2- f(x,y,u,v)= u+ e u+v g(x,y,u,v)= v+eu-v u= Ø(x,y) dz (1,1) dx v= Ø(x,y) dz (1,1) dy dz = du + dv , dz = du + dv dx dx dx dy dy dy du + e u+v du + dv – 1 = 0 dv + eu-v du – dv = 0 dx dx dx dx dx dx du + e u+v du + dv = 0 dv + eu-v du – dv -1 = 0 dy dy dy dy dy dy dz = 0+1 = 1 dx dz = 1+(-2)= -1 dy 3- f(x,y,u,v)= u-v+2x-2y=0 g(x,y,u,v)= 3u3+v3-5×2+y3=0 d(F,G) = d(u,v) 1 9u2 -1 3v2 = 3v2+9u2 du = 2v2-y2 dy v2+3u2 dv = 1 1 -2 = -y2+6u2 dy 3v2+ 9u2 9u2 3y2 v2 + 3u2 d2u = – 2v (y2+6u2) + 6u (2v2-y2)2- 2y (v2 + 3u2) dy 2 2 2 3 (v + 3u ) 4- f(x,y,u,v,w)= x+y+u+v+w= 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 En P(1,-1,1,-1,0) d(F,G,H) = d(u,y,w)
d(F,G,H) = d(u,y,w)
d(F,G,H) = d(u,y,w) 1 2u 4u3
1 2u 4u3
1 2 4 1 -4y -12v3
1 -2y 3y2
1 2 3 1 2w 32w2
1 2w 32w3
1 0 0 = -2 =8 dv (1,-1)= 1 dy 4
5- f(x,y,z)= 0 z= f(x,y)
df dx + df dy = 0 dx dy df dy = – df dx dy dx dy = dx – df dx df dy dy = – df dx df dy DERIVADA DIRECCIONAL 1- f(x,y,z) = 2×3+7y2+9z2 v = (a,b,c) x = xo+ as y = yo+bs z = zo+cs
gs = 2 (xo+ as)3+7(yo+bs)2+9(zo+cs)2 = 2 (xo3+3x2as+3xoa2s2+a3s3)+7(yo2+2 yobs+b2s2)+9(zo2+2 zocs+c2s2) = 2 xo3+6 x2as+6 xoa2s2+2 a3s3+7 yo2+14yobs+7 b2s2+9 zo2+18 zocs+9 c2s2 g’(s) = 6ax2+12a2xs+6a3s2+14byo+14b2s+18czo+18c2s = 6ax2+14byo+18czo 2- f(x,y) = 3x-2y v = (1/ 2 , 1/ 2 ) x = xo+ 1/ y = yo+ 1/ 2 s 2 s gs = 3(xo+ 1/ 2 s)-2(yo+ 1/ 2 s) = 3xo+3/ 2 s- 2 yo-2/ 2 s) g’(s) = 3/ 2 – 2/ 2 = 1/ 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 =0
2 df ( 4- f(x,y) = x y2+ x2y v=(1,0) x = xo+ s y = yo+0
gs = xo+ s (y o +0)2+( x o + s)2 y o +0 g’(s) = (y o +0) + y o +0 .2(x o + s) 2 = y o +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 = xo+ 1/3s y = yo-2/3s z = zo-2/3s
gs = (xo+ 1/3s)( y o -2/3s)( z o -2/3s) = xo yo – x o 2/3s+1/3s y o -2/9s2(zo-2/3s) = zo xo y o – z o x o 2/3s+ zo1/3s yo- zo2/9s2 -2/3s xo yo+4/9s2-2/9 yos3+4/27 s3 g’(s) = – z o x o 2/3+ zo1/3 yo-4/9s zo-8/9s-6/9syo+12/27s2 = -2 z o x o + yo zo- 2 xo yo 3
DERIVADA PARCIAL
1- f(x,y) = x2y3
x2(0)+ y32x = 2xy3 dx x2 3 y2+ y3(0) df dy 3 x2y2 2 – f(x,y) = Sen ( 2 x 3 y 2 ) Cos ( 2 x 3 y 2 ) . 1/2 ( 2 x 3 y 2 )-1/2 . 6×2 df dx = 3x2cos( 2 x 3
2 x 3 y 2 )
y 2 ) Cos ( 2 x 3 y 2 ).1/2 ( 2 x 3 y 2 )-1/2 . 2y df y Cos ( 2 x 3 y 2 ) dy ( 2 x 3 y 2 ) 3- f(x,y) = xy+yx
