Integrals and derivatives , velocity and displacement

[madah12]

This is a very basic question, but I just started wondering about it. First which is the original quantity and which is the derived one. I mean is the displacement by definition is integral from a to b v(t) dt or is it the distance from the starting point to the finishing point or change in position which is the fundamental definition. Second Is the instantaneous velocity of an object defined to be the derivative if so then is velocity more of a mathematically defined quantity instead of physically because I don't really understand the physical meaning of saying that at this second the velocity is equal to 2 m/s. but then back to number one is integration an approximate or an exact method of finding the area under the velocity curve and is that area only an approximation or an exact value for the displacement and why?

 

[Char. Limit]

Integration is indeed an exact method, as long as you have an integrable function for the velocity curve that is not itself an approximate.

 

[madah12]

Integration is indeed an exact method, as long as you have an integrable function for the velocity curve that is not itself an approximate.

ok even if it is an exact method why is that area (physically not mathematically) equal to the displacement or it is just is by experiments?

 

[sophiecentaur]

If you were starting to develop a system of units, it would be best (imo) to kick off with quantities that can be measured directly. This would be why SI units start with mass, length, time, etc. These can be measured and specified in terms of readily obtainable conditions. (wavelength and frequency of reproduceable em radiations, lumps of platinum kept in a safe in Paris etc etc)
This sort of determines which are the basic quantities and which are the derived ones - so it would seem that the choice is a matter of convenience. I don't think there's anything particularly fundamental about the actual choices; it's just a pragmatic matter.
As technology progresses, the choices have changed and will change again, I'm sure.

 

[madah12]

If you were starting to develop a system of units, it would be best (imo) to kick off with quantities that can be measured directly. This would be why SI units start with mass, length, time, etc. These can be measured and specified in terms of readily obtainable conditions. (wavelength and frequency of reproduceable em radiations, lumps of platinum kept in a safe in Paris etc etc)
This sort of determines which are the basic quantities and which are the derived ones - so it would seem that the choice is a matter of convenience. I don't think there's anything particularly fundamental about the actual choices; it's just a pragmatic matter.
As technology progresses, the choices have changed and will change again, I'm sure.

ok so we choose distance and displacement to be the more original quantities, but then is the definition v=dx/dt just a mathematical definition I mean i get it physically with the average velocity = delta x /delta t which is just the change of position divided by the change in time it is very physically straight forward however when saying at t=1, v=6m/s the only physical significance I see is that if at the velocity remained constant then it would have moves 6 meters in one second but since the velocity isn't constant what is the physical significance of the value of v at time =t? Also I still don't understand why does the definite integral of the velocity which is just a mathematical definition gives the physical change of distance? is it that we just find it to be like that by experiments?

 

[sophiecentaur]

I must say, I am not sure what, exactly, is your problem with this - although it is one of many examples of where, in an almost magical way, Maths gives you the right answer to a physical problem. Maths is a model of things and, on a Philosophical level, we sort of believe that 'turning the handle' of a mathematical process will yield the answer to the originally posed physical question
Expressing velocity as ds/dt is just a way of putting it in symbols. The differential is just a way of expressing a 'rate of change with time or some other variable'. The whole concept of talking in terms of the limiting value of a ratio δy/δx as δx approaches zero and the consequences of it are pretty sophisticated. I remember, at A level, finding it exciting.
If you go the other way then, in the physical world, it involves a Definite Integral and that involves specifying the initial situation - or else you have an indefinite Integral and you then need to add a "+c" to the answer.
Although you can easily verify the connection between calculus and the real world, I don't think it is actually necessary to do experiments to prove that calculus 'works'.

It all this relevant to what you have been asking?

 

[madah12]

I am saying how can we know that the mathematical method we devised to calculate displacement "the integral" agrees with the physical meassurements by a metric meassure is it by experiments? I mean just because an equation is mathematically valid doesn't mean the physical world has to obey it right?

 

[Pythagorean]

it makes a lot of sense if you do the dimensional analysis. The area you're calculating is (like the area of a rectangle) from two dimensions. In a rectangle, you're multiplying a spatial dimension by a spatial dimension. In this particular integral, you're multiplying the velocity (m/s) by the time (s). This gives you m(s/s). The seconds cancel and you have space.

But it makes sense physically too, doesn't it? If you go 15 meters per second for 1 second, you'll have gone 15 meters. I mean, since you're going 15 meters every second, you've had to have gone 15 meters in a second, otherwise you're not going 15 meters per second.

Yes, I've personally verified and helped students verify this experimentally. It's the first lab we do in the first required physics lab using data logger pro's motion sensor.

 

[madah12]

yes but what about a non constant velocity if in every instant you go with a different velocity then you can't say in this second because the change in velocity happens in less than a second right? I mean you specify at t=a v=b but then then at t=a+delta where delta is a very small number , v =b2 so with an instantaneously changing velocity and changing acceleration what is the physical meaning of v at time t

 

[Pythagorean]

Well, the assumption is that you have a known function that describes the position so that it's derivative is known too.

Can we ways find a function to describe the behavior we see? Usually, yes! Math is logical clay; we can manipulate terms until they match the behavior observed. That's modeling.

 

[Pythagorean]

Addendum: it's a statement about rates of change in general, position is just one example (velocity being the rate of change for position). It also assumes smooth spacetime.

 

[sophiecentaur]

I am saying how can we know that the mathematical method we devised to calculate displacement "the integral" agrees with the physical meassurements by a metric meassure is it by experiments? I mean just because an equation is mathematically valid doesn't mean the physical world has to obey it right?

Start with distance / displacement as being your fundamental measure. You could say that our definition of velocity is axiomatic, I suppose.
If you acknowledge / define that velocity is the time differential of distance, that will apply to non-constant velocity as well as constant velocity motion. What the Maths throws up as a result of that definition will be correct - particularly because the integral and differential are just alternative statements of the same thing - you might even call it 'commutative'. The only thing you must do, when integrating, is to get your limits right.

Of course, all this may only apply where SR is not considered. But I don't think your original question involves that.

I would agree, in general, that you are right to question the outcome of just any old mathematical tinkering. After all - even the quadratic equations, used to calculate motion, can yield a root which is not physically realisable - possibly implying that your journey started before you set off!

It's very easy to 'prove' 0=1 and all sorts of other things but, when the Maths is valid and nothing else, physical, creeps into the situation, then I think you can rely on the outcome without needing to measure it.

 

[JDługosz]

ok even if it is an exact method why is that area (physically not mathematically) equal to the displacement or it is just is by experiments?

The http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus" .

 

[sophiecentaur]

Why do so many people who post on these fora expect a simple answer to these questions? The fact is that you can't understand stuff until you've actually learned something about it. There is no easy answer. There is no simple model. It's just HARD.
I'm not being elitist; I'm just stating the way it is.