4

The volume of a cylinder is given by the expression

$V=\pi r^2 h$

The uncertainties for $V$ and $h$ are as shown below

$\begin{align} V&\pm 7\%\\ h&\pm 3\% \end{align}$

What is the uncertainty in $r$?

Now, the obvious answer would be $2\%$, from $$\frac{dV}{V}=\frac{dh}{h}+2\frac{dr}{r}$$

However, rearranging to $r^2=\frac{V}{\pi h}$ gives $$2\frac{dr}{r}=\frac{dV}{V}+\frac{dh}{h}$$ which gives a different answer of $5\%$. Thus, by simply rearranging the formula, we get different values for uncertainty in $r$.



How do you explain this?

(The mark scheme lists $5\%$ as the correct answer)

Helder Velez
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Milo Wielondek
  • 164

5 Answers5

8

You're confusing independent and dependent variables. When you propogate from uncertainties in the $x_{i}$ to some $f(x_{1},x_{2}...)$, the formula $\delta f(x_{1}...)=\sum \left|\frac{\partial f}{\partial x_{i}}\right|\delta x_{i}$ assumes that each of the $x_{i}$ is an independently measured variable and that $f$ is a dependent variable to be calculated from the $x_{i}$.

In the example you give, you have two independent measurements of $V$ and $h$ and are expected to calculate the uncertainty in $r$. Well, to use the above formula, you need to write $r$ as a dependent variable of $V$ and $h$. Therefore, it's only correct to solve for $r$ first, and then calculate the uncertainty.

Zo the Relativist
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8

Jerry Schirmer's right about why solving for $r$ first is the right procedure. One way to illustrate this is to notice that with the other procedure the uncertainty could go negative, which can't be right.

But the main thing I wanted to point out is that, if the measurements of $V$ and $h$ are independent, and if the "errors" mean standard deviations as usual, then the correct procedure is to add the errors in quadrature (i.e., to add the squares and take the square root): $$ {\delta r\over r}={1\over 2}\sqrt{\left(\delta V\over V\right)^2+\left(\delta h\over h\right)^2}. $$ See, e.g., any of the first few Google hits for "propagation of errors."

Ted Bunn
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2

This should be a comment to Jerry Schirmer's right answer.

The solutions you give solve different problems.

In the first calculation you measure V with its uncertainty $\Delta V$, you know it depends with r and h, you know the uncertainty you have in the measurement of h, $\Delta h$ and thus you infer the uncertainty you had associated with measuring r, $\Delta r$.

In the second calculation you want to measure r from V and h. Knowing the uncertainties in V and h you predict the uncertainty that you will have in r.

perplexity
  • 146
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1

Your first formula is correct, but it looks like you substituted in the incorrectly signed value for (dh/h). Your second formula is missing a sign change for the (dh/h) term.

In the first equation, substituting, to get the MAXIMUM (dr/r) --->

+7% = (-3%) + (2 x (+5%)) -OR- -7% = (+3%) + (2 x (-5%))

So, your text is correct: +/-5% is the (approximate) uncertainty for r.

If you solve this problem algebraically, you will find that the (corrected) differential equations you constructed gives a very close but not exact answer. The correct uncertainty for r is something like + SQRT(1.07/0.97) and - SQRT(0.93/1.03) = +5.0282% / -4.9783%. This is because the differential eqns are valid only for infinitesimal deltas; but we are using deltas on the order of +/-5% (hardly infinitesimal).

Vintage
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0

I will use two ways to derive the uncertainity on r
1 - very simple method: the 'interval arithmetic' present in the Euler math package (free)
2 - derive the formula of dr(V,h, dV,dh)
Both answers : 5% (for any value V or h that I choose for V and h) 

// 'using interval arithmetic'    
>dV=0.07 dh=0.03         // deltas  
>V=1000 h=40             //arbitrary choosen values  
>r0=sqrt(1/pi*V/h)       //central value for r    
 2.82094791774  
>r=sqrt(1/pi*((V±V*dV)/(h±h*dh))) //interval values for r   
 ~2.6,3~                          //    from 2.6 to 3  
>(r-r0)  
 ~-0.14,0.14~  
>(r-r0)/r0                           
 **~-0.05,0.05~                     // 5% for dr/r0 for any V, h**

//calc dr(V,h, dV,dh) (seeResistance_measurement Example and adapting the derivatives)

dr=1/2/sqrt(pi)*sqrt(dV^2/V/h+dh^2*V/h^3)
0.00106245302667
1-(r-dr)/r0
~-0.05,0.05~ // the result is again 5%

Helder Velez
  • 2,685