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Hence the possible relative error, that is, the ratio of the possible error to the product, is

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The relative error of ab is therefore equal to the sum of the relative errors of a and b.

The case for the quotient is proved similarly, assuming that the possible error is half the difference of the limits

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with the further assumption that (Ab)2 is small in comparison with Aa and Ab and may be neglected.

Since the possible relative error of a product is greater than that of either factor, there can be no more significant figures in the product than in the factor with the smaller number of significant figures.* In practice, factors are rounded off to the same number of significant figures before multiplying, and only that number of significant figures retained in the product. In division, the dividend, divisor, and quotient are treated similarly.

Abridged methods of multiplication and division, which do away with the labor of retaining non-significant figures during either operation, are illustrated in the examples following.

EXAMPLE 2. Find the circumference of a circle whose measured diameter is 45.38 inches.

Since the diameter is given to four significant figures, the value of π should be taken to four significant figures. This value is π = 3.142.

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The value of the circumference, πd, to four significant figures is therefore 142.6.

*There may be less. For example, if two factors with three significant figures are close to 100, such as 123 and 114, the possible relative

Using ordinary multiplication first we have (a), which gives the arithmetically correct product. The last partial product contributes most to the significant figures of the product, and may be written first, as in (b), where the order of all the partial products is reversed. The figures to the right of the vertical lines in (a) and (b) pertain to the non-significant figures of the product, and may be discarded, as in (c), which shows the abridged multiplication.

In (c), the partial product obtained from the first figure on the left of the multiplier, 3, is entered first. Then the last number on the right of the multiplicand, 8, is canceled, and the second figure of the multiplier, 1, is used, the amount carried over from the canceled figure being added in. The other partial products are rounded off similarly.

EXAMPLE 3. Find the density of a mass of 7.643 grams whose volume is 3.564 cubic centimeters.

To obtain the density divide the mass by the volume. As there are four significant figures in both dividend and divisor, neither need be rounded off before the division is performed, and the quotient should be obtained to four significant figures.

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Ordinary long division, given on the left, shows that the density is 2.144.

The abridged division is shown on the right. After the first figure in the quotient is obtained, instead of adding a cipher to the dividend, the last figure of the divisor, 4, is canceled. The next partial product is rounded off in accordance error in each is about 0.5% (by the table in the preceding section), and the possible relative error in the product is therefore approximately 1%, which indicates that only two significant figures should be retained in the product.

with the amount carried over from the multiplication of the canceled number by the second digit in the quotient, etc. Ordinarily, in finding the partial products it is sufficient to consider only the nearest canceled figure of the divisor, but it is sometimes necessary to consider two canceled figures to determine the amount to be carried over.

EXERCISES

1. Write the following numbers in standard form and determine the number of significant figures and the percentage of error in each: 3.1416, 0.00732, 259.34, 678943, 0.0020.

2. The value of π to nine figures is 3.14159265. Round off the value to one significant figure; two; three; four; five. Determine the number of significant figures and the percentage of error of the approximations 3 and. Correct such of the following rounded off values of π as are incorrect: 3.141, 3.15, 3.14160.

3. Add the following numbers obtained by measurements: 3.4785, 16.743, 253.78, 36.583.

4. The dimensions of a rectangle found by measurement are 24.78 inches and 19.8 inches. Find the area and determine the number of significant figures and the relative error in the area.

5. The diameter of a circle is found by measurement to be 151.6 millimeters. Find the circumference. Is 3 a sufficiently accurate approximation for π in this instance?

6. If one square meter is equivalent to 1.196 square yards, to how much are 5 square meters equivalent? If 5 square meters are equivalent to 5.98 square yards, to what is one square meter equivalent?

7. If 1 centimeter is equivalent to 0.3937 inches, convert a measurement of 3.85 inches to centimeters. Convert 42.83 centimeters to inches. 8. If the length of the year is 365.24 days and the average distance of the earth from the sun is 9.31 × 107 miles, find approximately the velocity of the earth in miles per hour, assuming that the orbit is a circle with the Isun at the center.

27. Empirical Data Problems. It is frequently possible to measure corresponding pairs of values of two related variables when the law connecting them is unknown. An important problem presented by such empirical data is to determine a law which represents approximately the relation between the two variables.

If the pairs of values be plotted, it may happen that the points so obtained lie very nearly on a straight line. If such is the case, it is reasonable to assume that the graph of the relation is a straight line, and hence that the relation connecting the two variables is a linear equation.

When graphical methods are being used, all that is needed is the graph of the relation. Any straight line which passes through or near to each of the points will serve approximately as the graph. A method frequently used by engineers to get the line giving the best approximation is to stretch a rubber band over two pins stuck in the drawing board, and move the pins about until the stretched band appears to make the average distance from the band to each of the points as small as possible. When algebraic methods are employed, a method of obtaining the equation of a line which answers well as the graph is illustrated in the following examples.

W

FIG. 44.

EXAMPLE 1. In an experiment dealing with friction of wood on wood, a block of wood with various weights on it was placed on a horizontal board. A string fastened to the block ran over a pulley at the end of the board, and a pan was tied to the hanging end. Weights were placed in the pan until the block was just on the point of moving. Using W to represent the combined weight of the block and

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the weights on it, and W' for the

weight of the pan and of the
weights in it, corresponding
values of W and W' were found
as indicated in the table,

W 10, 20, 30, 40
W'

3.1, 5.8, 8.1, 11.2'

the unit of weight being the gram. Determine approximately the relation between W and W'.

Plot the points whose coördinates are the pairs of values of The four points obtained lie

W and W', using values of W as abscissas.

very nearly on a straight line through the origin, and hence we assume that the graph of the relation is a straight line through the origin.

(That the graph ought to pass through the origin is clear, for if W = 0, so also must W' = 0.) The required relation must therefore be of the form W' = mW, where m is the slope of the line.

The slopes of the lines joining the origin to each of the four points are respectively

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From these values it appears that m may have any value between 0.27 and 0.31, and the line W' mW would be a fair approximation to the graph required. We shall choose as a good value of m the average of the four values of m, namely,

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In dividing 1.15 by 4, the second decimal figure is 8, but the quotient is nearer to 0.29 than to 0.28; hence the former value is chosen.

Hence the relation desired is represented approximately by the equation

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% of

error

(1)

The accuracy with which this equation represents the given data may be determined by constructing the table adjoined. The first two columns give the observed values of W and W', the third, the values of W' computed by means of equation (1) from the observed values of W; the fourth, the error in the com

- 6.4

W W'

0.29W Error

10

3.1

2.9

- 0.2

20

5.8

5.8

0.0

0

30

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puted values of W', which are obtained by subtracting the second column from the third; and the fifth, the percentage of error. The percentage of error is found by dividing the error by the observed value of W'. Thus 0.2/3.1 = 0.064, or 6.4%.

Note the part played by mathematics in this illustration of the scientific method. The given data are obtained by observation. The principles of graphic representation enable us to put the data in a form which makes reasonable the hypothesis that the graph of the law under investigation is a straight line, and that the law is represented by a linear equation. By deductive processes we determine the numerical values of the coefficients of the equation, and the accuracy of the representation of the given data by the equation found. The verification

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