3Blue1Brown Cloudy Argument
After watching 3blue1brown's video "Integration and the fundamental theorem of calculus," I got confused at around 11:00, where Grant says the derivative of the area function is exactly the function itself. The reason for this is since when dT approaches zero, the rectangular approximation $v(T)dT$ resembles $dS$ more and more, it makes sense then that at $dT=0$, the derivative of $s(T)$ is exactly $v(T)$. However, $dT=0$ invalidates the equality because a zero denominator is undefined. I think this contradiction possibly highlights a consequence of treating differentials as variables when they're actually not. Is my reasoning correct?
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$\begingroup$There are two things I want to say.
First of all, $ds = v(T)dT$ is literally true when you interpret what $ds$ and $dT$ stand for correctly (they are differential forms, but that's not important). However, this is a consequence of the fundamental theorem.
But, 3Blue1Brown doesn't treat $ds$ and $dT$ in this abstract sense, what they are actually doing is looking at $\Delta s = s(T+\Delta T) - s(T)$ and $\Delta T$. I suppose that for the sake of clarity of exposition, they conflate these things not to overburden viewers with too much notation. (Who said YouTube videos need to be formally correct?) However, their explanation really leads to formally correct proof that $s'(T) = v(T)$. Let's take a look.
$$\Delta s = s(T+\Delta T) - s(T) = \int_0^{T+\Delta T}v(t)\,dt - \int_0^{T}v(t)\,dt = \int_T^{T+\Delta T}v(T)\,dt.$$
By the mean value theorem for integrals, there exists some $T_1\in [T,T+\Delta T]$ such that $$\int_T^{T+\Delta T}v(T)\,dt = v(T_1)\Delta T$$
and this is formal explanation why 3Blue1Brown say $\Delta s \approx v(T)\Delta T$ when $\Delta T$ is small. This still needs more polishing to be formally correct, we need $v$ to be continuous. So far, we concluded that for some $T$ and $\Delta T$, there exists $T_1\in[T,T+\Delta T]$ such that $$\frac{\Delta s}{\Delta T} = v(T_1).$$ Now, in the last expression you obviously cannot let $\Delta T = 0$, as you say, but you can look at the limit $\lim_{\Delta T\to 0}$. We now have
$$s'(T) = \lim_{\Delta T\to 0} \frac{\Delta s}{\Delta T} = \lim_{\Delta T\to 0} v(T_1) = v(\lim_{\Delta T\to 0}T_1) = v(T),$$
where second to last equality is due to continuity of $v$ and the last equality is due to squeeze theorem, since $T\leq T_1 \leq T+\Delta T$. And this completes the proof.
The moral of the story is that YouTube videos can be great to build intuition and good ones provide sketches of formal proofs, however, they are not to be used as a substitute for textbooks when it comes to formal correctness.
$\endgroup$ 8 $\begingroup$I agree. You can use this calculation with differentials to reason and as a mental aid, but it is very dangerous to use them in formal proofs.
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