In summary, consider these three facts:
Explaining how convergence works – Presented by: Steve Patterson.
You can also see: Steve Patterson's related article which is linked to from the following page: Infinity Articles & Websites
The derivative of position with respect to time – Presented by: Grant Sanderson.
A technical insight into the limitations of asymptotic relationships in calculus.
See: 8:17 to 9:45 – Graphical illustration which emphasises the limitation of calculus with respect! to velocity.
Quote: “Notice what I’m not saying, I’m not saying that the derivative is whatever happens when dt is infinitely small, whatever that would mean...”
See: 10:25 to 16:38 – Derivation of the derivative for distance vs time, which also demonstrates how the remaining terms that contain “dt” are discarded (12:20 to 12:55) as part of the derivation – the result is an equation that produces the “best constant approximation” of the car's velocity.
Quote: “Instantaneous rate of change, a phrase which is intrinsically oxymoronic”, i.e., does not make any sense.
Explanation: In other words, by discarding the remaining “dt” terms (as stated above), any arguable infinitesimal existence has simply been dismissed – instead of any result given by the final equation representing an infinitesimal existence, it actually produces the "best constant approximation" of reality.
A complete example which shows all the algebraic steps
Calculus from graph: (2 + dt)³ - (2)³
Algebra now begins: (“2” + “dt”) (2 + dt) (2 + dt) - 2³... (- 2³ is added back in near the end)
1st term “2”: (2×2×2) + (2×2×dt) + (2×dt×2) + (2×dt×dt) = 2³ + 4(dt) + 4(dt) + 2(dt)²
2nd term “dt”: (dt×2×2) + (dt×2×dt) + (dt×dt×2) + (dt×dt×dt) = 4(dt) + 2(dt)² + 2(dt)² + (dt)³
1st + 2nd terms: 2³ + 4(dt) + 4(dt) + 2(dt)² + 4(dt) + 2(dt)² + 2(dt)² + (dt)³ = 2³ + 12(dt) + 6(dt)² + (dt)³
Translate to video: 2³ + 12(dt) + 6(dt)² + (dt)³ = 2³ + 3(2)²dt + 3(2)(dt)² + (dt)³
(2³ + 3(2)²dt + 3(2)(dt)² + (dt)³ - 2³) / dt... (video continues from here – starting with cancelling the “dt” terms)
= 3(2)² + 3(2)(dt) + (dt)²... (terms with “dt” remaining are now discarded)
= 3(2)²... or 3(t)²
The popular solution to Zeno's motion paradox is effectively a straight-line graph (linear), because 1/2 the distance is claimed to take 1/2 the time, 1/4 the distance takes 1/4 the time, and so on.
Unnecessary, but amusing: straight-line graph (linear) examples (equal scale X and Y axes):
Example 1:
1 to 1 = 45 degrees, time = 2 ... s(t) = 1t.
(1(2 + dt) – 1(2)) / dt ... 2 cancels, so we’re left with 1, because: 1(dt) / dt = 1 ... ds / dt (t) = 1(t).
Example 2:
4 to 1 = arctangent 4 = 76 degrees, time = 9 ... s(t) = 4t.
(4(9 + dt) – 4(9)) / dt ... 36 cancels, so we’re left with 4, because: 4(dt) / dt = 4 ... ds / dt (t) = 4(t).
Video: Zeno's Dichotomy Paradox (The Stadium): Zeno of Elea – Presented by: Professor Angie Hobbs ... also includes logical quantum movement examples: “The Arrow – A Clock for Zeno” and “Movement Logic of a Single Entity”.
Conclusion:
Calculus serves no purpose in the above linear examples...
The cancellation of the “dt” terms is necessary to complete the derivation, which means no “dt” terms ever existed that “tend to zero”.
Calculus as demonstrated on this page – either discards “dt” terms that “tend to zero” ... or never produces terms that “tend to zero”.