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## Homework Statement

I'm trying to solve the equations:

[tex] \ddot{\phi} + 2\left(\frac{\cos \theta}{\sin \theta}\right) \dot{\theta}\dot{\phi} =0 [/tex]

and

[tex] \ddot{\theta} - \sin \theta \cos \theta \dot{\phi^2} =0 [/tex]

for [itex] \theta(\lambda), \phi(\lambda) [/itex] where the dots represent differentiation w.r.t the parameter [itex] \lambda [/itex].

## Homework Equations

Chain rule, other differentiation tricks

## The Attempt at a Solution

I've tried to tackle the first equation by writing:[tex] \frac{\cos \theta}{\sin \theta} \dot{\theta} = \frac{1}{\sin \theta} \frac{d}{d\theta}(\sin \theta) \dot{\theta} = \frac{d}{d\lambda}[\ln (\sin \theta) ] [/tex]

I'm wondering if it's correct to do that. It seems to follow from the chain rule, but I'm not sure. If so, then the equation becomes:

[tex]\frac{\ddot{\phi}}{\dot{\phi}} = -2\frac{d}{d\lambda}[\ln (\sin \theta) ] [/tex]

[tex]\frac{1}{\dot{\phi}}\frac{d \dot{\phi}}{d\lambda} = -2\frac{d}{d\lambda}[\ln (\sin \theta) ] [/tex]

[tex]\frac{d \ln ( \dot{\phi})}{d\lambda} = -2\frac{d}{d\lambda}[\ln (\sin \theta) ] [/tex]

[tex]\frac{1}{\dot{\phi}}\frac{d \dot{\phi}}{d\lambda} = -2\frac{d}{d\lambda}[\ln (\sin \theta) ] [/tex]

[tex]\frac{d \ln ( \dot{\phi})}{d\lambda} = -2\frac{d}{d\lambda}[\ln (\sin \theta) ] [/tex]

Then I integrated both sides w.r.t. lambda:

[tex]\ln ( \dot{\phi}) = - 2\ln (\sin \theta) + \mathrm{~const.} [/tex]

[tex] \dot{\phi} \propto \frac{1}{\sin^2 \theta} [/tex]

[tex] \dot{\phi} \propto \frac{1}{\sin^2 \theta} [/tex]

Then I tried substituting this result into the second ODE, but it gave me:

[tex] \ddot{\theta} \propto \frac{\cos \theta}{\sin^3 \theta} [/tex]

which I don't know what to do with.

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