df dx
df dy = yxy-1 + yxlny
xy lnx + xy x-1 4- f(x,y) = (2y) x + 2 y si yxy-1 + yx lny df dx = (2y)x ln 2y si xylnx + xyx-1 df dy 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) df dx = lny – y x x(1/y)+(-y)(0)+lnx(-1) df dy x – lnx y
DERIVADA DIRECCIONAL GRADIENTE
1- f(x, y, z)= x2y3z4 df = 2xy3z4 dx
En el punto (1,1,1) son:
df (1,1,1)= 2 dx df = 3x2y2z4 dy
df (1,1,1)= 3 dy df = 4x2y3z3 dz
df (1,1,1)= 4 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 dx df = 3×2 –sen (xy) x dy df (1,1) = dx 5.982 df (1,1)= 2.98 dy
4 -1 grad f (1,1)= (5.982 , 2.98 )
3- f(x,y)= xy en p= (2,2) df = yxy-1 dx
df (2,2) = dx df = xy ln x dy df (2,2)= 4 ln 2 dy grad f (1,1)= (4, 4 ln 2 ) 4- f(x,y)= 3 x 2 y 2 df = -x df = -x dx 3 x 2 y 2 3 x 2 y 2 En el punto (1,1) df = -1 df = -1 1
df (1,1) = dx
grad f (1,1)= (-1, -1 )
5- f(x, y, z)= ln (x, y, z)
En el punto (1,1,1) 1
df (1,1)= -1 dy 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 dx df = dy 1 df = 1 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
) y= 1 dx 2 df = 8 – 2x = 0 dy
pc= ( 4, 3 2
2- f(x,y)= x2+x+ y2+1
df = 2x+1 = 0 dx
df = 2y = 0 dy
pc= ( -1 , 0 ) -2x= -8
2x = -1
2y= 0 x= 4
x= -1 2
y= 0 2
3- f(x,y)= x2 + 2x + y2 – 4y + 10 df = 2x+2 = 0 dx
df = 2y- 4 = 0 dy
pc= ( -1 , -2 ) 2x = -2
2y= -4 x= -1
y= -2 4- f(x,y)= 2×3 + 3×2 + 6x +y3 + 3y + 12 df = 6×2+6x +6 = 0 6×2= -6x – 6 x2 = -x -1 raiz negativa dx df = 3y2+3y = 0 dy 3y( y + 1) = 0 y = -1 y= 0 No hay puntos criticos
5- f(x,y)= x2y – x2 – 3xy + 3x + 2y -2 df = 2xy – 2x – 3y +3= 0 dx 2x (y -1 ) -3 (y -1) = 0 x= 1 df = dy
pc= ( x2 – 3x + 2 = 0
1, 1 ) x(x-3) +2 = 0 DERIVADA DIRECCIONAL GRADIENTE
1- f(x, y, z)= x2y3z4
4 dx x2+y2 x2+y2 df = 2xy3z4 dx
En el punto (1,1,1) son:
df (1,1,1)= 2 dx df = 3x2y2z4 dy
df (1,1,1)= 3 dy df = 4x2y3z3 dz
df (1,1,1)= 4 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 dx df = 3×2 –sen (xy) x dy df (1,1) = dx 5.982 df (1,1)= 2.98 dy grad f (1,1)= (5.982 , 2.98 )
3- f(x,y)= xy en p= (2,2) df = yxy-1 dx
df (2,2) = dx df = xy ln x dy
df (2,2)= 4 ln 2 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
d2f = d ( df ) = d (2x) = 2 dx2 dx dx
d2f = d ( df ) = d (2x) = 0 dy
d2f = d ( df ) = d (2y) = 0 dxdy dx dy dx
