = 0

dx dy

x= 1 y= 2

d²f d²f 4 0

dx² dydx = 0 6

dxdy dy²

(1,2 ) mínimo local

2 - f(x,y,z)= senx + sen y + sen z – sen (x+y+z)

Para ver las fórmulas seleccione la opción "Descargar" del menú superior 

Para ver las fórmulas seleccione la opción "Descargar" del menú superior 

El dominio del logaritmo mayor que cero.

Solo puede tomar valores positivos (x,y), entonces:

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1. Considrear el punto con funcion de posicion ס : t (1- cost, sent) hallar la velocidad, rapidez y longitud de arco

ס (t) =

2. Considerar una particula de masa m moviendose con rapidez constante S en una trayectoria circular de radio ro . ro = 5 s= 2 m= 3 t=1

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a (t)= r’’(t) = - s² cos ts, - s² sen ts = - S² r (t)

a (t)= r’’(t) =.8

P’T = (-2sen2t, 2 cos2t, )

L (p) = 

4. f (t) = (1- cost, sent)

f (t) =

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1. fF tiene componenete f F_i i= 1,2,3

div F = d (fF1) + d (fF2) + d(fF3) =

dx dy dz

f df₁ + df₂ + df₃ + F₁ df + F₂ df + F₃ df

dx dy dz dx dy dz

= f (∇.F)+ F. ∇f

∇_r = (x/r, y/r, z/r)

div (∇fx∇g) = ∇g. (∇fx∇) - ∇f. (∇gx∇) = 0

∇fx∇ = 0

∇gx∇= 0

3. v= (x+3y)i + (y-2z)j + (x+az) k

∇x V = = d (fF1) + d (fF2) + d(fF3) =

dx dy dz

∇x V = = d (x+3y) + d (y-2z) + d(x+az) =

dx dy dz

∇. V = 1 +1 + a = 0

2+a= 0+a= -2

. v= (x+3y)i + (y-2z)j + (x+az) k

4. V= (x+2y)i + (z+y)j + (x+az)

∇x V = = d (fF1) + d (fF2) + d(fF3) =

dx dy dz

∇x V = = d (x+2y) + d (y+z) + d(x+az) =

dx dy dz

2+a= 0+a= -2

5. F = x ²y i + z j + xyz k

div F = d (x ²y) + d (z) + d(xyz) = 2xy + 0 + xy = 3xy

dx dy dz

si x=1 y =2

1. Verificar que el campo vectorial es irrotacional en cada punto (x,y) ≢ 0

V(x,y) = yi - x j = y , - x

x²+y² x²+y² x²+y² x²+y²

d d d =

= -( x²+y² ) + 2x² + - ( x²+y²) + 2y² k

= 0

2. V(x,y,z) = yi – xj

rot(V) = i j k

3. f (xyz) = -wy i + wx j

rot = i j k

d d d = 2wk

-wy wx 0

4.. A(xyz) = x i + xy j + k

rot ( A) = i j k

d d d =(0-0)i- (0-0)j + (y-0)k

x xy 1

= 0 i + 0j + 0k

1- f ( x, y, z) = x² + y² +z²

sujeto a

x² + 1 y² + 1 z² = 1

4 9

4 9

dx

dy 2

dz 9

df = x² + 1 y² + 1 z² -1 = 0

p1( 1, 0, 0 ) p2 (0, 2, 0) p3(0, 0, 3) p4 ( -1, 0, 0 ) p5(0, -2, 0) p6(0, 0, -3)

x² + 1 y² + 1 z² = 1 p1( 1, 0, 0 ) = 1

4 9 p4 ( -1, 0, 0 ) =1

p2 (0, 2, 0 ) = 4

p3 (0, 0, 3) = 9

p6 (0, 0, -3) = 9

minimo ( 1, 0, 0 ) , ( -1, 0, 0 ) = 1

maximo (0, 0, 3) , (0, 0, -3) = 9

2- f ( x, y, z) = xyz

sujeto a

g₁( x, y, z) = x² + y² +z² -1 = 0 y g₂ = ( x, y, z) = x + y +z = 0

dx

dy

dz

df = x² + y² +z² -1 = 0 x² + 2 y² = 1

p1( -2, 1, 1 ) p2 (2, -1, -1) p3(1 , 1, -2 ) p4 (-1, -1 , -2 ) p5(-2 , 1 , 1 )

p6(2, -1, -1)

maximos p_2, p₄ , p₆ = 1 minimos p_1, p₃ , p₅ = - 1

3 3

3- f ( x, y, z) = 2xz+ 2yz + xy

sujeto a

g( x, y, z) = xyz- v

f ( x,y,z, λ) = 2xz + 2yz + xy + λ (xyz- v)

dx

dy 2

dz x

df = xyz - v = 0 x (x) ( x ) = V

dλ 2

minimo = 3 (2v)^2/3

4 - f ( x, y, z) = d² = (x-x_o)² + (y-y_o)² + (z-z_o)²

sujeto a

Ax + By +Cz = D

dx

dy

dz

df = D – Ax – By - Cz = 0

dλ

f ( x, y, z) = (x-x_o)² + (y-y_o)² + (z-z_o)² = (D –Ax_o – By_o – Cz_o)²

( A² +B² + C² )^1/2

minimo = Ax_O +By₀ + Cz_o - D

( A² +B² + C² )^1/2

5- f ( x, y, z) = Ax + By +Cz

sujeto a

x^ay^bz^c = N

dx

dy aB aC

dz

df = x^ay^bz^c –N = 0 x^a (bA ) x^b (cA)^c x^c = N

dλ aB aC

x = (bA ) ^-b (cA)^-c N ^1/a+b+c

aB aC

aB aC

aB aC

1- r (t) = (cost) i + (sent) j + t k

v= (-sent)i + (cost)j + k

v = (-sent)² + (cost)² + 1^{2 ½} = 2^1/2

T = v = -sent i + cost j + 1 k

2- r (t) = (cost + tsnet) i + (sent – tcost)j, t > 0

v = dr = ( -sent + sent + cost)i + ( cost- cost + tsent)j

dt

= (tcost)i + (tsent) j

2

v = ( t²cos²t + t²sen²t )^1/2 = (t²) ^½ = t

T = (cost)i + ( sent) j

3- r (t) = (cost) i + (sent) j

v= (-sent)i + (cost)j

es un vector unitario entonces

T = V

T= (-sent)i + (cost)j

4- r (t) = (2cost)i + (2sent)j + 5^1/2t k

v= (-2sent) i + (2cost) j + 5^1/2

v = (-2sent)² + (2cost)² +5 ^½ = 4sen²t + 4 cos²t +5 ^½ = 3

T= ( - 2 sent ) i + (2 cos t) j + 5^1/2 k

3 3

5- r (t) = ti + (sen2t)j + (cos 2t) k en p = (Π/4)

= i + 2 cos Π/2 j – 2 sen Π/2 k

= i + 2 (0) – 2 ( 1)

v = 5^1/2

T = 1 i - 2

5^1/2 5^1/2

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