Estimating the maximum possible percentage error

[benji123]

Homework Statement

Hi all! I am 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.

Homework Equations

n=AL/d^4

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 :)

 

[Mark44]

Homework Statement

Hi all! I am 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.

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.

Homework Equations

n=AL/d^4

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.,
$$\frac{\partial n}{\partial L} \text{and} \frac{\partial n}{\partial d}$$

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.

 

[benji123]

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?

 

[Mark44]

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

 

[benji123]

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" ?

 

[Mark44]

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.

 

[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?

 

[Mark44]

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.

 

[Dickfore]

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:

$$\left\{\begin{array}{l}<br /> n_{\mathrm{min}} = \bar{n} - \Delta n \\<br /> <br /> n_{\mathrm{max}} = \bar{n} + \Delta n<br /> \end{array}\right. \Leftrightarrow \left\{\begin{array}{l}<br /> \bar{n} = \frac{1}{2} \, (n_{\mathrm{min}} + n_{\mathrm{max}}) \\<br /> <br /> \bar{n} = \frac{1}{2} \, (n_{\mathrm{max}} - n_{\mathrm{min}}) \\<br /> \end{array}\right.$$

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 L_min and L_max by knowing $\bar{L}$ (the nominal value) and $\Delta L = \delta_{L}/{100 \%} \, \bar{L}$ (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 $\bar{n}$ (nominal value) and $\Delta n$ (absolute uncertainty) according to the above formulas;

5. Find the relative uncertainty $\delta_{n}$.