Estimating maximum percentage error using partial differentiation

In summary: And the maximum error would be:ΔD_max = 6,70208 * 10^-6 E / Eh^3/(12(1-v^2))ΔD_max = 6,70208 * 10^-6 * 12(1-v^2) / EhΔD_max = 8,0425 * 10^-5 / EhIs this correct?In summary, the maximum error in D can be expressed as 8,0425 * 10^-5 / Eh, where E is a constant and h and v are given values with uncertainties. The error is calculated using the partial derivatives of D with respect to h and v, and the values of Δh and
  • #1
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Homework Statement
1. In the formula D = Eh^3/(12(1-v^2)), h is given as 0,1 +- 0,002 and v as 0,3 +- 0,02. Express the approximate maximum error in D in terms of E.

2. The coefficient of rigidity (n) of a wire of length (L) and uniform diameter (d) is given by n = AL/d^4, where A is constant. If errors of +-0,25% and +-1% are possible in measuring L and d respectively, determine the maximum percentage error in the calculated value of n.
Relevant Equations
1.
- D = Eh^3/(12(1-v^2))
- ΔD = ∂D/∂h * Δh + ∂D/∂v * Δv

2.
- n = AL/d^4
- Δn = ∂n/∂L * ΔL + ∂n/∂d * Δd
no.1.png


Attempt at question No. 1:
ΔD = ∂D/∂h * Δh + ∂D/∂v * Δv

∂D/∂h = 3Eh^2/(12(1-v^2))
∂D/∂v = 2Eh^3/(12(1-v^2)^2)
Δh = +- 0,002
Δv = 0,02
h = 0,1
v = 0,3

ΔD = 3Eh^2/(12(1-v^2)) * Δh + 2Eh^3/(12(1-v^2)^2) * Δv

Because the problem asked for maximum percentage error then I decided to use the positive value of Δh = 0,002, because the value of Δv is positive, is this right?

ΔD = 3E(0,1)^2/(12(1-(0,3)^2)) * (0,002) + 2E(0,1)^3/(12(1-(0,3^2)^2) * (0,02)

Finish the equation and then the result will be:
ΔD = 4,5746 * 10^-6 E.

Really skeptical about this answer though.
no.2.png


Attempt at question No. 2:

Δn = ∂n/∂L * ΔL + ∂n/∂d * Δd

∂n/∂L = A/d^4
∂n/∂d = -4AL/d^5
ΔL = +-0,25% L
Δd = +-1% d

Δn = A/d^4 * (0,25%*L) + -4AL/d^5 * (1%*d)
Δn = AL/d^4 * (0,0025) + -4AL/d^4 * (0,01)
Δn = 0,0025n - 0,04n = -0,0375n = -3,75%n

I'm not confident in this one though, I think the answer should be 4,25%, but I don't know how to achieve that value.
 
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  • #2
manareus said:
because the value of Δv is positive, is this right?
Why would ##\Delta v## be positive? You simply do not know whether it is positive or negative.

Once you have arrived at the expression for the error, the expectation for the error should be zero, but you are interested in how large the typical error would be, i.e., the variance in the measurement, so the typical thing to do would be to take the expectation value of the square of the error. Now, this may or may not be what your book wants you to do, it should tell you when it discusses error estimates.
 
  • #3
Should you not be adding in quadrature? For instance,
\begin{align*}
\Delta D^2 = \left( \frac{\partial D}{\partial h} \Delta h \right)^2 + \left( \frac{\partial D}{\partial v} \Delta v \right)^2
\end{align*}
 
  • #4
The book wants "maximum error" it says. Bad book
 
  • #5
Orodruin said:
Why would ##\Delta v## be positive? You simply do not know whether it is positive or negative.

