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

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):
$\frac{{\mathrm{sin}}^{2}\theta {\varphi }^{\prime }}{\left(1+{\mathrm{sin}}^{2}\theta \left({\varphi }^{\prime }{\right)}^{2}{\right)}^{\frac{1}{2}}}=C$
where $C$ is an arbitrary constant and ${\varphi }^{\prime }$ denotes the derivative of $\varphi$ with respect to $\theta$.

My reasoning:
By setting $\varphi \left(0\right)=0$, 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
$\left({\varphi }^{\prime }{\right)}^{2}=\frac{{C}^{2}}{{\mathrm{sin}}^{4}\theta -{C}^{2}{\mathrm{sin}}^{2}\theta }$
From this I can see no way forward.
Where do i go from here/ what should I do instead?
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Elliott Gilmore
Let $u=\mathrm{cot}\theta$, then $1+{u}^{2}={\mathrm{csc}}^{2}\theta$ and $du=-{\mathrm{csc}}^{2}\theta \phantom{\rule{thinmathspace}{0ex}}d\theta$.
$\begin{array}{rl}\frac{d\varphi }{d\theta }& =\frac{d\theta }{\mathrm{sin}\theta \sqrt{{\mathrm{sin}}^{2}\theta -{C}^{2}}}\\ & =\frac{C{\mathrm{csc}}^{2}\theta }{\sqrt{1-{C}^{2}{\mathrm{csc}}^{2}\theta }}\\ d\varphi & =-\frac{C\phantom{\rule{thinmathspace}{0ex}}du}{\sqrt{1-{C}^{2}\left(1+{u}^{2}\right)}}\\ & =-\frac{C\phantom{\rule{thinmathspace}{0ex}}du}{\sqrt{\left(1-{C}^{2}\right)-{C}^{2}{u}^{2}}}\\ \text{(}C=\mathrm{cos}\alpha \text{)}& & =-\frac{du}{\sqrt{\mathrm{tan}{\alpha }^{2}-{u}^{2}}}\varphi & ={\mathrm{cos}}^{-1}\left(\frac{u}{\mathrm{tan}\alpha }\right)+\beta \\ \mathrm{cos}\left(\varphi -\beta \right)& =\frac{\mathrm{cot}\theta }{\mathrm{tan}\alpha }\\ \mathrm{cot}\theta & =\mathrm{tan}\alpha \mathrm{cos}\left(\varphi -\beta \right)\end{array}$
Rearrange,
$\left(\mathrm{sin}\theta \mathrm{cos}\varphi \right)\left(\mathrm{sin}\alpha \mathrm{cos}\beta \right)+\left(\mathrm{sin}\theta \mathrm{sin}\varphi \right)\left(\mathrm{sin}\alpha \mathrm{sin}\beta \right)=\left(\mathrm{cos}\theta \right)\left(\mathrm{cos}\alpha \right)$
which lies on the plane
$x\mathrm{sin}\alpha \mathrm{cos}\beta +y\mathrm{sin}\alpha \mathrm{sin}\beta -z\mathrm{cos}\alpha =0$

gorgeousgen9487
We have
${\varphi }^{\prime }=\frac{C{\mathrm{sin}}^{-2}\theta }{\sqrt{1-{C}^{2}{\mathrm{sin}}^{-2}\theta }}$
now changing variable
$u=C\mathrm{cot}\theta ⇒du=-{\mathrm{sin}}^{-2}\theta d\theta$
and then
$d\varphi =-\frac{du}{\sqrt{1-{u}^{2}}}$
and integrating
$\varphi ={C}_{1}-{\mathrm{sin}}^{-1}\left(u\right)={C}_{1}-{\mathrm{sin}}^{-1}\left(C\mathrm{cot}\theta \right)$
and then
$C\mathrm{cot}\theta =\mathrm{sin}\left({C}_{1}-\varphi \right)$
or
$C\mathrm{cos}\theta =\mathrm{sin}\left({C}_{1}\right)\mathrm{sin}\theta \mathrm{cos}\varphi -\mathrm{cos}\left({C}_{1}\right)\mathrm{sin}\theta \mathrm{sin}\varphi$
or changing to cartesian coordinates
$Cz-\mathrm{sin}\left({C}_{1}\right)x+\mathrm{cos}\left({C}_{1}\right)y=0$
which is the equation of the plane intersecting the sphere and containing the great circle.