UNIVERSIDAD NACIONAL DEL ALTIPLANO

Ingeniería mecánica eléctrica, electrónica y Sistemas.

ESCUELA PROFECIONAL INGENIERÍA DE SISTEMAS.

Estudiante: Leydi Ayled Fuentes Agramonte

III SEMESTRE“CALCULO AVANZADO”

------EJERCICIOS RESUELTOS-----

Lic. Esmer Monzón Astete

Puno-Perú15/06/2012

CURVAS EN EL ESPACIO, Ecuaciones vectoriales paramétricas

1- f (t) = 2i + 4 j – 2 k y que pasa por ( -2, 0. 4)para(-2, 0, 4) x = xo + tAAi + Bj + Ck = 2i + 4 j – 2 k y= yo + tB Z= zo + tCx =-2 + 2 ty = 4tz = 4 – 2 t

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

x= -3 + 4ty = 2 – 3tz = -3 + 7 t

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

x= 1 + 4ty = -1 – 3tz = 4 + 7t

4- Ai+ Bj + CkP ( -3, 3, -3) Q (1 , -1, 4)

x= -3 + 4ty = 2– 3tz = -3+7t

5- f (t) = a sent i + b cos t jx= a sen ty= b costz= 0

ARCO DE LONGITUD

1 - P’T = (-2sen2t, 2 cos2t, ) de (0,4П)

L(p)= (4 (sen2t)2 + 4 (cos2t)2 + 5 )1/2 = 91/2 = 3

L (p) =

2 – f (t) = (1- cost, sent)

f (t) =

2TT

0 = 8

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].

5- ( t, t -1/2, 0) en (-1,1) en R3

-1= to < t1 < ½= t2 < 1 = t3

(-1,0), (0,1/2) y (1/2,1)(-1,0)x(t) = -ty (t)= -tz (t) = 0

ds = 21/2

dt

(0,1/2)x(t) = ty (t)= -t + 1/2z (t) = 0ds = ( 21/2) 1/2dt

(1/2, 1)

x(t) = ty (t)= t – 1/2z (t) = 0

= 2 (2) 1/2

DERIVACION IMPLICITA

1- f(x,y,u,v)= xeu+v+uv-1= 0 g(x,y,u,v)= yeu-v-2uv-1=0

df = eu+v df= 0 df= xeu+v +v df= xeu+v+udx dy du dv

dg= 0 dg= eu-v dg= yeu-v-2u dg=-yeu-v -2udx dy du dv

d(F,G) = df df = xeu+v+u xeu+v +v du dv -yeu-v-2v -yeu-v -2u dg dg du dv

du = eu+v xeu+v +v = y e 2u + 2 u e u+v dx 0 -ye u-v - 2u 2xye2u+ 2(u-v)xeu+v+ (u+v)yeu-v

xeu+v+u xeu+v +v yeu-v-2v -yeu-v -2u

2- f(x,y,u,v)= u+ e u+v g(x,y,u,v)= v+eu-v

u= Ø(x,y) v= Ø(x,y)dz (1,1) dz (1,1)dx dy

dz = du + dv , dz = du + dvdx dx dx dy dy dy

du + e u+v du + dv - 1 = 0 dv + eu-v du - dv = 0dx dx dx dx dx dx

du + e u+v du + dv = 0 dv + eu-v du - dv -1 = 0dy dy dy dy dy dy

dz = 0+1 = 1 dz = 1+(-2)= -1dx dy

3- f(x,y,u,v)= u-v+2x-2y=0 g(x,y,u,v)= 3u3+v3-5x2+y3=0

d(F,G) = 1 -1 = 3v2+9u2

d(u,v) 9u2 3v2

du = 2v 2 -y 2 dy v2+3u2

dv = 1 1 -2 = -y 2 +6u 2 dy 3v2+ 9u2 9u2 3y2 v2 + 3u2

d 2 u = - 2v (y 2 +6u 2 ) + 6u (2v 2 -y 2 ) 2 - 2y (v 2 + 3u 2 ) dy2 (v2 + 3u2)3

4- f(x,y,u,v,w)= x+y+u+v+w= 0 En P(1,-1,1,-1,0) g(x,y,u,v,w)= x2-y2+u2-2v2+w2+1= 0 h(x,y,u,v,w)= x3+y3+u4-3v4+8w4+2= 0