Once you have arrived at the expression for the error, the expectation for the error should be zero, but you are interested in how large the typical error would be, i.e., the variance in the measurement, so the typical thing to do would be to take the expectation value of the square of the error. Now, this may or may not be what your book wants you to do, it should tell you when it discusses error estimates.
Apologies, my teacher changed the value of ##\Delta v##, from +-0,02 to +0,02.
ergospherical said:
Should you not be adding in quadrature? For instance,
\begin{align*}
\Delta D^2 = \left( \frac{\partial D}{\partial h} \Delta h \right)^2 + \left( \frac{\partial D}{\partial v} \Delta v \right)^2
\end{align*}
From examples, to solve the problem I don't need to be adding quadrature. Here is an example:
ex.png
 
  • #6
So it is a small deviation (calculus) problem, not an "error" (physics) problem per se.
 
  • #7
hutchphd said:
So it is a small deviation (calculus) problem, not an "error" (physics) problem per se.
Yes, correct.
 
  • #8
The point is that for most systems ICBS that the errors from an ensemble are symmetrically distributed about the mean value. The important measure is the RMS value i.e. the percentage Variance about the mean.
 
  • #9
hutchphd said:
The point is that for most systems ICBS that the errors from an ensemble are symmetrically distributed about the mean value. The important measure is the RMS value i.e. the percentage Variance about the mean.
I'm not sure I follow, but I kind of get what you mean.
 
  • #10
manareus said:
Homework Statement:: 1. In the formula D = Eh^3/(12(1-v^2)), h is given as 0,1 +- 0,002 and v as 0,3 +- 0,02. Express the approximate maximum error in D in terms of E.

Relevant Equations:: 1.
D = Eh^3/(12(1-v^2))
ΔD = ∂D/∂h * Δh + ∂D/∂v * Δv

View attachment 299254

Attempt at question No. 1:
ΔD = ∂D/∂h * Δh + ∂D/∂v * Δv

∂D/∂h = 3Eh^2/(12(1-v^2))
∂D/∂v = 2Eh^3/(12(1-v^2)^2)

ΔD = 3Eh^2/(12(1-v^2)) * Δh + 2Eh^3/(12(1-v^2)^2) * ΔvI'm not confident in this one though, I think the answer should be 4,25%, but I don't know how to achieve that value.
It appears that you have an error in differentiating ##D## w.r.t. ##v##.
You are missing a factor of ##v## .

Then if you are interested in percent error you should be looking at ##\dfrac{\Delta D} D ## .
 
  • #11
SammyS said:
It appears that you have an error in differentiating ##D## w.r.t. ##v##.
You are missing a factor of ##v## .

Then if you are interested in percent error you should be looking at ##\dfrac{\Delta D} D ## .
Yeah, I noticed that when I tried to solve the problem again, thank you for pointing it out though!

So the final result would be:

ΔD = 3Eh^2/(12(1-v^2)) * Δh + 2Eh^3v/(12(1-v^2)^2) * Δv

ΔD = 3E(0,1)^2/(12(1-(0,3)^2)) * (0,002) + 2E(0,1)^3(0,3)/(12(1-(0,3^2)^2) * (0,02)
ΔD = 5,4945 * 10^-6 E + 1,20758 * 10^-6 E
ΔD = 6,70208 * 10^-6 E
 

What is partial differentiation?

Partial differentiation is a mathematical concept used to calculate the rate of change of a function with respect to one of its variables while holding all other variables constant. It is commonly used in multivariate calculus and is an important tool in many scientific fields.

Why is partial differentiation used in estimating maximum percentage error?

Partial differentiation allows us to calculate the sensitivity of a function to changes in its variables. By using this technique, we can determine the maximum percentage error in a function's output based on the maximum percentage error in its input variables.

What is the process for estimating maximum percentage error using partial differentiation?

The process involves taking the partial derivative of the function with respect to each of its variables, multiplying these derivatives by the corresponding maximum percentage error in the input variables, and then adding these values together to get the estimated maximum percentage error in the function's output.

Can partial differentiation be used for any type of function?

Yes, partial differentiation can be used for any function that has multiple variables. However, the function must be differentiable, meaning that it must have a well-defined derivative at every point in its domain.

Are there any limitations to using partial differentiation for estimating maximum percentage error?

Partial differentiation relies on the assumption that the input variables are independent of each other. If this assumption is not met, the estimated maximum percentage error may not be accurate. Additionally, it is important to note that partial differentiation only provides an estimate and not an exact value for the maximum percentage error.

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