B Understanding the instantaneous velocity formula

[adjurovich]

If the formula for instantaneous velocity is:

$v = \lim_{\Delta t \to 0} \dfrac{\Delta s}{\Delta t}$

Why the result of equation isn’t infinity? It’s said that if we divide something by number very close to zero, it results in infinitely large number. But how does this equation work then? It obviously does work fine, but how?

 

[PeroK]

If the formula for instantaneous velocity is:

$v = \lim_{\Delta t \to 0} \dfrac{\Delta s}{\Delta t}$

Why the result of equation isn’t infinity? It’s said that if we divide something by number very close to zero, it results in infinitely large number. But how does this equation work then? It obviously does work fine, but how?

$\Delta s$ also tends to zero, as $\Delta t$ tends to zero. The limit can be finite in that case.

A purely mathematical example is:
$$\lim_{x \to 0}\frac{2x}{x} = 2$$

 

[fresh_42]

If the formula for instantaneous velocity is:

$v = \lim_{\Delta t \to 0} \dfrac{\Delta s}{\Delta t}$

Why the result of equation isn’t infinity? It’s said that if we divide something by number very close to zero, it results in infinitely large number. But how does this equation work then? It obviously does work fine, but how?

It depends on what you divide by $\Delta t.$ You do not divide by zero, only by a decreasing positive number. E.g. if we divide $\Delta s = (\Delta t)^2$ by $\Delta t$ then we get
$$
v = \lim_{\Delta t \to 0} \dfrac{\Delta s}{\Delta t}= \lim_{\Delta t \to 0} \dfrac{(\Delta t)^2}{\Delta t}=\Delta t =0
$$
which is far from being infinitely large.

 

[adjurovich]

$\Delta s$ also tends to zero, as $\Delta t$ tends to zero. The limit can be finite in that case.

A purely mathematical example is:
$$\lim_{x \to 0}\frac{2x}{x} = 2$$

Thanks. I would like to ask you one more thing because I’m struggling with understanding what derivative is.

I’ve seen everywhere that definition “derivative is instantaneous rate of change”. I am so confused by this definition I can’t even explain the part I don’t understand, it just sounds like a different language to me. So what does that mean actually?

 

[fresh_42]

Thanks. I would like to ask you one more thing because I’m struggling with understanding what derivative is.

I’ve seen everywhere that definition “derivative is instantaneous rate of change”. I am so confused by this definition I can’t even explain the part I don’t understand, it just sounds like a different language to me. So what does that mean actually?

I'm not sure whether this can help you, but I once have been asked here to write a walkthrough the maze of differentiation. It took me five articles in the end to cover at least the main aspects. Parts 2-5 may be too difficult as an introduction, but perhaps part 1 can help you:

https://www.physicsforums.com/insights/the-pantheon-of-derivatives-i/

and a picture. Those lines are possible tangents at a (curved) function. In the case of velocity, they reflect the straight line a car takes if it loses contact with the road for whatever reason.

If we consider $\dfrac{\Delta s}{\Delta t}=\dfrac{\Delta y}{\Delta x}$ then this represents a ratio. That ratio corresponds to the slope of the red line, a secant. A secant is a straight that intersects the black function twice. Now, if we make $\Delta y = \Delta s$ smaller and smaller, then we end up with the yellow tangent that intersects, better: touches the black function only once. The slope of the tangent corresponds to the value of the velocity. This number is not infinity.

 

[PeroK]

Thanks. I would like to ask you one more thing because I’m struggling with understanding what derivative is.

I’ve seen everywhere that definition “derivative is instantaneous rate of change”. I am so confused by this definition I can’t even explain the part I don’t understand, it just sounds like a different language to me. So what does that mean actually?

This is a key idea in physics. Let's take a particle moving in a circle. It's not moving in the same direction for any finite period of time. If it were, it would be moving in a polygon, rather than a circle. This leads us to the idea that it has some instantaneous velocity at each point in the circle. In fact, we can write its position as a function of time:
$$\vec r(t) = (R \cos \omega t) \hat x + (R \sin \omega t ) \hat y$$That's a particle moving in a circle of radius $R$, centred on the origin, at some angular velocity $\omega$.That means its velocity must be:
$$\vec v(t) = (-\omega R \sin \omega t) \hat x + (\omega R \cos \omega t ) \hat y$$It might be an idea to find some software that would show you the velocity for this particle. So, you could visually how the motion of the particle translates to that particular derivative.

 

[fresh_42]

This is a key idea in physics.

