I have to show that the solution of a differential equation is an arc of a great circle. The differential equation is as follows (in spherical coordinates): sin 2 ⁡ θ ϕ ′ ( 1 + sin 2 ⁡ θ ( ϕ ′ ) 2 ) 1 2 = Cwhere C is an arbitrary constant and ϕ ′ denotes the derivative of ϕ with respect to θ.My reasoning:By setting ϕ ( 0 ) = 0, any arc of a great circle will have no change in ϕ with respect to θ, so with this initial condition the answer follows by proving that ϕ ′ = 0. My issue is that upon working this round I end up with ( ϕ ′ ) 2 = C 2 sin 4 ⁡ θ − C 2 sin 2 ⁡ θ From this I can see no way forward.Where do i go from here/ what should I do instead?

Question

I have to show that the solution of a differential equation is an arc of a great circle. The differential equation is as follows (in spherical coordinates):                    sin        2            ⁡      θ              ϕ        ′                    (      1      +              sin        2            ⁡      θ      (              ϕ        ′                    )        2                    )                              1            2                                =  Cwhere   C is an arbitrary constant and       ϕ    ′   denotes the derivative of   ϕ with respect to   θ.My reasoning:By setting   ϕ  (  0  )  =  0, any arc of a great circle will have no change in   ϕ with respect to   θ, so with this initial condition the answer follows by proving that       ϕ    ′    =  0. My issue is that upon working this round I end up with  (      ϕ    ′        )    2    =            C      2                      sin        4            ⁡      θ      −              C        2                    sin        2            ⁡      θ      From this I can see no way forward.Where do i go from here/ what should I do instead?

poquetahr · Accepted Answer

Let   u  =  cot  ⁡  θ, then   1  +      u    2    =      csc    2    ⁡  θ and   d  u  =  −      csc    2    ⁡  θ    d  θ.                                          d            ϕ                                d            θ                                              =                              d            θ                                sin            ⁡            θ                                          sin                2                            ⁡              θ              −                              C                2                                                                                      =                              C                          csc              2                        ⁡            θ                                1            −                          C              2                                      csc              2                        ⁡            θ                                              d        ϕ                            =        −                              C                        d            u                                1            −                          C              2                        (            1            +                          u              2                        )                                                            =        −                              C                        d            u                                (            1            −                          C              2                        )            −                          C              2                                      u              2                                                                    (                      C            =            cos            ⁡            α                    )                                          =        −                              d            u                                tan            ⁡                          α              2                        −                          u              2                                                          ϕ                            =                  cos                      −            1                          ⁡                  (                      u                          tan              ⁡              α                                )                +        β                            cos        ⁡        (        ϕ        −        β        )                            =                              cot            ⁡            θ                                tan            ⁡            α                                              cot        ⁡        θ                            =        tan        ⁡        α        cos        ⁡        (        ϕ        −        β        )            Rearrange,  (  sin  ⁡  θ  cos  ⁡  ϕ  )  (  sin  ⁡  α  cos  ⁡  β  )  +  (  sin  ⁡  θ  sin  ⁡  ϕ  )  (  sin  ⁡  α  sin  ⁡  β  )  =  (  cos  ⁡  θ  )  (  cos  ⁡  α  )which lies on the plane  x  sin  ⁡  α  cos  ⁡  β  +  y  sin  ⁡  α  sin  ⁡  β  −  z  cos  ⁡  α  =  0

fythynwyrk0 · Accepted Answer

We have      ϕ    ′    =            C              sin                  −          2                    ⁡      θ              1      −              C        2                    sin                  −          2                    ⁡      θ      now changing variable  u  =  C  cot  ⁡  θ  ⇒  d  u  =  −      sin          −      2        ⁡  θ  d  θand then  d  ϕ  =  −            d      u              1      −              u        2            and integrating  ϕ  =      C    1    −      sin          −      1        ⁡  (  u  )  =      C    1    −      sin          −      1        ⁡  (  C  cot  ⁡  θ  )and then  C  cot  ⁡  θ  =  sin  ⁡  (      C    1    −  ϕ  )or  C  cos  ⁡  θ  =  sin  ⁡  (      C    1    )  sin  ⁡  θ  cos  ⁡  ϕ  −  cos  ⁡  (      C    1    )  sin  ⁡  θ  sin  ⁡  ϕor changing to cartesian coordinates  C  z  −  sin  ⁡  (      C    1    )  x  +  cos  ⁡  (      C    1    )  y  =  0which is the equation of the plane intersecting the sphere and containing the great circle.

I have to show that the solution of a differential equation is an arc of a great circle. The differe

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