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 dx df = x2e x2+y2 (2y) = 2x2ye x2+y2 dy (x3+x) d2f = d ( df ) = d 2x (x3+x) = 2xe dx2 dx dx dx (3×2+1)+4 x2e x2+y2 (x3+x)= 2 e x2+y2(2×4+5×2+)
x2+y2 x2+y2 x2+y2 d2f = d ( df ) = d (2x e x2+y2(x3+x)= 4y e (x3+x) dydx dy dx dy d2f = d ( df ) = d (2×2 ye x2+y2) = 2×2 ye dxdy dx dy dx (2x)+ 4x ye x2+y2= 4y e (x3+x) 2 2 d f = d ( df ) = d (2x ye x2+y2 2 ) = 2x ye x2+y2 2 (2y)+ 2x e x2+y2 2 = 2x e x2+y2 2 (2y +1) dy2 dy dy dy 3- f(x,y) = x3+6x2y4+7xy5+10x3y si df = 3×2+12xy4+7y5+30x2y dx df = 24y3x2+35xy4+10×3 dy d2f = d ( df ) = d (3×2+12xy4+7y5+30x2y ) = 6x+12y4+60xy dx2 dx dx dx d2f = d ( df ) = d (24y3x2+35xy4+10×3) = 48xy3+35y4+30x dxdy dx dy dx d2f = d ( df ) = d (3×2+12xy4+7y5+30x2y ) = 6x+12y4+60xy dydx dy dx dy d2f = d ( df ) = d (24y3x2+35xy4+10×3) = 72x2y2+140xy3 dy2 dy dy dy 4- f(x,y) = x+y Verificar que satisfaga lo siguiente: d2f + d2f = 0 2 2 df = y -2xy-x x2+y2 dx2 dy2 dx (x2+y2)2 d2f = d ( df ) = 2×3-2y3+6x2y-6xy2 dx2 dx dx
df = x2-2xy-y2 (x2+y2)3 dy (x2+y2)2 d2f = 2y3-2×3+6y2x-6xy2 dy2 (x2+y2)3 d2f + d2f = 2×3-2y3+6x2y-6xy2 + 2y3-2×3+6y2x-6xy2 = 0 dx2 dy2 (x2+y2)3 (x2+y2)3 5- f(x,y) = xy df = yxy-1 dy
d2f = xy-2(y2-y) df = xy lnx dx
d2f = xyln2x
dy2
d2f = xy-1(ylnx+1) dx dy dx2
d2f = xy-1(ylnx+1) dydx FUNCIONES DIFERENCIABLES 1- f(x,y) = xy2 ?f= (x+?x) (y+?y)2= x+?x (y2+2y ?y + (?y)2) – xy2= xy2+2xy?y+x(?y)2+ ?x y2+ ?x2y ?y+ ?x(?y)2- xy2= 2xy?y+x(?y)2+ ?x y2+ ?x2y ?y+ ?x(?y)2 Si es diferenciable
2-f(x,y) = x2+y2
?f= (x+?x)2+(y+ ?y)2 –( x2+y2) = x2+2x ?x+( ?x)2+ y2+2y ?y + (?y)2)- ( x2+y2)= 2x ?x+( ?x)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
?f= 3(x+?x)-3x = 3x+3 ?x-3x = 3 ?x Si es diferenciable
DIVERGENCIA Y ROTACIONAL
Y LAPLACIANO
1. f (xyz) = x i + xy j + k ? xF= i j k d dx x d dy xy d dz 1 =(0-0)i- (0-0)j + (y-0)k ? x F = yk
2. f (xyz) = -wy i + wx j rot = i j k d dx -wy d dy wx d dz 0 = 2wk 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 2xy + xy =
4. F = F1i + F2 j + F3 k
? 2 f = ? . (? f) = d2f + d2f + d2f dx 2 3xy
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
ECUACIONES DEL PLANO OSCULADOR, NORMAL Y RECTIFICANTE 1- F (s) = cos s , sen s , s p= f ( 2 ?) = (0 ,1, ?) 2 2 2
