martes, 16 de septiembre de 2008

tarea 3.....

Calcular la gradiente de la funcion:

\\f(x,y)= 4x^2 - 3xy + y^2\\ f(x)(xy)= 8x -3y\\ f(y)(xy)= -3x +2y\\ \\ \overline{V} = 8x - 3y i -3x + 2y j





2.- \\f(x,y,z)=\frac{x-y}{x-z}


\\f(x)(x,y,z)=\frac{(x+z)Dx(x-y)-(x-y)Dx(x+z)}{(x+z)^2}


\\=\frac{(x+z)(1)-(x-y)(1)}{(x+z)^2}


\\=\frac{x+z -x-y}{x+z^2}

\\= \frac{z+y}{x+z^2} i


\\f(y)(x,y,z) = \frac{(x+z)Dy (x-y)-(y-y)Dy (x+z)}{x+z^2}



\\=\frac{(x+z)(-1) - (x-y)(0)}{x+z^2}


\\=\frac{-x-z}{x+z^2} j



\\f(z)(x,y,z)=\frac{(x+z)Dz(x-y)(x-y)Dz(x+z)}{x+z^2}


\\= \frac{(x+z)(0)-(x-y)(1)}{x+z^2}

\\= \frac{-x+y}{(x+z)^2} k



\overline{V}= \frac{z+y}{(x+z^2}i - \frac{x+z}{(x+z)^2}j + \frac{x+y}{(x+z)^2} k





Calcular la divergencia y el rotacional del campo vectorial F


f(x,y,z) = 6x^2 j + xy^2j


m= 6x^2 \; n= xy^2 \; r=0

divergencia: 12x + 2xy


\frac{{\partial f}}{{\partial x}}= 12x ; \frac{{\partial f}}{{\partial y}}= 2xy ; \frac{{\partial f}}{{\partial z}}= o


Producto cruz.....

rotacional \overline{V} * F \begin{bmatrix}{\frac{{\partial f}}{{\partial x}}}&{\frac{{\partial f}}{{\partial y}}}&{\frac{{\partial f}}{{\partial z}}}\\{m}&{n}&{r}\\\end{bmatrix}


=\begin{bmatrix}\frac{{\partial f}}{{\partial y}}(R) - (N)\frac{{\partial f}}{{\partial z}}\end{bmatrix} - \begin{bmatrix}\frac{{\partial f}}{{\partial x}}(R) - (M)\frac{{\partial f}}{{\partial z}}\end{bmatrix} + \begin{bmatrix}\frac{{\partial f}}{{\partial x}}(N) - (M)\frac{{\partial f}}{{\partial y}}\end{bmatrix}

\\=(0-0)i -(0-0)j + 12x(xy^2) - 6x^2(2xy)k\\ \\=0i - 0j + (12X^2 - 6x^3y)k


F(x,y,z) = Sen x i + Cos y j + z^2 k

m= Sen x n= Cos y r = z^2

divergencia: Cos x -Sen y + 2z


\frac{{\partial f}}{{\partial x}}= Cos x \frac{{\partial f}}{{\partial y}}= -Sen y \frac{{\partial f}}{{\partial z}}= 2z


producto cruz......


rotacional \overline{V} * F \begin{bmatrix}{\frac{{\partial f}}{{\partial x}}}&{\frac{{\partial f}}{{\partial y}}}&{\frac{{\partial f}}{{\partial z}}}\\{m}&{n}&{r}\end{bmatrix}

=\begin{bmatrix}\frac{{\partial f}}{{\partial y}}(r)- (n)\frac{{\partial f}}{{\partial z}}\end{bmatrix} - \begin{bmatrix}\frac{{\partial f}}{{\partial x}}(r)- (m)\frac{{\partial f}}{{\partial z}}\end{bmatrix} + \begin{bmatrix}\frac{{\partial f}}{{\partial x}}(n)- (m)\frac{{\partial f}}{{\partial y}}\end{bmatrix}

\\=\begin{bmatrix}Sen y(z^2)-2z(Cos y)\end{bmatrix} - \begin{bmatrix}Cosx (z^2)-2z(Senx)\end{bmatrix} + \begin{bmatrix}Cosx(Cosy) - Senx(Seny)\end{bmatrix}
=(Sen y Z^2 - 2Cos y z) i - (Cos xz^2 - 2 Senxz) j + (Cos x Cos y - Sen x Sen y) k



Publicado por en 4 comentarios:

martes, 9 de septiembre de 2008

tarea 2.....

1.- Cambiar las coordenadas cilindricas dadas a coordenadas rectangulares

a) (5,\frac{ \Pi}{2} ,3)\\ b) (6,\frac{ \Pi}{3},-5)\\ \\ a) (5,\frac{ \Pi}{2} ,3)\\ x = 5 cos\Pi}{2}\\ x = 5(0)\\ x = 0


y = 5 sen \frac{ \Pi}{2}\\ =5(1)\\ \\ \\ z=3

(0, 5, 3)


B)


(6, \frac{ \Pi}{-5}, 3)\\ x= 6 cos \frac{ \Pi}{3}\\ x=6(\frac{1}{2})\\ x= 3\\ \\ \\ y = 6 sen \frac{ \Pi}{3} y = 6(.8660)\\ y = 5.2\\ \\ z = -5\\



