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Given the diferentials $$ \begin{equation} d'Q=K(x-x')d'x \end{equation} $$ $$ \begin{equation} d'Q'=K(x'-x)d'x' \end{equation} $$

where $K$ is a constant, I need to show that

$$ \begin{equation} \frac{d'Q}{dt}+\frac{d'Q'}{dt}=\frac{d}{dt}\left[\frac{K}{2}\left(x-x'\right)^{2}\right] \end{equation} $$

How can I do so?

Attempt:

$$ \begin{array}{ll} \frac{d}{dt}\left(d'Q+d'Q'\right) & =K\frac{d}{dt}\left[(x-x')d'x+(x'-x)d'x'\right]\\ & =K\left[(x-x')\frac{d}{dt}\left(d'x\right)+d'x\frac{d}{dt}(x-x')\right.\\ & \qquad\left.(x'-x)\frac{d}{dt}\left(d'x'\right)+d'x'\frac{d}{dt}(x'-x)\right]\\ & =K\left[(x-x')\frac{d}{dt}\left(d'x-d'x'\right)+\left(d'x-d'x'\right)\frac{d}{dt}(x-x')\right]\\ & =? \end{array} $$

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  • $\begingroup$ What is $d'$?.. $\endgroup$ – amsmath Aug 29 '18 at 16:21
  • $\begingroup$ I am sorry, but as long as you cannot explain the symbols you use, it is not possible to help you. $\endgroup$ – amsmath Aug 29 '18 at 18:20
  • $\begingroup$ @amsmath Ok, it's meant to represent an infinitesimal. From mathworld: "An infinitesimal is some quantity that is explicitly nonzero and yet smaller in absolute value than any real quantity". $\endgroup$ – Igaturtle Aug 29 '18 at 18:29
  • $\begingroup$ So, $d'Q$ is the same as $dQ$? $\endgroup$ – amsmath Aug 29 '18 at 18:30
  • $\begingroup$ @amsmath $d'Q$ is in the context of thermodynamics usually called a differential; here in particular it means the average heat transferred over some stochastic process between $t$ and $t+dt$. $\endgroup$ – Igaturtle Aug 29 '18 at 18:37
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In formal terms (I set $K=1$ and $d' = d$), \begin{align*} \frac 1 2\frac d{dt}(x-x')^2 &= (x-x')\left(\frac{dx}{dt}-\frac{dx'}{dt}\right)\\ &= \frac{1}{dt}\big((x-x')dx + (x'-x)dx'\big)\\ &= \frac{1}{dt}(dQ+dQ'). \end{align*}

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  • $\begingroup$ I tried that before but I was unsure if it was on good grounds to work the infinitesimals with such freedom... Since you have more of a formal knowledge maybe you can assure me it is? $\endgroup$ – Igaturtle Aug 29 '18 at 19:15
  • $\begingroup$ Since it's physics, it is. ;-) $\endgroup$ – amsmath Aug 29 '18 at 19:18
  • $\begingroup$ -_- ... Thanks! $\endgroup$ – Igaturtle Aug 29 '18 at 19:20

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