I would even say that it wasn't the apple, it was differentiation when modern physics began. And whether we call Newton or Leibniz, or even French mathematicians the ones who found it doesn't matter. It is simply one example out of so many in science of convergent developments. Physics without the infinitesimal change is unthinkable. I find it pitiful that everybody associates with Newton that silly (invented story) of the apple.

 

[gmax137]

I’ve seen everywhere that definition “derivative is instantaneous rate of change”. I am so confused by this definition I can’t even explain the part I don’t understand, it just sounds like a different language to me. So what does that mean actually?

Let's say you ride your bike from home to school. Also, say the trip is 4 km, and you leave home at noon. If you arrive at school at 12:20, you can easily find your average speed ($v=\frac {\Delta x}{\Delta t}=\frac {4 km}{0.33 hour}=12 \frac{km}{hr}$).

But during the ride you speed up, slow down, maybe even stop for a red light. What was your speed at 12:05? Or at 12:13? These would be "instantaneous velocities" at particular times.

EDIT (sorry!)
If you can define your velocity position (x) as a function of time (from 12:00 to 12:20) then you can use calculus (the derivative) to find your instantaneous velocity at each time along the route.

 

[adjurovich]

This is a key idea in physics. Let's take a particle moving in a circle. It's not moving in the same direction for any finite period of time. If it were, it would be moving in a polygon, rather than a circle. This leads us to the idea that it has some instantaneous velocity at each point in the circle. In fact, we can write its position as a function of time:
$$\vec r(t) = (R \cos \omega t) \hat x + (R \sin \omega t ) \hat y$$That's a particle moving in a circle of radius $R$, centred on the origin, at some angular velocity $\omega$.That means its velocity must be:
$$\vec v(t) = (-\omega R \sin \omega t) \hat x + (\omega R \cos \omega t ) \hat y$$It might be an idea to find some software that would show you the velocity for this particle. So, you could visually how the motion of the particle translates to that particular derivative.

I would even say that it wasn't the apple, it was differentiation when modern physics began. And whether we call Newton or Leibniz, or even French mathematicians the ones who found it doesn't matter. It is simply one example out of so many in science of convergent developments. Physics without the infinitesimal change is unthinkable. I find it pitiful that everybody associates with Newton that silly (invented story) of the apple.

Would it be accurate if I interpreted derivatives this way:

The rate of change of linear functions tells us how many times y changes if we increase x by 1.

So: $y = mx + b$
Where m is the initial value and we will set it to zero:

$y = mx$

So solving for m we get:

$m = \dfrac{y}{x}$

So in this case derivative (the unique function used to describe rate of change at any point) would be:

$y’ = m$

However, if we were to do the same for quadratic, we wouldn’t be able to because we can’t make a triangle out of curve. So we observe the rate of change at each point (it’s not the same at each point) so we just draw the tangent (I am still wondering if length of what is the length of that tangent determined by) and then we have to take the derivative to be able to find some global law of how value changes, and it’s:

$y’ = 2x$

That way, if we wanted to know “m” at some point, we have to plug in the coordinate of that point at x?

This all seems to me somehow confusing, maybe it needs to “cook for some time”? If I am wrong, please correct me.

Also, one more thing, does derivative represent how function would change if it became linear all of sudden at some point at curve and decided to ignore it’s previous non-linear path?

 

[Mark44]

Would it be accurate if I interpreted derivatives this way:

The rate of change of linear functions tells us how many times y changes if we increase x by 1.

So: $y = mx + b$
Where m is the initial value and we will set it to zero:

I think you're actually referring to b, which is the y-intercept of the line.

So solving for m we get:

$m = \dfrac{y}{x}$

So in this case derivative (the unique function used to describe rate of change at any point) would be:

$y’ = m$

Yes. The derivative of a function that represents a straight line is equal to the slope of that line.

However, if we were to do the same for quadratic, we wouldn’t be able to because we can’t make a triangle out of curve. So we observe the rate of change at each point (it’s not the same at each point) so we just draw the tangent (I am still wondering if length of what is the length of that tangent determined by) and then we have to take the derivative to be able to find some global law of how value changes, and it’s:

You don't really need to know the length of the tangent line. For an actual curve, as opposed to a straight line, the tangent line is relevant only for a relatively small neighborhood around the point of tangency.

$y’ = 2x$

That way, if we wanted to know “m” at some point, we have to plug in the coordinate of that point at x?

Yes.

This all seems to me somehow confusing, maybe it needs to “cook for some time”? If I am wrong, please correct me.

