2nalfq8

Answered

2022-06-30

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):

$\frac{{\mathrm{sin}}^{2}\theta {\varphi}^{\prime}}{(1+{\mathrm{sin}}^{2}\theta ({\varphi}^{\prime}{)}^{2}{)}^{\frac{1}{2}}}=C$

where $C$ is an arbitrary constant and ${\varphi}^{\prime}$ denotes the derivative of $\varphi $ with respect to $\theta $.

My reasoning:

By setting $\varphi (0)=0$, any arc of a great circle will have no change in $\varphi $ with respect to $\theta $, so with this initial condition the answer follows by proving that ${\varphi}^{\prime}=0$. My issue is that upon working this round I end up with

$({\varphi}^{\prime}{)}^{2}=\frac{{C}^{2}}{{\mathrm{sin}}^{4}\theta -{C}^{2}{\mathrm{sin}}^{2}\theta}$

From this I can see no way forward.

Where do i go from here/ what should I do instead?

$\frac{{\mathrm{sin}}^{2}\theta {\varphi}^{\prime}}{(1+{\mathrm{sin}}^{2}\theta ({\varphi}^{\prime}{)}^{2}{)}^{\frac{1}{2}}}=C$

where $C$ is an arbitrary constant and ${\varphi}^{\prime}$ denotes the derivative of $\varphi $ with respect to $\theta $.

My reasoning:

By setting $\varphi (0)=0$, any arc of a great circle will have no change in $\varphi $ with respect to $\theta $, so with this initial condition the answer follows by proving that ${\varphi}^{\prime}=0$. My issue is that upon working this round I end up with

$({\varphi}^{\prime}{)}^{2}=\frac{{C}^{2}}{{\mathrm{sin}}^{4}\theta -{C}^{2}{\mathrm{sin}}^{2}\theta}$

From this I can see no way forward.

Where do i go from here/ what should I do instead?

Answer & Explanation

poquetahr

Expert

2022-07-01Added 18 answers

Let $u=\mathrm{cot}\theta $, then $1+{u}^{2}={\mathrm{csc}}^{2}\theta $ and $du=-{\mathrm{csc}}^{2}\theta \phantom{\rule{thinmathspace}{0ex}}d\theta $.

$\begin{array}{rl}\frac{d\varphi}{d\theta}& =\frac{d\theta}{\mathrm{sin}\theta \sqrt{{\mathrm{sin}}^{2}\theta -{C}^{2}}}\\ & =\frac{C{\mathrm{csc}}^{2}\theta}{\sqrt{1-{C}^{2}{\mathrm{csc}}^{2}\theta}}\\ d\varphi & =-\frac{C\phantom{\rule{thinmathspace}{0ex}}du}{\sqrt{1-{C}^{2}(1+{u}^{2})}}\\ & =-\frac{C\phantom{\rule{thinmathspace}{0ex}}du}{\sqrt{(1-{C}^{2})-{C}^{2}{u}^{2}}}\\ \text{(}C=\mathrm{cos}\alpha \text{)}& & =-\frac{du}{\sqrt{\mathrm{tan}{\alpha}^{2}-{u}^{2}}}\varphi & ={\mathrm{cos}}^{-1}\left(\frac{u}{\mathrm{tan}\alpha}\right)+\beta \\ \mathrm{cos}(\varphi -\beta )& =\frac{\mathrm{cot}\theta}{\mathrm{tan}\alpha}\\ \mathrm{cot}\theta & =\mathrm{tan}\alpha \mathrm{cos}(\varphi -\beta )\end{array}$

Rearrange,

$(\mathrm{sin}\theta \mathrm{cos}\varphi )(\mathrm{sin}\alpha \mathrm{cos}\beta )+(\mathrm{sin}\theta \mathrm{sin}\varphi )(\mathrm{sin}\alpha \mathrm{sin}\beta )=(\mathrm{cos}\theta )(\mathrm{cos}\alpha )$

which lies on the plane

$x\mathrm{sin}\alpha \mathrm{cos}\beta +y\mathrm{sin}\alpha \mathrm{sin}\beta -z\mathrm{cos}\alpha =0$

