8. If a pipe whose diameter is 1.5 in. fill a cistern in 5 hours, in what time will a pipe whose diameter is 3 in. fill the same cistern? SUG.-It pours in 4 times as much, and fills it in 9. Required the distance between a lower corner and the opposite upper corner of a room 48 feet long, 20 feet wide, and 39 feet high. Ans. 65 feet. CUBE ROOT. Ans. 14 hours. of 5 hours. 48 3 39 644. There are Two Methods of explaining the general process of extracting the Cube Root, called the Analytic or Algebraic Method, and the Synthetic or Geometrical Methou. 645. The Analytic Method of cube root is so called because it analyzes the number into its elements, and derives the process from the law of involution. 646. The Geometrical Method of cube root is so called because it makes use of a cube to explain the process. 1. Extract the cube root of 91125. ANALYTIC SOLUTION. -Since the cube of a number consists of three times as many places as the number itself, or of three times as many less one or two, the cube root of 91125 will consist of two places, or of tens and units, and the number itself will consist of tenss + 3x tens2 X units+ 3 Xtens X units+ units. SHOWN BY LETTERS. ts+ 313 × u + 3t × u3 + u3 = 91125( 45 3tu+3t × u2 + u3 3ť × 3 × 402 3t X u= 3 × 40 × 5 = = 64000 = 27125 4800 600 = 25 The greatest number of tens whose cube is contained in 91125 is 4 tens. Cubing the tens and subtracting, we have 27125, which equals 3 X tens X units + 3x tens x units2 + units3. Now, since 3x tens units is much greater than 3xtens x units-units, 27125 consists principally of 3 times tens3 X units: hence, if we divide by 3 times tens, we can ascertain the units (3ť2 + 31 × u + u2) × u = 5425 × 5=27125 3 times tens2 equals 3 x 402 4800; dividing by 4800, we find the units to be 5. We then find 3 times tens X units equal to 3 x 40 x 5 = 600, and units=52=25, and adding these and multiplying by units we have (3 tens + 3 tens x units + units) X units, which equals 5425 x 5 = 27125; subtracting, nothing remains, hence the cube root of 91125 is 45 GEOMETRICAL SOLUTION.-Let Fig 1 represent the cube which contains 91125 cubic units, then our oject is to find the number of linear units in its edge. The number of terms in the root, found as before, is two. 3 x 4024800 3 X 40 X 5= 600 52- 251 5425 27125 The greatest number of tens whose cube is contained in the given number is 4 tens. Let A, Fig. 1, represent a cube whose sides are 40, its contents will be 403 Subtracting 64000 from 91125 we find a remainder of 27125 cubic units, which by removing the cube A from Fig. 1, leaves a solid represented by Fig. 2. 64000. Inspecting this solid, we perceive that the greater part of it consists of the three rectangular slabs, B, C, and D, each of which is 40 units in length and breadth, hence if we divide 27125 by the sum of the areas of one face of each regarded as a base, we can ascertain their thickness. The area of a face of one slab is 402-1600, and of the three, 3 x 1600, 4800, and dividing 27125 by 4800 we have a quotient of 5, hence the thickness of the slab is 5 units. = Removing the rectangular slabs, there remain three other rectangular solids, E, F, G, as shown in Fig. 3, each of which is 40 units long and 5 units thick, hence the surface of a face of each is 40 x 5=200 square units, and of the three it is 3 x 40 x 5600 square units. Finally removing E, F, and G, there remains only the little corner cube H, Fig. 4, whose sides are 5 units, and the surface of one of its faces 52-25 square units. We now take the sum of the surfaces of the solids remaining after the removal of the cube A, and multiply this by the common thickness, which is 5, and we have their solid contents equal to (4800+600+25) x 5 = 27125 cubic units, which, subtracted from the number of cubic units remaining after the removal of A, leaves no remainder. Hence the cube which contains 91125 cubic units is 40+ 5, or 45 units on a side. NOTE. This can also be explained by building up the cube instead of separating it into its parts, for which see Manual. 647. We will now solve a problem with three figures in the root, indicating the solution by means of letters, and abbreviating the operation as in practice. A point like a period indicates the multiplication of the letters. NOTES.