(3, 5.2, -5)


2.- cambiar las coordenadas rectangulares a coordenadas esfericas



a) ( 1,1,\sqrt[]{2})\\ b) (1,\sqrt[]{3},0)\\ \rho=\sqrt[]{x^2+y^2+z^2}\\ tan^-1 \phi=\frac{y}{z}\\ cos \theta = \frac{z}{\sqrt[]{x^2+y^2+z^2}}\\ \\ \\ a)(1,1,\sqrt[]{2})\\ \rho=\sqrt[]{1^2+1^2+2^2}\\ =\sqrt[]{1+1+2}\\ =\sqrt[]{4}\\ \rho= 2\\



tan^-1 \phi = \frac{y}{z}\\ = \frac{1}{1}\\ tan^-1 \phi = 1.45\circ{},\frac{ \Pi}{4}\\ cos \phi = \frac{\sqrt[]{2}}{\sqrt[]{(1)^2 + (1)^2 + (\sqrt[]{2}^2}})\\ = \frac{\sqrt[]{2}}{\sqrt[]{4}}\\ cos \phi = 1 = 45\circ{}= .79 rad


(2, 45`,45`)



b) (1, \sqrt[]{3}, 0)

\rho = \sqrt[]{(1)^2+(\sqrt[]{3})^2+(0)^2}

= \sqrt[]{1 + 3 + 0}

= \sqrt[]{4} = 2


tan^-1 \phi = \frac{y}{z}

= \frac{\sqrt[]{3}}{1} = 60 º


cos \theta = \frac{0}{\sqrt[]{(1)^2+(\sqrt[]{3})^2} + (0)^2 }

= \frac{0}{\sqrt[]{1 + 3}}

= \frac{0}{\sqrt[]{4}} = \frac{0}{2}

cos \theta = 0 = 1 = \frac{ \Pi}{2}

( 2, 60º, \frac{ \Pi}{2} )


3.- Convertir las coordenadas esfericas dadas a coordenadas cilindricas.

  • a) ( 4, \frac{ \Pi}{3}, \frac{ \Pi}{3} )
  • b) (2, \frac{5}{6} \Pi, \frac{ \Pi}{4} )

a) ( 4, \frac{ \Pi}{3}, \frac{ \Pi}{3} )

ESFERAS-CARTESIANAS

x = \rho sen \phi cos \theta

= 4 sen \frac{ \Pi}{3} cos \frac{ \Pi}{3}
= 4 (.87)(.5)
x = 1.74


y = \rho sen \phi sen \theta

= 4 sen \frac{ \Pi}{3} sen \frac{ \Pi}{3}

= 4 (.87)(.87)

y = 3.03


z = 4 cos \frac{ \Pi}{3}

= 2


( 1.74, 3.03, 2 )

cartesianas a cilindricas

r = \sqrt[]{x^2 + y^2}

= \sqrt[]{(1.74)^2 + (3)^2}

= \sqrt[]{3 + 9} = \sqrt[]{12}

r = 3.46


tan ^-1 \theta = \frac{3}{1.74}

\theta = 1 = 60º = \frac{ \Pi}{3}

z = 2

( 3,46, 60º, 2 )



b) (2, \frac{5}{6} \Pi, \frac{ \Pi}{4} )

esfericas a cartesianas

x = 2 sen \frac{5}{6} \Pi cos \frac{ \Pi}{4}

= 2 (.5)(.71)

x = .71



y = 2 sen \frac{5}{6} \Pi sen \frac{ \Pi}{4}

= 2 (.5)(.71)

y = .71


z = 2 cos \frac{5}{6} \Pi

= 2 (-.87)

z = -1.73


( .71, .71, -1.73 )

cartesianas a cilindricas

r = \sqrt[]{x^2 + y^2}

= \sqrt[]{(.71)^2 + (.71)^2}

= \sqrt[]{.5 + .5} = \sqrt[]{1}

r = 1


tan ^-1 \theta = \frac{.71}{.71}

\theta = 1 = 60º = \frac{ \Pi}{3}

z = -1.74

( 1, \frac{ \Pi}{3}, -1.73 )




4.- Describir la grafica de ecuacion en 3 dimenciones.

  • a) \phi = \frac{ \Pi}{6}
  • b) \rho = 4 cos \phi

b) \rho = 4 cos \phi

\rho^2 = 4 \rho cos \phi

Ecuacion cartesiana

x^2 + y^2 + z^2 = 4z

x^2 + y^2 + z^2 -4z = 0

Completar Trinomio cuadrado perfecto

x^2 + y^2 + z^2+(\frac{4}{2})^2 = (\frac{4}{2})^2

x^2 + y^2 + z^2+4=4

C ( 0, 2, 0 ) r = 2


5.- encontrar una ecuacion en coordenadas cilindricas una en coordenadas esfericas para la grafica de la ecuacion dada:


a) x^2 + y^2 + z^2 = 4\\ b)y^2 + z^2 = 9

a)





b)

\\y^2 + z^2 = 9\\ y = (r sen \theta)^2 + z^2 = 9\\ r^2sen \theta^2 + z^2 = 9\\ r^2= \frac{9 - z^2}{sen \theta2}\\ r = \sqrt[]{\frac{9 - z^2}{sen \theta2}\\}






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