(0, 5, 3)
B)
(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\\](http://rinconmatematico.com/latexrender/pictures/d4aabe78dd4c4ad8ca79d23ce7c21eea.png "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](http://rinconmatematico.com/latexrender/pictures/e4ddbfa689476a48afeadd50813aca07.png "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}](http://rinconmatematico.com/latexrender/pictures/b84ccc0f808c82dca2d7b0f887c64445.png "\sqrt[]{3}")
, 0)
= ![\sqrt[]{(1)^2+(\sqrt[]{3})^2+(0)^2}](http://rinconmatematico.com/latexrender/pictures/d02906b3951b1ec4bad80fa29433960f.png "\sqrt[]{(1)^2+(\sqrt[]{3})^2+(0)^2}")
=
![\sqrt[]{1 + 3 + 0}](http://rinconmatematico.com/latexrender/pictures/9c09fa7367fc29cdabeb3d3128b97b22.png "\sqrt[]{1 + 3 + 0}")
=
![\sqrt[]{4}](http://rinconmatematico.com/latexrender/pictures/c9134ba422cb8352ff186028fe0cbdfc.png "\sqrt[]{4}")
= 2tan
= 
=
![\frac{\sqrt[]{3}}{1}](http://rinconmatematico.com/latexrender/pictures/b373994e9f0fbd961f6bff1e983be23e.png "\frac{\sqrt[]{3}}{1}")
= 60 ºcos
= ![\frac{0}{\sqrt[]{(1)^2+(\sqrt[]{3})^2} + (0)^2 }](http://rinconmatematico.com/latexrender/pictures/abaa3c265f315dbbfc17ff681652777a.png "\frac{0}{\sqrt[]{(1)^2+(\sqrt[]{3})^2} + (0)^2 }")
=
![\frac{0}{\sqrt[]{1 + 3}}](http://rinconmatematico.com/latexrender/pictures/e65854deca22287fa54cb12b7f3d6b24.png "\frac{0}{\sqrt[]{1 + 3}}")
=
![\frac{0}{\sqrt[]{4}}](http://rinconmatematico.com/latexrender/pictures/fde7491a3d14387db314314f9eb1ea1f.png "\frac{0}{\sqrt[]{4}}")
= 
cos
= 0 = 1 = 
( 2, 60º,
)3.- Convertir las coordenadas esfericas dadas a coordenadas cilindricas.
- a) ( 4,
, )
- b) (2,
, 
)
a) ( 4,
, ) ESFERAS-CARTESIANAS
x =
sen cos = 4 sen
cos = 4 (.87)(.5)
x = 1.74
y =
sen sen = 4 sen
sen = 4 (.87)(.87)
y = 3.03
z = 4 cos
= 2
( 1.74, 3.03, 2 )
cartesianas a cilindricas
r =
![\sqrt[]{x^2 + y^2}](http://rinconmatematico.com/latexrender/pictures/daf63bf2a865860b88b0956aa5dc70a8.png "\sqrt[]{x^2 + y^2}")
=
![\sqrt[]{(1.74)^2 + (3)^2}](http://rinconmatematico.com/latexrender/pictures/83247e0c6b8f4ea3ffac011af5e378ca.png "\sqrt[]{(1.74)^2 + (3)^2}")
=
![\sqrt[]{3 + 9}](http://rinconmatematico.com/latexrender/pictures/84428fc9701695e8f354be3724247ee7.png "\sqrt[]{3 + 9}")
= ![\sqrt[]{12}](http://rinconmatematico.com/latexrender/pictures/7af2f08af843602622988af4f73c0473.png "\sqrt[]{12}")
r = 3.46
tan
= 
= 1 = 60º = z = 2
( 3,46, 60º, 2 )
b) (2,
, )esfericas a cartesianas
x = 2 sen
cos = 2 (.5)(.71)
x = .71
y = 2 sen
sen = 2 (.5)(.71)
y = .71
z = 2 cos
= 2 (-.87)
z = -1.73
( .71, .71, -1.73 )
cartesianas a cilindricas
r =
=
![\sqrt[]{(.71)^2 + (.71)^2}](http://rinconmatematico.com/latexrender/pictures/442f8748bf3efe286da1c5121239fcd9.png "\sqrt[]{(.71)^2 + (.71)^2}")
=
![\sqrt[]{.5 + .5}](http://rinconmatematico.com/latexrender/pictures/b3cf8c26ad64909154e4135172285459.png "\sqrt[]{.5 + .5}")
= ![\sqrt[]{1}](http://rinconmatematico.com/latexrender/pictures/c27c91cace800eaf0a3cd0936a90638a.png "\sqrt[]{1}")
r = 1
tan
= 
= 1 = 60º = z = -1.74
( 1,
, -1.73 )4.- Describir la grafica de ecuacion en 3 dimenciones.
- a) =
- b) = 4 cos
b)
= 4 cos 
= 4 cos Ecuacion cartesiana
= 4z
= 0Completar Trinomio cuadrado perfecto
= 
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)
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}\\}](http://rinconmatematico.com/latexrender/pictures/2b1a37463752664c4480c19cc2ed7055.png "\\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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