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If $r=r(t)$, why is $\frac{r'(t)}{(r(t))^2}$ = $\frac{1}{r(t)}$ where $'$ denotes the derivative? I saw it in a lecture.

Can you please explain?

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closed as too localized by Waffle's Crazy Peanut, zhermes, David Z May 11 '13 at 19:23

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You're missing a derivative on the right-hand-side of your equation. The equation you saw was likely $ \frac{r'(t)}{(r(t)^2)} = - \left(\frac{1}{r(t)}\right)'$. –  Peter Shor May 12 '13 at 12:26

2 Answers 2

It's not...not unless $r(t)$ happens to be $r_0e^{-t}$.

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It isn't true. What's true is that: $$\int \frac {r'(t)}{r^2(t)} dt = -\frac{1}{r(t)}$$

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Thank you. But I am confused a bit - is there a similar relation maybe? Maybe I have a typo. –  Guest11 May 11 '13 at 18:45
    
As @JLA said, that is a differential equation, that you may simply write as: $\frac{dr}{dt}=r$ and has unique solution $r=r_0 e^{-t}.$ If you could give some context maybe one could say where it comes from (if it is a typo). –  pppqqq May 11 '13 at 18:49

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