s s 0 1 , T (s) = f’’(s) = -1 sen s , 1 cos s , 2 2 2 2 2 f’’ (s) = -1 cos s , -1 sen s , 0 2 2 2 2 k ( s) = ½ N (s) = 1 f’’ (s) = 1 -1 cos s , -1 sen s , 0 k ( s) ½ 2 2 2 2 = – cos s , 2 – sen s 2 , 0 B (s) = T (s) x N (s) = det i j k -1 sen s 1 cos s 2 2 2 2 2 – cos s 2 – sen s 2 = 1 sen s , -1 cos s , 1 2 2 2 2 2 T ( 2 ?)= 0 , -1 , 2 2 N ( 2 ?) = ( 1, 0, 0) B( 2 ? ) = 0 , 1 2 2 1 0 ( x-0) + 1 ( y-1) + 1 ( z- ?) = 0 y+ z = ? +1 Osculador 2 2 0 ( x-0) – 1 ( y-1) + 1 ( z- ?) = 0 -y+ z = ? +1 Normal 2 2 1 ( x-0) + 0 ( y-1) + 0 ( z- ?) = 0
2 3 2- F (t) = ( t , t , t )
u = f’(t) = (1, 2t, 3t2 ) , x = 0
p= f (2) = ( 2, 4, 8 )
f’’’(t) =( 0, 2, 6t) Rectificante v= f’(t) x f´´(t) = det i j k 1 2t 3t2 = ( 6t2 – 6t, 2)
3 4 3 0 0 0 2 6t w = v x u = det i j k 6t 1 2 – 6t 2t 2 3t2 = ( 18t -4t, 3 – 18 t , 12 t + 6t) 24 ( x-2) – 12 ( y – 4) + 2 ( z – 8 ) = 0
12x – 6y + z = 8
1 ( x-2) + 4 ( y – 4) + 12 ( z – 8 ) = 0
x + 4y + 12 z = 114
-152 ( x-2) – 286 ( y – 4)+ 108 ( z – 8 ) = 0
76x + 143 y – 54z= 292 Osculador
Normal
Rectificante 3- f ( t) = ( 2 cost, 2 sent , 2 ) p= (1 , 1, 2 ) f’(t) x f´´(t) = (- 2 sent, 2 cost , 0 ) x (- 2 cost, – 2 sent , 0) = det i – 2 sent j 2 cost k = ( 0, 0, 2) – 2 cost
0 ( x-1) + 0 ( y-1) + 2 ( z-2) = 0 – 2 sent Osculador 4 – si T ( – 3/5, 0, 4/5) ; p ( 0, 3 , 2 ?)
-3 ( x-0) + 0 ( y-3) + 4 (z- 2 ?) = 0 5 5 -3 x- + 0 + 4 z- 8 ? = 0 5 5 5 – 3x + 4z – 8 ? = 0 3x – 4z + 8 ? = 0
4x + 3z – 6 ? =0 Normal
Osculador 5- x= t – cost y= 3 + sen 2t z= 1 + cos 3 t p= t= ? 2
lim x’ = 1 + sen t = 2
y’= 3 + sen2t = -2
z’= 1 + cos 3t = 3 x= t – cos t = ? – cos ? = ? 2 2 2 y= 3 + sen 2t = 3
z= 1 + cos 3t = 1
2x- ? – 2y + 6 + 3z – 3 = 0 2x- 2y + 3z + 3 – ? = 0
FUNCIONES VECTORIALES
1- F(t) = (t2 +1, 2t, t) Normal lim (t2 + 1) = 1 +1=2 , lim (2t) = 2 , lim (2t) = 2 t 1 t 1 t 1 lim t (t) = 1 1 lim t 1 = 2i + 2j + k 2- F(t) = (t3 +1, t2 – 2 t +1 ) lim (t3 +1)= 2 ( t2 – 2 t +1) = 0 t 1 t 1 lim = 2i t 1 3- F(t) = (t , t2 , sent ) t lim (t , t2 , sent ) = (0, 0 1) t 0 t lim t (t) = 0 0 t lim 0 (t2) = 0 t lim 0 (t) = t sent =1 lim t 0 = k
4- F(t) = (3t-1, t) lim t (3t-1) = 3- 1 = 2 t
lim t t
lim t
t 1
1
1 (t) = 1
= 2i , j 5- F(t) = (1- cos t , 1- cos2t ) t2 t2 lim t t
lim t t 1
1 =
= 1 . t
1 . t t
t lim t 1
lim t 1 1- cost t
1- cos2t t =
= 1.0=0
1. .01 = .01 lim t t 1 = .01 j MATRIZ HESIANA
1. f(x,y)= 2 (x-1)2 + 3(y-2)2 df = 4 (x-1) = 0 dx
