Fisica II: 09/04/08

1) Cambiar las coordenadas cilindricas dadas, a coordenadas rectangulares.

a) $(5, \frac{ \Pi}{2}, 3)$
x = r Cos $\theta$ ----> 5 Cos $\displaystyle\frac{\prod_{}^{}{}}{2}$ = 5 (0) = 0

y = r Sen $\theta$ ----> 5 Sen $\displaystyle\frac{\prod_{}^{}{}}{2}$ = 5 (1) = 5

z = z ----------------------------> = 3

P (0, 5, 3)

b) $(6, \frac{ \Pi}{3}, -5)$
x = r Cos $\theta$ ----> 6 Cos $\displaystyle\frac{\prod_{}^{}{}}{3}$ = 6 ( $\displaystyle\frac{1}{2}$ ) = 3

y = r Sen $\theta$ ----> 5 Sen $\displaystyle\frac{\prod_{}^{}{}}{3}$ = 6 $(\displaystyle\frac{1}{2}\sqrt{3})$ = $3\sqrt{3}$

z = z ----------------------------> = -5

P (3, $3\sqrt{3}$ , -5)

2) Cambiar las coordenadas rectangulares a coordenadas esfèricas.

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

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

$=\sqrt{(1)^2+(1)^2+(\sqrt{2})^2}$

$=\sqrt{1+1+2}$

$=\sqrt{4}$

=2

Arctan $\theta$ = $\displaystyle\frac{y}{x}$

$=\displaystyle\frac{1}{1}$
Arctg $\theta$ = 1
Arctg $\theta$ = 45º
$\theta$ = $\displaystyle\frac{\pi }{4}$

$\cos{ \phi} = \displaystyle\frac{z}{\sqrt{x^2+y^2+z^2}}$

$= \displaystyle\frac{\sqrt{2}}{\sqrt{(1)^2+(1)^2+(\sqrt{2})^2}}$

$= \displaystyle\frac{\sqrt{2}}{\sqrt{4}}$

Cos $\phi$ = $\displaystyle\frac{1}{2}\sqrt{2}$

$\phi = \displaystyle\frac{\pi }{4}$

P ( $2,\: \displaystyle\frac{\pi }{4}, \: \displaystyle\frac{\pi }{4}$ )

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

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

$= \sqrt{1^2+(\sqrt{3})^2+0^2}$

$= \sqrt{4}$

= 2

ArcTg $\theta$ = $\displaystyle\frac{y}{x}$

$=\displaystyle\frac{\sqrt{3}}{1}$

ArcTg $\theta$ = $\displaystyle\frac{\sqrt{3}}{1}$

$\theta$ = $\displaystyle\frac{\pi }{6}$

$\cos{ \phi} = \displaystyle\frac{z}{\sqrt{x^2+y^2+z^2}}$

$= \displaystyle\frac{0}{\sqrt{1^2+(\sqrt{3})^2+0^2}}$

$= \displaystyle\frac{0}{2}$

= 0

$\phi =\displaystyle\frac{\pi }{2}$

P(2, $\displaystyle\frac{\pi }{6}$ , $\displaystyle\frac{\pi }{2}$ )

3) Convertir las coordenadas esfèricas dadas a coordenadas cilìndricas.

a) $(4, \displaystyle\frac{\pi }{3}, \displaystyle\frac{\pi }{3})$ Esfericas-Cartesianas

x = $\rho Sen \phi Cos\theta$

= 4 Sen $\displaystyle\frac{\pi }{3}$ Cos $\displaystyle\frac{\pi }{3}$

=(4) ( $\displaystyle\frac{1}{2}\sqrt{3}$ ) ( $\displaystyle\frac{1}{2}$ )

= 1.73

y = $\rho Sen \phi Sen \theta$

= 4Sen $\displaystyle\frac{\pi }{3}$ Sen $\displaystyle\frac{\pi }{3}$

= (4) ( $\displaystyle\frac{1}{2}\sqrt{3}$ ) ( $\displaystyle\frac{1}{2}\sqrt{3}$ )

= 3

z = $\rho Cos \phi$

= 4 Cos $\displaystyle\frac{\pi }{3}$

= (4) ( $\displaystyle\frac{1}{2}$ )

= 2
P (1.79, 3, 2)

Cartesianas-Cilindricas

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

= $\sqrt{1.73^2+3^2}$

= $\sqrt{11.99}$

= 3.46

Arctg $\theta$ = $\frac{y}{x}$

= $\displaystyle\frac{3}{1.73}$

$\theta$ = 1.73

= $\sqrt{3}$

= $\displaystyle\frac{\pi }{3}$

z = z

= 2

P(3.46, $\displaystyle\frac{\pi }{3}$ , 2)

b) $(2, \displaystyle\frac{5}{6}\pi, \displaystyle\frac{\pi }{4} )$ Esfericas-Cartesianas

x = $\rho Sen \phi Cos\theta$

= 2 Sen $\displaystyle\frac{5}{6}\pi$ Cos $\displaystyle\frac{\pi }{4}$

= (2) ( $\displaystyle\frac{1}{2}$ ) ( $\displaystyle\frac{1}{2}\sqrt{2}$ )

= 0.70

y = $\rho Sen \phi Sen \theta$

= 2 Sen $\displaystyle\frac{5}{6}\pi$ Sen $\displaystyle\frac{\pi }{4}$

= (2) ( $\displaystyle\frac{1}{2}$ ) ( $\displaystyle\frac{1}{2}\sqrt{2}$ )

= 0.70

z = $\rho Cos \phi$

= 2 Cos $\displaystyle\frac{5}{6}\pi$

= (2) ( $-\displaystyle\frac{1}{2}\sqrt{3}$ )

= -1.73

P(0.70, 0.70, -1.73)

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

= $\sqrt{0.70^2+0.70^2}$

= $\sqrt{0.98}$

= 0.99

Arctg $\theta$ = $\frac{y}{x}$

= $\frac{0.70}{0.70}$

= 1

$\theta$ = 1

= $\displaystyle\frac{\pi }{4}$

z = z

= -1.73

P(0.99, $\displaystyle\frac{\pi }{4}$ , -1.73)

4) Describir la grafica de la ecuacion en 3 dimensiones.

a) $\phi = \displaystyle\frac{\pi }{6}$

= $\rho Cos\phi$

$=\rho \displaystyle\frac{\sqrt{3}}{2}$

z $=\rho \displaystyle\frac{\sqrt{3}}{2}$

$z = (\sqrt{x^2 + y^2 + z^2})^2 (\displaystyle\frac{\sqrt{3}}{2})^2$

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

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

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

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

$\displaystyle\frac{3}{2}x^2 + \displaystyle\frac{3}{2}y^2 + (\displaystyle\frac{\pi }{6}\pi-\displaystyle\frac{3}{4})^2$

P $(\displaystyle\frac{3}{2},\displaystyle\frac{3}{2},\displaystyle\frac{3}{4})$

b) $\rho = 4 Cos\phi$

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

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

$z =\rho Cos\phi$

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

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

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

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

$x^2 + y^2 +(z+2)^2$

C(0, 0, 2)

Fisica II

jueves, 4 de septiembre de 2008

Tarea 2

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