Question

If the relative errors of three quantities to be multiplied together are 0.009, 0.006 and 0.003, what is the relative error of the resulting quantity? Give your answer to three decimal places.

Answer 1

Answer on Question #83078, Chemistry/ Organic Chemistry

If the relative errors of three quantities to be multiplied together are 0.009, 0.006 and 0.003, what is the relative error of the resulting quantity? Give your answer to three decimal places.

Solution

$$\alpha = \alpha_1 \alpha_2 \cdot \alpha_3$$ $$\ln \alpha = \ln \alpha_1 + \ln \alpha_2 + \ln \alpha_3$$

For little quantities:

$$\frac{\delta \alpha}{\alpha} = \frac{\delta \alpha_1}{\alpha_1} + \frac{\delta \alpha_2}{\alpha_2} + \frac{\delta \alpha_3}{\alpha_3} = 0.003 + 0.006 + 0.009 = 0.018$$

Answer: 0.018

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