d(F,G,H) = 1 1 1 =8 d(u,y,w) 2u -4y 2w 4u3 -12v3 32w2

d(F,G,H) = 1 1 1d(u,y,w) 2u -2y 2w 4u3 3y2 32w3

d(F,G,H) = 1 1 1 = -2d(u,y,w) 2 2 0 4 3 0

dv (1,-1)= 1dy 4

5- f(x,y,z)= 0 z= f(x,y)

df dx + df dy = 0dx dy

df dy = - df dxdy dx

dy = - df dy = - dfdx dx dx df df dy dy

DERIVADA PARCIAL

1- f(x,y) = x2y3

x2(0)+ y32x

= 2xy3

x2 3 y2+ y3(0)

3 x2y2

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

Cos ( ) . 1/2 ( )-1/2 . 6x2

= 3x2cos( )

( )

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

y Cos ( )

( )

3- f(x,y) = xy+yx

= yxy-1 + yxlny

xylnx + xyx-1

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

siyxy-1 + yxlny

= (2y)x ln 2y

sixylnx + xyx-1

x(2y)x-1 + 2y ln 2

5- f(x,y) = x ln y – y ln x

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

= lny – y

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

x – lnx

y DERIVADA DIRECCIONAL GRADIENTE

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)= 4dx 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) xdx dy

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

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

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

df = yxy-1 df = xy ln xdx dy

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

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

PUNTOS CRITICOS DE UNA FUNCION

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 )

DERIVADA DIRECCIONAL GRADIENTE

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)= 4dx 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) xdx dy

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

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

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

df = yxy-1 df = xy ln xdx dy

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

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

DERIVADAS PARCIALES DE ORDEN SUPERIOR(Teorema de Schwarz)

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

d 2 f = d ( df ) = d (2x) = 2 d 2 f = d ( df ) = d (2y) = 0dx2 dx dx dx dxdy dx dy dx

d 2 f = d ( df ) = d (2x) = 0 d 2 f = d ( df ) = d (2y) = 2dydx 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

d 2 f = 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+)

d 2 f = d ( df ) = d (2x e x2+y2(x3+x)= 4y e x2+y2(x3+x)dydx dy dx dy

d 2 f = 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

d 2 f = 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

df = 24y3x2+35xy4+10x3

dy

d 2 f = d ( df ) = d (3x2+12xy4+7y5+30x2y ) = 6x+12y4+60xydx2 dx dx dx

d 2 f = d ( df ) = d (24y3x2+35xy4+10x3) = 48xy3+35y4+30xdxdy dx dy dx

d 2 f = d ( df ) = d (3x2+12xy4+7y5+30x2y ) = 6x+12y4+60xydydx dy dx dy

d 2 f = d ( df ) = d (24y3x2+35xy4+10x3) = 72x2y2+140xy3

dy2 dy dy dy

4- f(x,y) = x+y Verificar que satisfaga lo siguiente: d 2 f + d 2 f = 0 x2+y2 dx2 dy2

df = y 2 -2xy-x 2 dx (x2+y2)2

d 2 f = d ( df ) = 2x 3 -2y 3 +6x 2 y-6xy 2 dx2 dx dx (x2+y2)3

df = x 2 -2xy-y 2 dy (x2+y2)2

d 2 f = 2y 3 -2x 3 +6y 2 x-6xy 2 dy2 (x2+y2)3

d 2 f + d 2 f = 2x 3 -2y 3 +6x 2 y-6xy 2 + 2y 3 -2x 3 +6y 2 x-6xy 2 = 0 dx2 dy2 (x2+y2)3 (x2+y2)3

