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> How can i calculate the absolute uncertainty dt in the following case T=(P/A)^1/4 if P=(5±0.15) and A=(0.3±0.009)

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  • \$\begingroup\$ When uncertainty gets confusing just use min/max analysis. All other methods are really just shortcuts anyways. \$\endgroup\$ – caveman Jan 20 '15 at 1:59
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P has a relative uncertainty of ±0.15/5 = ±0.03

A has a relative uncertainty of ±0.009/0.3 = ±0.03

When you add/subtract numbers you add their absolute uncertainties.

When you multiply/divide numbers you add their relative uncertainties.

Therefore, the relative uncertainty in P/A is ±0.03 + ±0.03 = ±0.06. The nominal value of P/A is 5/.3 = 16.67, so its absolute uncertainty is 16.67 × ±0.06 = ±1.0.

You can verify this: The maximum value of P/A occurs when P is maximum and A is minimum: 5.15/0.291 = 17.70. Similarly, the minimum value of P/A occurs when P is minimum and A is maximum: 4.85/0.309 = 15.70.

When you raise a value to a power, you multiply its relative uncertainty by the power. Therefore T has a relative uncertainty of ±0.06 × 1/4 = ±0.015. The nominal value of T is 16.671/4 = 2.02, so its absolute uncertainty is 2.02 × ±0.015 = ±0.03.

Again, verify this: The minimum value is 15.701/4 = 1.99 and 17.701/4 = 2.05.

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