Fisica 2: Converciones de coordenadas

1> Cambiar las coordenadas cilindricas dadas a coordenadas rectangulares

a) ( 5, $\frac{\pi }{2}$ ,3)

x= r cos $\theta$

= 5 cos $\frac{\pi }{2}$
= 0

y= r sen $\theta$

= 5 sen $\frac{\pi }{2}$
= 5

z=z
= 3

C (0,5,3)

b) ( 6, $\frac{\pi }{3}$ ,-5)

x= r cos $\theta$

=6 cos $\frac{\pi }{3}$
=3

y= r sen $\theta$

= 6 sen $\frac{\pi }{3}$

= 5.2

z=z

=-5

c(3, 5.2,-5)

2> Cambiar las coordenadas rectangulares a coordenadas esfericas

a) ( 1,1, $\sqrt[2]{2}$ )

$\sigma$ = $\sqrt[2]{x^2+y^2+z^2}$

= $\sqrt[2]{1^2+1^2+\sqrt[2]{2}^2}$

$\theta$ = $tan^-1$ $\frac{y}{x}$

= $tan^-1$ 1

= 0.79

$\phi$ = $cos^-1$ $\frac{z}{\sqrt[2]{x^2+y^2+z^2}}$

= $cos^-1$ $\frac{2}{\sqrt[2]{1^2+1^2+\sqrt[2]{2}^2}}$

=0.79

c. esfericas ( 2, 0.79, 0. 79)

b) (1, $\sqrt[2]{3}$ , 0)

$\sigma$ = $\sqrt[2]{x^2+y^2+z^2}$

= $\sqrt[2]{1^2+\sqrt[2]{3}^2+0^2}$

=2

$\theta$ = $tan^-1$ $\frac{y}{x}$

= $tan^-1$ $\frac{\sqrt[2]{3}}{1}$

=1.05

$\phi$ = $cos^-1$ $\frac{z}{\sqrt[2]{x^2+y^2+z^2}}$

= $cos^-1$ $\frac{0}{\sqrt[2]{1^2+\sqrt[2]{3}^2+0^2}}$

=1.57

c esfericas (2,1.05, 1.57)

3> Convertir las coordenadas esfericas dadas a coordenadas cilindricas

a) ( 4, $\frac{\pi }{3}$ , $\frac{\pi }{3}$ )

x= 4 sen $\frac{\pi }{3}$ cos $\frac{\pi }{3}$

=4(.87)(.5)

=1.47 rad

y=4 sen $\frac{\pi }{3}$ sen $\frac{\pi }{3}$

= 4(.87) (.87)

= 3.02 rad

z= cos $\frac{\pi }{3}$

= 4(.5)

=2 rad

c ( 1.74,3.02 ,2)

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

= $\sqrt[2]{1.72^2+3.02^2}$

= 3.49

$\theta$ = $tan^-1$ $\frac{y}{x}$

= $tan^-1$ $\frac{3.02}{1.74}$

= 1.05

z=z

=2

C( 3.49, 1.05,2)

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

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

=2(.5)(.71)

= 0.71

y= 2 sen $\frac{5}{6}\pi$ sen $\frac{\pi }{4}$

=2(.5)(.71)

=0.71

z= 2 cos $\frac{5}{6}\pi$

= 2 (-.87)

=-1.74

c (.71,.71,-1.74)

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

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

=1

$\theta$ = $tan^-1$ $\frac{y}{x}$

= $tan^-1$ 1

= .78

z=z

=-1.74

c ( 1,0.78,-1.74)

4> Describir la grafica de la ecuacion en 3 dimenciones

a) $\phi$ = $\frac{\pi }{6}$
formula
z= $\rho$ cos $\phi$

* $\rho=\sqrt[2]{x^2+y^2+z^2}$

entonses:

cos $\pi$ 6 = $\frac{\sqrt[2]{3}}{2}$

z= $\rho$ $\frac{\sqrt[2]{3}}{2}$

* $z^2$ =( ${x^2+y^2+z^2}$ ) $\frac{3}{4}$

4 $z^2$ =( ${x^2+y^2+z^2}$ )3

${x^2+y^2-\frac{1}{3}z^2$ = 0

b) $\rho$ = 4 cos $\phi$

$\rho$ ( $\rho$ = 4 cos $\phi$ )

$\rho^2$ = 4 $\rho$ cos $\phi$

formulas utilizadas

$\rho^2$ = $x^2+y^2+z^2$

z = $\rho$ cos $\phi$

entonces:
$x^2+y^2+z^2$ =4z

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

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

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

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

ecuacion de esfera

5> Encontrar una ecuación en coordenadas cilíndricas y una en coordenadas esféricas para la gráfica de la ecuación dada:

a) $x^2+y^2+z^2$ = 4

Cilíndrica

Formulas

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

z=z

entonces:

$r^2$ + $z^2$ =4

Esféricas

formulas

$\rho$ = $x^2+y^2+z^2$

entonces:

$\rho^2$ = 4

b) $y^2+z^2$ =9

Cilindricas

Formulas

y= r sen $\theta$
z=z

$r^2$ $sen^2$ $\theta$ + $z^2$ =9

r sen $\theta$ +z = 3

Esfericas

Formulas

$\rho^2$ = $x^2+y^2+z^2$

x= $\rho$ sen $\theta$ cos $\theta$

entonses
$\rho^2$ - $x^2$ = $y^2+z^2$

$\rho^2$ - $x^2$ = 9

$\rho^2$ = $\frac{x^2}{sen^2 \phi cos^2 \theta}$

$\frac{x^2}{sen^2 \phi cos^2 \theta}$ - $x^2$ =9

$\frac{0}{sen^2 \phi cos^2 \theta}$ =9

9 ${sen^2 \phi cos^2 \theta}$ = 0

11 de septiembre de 2008

Converciones de coordenadas

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