Answer to Question #151520 in Differential Geometry | Topology for Dolly

Question #151520
Find the rectangular coordinates of the point whose spherical coordinates are
1) (5, π/2,π/2)
2) (4, π/3, 2π/3)
3) (0,π/11,π/5)
4) (2, 5π/3, 3π/4)
1
Expert's answer
2020-12-21T19:06:49-0500

(1) Given spherical coordinates: "(5,\\frac{\\pi }{2},\\frac{\\pi }{2})"

Require to find the rectangular coordinates.

Now "(\\rho ,\\theta ,\\varphi )=(5,\\frac{\\pi }{2},\\frac{\\pi }{2})"

"\\Rightarrow \\rho=5,\\theta=\\frac{\\pi }{2},\\varphi =\\frac{\\pi }{2}"

The relation between the rctangular coordinates and spherical coordinates are given by the equaitons

"x=\\rho sin\\varphi cos\\theta ,y=\\rho sin\\varphi sin\\theta,z=\\rho cos\\varphi"

Subtituting the given values, we get

"x=5sin(\\frac{\\pi }{2})cos(\\frac{\\pi }{2})=5(1)(0)=0"

"y=5sin(\\frac{\\pi }{2})sin(\\frac{\\pi }{2})=5(1)(1)=5"

"z=5cos(\\frac{\\pi }{2})=5(0)=0"

Therefore, "(x,y,z)=(0,5,0)"

(2) Gievn: "(4,\\frac{\\pi }{3},\\frac{2pi }{3})"

"x=4sin(\\frac{2pi }{3})cos(\\frac{pi }{3})=4(\\frac{\\sqrt{3}}{2})(\\frac{1}{2})=\\sqrt{3}"

"y=4sin(\\frac{2pi }{3})sin(\\frac{pi }{3})=4(\\frac{\\sqrt{3}}{2})(\\frac{\\sqrt{3}}{2})=3"

"z=4cos(\\frac{2pi }{3})=4(-\\frac{1}{2})=-2"

Therefore, "(x,y,z)=(\\sqrt{3},3,-2)"

(3)Given: "(0,\\frac{\\pi }{11},\\frac{\\pi }{5})"

"\\Rightarrow \\rho=0,\\theta=\\frac{\\pi }{11},\\varphi =\\frac{\\pi }{5}"

"x=0sin(\\frac{pi }{5})cos(\\frac{pi }{11})=0"

"y=0sin(\\frac{pi }{5})sin(\\frac{pi }{11})=0"

"z=0cos(\\frac{pi }{5})=0"

Therefore, "(x,y,z)=(0,0,0)"

(4) Given: "(2,\\frac{5pi }{3},\\frac{3pi }{4})"

"\\Rightarrow \\rho=2,\\theta=\\frac{5pi }{3},\\varphi =\\frac{3pi }{4}"

"x=2sin(\\frac{3pi }{4})cos(\\frac{5pi }{3})=2(\\frac{1}{\\sqrt2})(\\frac{1}{2})=\\frac{1}{\\sqrt2}"

"y=2sin(\\frac{3pi }{4})sin(\\frac{5pi }{3})=2(\\frac{1}{\\sqrt2})(\\frac{-\\sqrt3}{2})=-\\frac{\\sqrt3}{\\sqrt2}"

"z=2cos(\\frac{3pi }{4})=2(-\\frac{1}{\\sqrt2})=-\\sqrt2"

Therefore, "(x,y,z)=(\\frac{1}{\\sqrt2},-\\frac{\\sqrt3}{\\sqrt2},-\\sqrt2)"



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