Respuesta :
Answer and Step-by-step explanation: Spherical coordinate describes a location of a point in space: one distance (ρ) and two angles (Ф,θ).To transform cartesian coordinates into spherical coordinates:
[tex]\rho = \sqrt{x^{2}+y^{2}+z^{2}}[/tex]
[tex]\phi = cos^{-1}\frac{z}{\rho}[/tex]
For angle θ:
- If x > 0 and y > 0: [tex]\theta = tan^{-1}\frac{y}{x}[/tex];
- If x < 0: [tex]\theta = \pi + tan^{-1}\frac{y}{x}[/tex];
- If x > 0 and y < 0: [tex]\theta = 2\pi + tan^{-1}\frac{y}{x}[/tex];
Calculating:
a) (4,2,-4)
[tex]\rho = \sqrt{4^{2}+2^{2}+(-4)^{2}}[/tex] = 6
[tex]\phi = cos^{-1}(\frac{-4}{6})[/tex]
[tex]\phi = cos^{-1}(\frac{-2}{3})[/tex]
For θ, choose 1st option:
[tex]\theta = tan^{-1}(\frac{2}{4})[/tex]
[tex]\theta = tan^{-1}(\frac{1}{2})[/tex]
b) (0,8,15)
[tex]\rho = \sqrt{0^{2}+8^{2}+(15)^{2}}[/tex] = 17
[tex]\phi = cos^{-1}(\frac{15}{17})[/tex]
[tex]\theta = tan^{-1}\frac{y}{x}[/tex]
The angle θ gives a tangent that doesn't exist. Analysing table of sine, cosine and tangent: θ = [tex]\frac{\pi}{2}[/tex]
c) (√2,1,1)
[tex]\rho = \sqrt{(\sqrt{2} )^{2}+1^{2}+1^{2}}[/tex] = 2
[tex]\phi = cos^{-1}(\frac{1}{2})[/tex]
[tex]\phi[/tex] = [tex]\frac{\pi}{3}[/tex]
[tex]\theta = tan^{-1}\frac{1}{\sqrt{2} }[/tex]
d) (−2√3,−2,3)
[tex]\rho = \sqrt{(-2\sqrt{3} )^{2}+(-2)^{2}+3^{2}}[/tex] = 5
[tex]\phi = cos^{-1}(\frac{3}{5})[/tex]
Since x < 0, use 2nd option:
[tex]\theta = \pi + tan^{-1}\frac{1}{\sqrt{3} }[/tex]
[tex]\theta = \pi + \frac{\pi}{6}[/tex]
[tex]\theta = \frac{7\pi}{6}[/tex]
Cilindrical coordinate describes a 3 dimension space: 2 distances (r and z) and 1 angle (θ). To express cartesian coordinates into cilindrical:
[tex]r=\sqrt{x^{2}+y^{2}}[/tex]
Angle θ is the same as spherical coordinate;
z = z
Calculating:
a) (4,2,-4)
[tex]r=\sqrt{4^{2}+2^{2}}[/tex] = [tex]\sqrt{20}[/tex]
[tex]\theta = tan^{-1}\frac{1}{2}[/tex]
z = -4
b) (0, 8, 15)
[tex]r=\sqrt{0^{2}+8^{2}}[/tex] = 8
[tex]\theta = \frac{\pi}{2}[/tex]
z = 15
c) (√2,1,1)
[tex]r=\sqrt{(\sqrt{2} )^{2}+1^{2}}[/tex] = [tex]\sqrt{3}[/tex]
[tex]\theta = \frac{\pi}{3}[/tex]
z = 1
d) (−2√3,−2,3)
[tex]r=\sqrt{(-2\sqrt{3} )^{2}+(-2)^{2}}[/tex] = 4
[tex]\theta = \frac{7\pi}{6}[/tex]
z = 3
