Estimating the maximum possible percentage errorby benji123 Tags: error, estimating, maximum, percentage 

#1
Aug1010, 01:53 PM

P: 4

1. The problem statement, all variables and given/known data
Hi all! Im new to the forum, came across it from googling the problem which I had and someone was having difficuilty with a question like this on here! Here's the question I have... The coefficient of rigidity, n, of a wire of length L and uniform diameter d is given by n=AL/d^4 If A is a constant and the parameters L and d are known to within ± 2% of their correct values, estimate the maximum possible percentage error in the calculated value of n. 2. Relevant equations n=AL/d^4 3. The attempt at a solution dn = dn/dL x dL + dn/dA x dA dn = A/d^4 dL + L/d^4 dA n = A/d^4 x ΔL + L/d^4 x ΔA n = ΔL/L + ΔA/A Where do I go from here? Any help will be greatly appreciated :) 



#2
Aug1010, 04:32 PM

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P: 20,933

[tex]\frac{\partial n}{\partial L} \text{and} \frac{\partial n}{\partial d}[/tex] 



#3
Aug1010, 05:02 PM

P: 4

I was wondering, how do I get the estimation when I have no values to work from for L and d? 



#4
Aug1010, 05:23 PM

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P: 20,933

Estimating the maximum possible percentage error
You don't have L and d, but you have their relative errors: ΔL/L and Δd/d.




#5
Aug1010, 06:04 PM

P: 4





#6
Aug1010, 08:02 PM

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P: 20,933

No, Δd is not a percent value; it is the error in determining d. The relative error, which is what you are given, is Δd/d, and that is .02. Actually, what is given is that Δd/d <= .02. Same with L and ΔL.




#7
Aug1110, 05:31 AM

P: 4

So would it be...
% change in n = % change in L  (4 times % change in d) % change in n = 2  (4 times 2) % change in n = 2 +8 % change in n = + 6% Therefore, ±2% change in L and d = ±6% change in n? 



#8
Aug1110, 09:01 AM

Mentor
P: 20,933

You started off in the right direction, but didn't fix the errors I pointed out earlier, so let's start from the beginning. n = AL/d^{4} [tex]dn = A(\frac{\partial n}{\partial L}~dL + \frac{\partial n}{\partial d}~dd)[/tex] The percent change in n is Δn/n. Fill in the partial derivatives in the equation above and divide both sides by n. 



#9
Aug1110, 09:20 AM

P: 3,015

There is a more elementary way.
[tex] n \propto L [/tex] means that n increases with increasing L, while [tex] n \propto d^{4} [/tex] means that n decreases with increasing d. Therefore, the smallest value for n is obtained by taking the smallest value for L and the largest value for d: [tex] n_{\mathrm{min}} = A \, \frac{L_{\mathrm{min}}}{d^{4}_{\mathrm{max}}} [/tex] The largest value for n, on the other hand, is obtained by taking the largest value for L and the smallest value for d: [tex] n_{\mathrm{max}} = A \, \frac{L_{\mathrm{max}}}{d^{4}_{\mathrm{min}}} [/tex] In this way, you obtain an interval for the possible values of n: [tex] n \in [n_{\mathrm{min}}, n_{\mathrm{max}}] [/tex] Instead of the interval notation, one usually uses the "techincal notation": [tex] n = \bar{n} \pm \Delta n [/tex] which actually means: [tex] \left\{\begin{array}{l} n_{\mathrm{min}} = \bar{n}  \Delta n \\ n_{\mathrm{max}} = \bar{n} + \Delta n \end{array}\right. \Leftrightarrow \left\{\begin{array}{l} \bar{n} = \frac{1}{2} \, (n_{\mathrm{min}} + n_{\mathrm{max}}) \\ \bar{n} = \frac{1}{2} \, (n_{\mathrm{max}}  n_{\mathrm{min}}) \\ \end{array}\right. [/tex] Then, of course, the relative uncertainty, expressed in percent, is defined as: [tex] \delta_{n} \equiv \frac{\Delta n}{\bar{n}} \cdot 100\% [/tex] It is up to you to: 1. Find L_{min} and L_{max} by knowing [itex]\bar{L}[/itex] (the nominal value) and [itex]\Delta L = \delta_{L}/{100 \%} \, \bar{L}[/itex] (the absolute uncertainty); 2. Do the same for d_{min} and d_{max}; 3. Find n_{min} and n_{max} according to the above formulas; 4. Find [itex]\bar{n}[/itex] (nominal value) and [itex]\Delta n[/itex] (absolute uncertainty) according to the above formulas; 5. Find the relative uncertainty [itex]\delta_{n}[/itex]. 


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