Also, one more thing, does derivative represent how function would change if it became linear all of sudden at some point at curve and decided to ignore it’s previous non-linear path?

Yes. The derivative "linearizes" a curve in a relatively small region around the point of tangency. For some curves with very little curvature, the tangent line lies pretty close to the curve. For other curves with greater curvature, the tangent line can diverge from the curve very quickly.

 

[fresh_42]

Would it be accurate if I interpreted derivatives this way:

The rate of change of linear functions tells us how many times y changes if we increase x by 1.

Yes. But linear functions have a constant rate of change, its steepness. If we have a non-linear function, then its rate of change is a local approximation by a linear function, the tangent. We then consider the steepness of the function at that point as the steepness of that tangent.

So: $y = mx + b$
Where m is the initial value and we will set it to zero:

$y = mx$

I'd rather say that $y(0)=b$ is the initial value, and $m$ is the constant steepness, the slope.

So solving for m we get:

$m = \dfrac{y}{x}$

So in this case derivative (the unique function used to describe rate of change at any point) would be:

$y’ = m$

Differentiation is a local property. It happens at a certain point, e.g. at $x=0,$ i.e. at point $P=(y,x)=(b,0).$

If we write $y'=mx$ then this is the equation of the tangent line at $P.$ The tangent is vector space, in this case, a vector space of dimension one, a straight. The origin of this vector space is $P.$ So we have $P=(b,0)$ in the coordinate system of the function, and $P=(0,0)$ in the coordinate system of the tangent.

However, if we were to do the same for quadratic, we wouldn’t be able to because we can’t make a triangle out of curve.

We can draw secants through two points $P$ and $P'.$ Now we let $P'$ approach $P$ until the secants become a tangent. All secants and the resulting tangent have their own slopes. The tangents however have different slopes at different points $P$, other than in the linear case.

So we observe the rate of change at each point (it’s not the same at each point) so we just draw the tangent (I am still wondering if length of what is the length of that tangent determined by) and then we have to take the derivative to be able to find some global law of how value changes, and it’s:

$y’ = 2x$

This common notation hides a few things. It is the equation of a function, and the long notation would be
$$
y'= (x_0 \mapsto y'=y'(x_0)=2\cdot x_0 = \left. \dfrac{d}{dx}\right|_{x_0} y(x)=\lim_{\Delta y \to 0}\dfrac{\Delta x}{\Delta x}=\lim_{\Delta x \to 0}\dfrac{y(x_0+\Delta x)-y(x_0)}{\Delta x}\quad (*)
$$
where $x_0$ is the $x$-coordinate of the point where we take the differentiation. $y'$ is a notation that does not say which and therefore covers all meanings in one letter:

• the function $x\longmapsto y'(x)$ that maps a location $x$ to the steepness of the function $y$ at that point
• the steepness itself, i.e. a certain real number $y'(x_0)$
• the limit $y'=\lim_{\Delta x \to 0}\dfrac{\Delta y}{\Delta x}$
• the tangent $T(x)=y'\cdot x$

This all can be hidden behind the notation $y'.$ Usually, it means the first point on the list
$$
y'(p) = \left(p\longmapsto \left. \dfrac{d}{dx}\right|_{x=p}y(p)\right)
$$
I have written $p$ instead of $x,$ simply to emphasize that it is what happens at a certain point.

That way, if we wanted to know “m” at some point, we have to plug in the coordinate of that point at x?

Correct.

This all seems to me somehow confusing, maybe it needs to “cook for some time”? If I am wrong, please correct me.

It is confusing, see my list. And people usually do not say which item on the list they refer to if they write $y'.$ It is highly context-sensitive. And we have only discussed the cases $y(x)=x^n$ and $y'(x)=nx^{n-1}.$ You will have taken a big step of understanding if you read and understand the long notation (*) above.

The best way to get through this confusion is to answer the following question: What is your variable? The function $y$, the location $P$, the variable of the derivative (a point on the tangent), the secants that approach a tangent, the steepness (slope) of the tangent. Since differentiation can be seen as a procedure (secants approaching a tangent, calculation of a limit, applying differentiation rules to a function), it automatically raises the question of which part of the procedure should be in the focus.

Also, one more thing, does derivative represent how function would change if it became linear all of sudden at some point at curve and decided to ignore it’s previous non-linear path?

Yes. The process can best be considered as those secants $\overline{PP'}$ that become a tangent at $P$ when $P'$ approaches $P.$