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Can Absolute Velocity be Measured? 
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#55
May512, 07:41 AM

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Hi Austin, you're right about nx^{n1}.
dx(t)/dt = h. I have to say that this is probably not the place for coaching in calculus. I was hoping you could see how straightforward it is and get a textbook ... 


#56
May512, 12:21 PM

P: 255

Austin0, if you are asking if the final "dx/dt" represents the division of a value dx by another value dt, the answer is no. The bit that gives you the value of v is [tex]v=\lim_{\Delta t\to 0} \, \ ...[/tex] The notation "d/dt" is a way of writing that limit so as Menzt114 says, that is the definition of dx/dt. 


#57
May512, 03:35 PM

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P: 17,271

Sorry I have been delayed. I was travelling for business very heavily. Some of your questions may have already been resolved, so I apologize for any redundancy.
My whole involvement in this thread was to correct the misunderstanding that the velocity is an approximation i.e. inexact. Your statements indicated to me that you still felt that it was inexace because it was a derivative. So I explained. That is all, no insult was provided nor implied. [tex]f'(x)=\lim_{\Delta x\to 0} \, \frac{f(\Delta x+x)f(x)}{\Delta x}[/tex] Now, look at the limit you are trying to evaluate: [tex]\lim_{t \to \infty}\,v(t)[/tex] There is no Δt which is approaching 0, so the limit cannot be expressed in the form of the above limit. Therefore it is not a derivative. Now, v(t) itself is a derivative v(t)=x'(t) so we can write the limit as [tex]\lim_{t \to \infty}\, \left( \lim_{\Delta t\to 0} \, \frac{x(\Delta t+t)x(t)}{\Delta t} \right)[/tex] This makes it clear that the overall expression is not a derivative, but a limit of a derivative. One of the key things that makes it not a derivative is the fact that the limit is taken to infinity. In a derivative the limit is always taken to 0. This is important, because a limit to infinity is fundamentally a different thing than a limit to some finite value. A limit to some finite value means "as I get arbitrarily close to X ..." but you cannot get arbitrarily close to infinity. For instance, the real number .0001 is within .001 of 0. Can you give me a real number which is within .001 of infinity? So a limit to infinity is a fundamentally different thing, it means what does f(x) tend to asymptotically as x increases without bound. In this case, that number is c. So the limit is c meaning that as t increases without bound v asymptotically approaches c. 


#58
May812, 09:42 PM

P: 1,162

Hi GeorgeDishman Thank you for your patience.There has been much confussion in this discussion and I have no doubt a certain frustration on both sides.
In the velocity equation [tex]v=\lim_{\Delta t\to 0} \, \frac{ x(t+\Delta t ) x(t)}{\Delta t}=dx/dt[/tex] the final dx/dt is the result? [QUOTE=GeorgeDishman;3899076]I may be overanalysing but it may avoid some confusion if I check something here. Austin0, if you are asking if the final "dx/dt" represents the division of a value dx by another value dt, the answer is no. The bit that gives you the value of v is [\QUOTE] I thought the final dx/dt represented the answer .A ratio like 20m/s not an operation of division to obtain the answer. Is this incorrect? The term delta itself seems to have another intrinsic meaning , that being; interval. Wrt dynamic systems it is found by subtracting one value of some variable at a point from the value at some other point to get a difference , an interval ..Correct? Now delta may have a more specific meaning as an operator in the context of derivation but it is certainy widely used outside that context as meaning interval with no implied reference to the operation of derivation,limit etc. Example Is there some distinction between small d delta and capitol D delta you are referring to??? Isn't the fundamental definition of velocity; rate of change of position as a function of time?? The difference in position with respect to the difference in time??? Are you saying this definition is invalid or has no relevance to a velocity as a description of motion in the world.?? Isn't this essential meaning validly expressed with symbols using the fundamental definition of delta as v=dx/dt ??? So any explicit velocity value , independent of the method of arriving at it , is expressed as a ratio of intervals of change eg. 20m/s This value has an explicit literal meaning under the fundamental definition of velocity which is: the ratio of a specific interval/change of position with respect to a specific interval/change of time. Eg. 20m/s I am not talking about the interpretation here but the literal meaning. Would you agree??? [QUOTE=GeorgeDishman;3894814] No, the meaning is that, at that instant, the rate at which the distance is changing is a slope of 20m/s but since the acceleration is constant, it has that value only at that instant. That is what is called velocity in physics. [\QUOTE] I was of , course, aware of this as it is the point of this discussion but: Here you are talking about interpretation which is fine.. So what is your interpretation of this value as a description of the real world motion of the particle?? Thanks 