x= 1 df = 6(y-2) = 0 dy
y= 2 d2f d2f 4 0 dx2 dydx = 0 6 d2f dxdy
(1,2 ) d2f dy2
mínimo local 2 – f(x,y,z)= senx + sen y + sen z – sen (x+y+z) en P(? , ? , ? ) 2 2 2 H (? , ? , ? ) = -2 -1 -1
2 2 2 -1
-1 -2
-1 -1
-2 A1 = -2 A 2 = -2 -1 = 3 A 3 = -4 -1 -2 H(p) – ?I) = -?3 – 6 ?2 – 9? – 4 = – (? + 1 )2 (? + 4 ) ? = -1 ? = -4 máximo local de p = 4 3- Se f: R4 – {( 0, 0, 0, 0)}
f ( x, y, z, u ) = x + y + z +u + 1 x y z u dada por df = 1 – y = 0 df = 1 – z = 0 df = 1 – u = 0 df = 1 – 1 = 0 dx x2 dy x y2 dz y z2 du z u2 H (x, y, z, u ) = 2y x3 -1 0 x2 0 0 -1 x2 -1 2z y3 -1 y2
2u -1 0 y2 z3 z2 0 0 -1 2 z2 u3
En el punto (1, 1,1 ,1 ) H ( 1, 1,1 ,1 ) = 2 -1 0 0 -1 2 -1 0 0 -1 2 -1 0 0 -1 2 A1 = 2 A2 = 2 -1 = 3 A3= 2 -1 0 = 2 A4 = 2 -1 0 0 =5 -1 2 -1 2 -1 -1 2 -1 0 0 -1 2 0 -1 2 -1 0 0 -1 2
mínimo local en ( 1, 1,1 ,1 ) = 5
x 4- f(x,y,z)= e-x2 + e –y2 + z2 df = -2xe-x2 dx df = -2y-y2 dy df = 2z dz H= (4×2-2)e-x2 0 0 0 (4y2-2)e-y2 0 0 0 2 H ( 0,0, 0)=
?1 = -2 -2 0 0
? 2 = 2 0 -2 0 0 0 2 Tiene punto de silla
5- f(x,y)= ax2 +by4 df = 2ax dx df = 4by3 dy H (x,y) = 2a 0 0 12by2 H (0,0) = 2a 0 0 0 ? = 0 Cualquier cosa puede pasar
LIMITES 1) lim (x,y) 3x2y = (0,0) x4+y2 si y(x=0) lim y 0
si y=x 0 = 0 y2 0 lim 0 3x2x = lim x4+x 2 3 x3 = x2(x2+1) lim 3x = x2+1 0= 0 0+1 si y=x2 lim x 0 3x2x 2 = lim x4+x 4 3 x4 = x2(x2+x2) lim 3×2 2×2 = 3 2
) ) ) 2 2 x x = 2) lim sen(x2+y2) = (x,y) (0,0 x2+y2 De acuerdo a una propiedad de limites de funciones trigonometricas especifica por definición que : lim x 0
lim (x,y)
3) lim (x,y) sen x = 1 x
sen(x2+y2) = 1 (0,0
x2 = (0,0 por lo tanto
x2+y2
( x2+y2 ) si y(x=0) lim (x,y)
si y=x 0 = 0 = 1 0 0+y2 lim 0 x2 = lim x2 = x + x x 0 2 x2 0 = 0 indeterminacion si y=x2 lim x 0 x2 = lim x2 +x4 x2 = x2(x+x2) 0 = 0 indeterminacion 4) lim (x,y) x4y = (0,0) x4+y4 si y(x=0) lim y 0
si y=x 0 = 0 x4 lim 0 x4x = lim x5 = x4+x 4 2×4 0 = 0 indeterminacion 5) lim
x
0= 0 x 1 x 1 1
indeterminacion saco el conjugado x 1 x 1 . x 1 x 1 x1/2 +1 implica que x=1 DOMINIOS
1) f(x,y)= 1 X 2 Y 2 = 1- X2 –y2> 0 despejo el uno X2 +y2 x < x+y =>x2 0 y> 0 y< 0
3) f(x,y,z)= ln(1-x2-y2+z)
El dominio del logaritmo debe ser mayor que cero. entoces: 2 2 1-x -y +z > 0 -x2-y2+z > -1 x2+ y2+z
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