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,π2,π2)(5,\frac{\pi }{2},\frac{\pi }{2})

Require to find the rectangular coordinates.

Now (ρ,θ,φ)=(5,π2,π2)(\rho ,\theta ,\varphi )=(5,\frac{\pi }{2},\frac{\pi }{2})

ρ=5,θ=π2,φ=π2\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=ρsinφcosθ,y=ρsinφsinθ,z=ρcosφx=\rho sin\varphi cos\theta ,y=\rho sin\varphi sin\theta,z=\rho cos\varphi

Subtituting the given values, we get

x=5sin(π2)cos(π2)=5(1)(0)=0x=5sin(\frac{\pi }{2})cos(\frac{\pi }{2})=5(1)(0)=0

y=5sin(π2)sin(π2)=5(1)(1)=5y=5sin(\frac{\pi }{2})sin(\frac{\pi }{2})=5(1)(1)=5

z=5cos(π2)=5(0)=0z=5cos(\frac{\pi }{2})=5(0)=0

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

(2) Gievn: (4,π3,2pi3)(4,\frac{\pi }{3},\frac{2pi }{3})

x=4sin(2pi3)cos(pi3)=4(32)(12)=3x=4sin(\frac{2pi }{3})cos(\frac{pi }{3})=4(\frac{\sqrt{3}}{2})(\frac{1}{2})=\sqrt{3}

y=4sin(2pi3)sin(pi3)=4(32)(32)=3y=4sin(\frac{2pi }{3})sin(\frac{pi }{3})=4(\frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2})=3

z=4cos(2pi3)=4(12)=2z=4cos(\frac{2pi }{3})=4(-\frac{1}{2})=-2

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

(3)Given: (0,π11,π5)(0,\frac{\pi }{11},\frac{\pi }{5})

ρ=0,θ=π11,φ=π5\Rightarrow \rho=0,\theta=\frac{\pi }{11},\varphi =\frac{\pi }{5}

x=0sin(pi5)cos(pi11)=0x=0sin(\frac{pi }{5})cos(\frac{pi }{11})=0

y=0sin(pi5)sin(pi11)=0y=0sin(\frac{pi }{5})sin(\frac{pi }{11})=0

z=0cos(pi5)=0z=0cos(\frac{pi }{5})=0

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

(4) Given: (2,5pi3,3pi4)(2,\frac{5pi }{3},\frac{3pi }{4})

ρ=2,θ=5pi3,φ=3pi4\Rightarrow \rho=2,\theta=\frac{5pi }{3},\varphi =\frac{3pi }{4}

x=2sin(3pi4)cos(5pi3)=2(12)(12)=12x=2sin(\frac{3pi }{4})cos(\frac{5pi }{3})=2(\frac{1}{\sqrt2})(\frac{1}{2})=\frac{1}{\sqrt2}

y=2sin(3pi4)sin(5pi3)=2(12)(32)=32y=2sin(\frac{3pi }{4})sin(\frac{5pi }{3})=2(\frac{1}{\sqrt2})(\frac{-\sqrt3}{2})=-\frac{\sqrt3}{\sqrt2}

z=2cos(3pi4)=2(12)=2z=2cos(\frac{3pi }{4})=2(-\frac{1}{\sqrt2})=-\sqrt2

Therefore, (x,y,z)=(12,32,2)(x,y,z)=(\frac{1}{\sqrt2},-\frac{\sqrt3}{\sqrt2},-\sqrt2)



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