#59
May812, 10:05 PM

P: 1,162

Everything i have said has been reinterpreted into a statement or argument directed to the proposition that a derivative or instantaneous velocity it not exact. This is in spite of the fact that right from the beginning and repeatedly after, I have clearly stated this was not the question , that it was given that a velocity was exact at an instant. The result is that my communication has not been understood and just as in this case , my arguments and questions have not ever been addressed. I thought that the inclusion of a cartoon face would make it clear enough I was not serious. But on the other hand your response here almost has me wondering if it was I and not you that made the incorrect conclusion about a brand new concept ;) But seriously, you have been nothing but helpful as long as I have been in the forum and I consider you ,with DrGreg, as being among the most considered and considerate members, so I never assumed any negative intent on your part whatever. post 32 The post above is #32 we are now up to #57 and it is apparently still unclear that I never felt or said the value of an instantaneous velocity was inexact , I have been trying to talk about the meaning and interpretation of the derived value itself, independent of its method of determination. What that value meant as a description of the motion of a particle in the real world. But somehow this is not possible and it always, circularly, ends up back focused on derivatives. This is also a rational interpretation of the statement ;" the tangent at a point is a linear approximation of the values of the curve near to that point of exact value." As the curve represents the motion in the real world this means within a short distance on the actual path. If you disagree with this interpretation of the tangent could you explain your alternative??? Given two measurements, the accuracy of the derived value is inversely dependent on the spatial separation between the two events. Increasing as the interval decreases. In this case derivation is no more accurate at interpolation than the simple dx/dt and would return the same value . Both are approximations in this circumstance. Is this correct?? This is irrelevant to the initial question but there is another question. Given an exact value at a point (ideally close measurements) what would you say about this value as a predictor of measurements of position or time at other points? So what is your interpretation of an instantaneous velocity as a description of the motion of an accelerating particle.??? Thanks for your patience and help .[/QUOTE] 


#60
May912, 12:53 AM

P: 255

The definition of that operation is the derivative. 


#61
May912, 04:53 PM

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P: 17,271

I am sorry, but I don't think I can help. I have given you the mathematical definition of the derivative and explained it in english to the best of my ability. The physical position can be represented as a vectorvalued function of time, x(t). The velocity is defined as the derivative of the position with respect to time, x'(t) = dx/dt. It seems to me that this is a complete description of the topic, so I don't see what else can be added and I don't understand where you think there is some ambiguity in physical interpretation. 


#62
May912, 10:10 PM

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#63
May1012, 06:55 AM

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#64
May1012, 09:08 PM

P: 1,162

I did look up the Taylor expansion and it is exacty as you said but is a different thing than the sources I quoted. Every source and also GeorgeDishman and Russ_Waters seem to agree that the tangent comes directly from the derivation to the limit and is exact. I may be mistaken but in a way it doesn't matter as I came into this with the understanding I had gotten from Minkowski diagrams of accelerating systems that the tangent was exact so that was never a question. So my interpretation of the statement that "the tangent is an approximation of the values of the curve near the exact point" , is with the understanding that the line is exact and the point of congruency is exact. I explicitely stated this in an earlier post. And that the tangent was also congruent with the ICMIF at that point. Thanks 


#65
May1112, 06:24 AM

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Where did I say that?!!
Perhaps the issue here is you are confusing "a line" with "the slope of a line"? 


#66
May1112, 07:54 AM

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[tex]f(x_0)+f'(x_0)(xx_0)\ne f'(x)[/tex] Since you are now saying that you believe that the tangent line is also "exact", I really have no idea what you think is inexact. Perhaps you can clarify, and also perhaps clarify what you mean by "exact" (eg exactly what). 


#67
May1112, 08:42 AM

P: 643

What I'm saying is there's no uncertainty in what the tangent line is, but it can be used to approximate stuff with some uncertainty. 


#68
May1112, 08:59 AM

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http://en.wikipedia.org/wiki/Taylor_series In a post some time back that I can't find at the moment, there was a comment to the effect that if you wanted to approximate a curve with a straight line, the best value for the slope of that line was to make it equal to the value of the derivative at the desired point. There was no implication in the cited source that a derivative was an approximation but possibly some readers might have mistakenly taken it that way. 


#69
May1112, 10:30 AM

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I think that part of the problem here is the ambiguous use of the words exact and approximate. Austin0 should clarify. 


#70
May1212, 05:09 AM

P: 512

When the tangent line is used to plot a point on a curved line in the neighbourhood of a given point, this point becomes an approximation because the tangent line is an approximation of the curvature in the vicinity of the given point.
The velocity at this point lies on the tangent, is not an approximation because we use the limit delta t tends to zero; in the domain of this limit, the infinitesimal displacement is almost an exact straight line(no room for curvature), hence the velocity is said to be exact. 


#71
May1312, 03:47 AM

P: 1,162

Apparently I was again mistaken as you seem to be saying it is necessary to resort to other processes to define the line. Is this the case?? SO is the tangent line referred to here not a direct product of the derivative?? Is it not to be taken, as I did, as being exact?? So is the approximation that is mentioned here referring to the line itself or is it referring to the relationship of the line to points on the curve. It seems to me to clearly be the latter , am I wrong?? Thanks 


#72
May1312, 03:58 AM

P: 1,162

Also I use the word ratio interchangebly with rate . I am beginning to suspect this may be semantically wrong in this context also? This statement was directed to the limits I was familiar with: asymptotic approach to c with constant acceleration, the "Slow clock transport" as velocity approaches 0, etc. When this was misinterpreted to apply to instantaneous velocity , tangents etc I immediately responded (below) that this was definitely not the case post 32 I was not completey wrong. Now I would say "like some limits and the Taylor series expansion" provide exceedingly useful and accurate approximations Just kidding. I plan to carefully avoid any such provocative statements in the future. Thanks 


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