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So for the proof of Kepler's first law they make a simplification of a differential equation, but I don't get how they come up with that simplification. It goes like this

$$2\frac{dr}{dt}\frac{d\theta}{dt}+r\frac{d^2\theta}{dt^2}=0$$ once simplified (which I don't understand) they came up with this answer $$\frac{1}{2r}\frac{d}{dt}\left(r\left(\frac{d\theta}{dt}\right)^2\right)=0$$ could someone please explain to me what they did?

thanks in advance.

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    $\begingroup$ Apply the chain rule to the derivative in your second equation. $\endgroup$
    – Karl
    Dec 23, 2020 at 21:59
  • $\begingroup$ Possibly related. $\endgroup$
    – K.defaoite
    Dec 23, 2020 at 22:00
  • $\begingroup$ @Karl thanks for your reply but what I actually want to know is how would you come up with this equation? Are there any steps I can apply on the first equation? $\endgroup$
    – TheCreator
    Dec 23, 2020 at 22:04
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    $\begingroup$ I'd say it's more of a "trick". There's no set of steps that always works for solving differential equations - you sometimes have to just "notice" that an expression can be rewritten in a useful way. The more examples you see, the better you'll get at it. $\endgroup$
    – Karl
    Dec 23, 2020 at 22:16
  • $\begingroup$ ok thank you :) $\endgroup$
    – TheCreator
    Dec 23, 2020 at 22:20

1 Answer 1

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$$\frac{dr}{dt}\frac{d\theta}{dt}+\color {red} {2r}\frac{d^2\theta}{dt^2}=0$$ More simply: $$r'\theta '+2r\theta ''=0$$ Multiply by $\theta ':$ $$r'(\theta ')^2+2 r \theta '\theta ''=0$$ Then you have: $$r' (\theta ')^2+r ((\theta ')^2)'$$ Finally: $$(r((\theta ')^2))'=0$$

But this $$2\frac{dr}{dt}\frac{d\theta}{dt}+r\frac{d^2\theta}{dt^2}=0$$ Is : $$2r'\theta ' +r \theta ''=0$$ $$2r'r\theta ' +r^2 \theta ''=0$$ $$(r^2 \theta ')'=0$$ The two equations you wrote are not giving the same result.

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