Estimating maximum percentage error using partial differentiation

[manareus]

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

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.

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.

 

[Orodruin]

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.

 

[ergospherical]

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*}

 

[hutchphd]

The book wants "maximum error" it says. Bad book

 

[manareus]

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.

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:

 

[hutchphd]

So it is a small deviation (calculus) problem, not an "error" (physics) problem per se.

 

[manareus]

So it is a small deviation (calculus) problem, not an "error" (physics) problem per se.

Yes, correct.

 

[hutchphd]

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.

 

[manareus]

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.

 

[SammyS]

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 $\dfrac{\Delta D} D$ .

 

[manareus]

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