calculoasd
TRANSCRIPT
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