Let the capacitor be charged to voltage $ V(t_0) $ at $ t=t_0 $ and then start discharging. At time $ t \, (>t_0)$, the delayed capacitor voltage is $$ v(t)=V(t_0)\operatorname{exp}(- \frac{t-t_0}{RC}). \tag 1$$

The rate of change of $ v(t) $ AT $ t=t_0 $ is $$\frac{dv}{d(t-t_0)}|_{t=t_0}=-\frac{V(t_0)}{RC}. \tag 2$$

My questions:

  1. When they are saying rate of change of $v(t)$ , why have they differentiated $ v(t) $ with respect to $ (t-t_0)$, why haven't they differentiated with respect to $t$ and then subtituted $ t $ with $ t_0 $ , which gives the same answer as shown: \begin{align} \frac{dv}{dt} &= V(t_0)e^{\frac{t_0}{R_C}}\frac{d}{dt}(e^{-\frac{t}{RC}}) \\ &=V(t_0)e^{\frac{t_0}{RC}}e^{-\frac{t}{RC}}(-\frac{1}{RC}) \\ &= >\frac{dv}{dt}|_{t=t_0}=-\frac{V(t_0)}{RC} \end{align}

  2. How to differentiate $ v(t)$ w.r.t $ (t-t_0)$. I am particularly confused about this because of the component $ V(t_0) $ of eq $ (1) $. Should I consider it a constant or should I consider it a variable as I am differentiating $ v(t) $ with respect to $ (t-t_0)$. In other words how to perform this differentiation?

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    $\begingroup$ Isn't $d(t-t_0) = dt$ since $t_0$ is a constant? $\endgroup$ – Atul Ingle Jul 20 '17 at 20:32
  • $\begingroup$ Yes thats true!!! $\endgroup$ – Soumee Jul 20 '17 at 20:35
  • $\begingroup$ @AtulIngle Then why are they complicating such a simple thing? Should I differentiate it just with.r.t $t$ instead of $t-t_0$. I wanted to know if anything in mathematics is there wherein they differentiate a function wrt $t-t_0$, but as you pointed out both are just the same in case $t_0$ is a constant, I would like to know what will be the differentiating procedure if $t_0$ is a variable. $\endgroup$ – Soumee Jul 20 '17 at 20:41
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    $\begingroup$ I have no idea what the physical meaning of "variable $t_0$" would be in this example. That aside, if $t_0$ were a function of $t$ you can work out its differential in the usual way. Eg. if $t_0=t^2$, then $dt_0 = 2t dt$. $\endgroup$ – Atul Ingle Jul 20 '17 at 21:37
  • $\begingroup$ No answer as to why except to point out that the decay is proportional to the time difference; but for calculations, you just apply the chain rule from calculus. $\endgroup$ – rrogers Jul 25 '17 at 19:36

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