# Interpretation of Velocity as a time derivative of position

Going by the Wikipedia explanation, a derivative measures the 'sensitivity' of a function to tiny nudges in its input.

How well does this fit with the velocity being the derivative of position? I can't see an intuitive relation between the two concepts. But I don't understand what the 'order' of smallness must be? How small is the time period? Because I can just change the order by changing the units I use

Further how would this 'sensitivity' definition extend to the standard approach given below?

$$v = \lim_{\Delta t \to 0} \frac{x_2 - x_1}{\Delta t}$$

where $$\Delta t = t_2 - t_1$$

• The smaller , the better . Commented Jul 19, 2020 at 9:02
• Are you trying to calculate average velocity or instantaneous velocity? Commented Jul 20, 2020 at 19:48
• @DavidWhite, I'm trying to calculate instantaneous velocity. I've made an edit to the mathematical statement. Commented Jul 21, 2020 at 1:00

The difference and the derivative is the same thing, but in the limit that the difference becomes "very small". So here is what I like you to do:

1. Let's assume that the position changes with time according to $$s = s(t) = t^3$$. Draw this function. [Sidemark: Let's assume that we always use SI units. Including the units in the defining equations makes them messy.]
2. Now, ask yourself, what is the velocity at time $$t=2$$?

Since the velocity is the change of position within a time interval, we could estimate it by considering differences. E.g. by taking the points $$(t_1, s_1) = (1.5, 1.5^3)$$ and $$(t_2, s_2) = (2.5, 2.5^3)$$, the velocity in the interval $$t=[1.5, 2.5]$$ can be approximated by $$\Delta s/\Delta t = 12.25$$, which is shown as red line in the following plot.

The chosen interval around the target value $$t=2$$ are arbitrary. Hence, I could have chosen to use different points, e.g. $$(t_1, s_1) = (1.8, 1.8^3)$$ and $$(t_2, s_2) = (2.2, 2.2^3)$$, which yields $$\Delta s/\Delta t = 12.04$$, or I could have chosen a non-symmetric interval around $$t=2$$.

So it seams to be natural to ask how we could improve this result of differences? And also, how can we obtain a result which is independent of the chosen interval? One possible answer to this question is to take smaller and smaller intervals. By doing so we restrict the variability of our choices. This leads to limit $$\Delta t \to 0$$ and hence to the derivative.

Nowm coming back to your first statement

[...] a derivative measures the 'sensitivity' of a function to tiny nudges in its input.

we can see how this is the case for the velocity: The velocity is per definition the change of the position with respect to time. Therefore, the input variable is the time, and the output variable is the position. If a car has a "large" velocity, it changes its position "fast". In contrast, if the car has a "small" velocity, it changes its position "slow".

Does this makes sense?

A simple intuitive explanation of sensitivity is that it is a measure of how 'volatile' the function is to increments in it's inputs. For example, consider the square function

$$f(x) = x^2$$

Suppose, I nudge the input by some quantity $$'h'$$

$$f(x+h) = (x+h)^2 = x^2 +2xh + h^2$$

The nudge in the out put due the corresponding nudge in input is exactly given by the follow expression:

$$f(x+h) - f(x) = 2xh + h^2$$

Now, going back to the 'rise over run' idea from the concept slopes we used in straight lines and all. The rise in our function is the quantity mentioned above, to find the 'gradient' as in how much the function is 'sloping', we need to divide this by our 'run' which is h.

$$\frac{ f(x+h) - f(x)}{h} = 2x +h$$

To turn this into a derivative, by skipping a lot of formal steps, we take the instantons slopes. That is we bring make the 'nudge' amount so small that the $$f(x+h)$$ and $$f(x)$$ are very close to each other but not 'coincident', we denote this procedure by using the limit notation.

$$\lim_{h \to 0} \frac{ f(x+h) -f(x)}{h} = \lim_{ h \to 0} (2x+h)$$

Now, as we shrink 'h' more and more the second term our expression becomes zero and we are left with,

$$\lim_{h \to 0} \frac{ f(x+h) -f(x)}{h} = 2x$$

Which is precisely the derivative of the $$x^2$$ function.

Now, how does this differ from the algebraic quantities which you might be familiar with? Well notice that we used a function here, when you did the algebraic stuff that you are all so familiar about, you probably never even thought of having a general relations which specifies the quantities of motion as a function of time. That is you're only consider between finding the change between two particular states when you do the algebraic manipulations $$\Delta$$ change procedures.

Now, let's say you model the motion of a car, and let's say you get a graph which looks sort of like this,

Note on the graph: At each point on the 't' (time-axis) the height of the curve corresponding to it gives position of the car at that point in time. For example, we can see that at t=0, the curve has no height and that means that the car is at a starting at t=0 with the position function evaluating to 0.

If you've seen a lot of function graphs, you may go like hmm this looks sort of like the graph of the square function. And, you would write the most general form of the square function which is given as:

$$f(t) = at^2 +bt +c$$

Now, that we have this, we can evaluate the function at a few points to figure out the coefficients. For example $$f(0) = C$$ but notice that at $$t=0$$ displacement is $$0$$ , so the functions value is zero and hence the constant term is zero.

Once, we figure out all the coefficients we could take the derivative of this function and find the velocity at any point of time. Like this,

$$f'(t) = v(t) = 2at + b$$

And, this is great because this tells us the velocity at any point of time while using the regular algebraic stuff we could only get the velocity to move between two points in time. And further, we could generalize the regular displacement

$$S= ut + \frac{1}{2} at^2$$

formula to account for acceleration ( yes, this formula doesn't hold for changing acceleration)

The final point, is that assuming this car follows this parabolic trajectory forever you could also find the time point where the velocity is zero! In essence, you can derive more information about the motion if you model it as a funciton.

Illustration: our previous velocity function was, $$v(t) = 2at+b$$

Now, if we impose the condition that $$v(t_o) = 0$$ for some $$t_o$$, then,

$$0 = 2at+b$$

$$\frac{-b}{2a} = t$$

So, notice that this condition occurs physically when the car is either at a start or stop, in essence its minimum or maximum position. Because like say you keep increasing after the maximum position , then by definition it's no longer the maximum position. Similar argument for minimum position. So, at this point you should velocity should switch signs, to switch sign the velocity must accelerate across 0 and become precisely 0 at the 'turning point'. So, this time of velocity being zero is also maximum of parabola

Let's plug it back in and see what happens...

I get

$$f(t) = -(\frac{b^2 -4ac}{4a})$$

oops... did I just derive the formula for vertex of a parabola when talking about kinematics?

Edit: how small should it be? as small as you can take it! Look back on how we defined the derivative

Derivative has a formal and exact meaning and relies on continuity of your position with respect to time for start. This is why it is exact. If you look close enough the deltas in locations are straight lines if you have continuity.

Regarding sensitivity: Picture a footrace that goes to a photo finish. The judges pore over the image and try to decide who won, but the camera exposure was a little too long and the final centimeter is just a blur. The cameraman next to the track, on the other hand, was walking slowly and they know almost exactly where he was.

The faster the runner, the longer a streak you see between x1 and x2, while the whole photograph has the same t2 - t1. And the ratio between how long the streak of the fast runners is relative to how long the streak of the slow cameraman will always be about the same for a photo taken at "the same time" with different lengths of exposure. With a camera of finite speed, you can only give that speed difference as an average over some interval of time, but with an infinitely fast camera you could give everyone's speed "at" one single moment in time. That's the limit (t2-t1 = 0) that makes the derivative (dx/dt).

Note: the derivative is not what wins the race, since the person going fastest at the precise moment the photo snapped may not be the one who was in the lead.