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In 3B1B's video on the paradoxes of derivatives, he mentions that the definition “instantaneous rate of change” is nonsense. Why is it no good, isn't the value that we are approaching as dx approaches 0 not actually the rate of change of a line tangent at a point? Computationally, we are just not measuring the rate of change at an instant is all, doesn’t mean our approximation isn’t the rate of change in an instant. Instead we are measuring constant approximation of rate of change or the rate of change that is being approached as the time becomes smaller and smaller and approaches zero but the value that we approach is rate of change at an instant (so I thought).

CalculusisCONFUSING
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    He’s using the term “nonsense” a bit loosely. As long as we define “instantaneous rate of change” precisely, it makes sense. (And though it took time for mathematicians to formulate the modern definition, now that we have the modern definition it is perfectly clear.) – littleO Feb 28 '23 at 21:48
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    The point is that there is no change happening in an instant, so "instantaneous rate of change" as a definition doesn't strictly make sense. What we mean by this phrase is the derivative, which is not actually defined in terms of an amount of change in an instant, but rather in terms of the amount of change in an arbitrarily but not infinitely small amount of time. – Ian Feb 28 '23 at 21:50
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    I suspect that they are talking about the limits of the definitions in the real world, where the lack of certainty of position and momentum (uncertainty) makes the idea nonsense. In mathematics, it is not nonsense, but the mathematics where it makes sense does not apply to the real world of physics. – Thomas Andrews Feb 28 '23 at 21:56
  • Since the origin of these ideas is Newton and the physical world, quantum behaviors of particles do not make sense, so the entire source of the idea is "nonsense." But the idea is not mathematical nonsense. – Thomas Andrews Feb 28 '23 at 21:58
  • Here's my answer: Why does 3blue1brown use the "around a point" to describe a derivative? – ryang Mar 01 '23 at 13:56

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Why is it no good, isn't the value that we are approaching as dx approaches 0 not actually the rate of change of a line tangent at a point?

But does that cleanly translate to the notion of "rate of change"? What is changing at an instant in time?

3Blue1Brown's comment about "nonsense" is a bit loose and I think you're overanalyzing it. What he is pointing out is that the average rate of change of a function $f$ over an interval $[a,b]$, quantified by

$$\text{average rate of change} = \frac{f(b) - f(a)}{b-a}$$

a priori (from a precalculus student's perspective) has no sensible definition "at a point", when $b=a$. If one carefully defines what that notion means, though (say, a limit as $b \to a$), then sure you get something that makes sense. (Remember, a lot of math is about definitions -- strict, unambiguous statements that mathematically describe a phenomenon.)

That is to say, nothing can change across an interval of zero width. A physical system measured across time has no change when you start and stop your observations at a time $t=1$ second. (If I take a picture of a car, for instance, you'd have trouble ascertaining its velocity.) So we need to do some work to qualify the notion of the derivative as "instantaneous rate of change" - and in particular, have it somewhat generalize and agree with more familiar notions of rate of change (e.g. slopes of lines).

PrincessEev
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