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I don't understand the following concept. It was given by the textbook.

The dimensions of a cylinder are measured to the nearest millimeter using a measuring tape. The circumference is measured to be 22.0 cm and height measured to be 15.0 cm. Using these measurements to (a) estimate the volume of the cylinder, and (b) estimate an upperbound for the percentage error in your answer.

I understand part a). We simply substitute the values given to 'estimate' the volume of the cylinder.

For part b) the working out was,

$$\Delta V \approx \frac{\partial V}{\partial C} \Delta C + \frac{\partial V}{\partial h} \Delta h$$

$$= \frac{Ch}{2\pi}\Delta C + \frac{C^2}{4\pi}\Delta h $$

$$= \frac{165}{\pi}\Delta C + \frac{121}{\pi}\Delta h$$

Taking the absolute value of both sides

$$|\Delta V |\approx |\frac{165}{\pi}\Delta C + \frac{121}{\pi}\Delta h|$$

$$\leq\frac{165}{\pi}|\Delta C| + \frac{121}{\pi}|\Delta h|$$

I understand this because the triangular inequality was applied. But for the next step is what I don't understand.

It says The absolute error in each measurement is at most 0.5 mm, which is 0.05 cm. Then,

$$\Delta C \leq 0.05, \Delta h \leq 0.05$$

Where did the 0.05 cm come from? It seems to me it was pulled from the air, am I missing something? Thanks.

Bobby
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  • Are you confused about how they went from $0.5$mm to $0.05$cm? – Michael Albanese Aug 25 '13 at 12:25
  • No, I'm confused on how 0.5 mm was obtained in the first place? How did they know $\Delta C \leq 0.05$ and $\Delta h \leq 0.05$ – Bobby Aug 25 '13 at 12:26
  • Well, if you take a measurement to the nearest millimetre, what is the largest error you can have? – Michael Albanese Aug 25 '13 at 12:26
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    So are you saying, that for example if I measure an object and I got 10.6 mm as my measurement. I round it up to 11 mm. Then I get an 'absolute error' of 0.4 mm. So the largest error in this case is if I were to measure 10.5 mm and round it up to 11 mm. Making the absolute error to 0.5mm. Hence that is the largest error I can have? Is this a right way of visualising it? – Bobby Aug 25 '13 at 12:31
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    That's correct. You can't have an absolute error of more than $0.5$mm; if you did, the measurement would not have been rounded to the nearest millimetre. – Michael Albanese Aug 25 '13 at 12:32
  • I'll post an answer to this question. Would you mind accepting it so this question doesn't remain open? – Michael Albanese Aug 25 '13 at 12:35

1 Answers1

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As the measurements are taken to the nearest millimetre, the largest absolute error that can be made in the measurement is $0.5$mm. If there was a larger absolute error, the measurement would not have been rounded to the nearest millimetre.

Explanation: Suppose an object is measured and, to the nearest millimetre, it has length $k$mm where $k \in \mathbb{N}$. If the absolute error is $\varepsilon$, then the true length of the object is $k-\varepsilon$ or $k + \varepsilon$. If $\varepsilon > 0.5$, then $|(k - 1)-(k-\varepsilon)| = |\varepsilon - 1| < 0.5$, so $k - \varepsilon$, to the nearest millimetre, is $k - 1$, not $k$. A similar calculation shows that if $\varepsilon > 0.5$, $k + \varepsilon$ to the nearest millimitre is $k + 1$, not $k$. Therefore, if measurements are made to the nearest millimetre, the absolute error in the measurement (i.e. $\varepsilon$) is at most $0.5$mm.

Note, this has nothing to do with millimetres at all. If a measurement is made to the nearest unit, then the absolute error in that measurement is at most half of a unit.

Michael Albanese
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