Given point P(-2, 6, 3) and vector B=yax+(x+z)ay, express P and B in cylindrical and spherical coordinates. Evaluate A at P in the Cartesian, cylindrical and spherical systems.

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dieseisB · Accepted Answer

Step 1 The objective is to express the point P(−2,6,3) and vector B=yax+(x+z)ay in cylindrical and spherical coordinates. Step 2 To convert Cartesian coordinates (x,y,z) to cylindrical coordinates (r,θ,z) r=x2+y2 θ=tan−1(yx) z=z The point P(−2,6,3) in cylindrical coordinates is, r=4+36 =40 =210 And θ=tan−1(6−2) =−1.25 Cylindrical coordinates is (210,−1.25,3) To convert Cartesian coordinates (x,y,z) to spherical coordinates (ρ,θ,ϕ) p=x2+y2+z2 θ=tan−1(yx) ϕ=tan−1(x2+y2z) The point P(−2,6,3) in spherical coordinates is, p=x2+y2+z2 4+36+9 =49 =7 And θ=tan−1(6−2) =−1.25 And ϕ=tan−1(4+363) =tan−1(403) =1.3034 Spherical coordinates is (7,−1.25,1.3034) Step 3 The vector B=yax+(x+z)ay in cylindrical and spherical coordinates. In the cartesian system B at P is B=6ax+ay For vector B, Bx=y By=x+z Bz=0 In cylindrical system Ar=ycos⁡θ+(x+z)sin⁡θ Aθ=−ysin⁡θ+(x+z)cos⁡θ

fudzisako · Accepted Answer

To express point P(-2, 6, 3) and vector 𝐁=y𝐚x+(x+z)𝐚y in cylindrical and spherical coordinates, we use the following transformations:1. Cylindrical Coordinates:P:(r,θ,z)wherer&amp;amp;=x2+y2=(−2)2+62=40=210,θ&amp;amp;=arctan(yx)=arctan(6−2)=arctan(−3)=−arctan(3),z&amp;amp;=z=3.𝐁 in cylindrical coordinates:𝐁:(Br,Bθ,Bz)whereBr&amp;amp;=Bx=0,Bθ&amp;amp;=By=x+z=−2+3=1,Bz&amp;amp;=Bz=0.2. Spherical Coordinates:P:(r,θ,ϕ)wherer&amp;amp;=x2+y2+z2=(−2)2+62+32=49=7,θ&amp;amp;=arctan(yx)=arctan(6−2)=arctan(−3)=−arctan(3),ϕ&amp;amp;=arctan(x2+y2z)=arctan(403)=arctan(2103).𝐁 in spherical coordinates:𝐁:(Br,Bθ,Bϕ)whereBr&amp;amp;=Bx=0,Bθ&amp;amp;=By=x+z=−2+3=1,Bϕ&amp;amp;=Bz=0.To evaluate 𝐀 at point P in different coordinate systems:1. Cartesian Coordinates:𝐀 at P: 𝐀(−2,6,3).2. Cylindrical Coordinates:𝐀 at P: 𝐀(210,−arctan(3),3).3. Spherical Coordinates:𝐀 at P: 𝐀(7,−arctan(3),arctan(210/3)).

xleb123 · Accepted Answer

1. Expressing point P (-2, 6, 3) in cylindrical coordinates:In cylindrical coordinates, a point is represented as (ρ,ϕ,z), where ρ is the distance from the origin to the point in the xy-plane, ϕ is the angle measured from the positive x-axis to the line segment connecting the origin and the point, and z is the height of the point.To find ρ and ϕ, we can use the following formulas:ρ=x2+y2ϕ=tan−1(yx)Given that P(−2,6,3), we can calculate:ρ=(−2)2+62=40=210ϕ=tan−1(6−2)=tan−1(−3)=−1.249Therefore, the cylindrical coordinates of point P are (210,−1.249,3).2. Expressing vector B as B=yax+(x+z)ay in cylindrical coordinates:To express vector B in cylindrical coordinates, we need to convert the Cartesian unit vectors ax and ay into cylindrical unit vectors. The cylindrical unit vectors are denoted as (ρ^,ϕ^,z^).The conversion formulas are:ρ^=cos(ϕ)ax+sin(ϕ)ayϕ^=−sin(ϕ)ax+cos(ϕ)ayz^=azGiven that B=yax+(x+z)ay, we can substitute the conversion formulas:B=y(cos(ϕ)ax+sin(ϕ)ay)+(x+z)(−sin(ϕ)ax+cos(ϕ)ay)Simplifying, we get:B=ycos(ϕ)ρ^−(x+z)sin(ϕ)ρ^+ysin(ϕ)ϕ^+(x+z)cos(ϕ)ϕ^Therefore, the expression of vector B in cylindrical coordinates is:B=(ycos(ϕ)−(x+z)sin(ϕ))ρ^+(ysin(ϕ)+(x+z)cos(ϕ))ϕ^.3. Expressing point P (-2, 6, 3) in spherical coordinates:In spherical coordinates, a point is represented as (r,θ,ϕ), where r is the distance from the origin to the point, θ is the angle measured from the positive z-axis to the line segment connecting the origin and the point, and ϕ is the angle measured from the positive x-axis to the projection of the line segment on the xy-plane.To find r, θ, and ϕ, we can use the following formulas:r=x2+y2+z2θ=cos−1(zr)ϕ=tan−1(yx)Given that &amp;lt;br&amp;gt;P(−2,6,3), we can calculate:r=(−2)2+62+32=49=7θ=cos−1(37)≈1.230ϕ=tan−1(6−2)=tan−1(−3)=−1.249Therefore, the spherical coordinates of point P are (7,1.230,−1.249).4. Evaluating vector A at point P in different coordinate systems:To evaluate vector A at point P, we can simply substitute the coordinates of P into the expression of vector A in each coordinate system.a) In Cartesian coordinates:A=xax+yay+zazSubstituting P(−2,6,3), we get:A=−2ax+6ay+3azb) In cylindrical coordinates:A=Aρρ^+Aϕϕ^+Azz^Substituting the cylindrical coordinates of P (210,−1.249,3), we get:A=Aρ(210,−1.249,3)+Aϕ(−sin(−1.249),cos(−1.249),0)+Az(0,0,1)c) In spherical coordinates:A=Arr^+Aθθ^+Aϕϕ^Substituting the spherical coordinates of P (7,1.230,−1.249), we get:A=Ar(7,1.230,−1.249)+Aθ(−sin(1.230)cos(−1.249),−sin(1.230)sin(−1.249),cos(1.230))+Aϕ(−sin(−1.249),cos(−1.249),0).

Given point P(-2, 6, 3) and vector B=ya_{x}+(x+z)a_{y}, express P and B in cylindrical and spherical coordinates. Evaluate A at P in the Cartesian, cylindrical and spherical systems.

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