Differentiation: small increments

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Homework Help Overview

The discussion revolves around the calculation of centripetal acceleration, specifically focusing on the maximum percentage error in the acceleration due to uncertainties in velocity and radius measurements. The subject area includes concepts from calculus and physics related to differentiation and error analysis.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the formula for maximum percentage error in centripetal acceleration and explore the implications of small increments in measurements. There are attempts to clarify the meaning of Δv and Δr, as well as the rationale behind the error propagation formula.

Discussion Status

The discussion is ongoing, with participants expressing confusion about the derivation and application of the error formula. Some have offered insights into the mathematical reasoning, while others are seeking further clarification on specific terms and concepts. There is no explicit consensus yet, but various interpretations and approaches are being explored.

Contextual Notes

Participants are navigating the complexities of error analysis in experimental physics, with specific attention to how measurement uncertainties affect calculations. There are mentions of homework guidelines and preferences for how solutions should be presented, which may influence the flow of the discussion.

DevonZA
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Homework Statement

Homework Equations

The Attempt at a Solution



1. The centripetal acceleration of a particle moving in a circle is ##a = \frac{v^2}{r}##, where v is the velocity and r is the radius of the circle. Approximate the maximum percentage error in the calculation of the acceleration if the error in experimental measurements of v is ± 0,2% and in r is ± 0,8%.

2. Differentiation: small increments

3. See attached.
 

Attachments

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You know , there is a simple way -
Δa/a = 2Δv/v + Δr/r , for small Δv and Δr .
The reason the two quantities on the right are added is because they both have a possible plus - minus error .

Hope this helps .
 
Qwertywerty said:
You know , there is a simple way -
Δa/a = 2Δv/v + Δr/r , for small Δv and Δr .
The reason the two quantities on the right are added is because they both have a possible plus - minus error .

Hope this helps .

##\frac {Δa}{a} = \frac {2(0.2-(-0.2))}{0.2} + \frac {(0.8-(-0.8))}{0.8}##
 
DevonZA said:
##\frac {Δa}{a} = \frac {2(0.2-(-0.2))}{0.2} + \frac {(0.8-(-0.8))}{0.8}##
Wrong .
Okay , perhaps I should have mentioned this earlier - Δx ( x is just some random variable ) represents uncertainty / error in the measurements of x .

So , now , what is Δv , and what is Δr ?
 
No, I'm lost sorry?
 
DevonZA said:
Approximate the maximum percentage error in the calculation of the acceleration if the error in experimental measurements of v is ± 0,2% and in r is ± 0,8%.
These values seem like Δv/v and ...
 
Qwertywerty said:
These values seem like Δv/v and ...

...Δr/r. They are the change in v and r because both can increase or decrease but I still don't understand what you are getting at and surely Δv = +-0.2 and Δr = +-0.8? Why put Δv over v and Δr over r?
 
DevonZA said:
Why put Δv over v and Δr over r?
Over here ? -
Qwertywerty said:
You know , there is a simple way -
Δa/a = 2Δv/v + Δr/r , for small Δv and Δr .

This is derived from -
Suppose error in a is da , and in some random variable x is dx .
Then , error in a w.r.t - x is
da/dx .
Also , a = v2/r .
Hence differentiating w.r.t x on both sides ,
da/dx = 2vrdv/dx - v2dr/r2.dx - Chain rule
⇒da = 2vrdv - (v/r)2 . - Cancelling dx
⇒da/a = 2dv/v - dr/r - Dividing by a

But error is plus - minus , and hence it would become
da/a = 2dv/v + dr/r .

I hope that was your doubt .
 
I am so confused but thank you for attempting to get me to understand
 
  • #10
DevonZA said:
I am so confused but thank you for attempting to get me to understand
What part ?
 
  • #11
All of it to be honest. Maybe it is the way it is written that is confusing me
 
  • #12
DevonZA said:

Homework Statement

Homework Equations

The Attempt at a Solution



1. The centripetal acceleration of a particle moving in a circle is ##a = \frac{v^2}{r}##, where v is the velocity and r is the radius of the circle. Approximate the maximum percentage error in the calculation of the acceleration if the error in experimental measurements of v is ± 0,2% and in r is ± 0,8%.

2. Differentiation: small increments

3. See attached.
Whoever gave you this problem expects you to know that an approximation to the error in a is da= \frac{2v}{r}dv- \frac{v^2}{r^2}dr where dv and dr are the respective errors.
 
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  • #13
Final answer attached. Thanks to all who helped.
 

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  • #14
DevonZA said:
Final answer attached. Thanks to all who helped.

Please type out your solution; I cannot open the pdf file on my new i-Phone, and you should not make me install an App just to read your work.
 
  • #15
Ray Vickson said:
Please type out your solution; I cannot open the pdf file on my new i-Phone, and you should not make me install an App just to read your work.
I agree with Ray. Posting hand-written work, which is often difficult to read, makes it harder for us as helpers to give you assistance. For one thing, if we spot a mistake, we have to provide context to indicate which line is in error.

For another thing, having the work in a separate window is a hassle that many homework helpers just won't bother with.
 
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  • #16
You can view it easily on a computer. The writing is pretty clear and I am satisfied with the answer, I simply put it here as information.
 
  • #17
DevonZA said:
You can view it easily on a computer. The writing is pretty clear and I am satisfied with the answer, I simply put it here as information.

You should have left out the first sentence above; the correct statement replacing it should be something like "I am sorry, and won't do that in future".

I suggest you read the "pinned" post "Guidelines for students and helpers", by Vela (at the start of the list of messages). Many helpers will not bother to read attachments, even on a computer, so will not-- freely and without pay-- offer their time to help.
 
  • #18
I will type the answer out when I get some spare time. Thanks.
 
  • #19
DevonZA said:
I will type the answer out when I get some spare time. Thanks.
No need to do that for this thread, but keep it in mind for future posts.
 
  • #20
Mark44 said:
No need to do that for this thread, but keep it in mind for future posts.

If there is no need to do it in this thread then why is Ray going on about it as if I have committed a crime? I did say that it is only there for information. I like to type my responses/answers/questions in correct latex form but for me to do this when I am satisfied with the answer and not actually asking anything seems a little extreme. Anyway I apologize if I upset anyone, I posted my final answer in PDF not knowing that it isn't a preferred method so it was unintentional.
 
  • #21
DevonZA said:
If there is no need to do it in this thread then why is Ray going on about it as if I have committed a crime? I did say that it is only there for information. I like to type my responses/answers/questions in correct latex form but for me to do this when I am satisfied with the answer and not actually asking anything seems a little extreme. Anyway I apologize if I upset anyone, I posted my final answer in PDF not knowing that it isn't a preferred method so it was unintentional.

I did mention it so that you would come to realize it, that's all. Everything I said was 100% true: many helpers will never bother to look at solution attachments (unless they are something like diagrams, etc.), and so by posting such attachments you are dismissing a significant portion of the PF community of helpers.

BTW: I did look at your pdf on my laptop, but still found it very hard to read; parts of it I could not make out at all.
 

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