5- f(x,y) = xy

df = yxy-1 df = xy lnxdy dx

d 2 f = xy-2(y2-y) d 2 f = xyln2xdy2 dx2

d 2 f = xy-1(ylnx+1) d 2 f = xy-1(ylnx+1) dx dy dydx

FUNCIONES DIFERENCIABLES

1- f(x,y) = xy2

Δf= (x+Δx) (y+Δy)2= x+Δx (y2+2y ∆y + (∆y)2) - xy2=xy2+2xy∆y+x(∆y)2+ Δx y2+ Δx2y ∆y+ Δx(∆y)2- xy2= 2xy∆y+x(∆y)2+ Δx y2+ Δx2y ∆y+ Δx(∆y)2

Si es diferenciable

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

Δf= (x+Δx)2+(y+ Δy)2 –( x2+y2) = x2+2x Δx+( Δx)2+ y2+2y ∆y + (∆y)2)- ( x2+y2)=2x Δx+( Δx)2+2y ∆y + (∆y)2

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

ECUACIONES DEL PLANO OSCULADOR, NORMAL Y RECTIFICANTE

1- F (s) = cos s , sen s , s p= f ( П) = (0 ,1, П)

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

T ( П)= 0 , -1 , 1 N ( П) = ( 1, 0, 0)

B ( П ) = 0 , 1 , 1

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

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

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)

0 2 6t

w = v x u = det i j k 6 t 2 – 6t 2 = ( 18t3-4t, 3 – 18 t4, 12 t3 + 6t)1 2t 3t2

24 ( x-2) – 12 ( y – 4) + 2 ( z – 8 ) = 0 Osculador

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)

- cost - sent 0

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

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

-3 ( x-0) + 0 ( y-3) + 4 (z- 2 П) = 0 5 5

-3 x- + 0 + 4 z- 8 П = 0 5 5 5

- 3x + 4z – 8 П = 0

3x – 4z + 8 П = 0 Normal

4x + 3z – 6 П =0 Osculador

5- x= t – cost y= 3 + sen 2t z= 1 + cos 3 t p= t= П 2

x’ = 1 + sen t = 2

y’= 3 + sen2t = -2

z’= 1 + cos 3t = 3

x= t – cos t = П - cos П = П 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

LIMITES

1) lim 3x 2 y =

(x,y) (0,0) x4+y2

si y(x=0) lim 0 = 0 y 0 y2 0 si y=xlim 3x 2 x = lim 3 x 3 = lim 3x = 0 = 0 x 0 x4+x 2 x2(x2+1) x2+1 0+1

si y=x2

lim 3x 2 x 2

= lim 3 x 4 = lim 3x 2 = 3 x 0 x4+x 4 x2(x2+x2) 2x2 2

2) lim sen(x 2 +y 2 ) = (x,y) (0,0 ) x2+y2

De acuerdo a una propiedad de limites de funciones trigonometricas especifica por definición que :

lim sen x = 1 por lo tantox 0 x

lim sen(x 2 +y 2 ) = 1 (x,y) (0,0 ) x2+y2

3) lim x 2 = (x,y) (0,0 ) ( x2+y2 )

si y(x=0) lim 0 = 0 = 1 (x,y) 0 0+y2

si y=xlim x 2 = lim x 2 = 0 = indeterminacion x 0 x2 + x2 x 0 2 x2 0

si y=x2

lim x 2 = lim x 2 = 0 = indeterminacion x 0 x2 +x4 x2(x+x2) 0

4) lim x 4 y =

(x,y) (0,0) x4+y4

si y(x=0) lim 0 = 0

y 0 x4

si y=xlim x 4 x = lim x 5 = 0 = indeterminacion x 0 x4+x 4 2x4 0

5) lim =

x 1

0 = indeterminacion0

saco el conjugado .

x1/2 +1 implica que x=1