## Estimating the maximum possible percentage error

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 :)
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Mentor
 Quote by benji123 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...
Welcome, benji. You have come to a good place for help.
 Quote by benji123 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
A is a constant, so dn/dA doesn't make any sense. Instead, you want dn/dd. For the record, dn/dL and dn/dd are really the partial derivatives of n. I.e.,
$$\frac{\partial n}{\partial L} \text{and} \frac{\partial n}{\partial d}$$
 Quote by benji123 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 :)
For the percent change in n, you want dn/n.

 Quote by Mark44 For the percent change in n, you want dn/n.
Thanks for the reply Mark :) and correcting me with the dn/dA and dn/dd situation!

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

Mentor

## Estimating the maximum possible percentage error

You don't have L and d, but you have their relative errors: ΔL/L and Δd/d.

 Quote by Mark44 You don't have L and d, but you have their relative errors: ΔL/L and Δd/d.
What do I do with the relative error? As "Δd" will be 2% what will I use for "d" ?
 Mentor 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.
 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?

Mentor
 Quote by benji123 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?
No. Where are you getting the -2 numbers?

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/d4
$$dn = A(\frac{\partial n}{\partial L}~dL + \frac{\partial n}{\partial d}~dd)$$

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

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