Approximation and Estimation

Chapter 3 Study Notes

Section 5 of 7
Contents

3.4Approximation Errors

When calculations have multiple steps, rounding too early can cause errors to accumulate. These are called premature rounding errors.

Avoiding premature rounding errors

The area of a circle is 54 cm². Find its circumference, correct to 3 s.f.

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A=πr2=54A = \pi r^2 = 54, so r=54π=4.14593r = \sqrt{\frac{54}{\pi}} = 4.14593\ldots cm
Keep that in the calculator. Circumference =2πr=26.0= 2\pi r = 26.0 cm (to 3 s.f.)
Now round rr to 4.154.15 first: 2π×4.15=26.072\pi \times 4.15 = 26.07\ldots, which gives 26.126.1 cm.
26.026.0 and 26.126.1 are different answers, and only the first is right.
Round once, at the end, to the accuracy the question asks for.

Reverse problem: error bounds on a derived quantity

The area of a square is given as 850850 cm² correct to 2 significant figures. Find the range of possible perimeters.

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Since 850850 cm² is correct to 2 s.f., the area AA satisfies 845A<855845 \leq A < 855.
The side length s=As = \sqrt{A}, so ss ranges from 845\sqrt{845} to 855\sqrt{855}.
845=29.0688\sqrt{845} = 29.0688\ldots cm and 855=29.2403\sqrt{855} = 29.2403\ldots cm.
Perimeter =4s= 4s, so the perimeter PP satisfies 4845P<48554\sqrt{845} \leq P < 4\sqrt{855}.
4845=116.27554\sqrt{845} = 116.2755\ldots cm and 4855=116.96154\sqrt{855} = 116.9615\ldots cm.
Therefore 116.27 cmP<116.97 cm116.27 \text{ cm} \leq P < 116.97 \text{ cm}, rounding each bound outwards so that no possible perimeter is left out.

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