30. Two men start from the same place, one goes exactly south 40 miles a day, the other goes exactly west 30 miles a day: how far apart will they be at the close of the first day? 31. How far apart will the same travelers be at the end of 4 days? 32. A line 75 feet long fastened to the top of a flag staff reaches the ground 45 feet from its base: what is the height of the flag staff? 33. Suppose a house is 40 feet wide, and the length of the rafters is 32 feet: what is the distance from the beam to the ridge pole? 34. The side of a square field is 30 rods: how far is it between its opposite corners ? 35. If a square field contains 10 acres, what is the length of its side, and how far apart are its opposite corners? EXTRACTION OF THE CUBE ROOT. 360. To extract the cube root, is to resolve a given number into three equal factors; or, to find a number which being multiplied into itself twice, will produce the given number. (Art. 345.) 1. What is the side of a cubical block containing 27 solid feet? Solution. Let the given block be represented by the adjoining cubical figure, each side of which is divided into 9 equal squares which we will call square feet. Now since the length of a side is 3 feet, if we multiply 3 into 3 into 3, the product 27 will be the solid contents of the cube. (Art. 164.) Hence, if we reverse the process, i. 3X3X3=27. into three equal factors, one of these factors will be the side of the cube. (Art. 344. Obs.) Ans. 3 ft. 2. A man wishes to form a cubical mound containing 15625 solid feet of earth: what is the length of its side? QUEST.-360. What is it to extract the cube root? Operation. 1. We first separate the given num15625(25 ber into periods of three figures each, by placing a point over the units' figure, then over thousands. This shows us that the root must have two figures, (Art. 342. a. N. 3,) and thus enables us to find part of it at a time. 8 Divisor. Dividend. 1200 7625 300 25 1525 7625 2. Beginning with the left hand period, we find the greatest cube of 15 is 8, the root of which is 2. Placing the 2 on the right of the given number for the first figure in the root, we subtract its cube from the period, and to the remainder bring down the next period for a dividend. This shows that we have 7625 solid feet to be added to the cubical mound already found. 3. We square the root already found, which in reality is 20; for since there is to be another figure annexed to it, the 2 is tens, and multiplying its square 400 by 3, we write the product on the right of the dividend for a divisor; and finding it is contained in the dividend 5 times, we place the 5 in the root. 4. We next multiply 20, the root already found, by 5, the last figure placed in the root; then multiply this product by 3 and place it under the divisor. We also place the square of 5, the last figure placed in the root, under the divisor, and adding these three results together, multiply their sum 1525 by 5, and subtract the product from the dividend. The answer is 25. PROOF. (25)325x25x25=15625, the given number. DEMONSTRATION BY CUBICAL BLOCKS. 361. The simplest method of illustrating the process of extracting the cube root to those unacquainted with algebra and geometry, is by means of cubical blocks.* 1. Place the large cube upon a table or stand. Let this represent * A set of these blocks contains 1st, a cube, the side of which is usually about 1 in. square; 2d, three side pieces about in. thick, the upper and lower base of which is just the size of a side of the cube; 3d, three corner pieces, whose ends are in. square, and whose length is the same as that of the side pieces: 4th, a small cube, the side of which is equal to the end of the corner pieces. It is desirable for every teacher and pupil to have a set. If not conveniently procured at the shops, any one can easily make them for himself. the greatest cube in the left hand period, which in the example above is 8, the root of which is 2. By subtracting 8 from the period, we find we have 7625 feet of earth to be added to this cube. In making this addition, it is plain the cube must be equally increased on three sides; otherwise its sides will become unequal, and it will then cease to be a cube. (Art. 154. Obs. 2.) 2. The object of squaring the root already found is to find the area of one side of this cube; (Art. 163;) we then multiply its square by 3, because the additions are to be made to three of its sides; and divide the dividend by this product to find the thickness of these additions, which is 5 feet. Now placing one of the side pieces on the top, and the other two on two adjacent sides of the cube, they will represent these additions. 