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I am preparing to take calculus-based physics. I have experience in algebra-based physics, in which the following formula was heavily emphasized:

$$v_{avg} = \frac{\Delta x}{\Delta t}$$

I assumed that this was a definition, but now that I have a better understanding of calculus I am wondering if this is really a derived quantity. From my reading, I have seen that the true definition of velocity is:

$$v = \frac{dx}{dt}$$

I know that you can take the average value of a function, like velocity, as follows:

$$v_{avg} = \frac{1}{t - 0}\int_0^t{v}\,dt = \frac{1}{t}\int_0^t{\frac{dx}{dt}}dt=\frac{\Delta x}{t}$$

So, is the average velocity equation really a derived equation? I apologize if this question is too simplistic: it has been bothering me for the past few days that there could be two definitions of velocity (albeit instantaneous vs. average velocity) that work together so well, but if average velocity is really just a derived quantity it would make a lot more sense. Was the calculus just "hidden" from me in the algebra-based class I took?

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  • $\begingroup$ Is there really a difference between a definition and a derived quantity? I don't see the point in drawing a distinction. Regardless, your proof that $v_{avg} = \frac{\Delta x}{t}$ is basically an answer to your own question; you can derive that equation from calculus principles, as you did. $\endgroup$
    – iammax
    Commented Aug 8, 2017 at 2:50
  • $\begingroup$ When I define x=1 or derive it from x-1=0, what difference does it make? None. Both are different ways of saying the same. Your question is moot, because your choices are not alternatives. They are the same definition stated differently, but meaning the same, as you clearly have shown with your integral. $\endgroup$
    – safesphere
    Commented Aug 8, 2017 at 2:51
  • $\begingroup$ Additionally, you can further ask this question, if we're going along these lines: $$f(x)_{avg} = \frac{1}{b-a} \int_{a}^{b} f(x) dx $$ Is the above equation a DEFINITION of the average, or just a useful equation that happens to be true? Really it doesn't matter, though. $\endgroup$
    – iammax
    Commented Aug 8, 2017 at 2:53
  • $\begingroup$ By the word derived quantity, do you really mean differentiated quantity? $\endgroup$
    – Qmechanic ♦
    Commented Aug 8, 2017 at 5:33
  • $\begingroup$ If you were new to physics, and someone told you that average velocity was defined to be $\Delta x / \Delta t$, you might ask them what this expression had to do with the usual meaning of the word "average", since on it's own, it's not obvious. In the last integral in the question before the last equals sign, you can see that you're adding up an infinite number of velocity values, each having a weighting of $dt$, and dividing that sum by a sum of the weights ($t$). It retains the usual meaning of "average", so both are acceptable as definitions. But the integral itself is equal to $\Delta x$ $\endgroup$
    – Apollonius
    Commented Aug 8, 2017 at 9:48

1 Answer 1

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In a sense you are questioning the region of validity of the symbol Δ . The d(x) symbol is the limit of taking the interval defined by Δ(x) to zero. Thus the d symbol defines a subset of the Δ symbol validity in the x space.

This distinction becomes clear in the usage of ΔxΔp in the Heisenberg uncertainty principle, where it is physical intervals that are defined and not the limiting values.

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