[Math] How does the concept of a derivative solve the problem of instantaneous velocity

calculusderivativeslimits

$$ \color{darkcyan}{\frac{dy}{dx}}
= \lim_{h \to 0} \frac{f(x+h) – f(x)}{h}$$
$$ \color{darkcyan}{m}
= \lim_{x \to a} \frac{f(x) – f(a)}{x-a} $$
Text source: https://i.stack.imgur.com/Kn3Bm.png

I think I have a fairly solid understanding of the derivative, but I don't get how it helps us find instantaneous velocity at a point. It only gives us the velocity that we can get infinitely close to, but that's not the velocity at the point. The velocity at the point is undefined as x-x in the denominator = 0.

I get the following about limits and derivatives:

  1. That the limit is an actual value, not an approximation. The limit is the actual value that we are getting infinitely closer to.

  2. That the derivative is the limit of the slope of x and a, as a is moved infinitely closer to a. It is the slope that is being approached, as a gets infinitely close to x.

But while this lets us know what the velocity is between two points as they get infinitely close to each other, that still doesn't give the actual instantaneous velocity at that point, because to find the actual velocity at that single instant, you have to do f(x)-f(x)/x-x= 0/0 = undefined. So how does the concept of the derivative give us instantaneous velocity?

How can this be explained without epsilon delta proofs, at the level of someone learning Khan Academy calculus?

Best Answer

Instantaneous rate of change of a differentiable function at a point is by definition the change in value of the function when the point is infinitesimally perturbed. The definition of instantaneous velocity at any point itself is the rate of change of position at that point, and is the velocity "at that point".

If $t$ denotes time and $f(t)$ denotes position at time $t$, then the velocity at time $t_0$ is defined as $$v_0 = f'(t_0)=\lim_{t\to t_0} \frac{f(t) - f(t_0)}{t-t_0}$$

As you mention, the limit is an "actual value", and may be defined based on $f$. When it is defined, it gives the exact velocity at time $t_0$. It is true that when you substitute $t_0$ in the limit, you get $\frac{0}{0}$, which is undefined. But the fact that a function is undefined at a point does not mean that the limit on approaching the point is undefined (take, for example, $\lim_\limits{x\to 2}\frac{x^2-4}{x-2}$). Moreover, substituting $t_0$ represents no change in time, whereas the velocity is defined for an infinitesimal change in time, for which the change in position is given by the limit, and is often well defined.

The key is that taking the limit allows you to exactly compute the change of the position for an infinitesimal change in time. Substituting $t_0$ represents no change in time, and substituting any non-infinitesimal change in $t_0$, say, $t_0+t_1$ gives the average velocity over the time period $t_1$.