Understanding the Concept of 'Cancelling dt's' in Derivatives and Integrals

[Abdullah Almosalami]

So I know that sometimes derivatives can behave in a multiplicative way but that is definitely not necessarily the case, and one should be rigorous to know for sure. So here's where my question kinda comes up. I'm going through my Circuits textbook, and to find the instantaneous power in a capacitor, you go as follows:

$p(t) = \frac {dw} {dt} = v(t) * i(t) = [v(t)] * (C * \frac {dv} {dt} ) = C * v(t) * \frac {dv} {dt}$

To find the instantaneous energy, assuming the energy is initially zero, and having t' as a dummy variable in the integral,

$w(t) = \int_0^t (C * v(t) * \frac {dv} {dt'}) \, dt' = C * \int_0^t (v(t) * \frac {dv} {dt'}) \, dt'$

So my question concerns the integral there ^. Here you can supposedly treat the $dt'$ as if they multiply and "cancel each other out" and you end up with $\int_0^v v(t) \,dv$ (since the initial energy is assumed to be zero in the capacitor, the initial voltage must've also been zero), which is trivial to integrate, and in the end you get, $w(t) = \frac {1} {2} * C * v^2(t)$. So what theorems are involved here? This kinda detail has never really been addressed in my Calc and Diff Eq. courses, nor in any of my engineering courses, so here I am asking.

My thinking of it is as follows. If I "expand" the derivatives and integrals to their limit definitions, I think you'd get the following:

$\int_0^t (v(t) * \frac {dv} {dt'}) \, dt'$
$= \lim_{n \rightarrow \infty} {\sum_{i=0}^n [ v(t_i) * \frac {dv} {dt} (t_i) ] * \Delta t }$ assuming equal partitions on the interval $[0, t],$ $\Delta t = \frac {t - 0} {n} = \frac {t} {n}$ and choosing arbitrary points $t_i$ in each partition $i$
$= \lim_{n \rightarrow \infty} {\sum_{i=0}^n [ v(t_i) * [\lim_{h \rightarrow 0} {(\frac {v(t_i + h) - v(t_i)} {h} )} ] ] * \Delta t }$

In order for the idea of the $dt's$ cancelling out in the original integral to be true, we'd need $\Delta t$ and $h$ to be the same variables (I think?). Now without taking the limits - as in, $\sum_{i=0}^n [ v(t_i) * \frac {v(t_i + h) - v(t_i)} {h} * \Delta t ]$ - and if we took say n=10 and h = 0.1, then $\Delta t$ and $h$ are definitely not necessarily the same (if the upper bound of the integral was 50 let's say, then $\Delta t = \frac {t} {n} = \frac {50} {10} = 5$, which is not the same as h; here we'd need n to be 500 in order for $\Delta t$ and $h$ to be the same and "cancel out"). However, and this is my reasoning, at the limit as n approaches $\infty$ and h approaches 0, they begin to behave as if they "cancel out" and there is where we get that property, though I am not sure how you continue simplifying and reducing. My guess would be as follows:

$= \lim_{n \rightarrow \infty} {\sum_{i=0}^n [ v(t_i) * [\lim_{h \rightarrow 0} {(v(t_i + h) - v(t_i))} ] ]}$

The integral $\int_0^v v(t) \,dv$ is equal to $\lim_{n \rightarrow \infty} {\sum_{i=0}^n ( v(t_i) * \Delta v )}$, with $\Delta v = \frac {v(t) - 0} {n} = \frac {v(t)} {n}$, which must mean that in the limit, you can say that $\lim_{h \rightarrow 0} {(v(t_i + h) - v(t_i))}$ behaves the same as $\Delta v$ as n approaches $\infty$, and so we arrive at,

$= \lim_{n \rightarrow \infty} {\sum_{i=0}^n [ v(t_i) * [\lim_{h \rightarrow 0} {(v(t_i + h) - v(t_i))} ] ]} = \int_0^v v(t) \,dv$

Yikes. I'm feeling the cringe from the fellow mathematicians here with my logic but that's the best I got.

 

[BvU]

and so we arrive at $C \int_0^v v(t) \,dv$

which is where we would be in one single step if we consider the work needed to bring an infinitesimal charge to a potential v : $$ dW = v \; dQ = C\ v \;dv \Rightarrow W = C \int_0^v v(t) \,dv $$

Deetail: your notation in $$
w(t) = \int_0^t (C * v(t) * \frac {dv} {dt'}) \, dt' = C * \int_0^t (v(t) * \frac {dv} {dt'}) \, dt'
$$ is not correct. There should be quotes in $v(t)$:$$
w(t) = \int_0^t (C * v(t') * \frac {dv} {dt'}) \, dt' = C * \int_0^t (v(t') * \frac {dv} {dt'}) \, dt'$$ but I think you understood that and treat it OK.

what theorems are involved here?

I don't think it's more than
If $F$ is a primitive of $f$ then $$\int_a^b f\,dx = F(b) - F(a)$$

 

[Abdullah Almosalami]

which is where we would be in one single step if we consider the work needed to bring an infinitesimal charge to a potential v : $$ dW = v \; dQ = C\ v \;dv \Rightarrow W = C \int_0^v v(t) \,dv $$

I was just using this as an example. I am aware of the other ways to arrive at the equation for energy. That was not my question, though I appreciate the reminder.

I don't think it's more than
If $F$ is a primitive of $f$ then $$\int_a^b f\,dx = F(b) - F(a)$$

I had a feeling perhaps the fundamental theorem was involved here, but I was thinking for something more rigorous, especially with the $v(t) * \frac {dv} {dt}$ in the integrand integrated with respect to t. Perhaps even a geometric interpretation with the switch from integrating with respect to t to integrating with respect to v.

 

[fresh_42]

If you like, then you can replaced all infinitesimals by their limits and Riemann sums for integration. You will probably have to swap some limits and need to prove why it is allowed. I predict a nasty calculation.

 

[Abdullah Almosalami]

If you like, then you can replaced all infinitesimals by their limits and Riemann sums for integration.

Well that's what I did. I am just unsure of whether it was correct or not or my reasoning was correct or not.

 

[fresh_42]

Well that's what I did. I am just unsure of whether it was correct or not or my reasoning was correct or not.

Looks ok as fas ar I could see on a quick view.