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

I think this is a simple question. If I have that $E(L)=\tau L$ and we are told that $\tau=BTL$ would this mean that $E=BTL^2$ implies $dE=(2BTL)dL$ or should I sub $\tau$ straight into the second law giving $dE=\tau dL=(BTL)dL$?

$\tau=tension \ L=length \ T=temperature \ B= constant$

share|cite|improve this question
Please explain your notations as well. What all $B,L...$ represent. – ABC May 30 '13 at 8:06

Both ways give you the same results:

(1) $E(L)=\tau L$ , $dE(L)=d\tau L + \tau dL = BTLdL + BTLdL = 2BTLdL$

(2) $E(L)=BTL^2$ and then $dE=2BTLdL$

Hope that helps,

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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