the operation of the lule. They may be made in a few minutes, from a small strip of a pine board, with a common penknife, at the longest, in less time than the teacher can make the pupil comprehend the reason, from merely seeing the picture on paper, In demonstrating the rule in this way, it will be an amusing and instructive exercise, to both teacher and pupil, and may be comprehended by any pupil, however young, who is so fortunate as to have progressed as far as this rule. It will give him distinct ideas respecting the different dimensions of square and cubic measures, and indelibly fix on his mind the reason of the rule, consequently the rule itself. But for the convenience of teachers, blocks, illustrative of the operation of the foregoing example, will accompany this work. The following are the supposed proportional dimensions of the several blocks used in the demonstration of the above example, which, when put together, ought to make an exact cube, containing 13824 cubic feet. One block, 20 feet long, 20 feet wide, and 20 feet thick; this we will call A. Three small blocks, each 20 feet long, 20 feet wide, and 4 feet thick; each of these we will call B. Three smaller blocks, each 20 feet long, 4 feet wide, and 4 feet thick; each of these we will call C. One block, and the smallest, 4 feet long, 4 feet wide, and 4 feet thick; this we will call D. We are now prepared to solve the preceding example. In this example, you recollect, we were to find the length of one side of the cube, containing 13824 cubic feet. In this example, we know that one side cannot be 30 feet, for 303: 27000 solid feet, being more than 13824, the given sum; therefore, we will take 20 for the length of one side of the cube. Then, 20 X 20 X 20 = 8000 solid feet, which we must, of course, deduct from 13824, leaving 5824. (See operation 1st.) These 8000 solid feet, the pupil will perceive, are the solid coutents of the cubical block, mark The same operation, by neglecting the ci- ed A. This corresponds with phers, may be performed thus: the operation; for we write 20 feet, the length of the cube A, at the right of 13824, in the form Ans. of a quotient; and its square, 8000, under 13894; from which subtracting 8000, leaves 5824, as before. As we have 5824 cubic feet romaining, we find the sides of the cube A are not so long as they ought to be; consequently we must enlarge A; but in doing this, we must enlarge the three sides of A, in order that we may preserve the cubical form of the block. We will now place the three blocks, each of which is marked B, on the three sides of A. Each of these blocks, in order to fit, must be as long and as wide as A; and, by examining them, you will see that this is the case; that is, 20 feet long and 20 feet wide; then 20 X 20= 400, the square contents in one B, and 3400 1200, square contents in 3 Bs; then it is plain, that 5824 solid contents, divided by 1200, the square contents, will give the thickness of each block. But an easier method is, to square the 2, (tens,). in the root 20, making 4, and multiply the product, 4, by 300, making 1200, a divisor, the same as before. We do the same in the operation, (which see ;) that is, we multiply the square of the quotient figure, 2, by 300, thus, 2X24 X 300 = 1200; then the divisor, 1200, (the square contents,) is contained in 5824 (solid contents) 4 times; that is, 4 feet is the thickness of each block marked B. This quotient figure, 4, we place at the right of 5824, and then 1200 square feet X 4 feet, the thickness, 4800 solid feet. If we now examine the block, thus increased by the addition of the 3 Bs, we shall see that there are yet 3 corners not filled up: these are represented by the 3 blocks, each marked C, and each of which, you will perceive, is as long as either of the Bs, that is, 20 feet, being the length of A, which is the 20 in the quotient. Their thickness and breadth are the same as the thickness of the Bs, which we found, by dividing, to be 4 feet, the last quotient figure. Now, to get the solid contents of each of these Cs, we multiply their thickness (4 feet) by their breadth (4 feet), 16 square feet; that is, the square of the last quotient figure, 4, 16; these 16 square contents must be multiplied by the length of each, (20 feet,) or, as there are 3, by 3 X 20 = 60; or, which is easier in practice, we may multiply the 2, (tens,) in the root, 20, by 30, making, 60, and this product by 4216, the square contents 960 solid feet. We do the same in the operation, by multiplying the 2 in 20 by 30=60X4 X4960 solid feet, as before; this 960 we write under the 4800, for we must add the several products together by and by, to know if our cube will contai all the required feet. By turning over the block, with all the additions of the blocks marked and C, which are now made to A, we shall spy a little square space, which prevents the figure from becoming a complete cube. The little block for this corner is marked D, which the pupil will find, by fitting it in, to exactly fill up this space. This block, D, is exactly square, and its length, breadth, and thickness are alike, and, of course, equal to the thickness and width of the Cs, that is, 4 feet, the last quotient figure; hence, 4 ft. X4 ft. X 4 ft. = 64 solid feet in the block D; or, in other words, the cube of 4, (the quotient figure,) which is the same as 43: 64, as in the operation. We now write the 64 under the 960, that this may be reckoned in with the other additions. We next proceed to add the solid contents of the Bs, Cs and D together, thus, 4500+960 + 645824, precisely the number of solid feet which we had remaining after we deducted 8000 feet, the solid contents of the cube A. If, in the operation, we subtract the amount, 5824, from the remainder or dividend, 5824, we shall see that our additions have taken all that remained after the first cube was deducted, there being no remainder. The last little block, when fitted in, as you saw, rendered the cube complete, each side of which we have now found to be 20+4=24 feet long, which is the cube root of 13824 (solid feet); but let us see if our cube contains the required number of solid feet. In operation 2d, we see, by neglecting the ciphers at the right of 