-1. By the geometric method, when there are more than two figures we remove the first cube, rectangular slabs and solids, and small cube, and we have remaining three slabs, three solids, and a small cube, as before. 2. The method employed in actual practice is derived from the other by omitting ciphers, using parts of the number instead of the whole number each time we obtain a figure of the root, etc. It will also be seen that by separating the number into periods of 3 figures each, we have the number of places in the root, the part of the number used in obtaining each figure of the root, etc. Rule.-I. Begin at units and separate the number into periods of three figures each. II. Find the greatest number whose cube is contained in the left hand period, write it for the first term of the root, subtract its cube from the left hand period, and annex the next period to this remainder for a dividend. III. Multiply the square of the first term of the root by 300 for a TRIAL DIVISOR; divide the dividend by it, and the result will be the second term of the root. IV. To the trial divisor add 30 times the product of the second term of the root by the first term, and also the square of the second term; their sum will be the TRUE DIVISOR. V. Multiply the true divisor by the second term of the root, subtract the product from the dividend, and annex the next period for another dividend. Square the root now found, mul tiply by 300, and find the third figure as before, and thus con tinue until all the periods have been used. NOTES.-1. If the product of the true divisor by the term of the root exceeds the dividend, the root must be diminished by a unit. 2. When a dividend will not contain a trial divisor, place a cipher in the root and two ciphers at the right of the trial divisor, bring down the next period, and proceed as before. 3. To find the cube root of a common fraction, extract the cube root of both terms. When these are not perfect cubes, reduce to a decimal and then extract the root. 4. By cubing 1, .1, .01, etc., we see that the cube of a decimal contains three times as many decimal places as the decimal; hence, to extract the cube root of a decimal, we point off the decimal in periods of three figures each, counting from the decimal point. WRITTEN EXERCISES. 13=1 .13.001 .013.000001 SHORT METHOD OF CUBE ROOT. 648. A Short Method of extracting the cube root is presented in the following modification of the ordinary method previously explained. The abbreviation consists in obtaining the successive trial divisors by a law which enables us to use our previous work. NOTE.-This method originated with Prof. Frank Albert, who has taught it to his classes for many years. 1. Extract the cube root of 14706125. SOLUTION.-We find as before the number of figures in the root, and the first term of the root, cube, subtract and bring down the first period. We then find as before the trial divisor, 12, by taking three times the square of the first term, and dividing find the second term of the root to be 4. We then, as before, take three times the product of the first and second terms, and the square of the second term, and add these to the trial divisor as 1456 OPERATION. 14.706.125(245 8 12 24 16 T. D. 16. 15824 1728 360 25 176425 T. D. 1882125 a correction to obtain the true divisor, 1456. We then multiply 1456 by 4, and subtract and bring down the next period. We then, to find the next til divisor, take the square of the last term, which is 16, and add it the previous true divisor and the two corrections (which were added to the previous trial divisor), and we have 1728 as the next trial divisor. Then to find the true divisor, we add 3 times the product of the last term of the root into the previous part of the root, and also the square of the last term, and have 176425 for the true divisor. Multiplying by 5 we have 882125. NOTE.-No rule need be given for the method, as it is merely a modifica tion of the previous method. The true divisor is obtained exactly as before; the abbreviation in obtaining the trial divisor is easily remembered by noticing that it equals the square of the last term, plus the true divisor, plus the corrections used in finding that divisor. The method is indicated in the following formula: 1. TRUE DIVISOR TRIAL DIVISOR+PRODUCT+SQUARE. 2. TRIAL DIVISOR SQUARE+TRUE DIVISOR+CORRECTIONS. The method is readily explained either by the blocks or the algebraic formula. 2. Extract the cube root of 105154048. |