Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

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

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

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

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

share|cite|improve this answer

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

share|cite|improve this answer
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

Not the answer you're looking for? Browse other questions tagged or ask your own question.