3. But we perceive there is a vacancy at three corners, each of which is of the same length as the side of the cube, viz: 20 ft., and the breadth and thickness of each is 5 ft., the thickness of the side additions. Placing the corner pieces in these vacancies, they will represent the additions necessary to fill them. The object of multiplying 20 ft. their common length, by 5, their common width, is to obtain the area of a side of one of them; we then multiply this area by 3, to find the area of a side of each of them. 4. We find also another vacancy at one corner, whose length, breadth, and thickness are each 5 ft., the same as the thickness of the side additions. This vacancy therefore is cubical. It is represented by the small cube, which being placed in it, will render the mound an exact cube again. The object of squaring 5, the figure last placed in the root, is to find the area of a side of this cubical vacancy. We now have the area of one side of each of the side additions, viz: 20X20X3=1200, the divisor; also the area of one side of each of the corner additions, viz: 20×5×3=300; and the area of one side of the cubical vacancy, viz: 5×5-25. We next multiply the sum of these areas, 1525, by 5, the thickness of each, in order to find the cubical contents of the several additions. (Art. 164.) These areas are added together, and their sum multiplied by the last figure placed in the root, for the sake of finding the solidity of all the additions at once. The result would obviously be the same, if we multiplied them separately, and then subtracted the sum of their products from the dividend. 362. From the preceding illustrations we derive the following general QUEST.-362. What is the first step in extracting the cube root? The second? Third? Fourth? Fifth? How is the cube root proved? Why point the given number into periods of three figures each? Why subtract the greatest cube from the left hand period? In finding the divisor. what does the square of the root show? Why multiply its square by 3? What does the quotient figure obtained by the divisor show? Why multiply the root previously found by the last figure placed in it? Why multiply this product by 3? What does the square of this last figure in the root show? What does the sum of these results show? Why multiply this sum by the last figure placed in the root? RULE FOR EXTRACTING THE CUBE ROOT. I. Separate the given number into periods of three figures each, placing a point over units, then over every third figure towards the left in whole numbers, and over every third figure towards the right in decimals. II. Find the greatest cube in the first period on the left hand; then placing its root on the right of the number for the first figure of the root, subtract its cube from the period, and to the remainder bring down the next period for a dividend. III. Square the root already found, giving it its true local value; multiply this square by 3, and place the product on the right of the dividend for a divisor; find how many times it is contained in the dividend, and place the result in the root. IV. Multiply the root already found, regarding its local value by this last figure added to it, then multiply this product by 3, and place the result on the right of the dividend under the divisor; under this result write also the square of the last figure placed in the root. V. Finally, add these results to the divisor; multiply the sum by the last figure placed in the root, and subtract the product from the dividend. To the right of the remainder bring down the next period for a new dividend ; find a new divisor, and proceed with the operation as above. PROOF.-Multiply the root into itself twice, and if the last product is equal to the given number, the work is right. OBS. 1. When there is a remainder, periods of ciphers may be added, as in square root. 2. If the right hand period of decimals is deficient, this deficiency must be supplied by ciphers. The root must contain as many decimals as there are periods of decimals in the given number. 3. What is the cube root of 1728? QUEST. Obs. When there is a remainder, how proceed? When the right hand period of decimals is deficient, what must be done? How many decimals must the root have? 11. What is the cube root of 91.125 ? 64 29 SECTION XV. EQUATION OF PAYMENTS. 363. Equation of Payments is the process of finding the equalized or average time when two or more payments due at different times, may be made at once, without loss to either party. OBS. The equalized or average time for the payment of several debts, due at different times, is often called the mean time. 364. From principles already explained, it is manifest, when the rate is fixed, the interest depends both upon the principal and the time. (Arts. 237, 238.) Thus, if a given principal produces a certain interest in a given time, Double that principal will produce twice that interest; Half that principal will produce half that interest; &c. In double that time the same principal will produce twice that interest; In half that time the same principal will produce half that interest; &c. 365. Hence, it is evident that any given principal will produce the same interest in any given time, as QUEST.-363. What is Equation of Payments? Obs. What is the average time for the payment of several debts sometimes called? 364. When the rate is fixed, upon what does the interest depend? Give an illustration. |