$\begin{array}{rl}\frac{d\varphi}{d\theta}& =\frac{d\theta}{\mathrm{sin}\theta \sqrt{{\mathrm{sin}}^{2}\theta -{C}^{2}}}\\ & =\frac{C{\mathrm{csc}}^{2}\theta}{\sqrt{1-{C}^{2}{\mathrm{csc}}^{2}\theta}}\\ d\varphi & =-\frac{C\phantom{\rule{thinmathspace}{0ex}}du}{\sqrt{1-{C}^{2}(1+{u}^{2})}}\\ & =-\frac{C\phantom{\rule{thinmathspace}{0ex}}du}{\sqrt{(1-{C}^{2})-{C}^{2}{u}^{2}}}\\ \text{(}C=\mathrm{cos}\alpha \text{)}& & =-\frac{du}{\sqrt{\mathrm{tan}{\alpha}^{2}-{u}^{2}}}\varphi & ={\mathrm{cos}}^{-1}\left(\frac{u}{\mathrm{tan}\alpha}\right)+\beta \\ \mathrm{cos}(\varphi -\beta )& =\frac{\mathrm{cot}\theta}{\mathrm{tan}\alpha}\\ \mathrm{cot}\theta & =\mathrm{tan}\alpha \mathrm{cos}(\varphi -\beta )\end{array}$

Rearrange,

$(\mathrm{sin}\theta \mathrm{cos}\varphi )(\mathrm{sin}\alpha \mathrm{cos}\beta )+(\mathrm{sin}\theta \mathrm{sin}\varphi )(\mathrm{sin}\alpha \mathrm{sin}\beta )=(\mathrm{cos}\theta )(\mathrm{cos}\alpha )$

which lies on the plane

$x\mathrm{sin}\alpha \mathrm{cos}\beta +y\mathrm{sin}\alpha \mathrm{sin}\beta -z\mathrm{cos}\alpha =0$

fythynwyrk0

Expert

2022-07-02Added 7 answers

We have

${\varphi}^{\prime}=\frac{C{\mathrm{sin}}^{-2}\theta}{\sqrt{1-{C}^{2}{\mathrm{sin}}^{-2}\theta}}$

now changing variable

$u=C\mathrm{cot}\theta \Rightarrow du=-{\mathrm{sin}}^{-2}\theta d\theta $

and then

$d\varphi =-\frac{du}{\sqrt{1-{u}^{2}}}$

and integrating

$\varphi ={C}_{1}-{\mathrm{sin}}^{-1}(u)={C}_{1}-{\mathrm{sin}}^{-1}(C\mathrm{cot}\theta )$

and then

$C\mathrm{cot}\theta =\mathrm{sin}({C}_{1}-\varphi )$

or

$C\mathrm{cos}\theta =\mathrm{sin}({C}_{1})\mathrm{sin}\theta \mathrm{cos}\varphi -\mathrm{cos}({C}_{1})\mathrm{sin}\theta \mathrm{sin}\varphi $

or changing to cartesian coordinates

$Cz-\mathrm{sin}({C}_{1})x+\mathrm{cos}({C}_{1})y=0$

which is the equation of the plane intersecting the sphere and containing the great circle.

${\varphi}^{\prime}=\frac{C{\mathrm{sin}}^{-2}\theta}{\sqrt{1-{C}^{2}{\mathrm{sin}}^{-2}\theta}}$

now changing variable

$u=C\mathrm{cot}\theta \Rightarrow du=-{\mathrm{sin}}^{-2}\theta d\theta $

and then

$d\varphi =-\frac{du}{\sqrt{1-{u}^{2}}}$

and integrating

$\varphi ={C}_{1}-{\mathrm{sin}}^{-1}(u)={C}_{1}-{\mathrm{sin}}^{-1}(C\mathrm{cot}\theta )$

and then

$C\mathrm{cot}\theta =\mathrm{sin}({C}_{1}-\varphi )$

or

$C\mathrm{cos}\theta =\mathrm{sin}({C}_{1})\mathrm{sin}\theta \mathrm{cos}\varphi -\mathrm{cos}({C}_{1})\mathrm{sin}\theta \mathrm{sin}\varphi $

or changing to cartesian coordinates

$Cz-\mathrm{sin}({C}_{1})x+\mathrm{cos}({C}_{1})y=0$

which is the equation of the plane intersecting the sphere and containing the great circle.

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