2

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?

Maximilian Janisch
  • 13,536
  • This is a good example of "missing the forest for the trees". If you want to be very formal, the rigorous definition of a limit does not need the value to be reached, so there is nothing wrong with the statement because while $dT\to 0$ at no point can we ever say $dT=0$ (I'm using $dT$ here but you can just replace the symbol with something that doesn't look like a differential, like the $h$ in the definition of the derivative). – Ninad Munshi Oct 08 '19 at 09:50
  • You can always replace $dT$ with $\Delta T$ and set it to be some small but nonzero value, and go from there (sort of like how epsilon-delta proofs set $\epsilon$ to be some usually small but nonzero positive number). It is possible to make the claims made by the YouTuber formal in this way. – YiFan Tey Oct 08 '19 at 10:24
  • This question and answer my be helpful: https://math.stackexchange.com/questions/1991575/why-cant-the-second-fundamental-theorem-of-calculus-be-proved-in-just-two-lines/1991585#1991585 – Ethan Bolker Oct 11 '19 at 15:41

2 Answers2

4

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.

Ennar
  • 23,082
  • +1, but I slightly disagree with regards to the last point, especially for topics like calculus: I think the YouTuber does an excellent job of providing a proof sketch, and as long as the viewers actually work through formalising the ideas presented this can easily replace a textbook. (Whether or not viewers actually do this is another story...) – YiFan Tey Oct 08 '19 at 10:26
  • @YiFan, better students can probably do this without problems, but if you are preparing for exam, I'd still advise using literature. As someone working as TA, I had to add this because I can easily imagine beginning students mistaking a sketch for a proof. – Ennar Oct 08 '19 at 10:39
  • Good point. It takes quite a bit of experience to be able to tell when a logical implication is too hand-wavy and when it is fine. – YiFan Tey Oct 08 '19 at 10:41
  • I understand almost all of what you said. I still have a few questions. In the proof of the MVT, function $F$ (which is in this case $\Delta s$) is equal to a integral with the upper bound being a variable and the lower bound being fixed. In this case however the upper bound is the variable $\Delta T$+ a fixed value $T$. –  Oct 08 '19 at 10:45
  • @user532874, to be clear, in MVT you have fixed bounds, nothing is variable. I have suppressed notation, but $\Delta s$ is a function of both $T$ and $\Delta T$. For fixed $T$ and $\Delta T$, there exists $T_1$ that satisfies what we want. Therefore, $T_1$ is also a function of both $T$ and $\Delta T$. So, what you can do is say at the beginning, ok, let me fix some $T$ and calculate $s'(T)$. Then, for some fixed $\Delta T$, I can do this and this and arrive at two functions of $\Delta T$, namely $\Delta s$ and $T_1$. I can apply $\lim_{\Delta T\to 0}$, then. Does that clear things? – Ennar Oct 08 '19 at 10:54
  • If you look at the proof part of the MVT on this page http://people.math.sc.edu/meade/Bb-CalcI-WMI/Unit5/HTML-GIF/MVTIntegral.html you will see that you have defined $a$ and $x$ but not $b$. You are saying that some $T_1$ exists on $[a,x]$ when it should be that some $T_1$ exists on $[a,b]$. $x$ is variable so in the MVT there is something variable. –  Oct 08 '19 at 11:09
  • My second question is that $T$, $T_1$, and $T+\Delta T$ are not functions so how can you apply the squeeze theorem? –  Oct 08 '19 at 11:13
  • @user532874, I don't use any proof of MVT, I use the statement that I gave you a link to. Saying that $f(\Delta T) = T$ and $g(\Delta T) = T+\Delta T$ are not functions of $\Delta T$ is like saying $f(x) = a$ and $g(x) = a + x$ are not functions of $x$. – Ennar Oct 08 '19 at 11:20
0

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.

Kaligule
  • 1,514
  • 10
  • 12