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Fig. 1.

The cubing of a number may be illustrated geometrically as fol. lows:

Let it be required to cube 45, the number before employed. To simplify the illustration, suppose we are required to find the number of cubic inches in a cube whose side is 45 inches. Separating 45 into 40+5, we will suppose the cube, (fig. 1,) to be 40 inches on a side; then 40 x

403=40 x 40 x 40 40 x 40 will give the

=64000 solid contents of this cube, represented by 40%. Let fig. 2 represent

Fig. 2. che cube increased by three equal slabs ;

40'=40 X 40 then 3 (the number

=1600 of slabs) times 40%

x by 3 (the surface of one of the slabs,)multiplied by 5, the thickness of

X by 5 a slab, will give the

24000 solid contents of the slabs, represented by 3 x 402.5. Let fig. 3 represent

Fig. 3. the solid, (as in fig. 2 ) further increased by three equal corner

3x40=120 pieces; then 3 (the

x by 5%= 25 number of corner

600 pieces) times 40(the

240 length of one cor

3000 per piece) multiplied into 52, the surface of an end of a comer piece, will give the solid contents of the corner pieces, represented by 3x 40.5.

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Let fig. 4 represent

Fig. 4. the solid (as in fig. 3) further increased by a little corner cube, each side of which is 5

53=5X5X5 inches; then 5x5x5

= 125 will give the solid contents of this cube, represented by 5%.

Then the whole cube thus increased will be represented by 453=403+3 x 402.5+3 x 40.52 +53

=64000+24000+3000+125=91125. 135. We will now endeavor to deduce a rule for the extraction of the Cube Root.

Let it be required to find the cube root of 382657176.

For the sake of simplicity, we will suppose 382657176 to denote the number of cubic feet in a geometrical cube; we are required to find the number of linear feet in a side of this cube, that is, the length of one of its sides.

We will first inquire how many figures the root will have.

The smallest number, consisting of two figures, which is 10, becomes, when cubed, 1000, having more than three figures. Again, the largest number, 99, which consists of two figures, becomes, when cubed, 970299, which consists of six figures. Hence, when a number consists of more than three figures, and not of more than six, its cube root will consist of two figures. By a similar method it may be shown, that when a number consists of more than six, and of not more than nine figures, its cube root will consist of three figures. Therefore, if we separate a number into groups of three figures each, the number of groups will denote the number of figures in the cube root of that number.. .

In the present example, we know that there must be three figures in the root.

We know that the side of the cube sought must exceed 700 lineam feet, since the cube of 700 is 343000000, which is less than 382657176, we also know that the side of this cube must be less than 800 linear feet, since the cube of 800 is 512000000, which is greater than 382657176. Hence the first figure of our root, or the figure in the hundred's place is 7; whose cube, 343, is the greatest cube con tained in 382, the first, or left-hand

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Fig. 1. period. If we suppose each side of the cube, represented by figure 1, to be 700 linear feet, one of the equal faces, as the upper face DEFG, will be denoted by 700 x 700=490000 square feet. The solid contents of the cube will be represented by 700 x 700=490000 x 700 =343000000 cubic feet. Subtracting 343000000 cubic feet from 382657176 cubic feet, we find 39657176 cubic feet for a remainder.

Hence it is necessary to increase the cube, figure 1, by 39657176 cubic feet. We have seen (Art. 134) that such increase is effected by the addition of three equal slabs, three equal corner pieces, and an additional cube ; and that the contents of the three slabs will make by far the largest portion of the whole increase. The number of square feet in the

Fig. 2. face of one of these slabs will be the same as the number of square feet in the face of the cube, figure 1, which has already been shown to be 490000 square feet. The surface of the three slabs will be three times 490000 square feet; or, which would be the same thing, twice 490000 square feet, added to 490000 square feet.* If to AB, (fig. 1,) which is 700 linear feet, we add BC, which is also 700 linear feet, we shall have AB+BC equal to 1400 linear feet, which, multiplied by DB, equal to 700 linear feet, will give 980000 square feet, for the area ABDG+BCED,which, added to DEFG, which is 490000 SqUzac feet,

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* It will be noticed that the peculiar steps throughout this demonstration, have reference to the mode of extracting the Cube Root which follows. The object of these processes is, to make use of what has been obtained in one stage of the work for the stage next succeeding; to obtain a new quantity by adding to one already in hand, instead of multiplying an original quantity; thereby saving much time and labor.

will give 1470000 square feet, for the area of three faces of the cube, figure 1, which is the same as the area of the three slabs. Were we to multiply 1470000 by the thickness of the slabs, we should obtain the cubic feet in these slabs. And since the contents of the slabs make nearly the whole amount added, it follows that 1470000 multiplied by the thickness of slabs, will give nearly 39657176 cubic feet. Consequently, if we divide 39657176 by 1470000, the quotient will give the approximate thickness of the slabs. Using 1470000 as a trial divisor, we find it to be contained between 20 and 30 times in 39657176; hence the second or tens' figure of the root is 2.

We have already remarked that 1470000 multiplied by 20, the thickness of the slabs, will give their solid contents. But besides the slabs there must be added three corner pieces, each of which is 700 feet long, and of the same thickness as the slabs, that is, 20 feet. Since each corner piece is the same length as a side of the cube, figure 1, it follows that adding 700 to 1400 or 700+700, the sum 2100 will represent the total length of the three corner pieces. Were we to multiply 2100 by 20, we should obtain the area of the three corner pieces, which might be added to 1470000, the area of

Fig. 3. the three slabs. But, since there is also to be added a little cube, each of whose sides is 20 linear feet, we will add 20 to 2100, and thus obtain 2120 for the total length of the three corner pieces, and of a side of the little cube. Now, multiplying 2120 by 20, we obtain 42400 square feet for the surface of the three corner pieces and a face of the little cube; which, added to 147000, the number of square feet in the faces of the three slabs, will give 1512400 square feet in all the additions. If we multiply 1512400 by 20, the thickness of these additions, we shall obtain 30248000 cubic feet for all the additions, which, subtracted from 39657176, leaves 9409176 cubro feet. The cube thus completed is 720 feet on a side, and is represented by figure 4.

Figure a.

The surfaces now obtained may be represented (figure a,) by the parts included within the heavy lines. The three divisions of the figure, including the dotted lines, may be supposed to be three entire faces of the cube, figure 4. But this cube is to be further

Fig. 4. increased by 9409176 cubic feet. And as before, the parts added will consist of three equal slabs, three equal corner pieces, and a litlle cube. The trial divisor, which is the area of the faces of the three slabs, is the same as three times the area of a face of the cube, figure 4, each of whose sides is 720 feet.

Now to obtain this area, we have only to add to the surfaces already obtained, and represented within the heavy lines, (figure a,) three rectangles, each 700 feet by 20, and two little squares 20 feet by 20 feet.

If to 2120, a number which we already have, we add 20, we shall obtain 2140, the linear extent of the rectangles and squares desired, as in the dotted portions, (figure a.) And as these dotted portions have all the same width of 20 feet, if we multiply 2140 by 20, we shall obtain 42800 square feet for the area of the dotted portion, (figure a,) which, added to 1512400, the area of the parts included within the heavy lines, will give 1555200 square feet for the area of three slabs, each equal to one face of the cube, (figure 4.) This will be a second trial divisor. We find this divisor contained between 6 and 7 times in 9409176; hence our third figure of the root, or the figure in the units' place, is 6. Were we to multiply 1555200 by 6, it would give the cubic feet in the second set of

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