8, the 8 is still 8000, by its standing under 3 (thousands); hence, we may point off three figures by placing a dot over the units, and another over the thousands, and so on. From the preceding example and illustrations we derive the following RULE. I. Divide the given number into periods of three figures each, by placing a point over the unit figure, and every third figure from the place of units to the left, in whole numbers, and to the right in decimals. II. Find the greatest cube in the left hand period, and place its root in the quotient. III. Subtract the cube thus found from the said period, and to the remainder bring down the next period, and call this the dividend. IV. Multiply the square of the quotient by 300, calling it the divisor. V. Find how many times the divisor is contained in the divi dend, and place the result in the root (quotient); then multiply the divisor by this quotient figure, placing the product under the dividend. VI. Multiply the former quotient figure, or figures, by 30, and this product by the square of the last quotient figure, and place the product under the last; under these two products place the cube of the last quotient figure, and call their amount the subtrahend. VII. Subtract the subtrahend from the dividend, and to the remainder bring down the next period for a new dividend, with which proceed as before, and so on, until the whole is finished. Note 1. When the subtrahend happens to be larger than the dividend, the quotient figure must be made one less, and we must find a new subtrahend. The reason why the quotient figure will be sometimes too large is, because this quotient figure merely shows the width of the three first additions to the original cube; consequently, when the subsequent additions are made, the width (quotient figure) may make the solid contents of all the additions more than the cubic feet in the dividend, which remain after the solid contents of the original cube are deducted. 2. When we have a remainder, after all the periods are brought down, we may continue the operation by annexing periods of ciphers, as in the square root. 3. When it happens that the divisor is not contained in the dividend, a cipher must be written in the quotient, (root,) and a new dividend formed by bringing down the next period in the given sum More Exercises for the Slate. 8. What is the cube root of 9663597 ? 9. What is the cube root of 17576? A. 26. 64 19, What is the cube root of 1? A. G. 20. What is the cube root of? A. 17. If the root be a surd, reduce it to a decimal before its root is extracted, as in the square root. ? 21. What is the cube root of A.,13+. 22. What is the cube root of 25 ? A.,34 +. 23. What is the length of one side of a cubical block, which contains 1728 solid or cubic inches? A. 12. 24. What will be the length of one side of a cubical block, whose contents shall be equal to another block 32 feet long, 16 feet wide, and 8 feet thick? 32X16X8= = 16 feet, Ans. 25. There is a cellar dug, which is 16 feet long, 12 feet wide, and 12 feet deep; and another, 63 feet long, 8 feet wide, and 7 feet deep; how many solid or cu bic feet of earth were thrown out, and what will be the length of one side of a cubical mound, which may be formed from said earth? A. 5832; 18. 26. How many solid inches in a cubical block, which measures 1 inch on each side? How many in one measuring 2 inches on each side? 3 inches on each side? 4 inches on each side? 6 inches on each side? 10 inches on each side? 20 inches on each side? A. 1, 8, 27, 64, 216, 1000, 8000. 27. What is the length of one side of a cubical block, which contains 1 solid or cubic inch? 8 solid inches? 27 solid inches? 64 solid inches? 125 solid inches? 216 solid inches? 1000 solid inches? 8000 solid inches? A. 1, 2, 3, 4, 5, 6, 10, 20. By the two preceding examples we see that the sides of the cube are as the cube roots of their solid contents, and their solid contents as the cubes of their sides. It is likewise true, that the solid contents of all similar figures are in proportion to each other as the cubes of their several sides or diameters. Note. The relative length of the sides of cubes, when compared with their solid contents, will be best illustrated by reference to the cubical blocks, accompanying this work. 28. If a ball, 3 inches in diameter, weigh 4 pounds, what will a ball of the same metal weigh, whose diameter is 6 inches? 3634:32: Ratio, 23 432 lbs., Ans. 29. If a globe of silver, 3 inches in diameter, be worth $160, what is the value of one 6 inches in diameter ? 3363 $160: $1280, Ans. 30. There are two little globes; one of them is 1 inch in diameter, and the other 2 inches; how many of the smaller globes will make one of the larger? A. 8. 31. If the diameter of the planet Jupiter is 12 times as much as the diameter of the earth, how many globes of the earth would it take to make one as large as Jupiter? A. 1728. 32. If the sun is 1000000 times as large as the earth, and the earth is 8000 miles in diameter, what is the diameter of the sun? A. 800000 miles. Note. The roots of most powers may be found by the square and cube roots only; thus the square root of the square root is the biquadrate, or 4th root, and the sixth root is the cube of this square root ARITHMETICAL PROGRESSION. ↑ LXXXVIII. Any rank or series of numbers more than 2, increasing by a constant addition, or decreasing by a constant subtraction of some given number, is called an Arithmetical Series, or Progression. The number which is added or subtracted continually is called the common difference. "When the series is formed by a continual addition of the common difference, it is called an ascending series; thus, 2, 4, 6, 8, 10, &c., is an ascending arithmetical series; but 10, 6, 4, 2, &c., is called a descending arithmetical series, because it is formed by a continual subtraction of the common difference, 2. The numbers which form the series are called the terms of the series or progression. The first and last terms are called the extremes, and the other terms the means. In Arithmetical Progression there are reckoned 5 terms, any three of which being given, the remaining two may be found, viz. 1. The first term. 2. The last term. 3. The number of terms. |