as often as necessary, and the root discovered to any assigned exact ness. Note. In order to find the value of the first assumed root, in this or any other power, divide the resolvend into periods by beginning at the place of units, and including in each period, so many figures as there are units in the exponent of the root; viz. 3 figures in the cube root; 4 for the biquadrate, and so on; then, by a table of powers, or otherwise, find a figure, which (being involved to the power whose exponent is the same with that of the required root) is the nearest to the value of the first period of the resolvend at the left hand, and to that figure annex so many cyphers as there are periods remaining in the integral part of the resolvend; this figure, with the cyphers annexed, will be the assumed root, and equal to r in the theorem; and it is of no importance whether the figure thus chosen be, when involved, greater or less than the left hand period, as the theorem is the same in both cases. 1st. What is the cube root of 436036824287? 7000 assumed root. 3 21 000)436036824287(20763658-2994 Subtract 7000x7000÷÷12-4083333.3333 16680324.9661-4084.15 Add the assumed root=3500 And it gives the approximated root=7584.15 For the second operation, use the approximated root as the assumed one, and proceed as above. THIRD METHOD BY APPROXIMATION. 1. Assume the root in the usual way, then multiply the square of the assumed root, by 3, and divide the resolvend by this product; to this quotient add of the assumed root, and the sum will be the true root, or an approximation to it.. 2. For each succeeding operation let the last approximated root be the assumed root, and proceeding in this manner, the root may be extracted to any assigned exa&ness. 1st. What is the cube root of 7? Let the assumed root be 2. Then, 2×2×3-12 the divisor. 12)7-0(583 to this add of 2=1·333, &c. that is, •583+1·333— 1.916 approximated root. Now assume 1916 for the root. Then, by the second process, the 2d. What is the cube root of 9? Let 2 be the assumed root as before. Then, +32=2.08 the approximated root. Now assume 2:08. Then, 9 2+3x2·08=2·08008, &c. 3d. What 2d. What is the cube root of 282? Let 6 be the assumed root.Then, 6x6x3=108)282(2·611, &c. and 2·611+3 of 6=6·611 approximated root. Now assume 6-611, and it will be 6.611x6·611x3= 131.116)282(2·1507, &c. and 2-1507+ of 6.611-6.558 a farther approximated root. 4th. What is the cube root of 1728 ?-Here the assumed root is 10. Then, 10x10x3=300)1728(5·76, and 5·76+3 of 10=12 426. Now assume 12 426, then 12 426×12·426×3=463-216428)1728 (3-732, and 3-732+2 of 12·426=12.014 a farther approximated root, and so on. APPLICATION AND USE OF THE CUBE ROOT. 1. To find two mean proportionals between any two given numbers. RULE.-1. Divide the greater by the less, and extract the cube root of the quotient. 2. Multiply the root, so found, by the least of the given numbers, and the product will be the least. 3. Multiply this product by the same root, and it will give the greatest. EXAMPLES. 1st. What are the two mean proportionals between 6 and 750? Then. 5×6=30=least, and 30X5= 3 750-÷6=125, and ✔ 125=5. 130 greatest. Answer 30 and 150. Proof. As 6: 30 :: 150: 750. 2d. What are the two mean proportionals between 56 and 12096? Answer 336 and 2016. Note. The solid contents of similar figures are in proportion to each other, as the cubes of their similar sides or diameters. 3d. If a bullet 6 inches diameter weigh 32lb.; What will a bullet of the same metal weigh, whose diameter is 3 inches? 6×6×6=216. 3x3x3=27. As 216 32lb. :: 27: 4lb. Ans. 4th. If a globe of silver of 3 inches diameter, be worth £.45, What is the value of another globe, of a foot diameter ? 3x3x3=27 12x12×12=1728. As 27 : 45 :: 1728 : £ 2880 Ans. The side of a cube being given, to find the side of that cube which shall be double, triple, &c. in quantity to the given cube. RULE. Cube your given side, and multiply it by the given proportion between the given and required cube, and the cube root of the product will be the side sought. 5th. If a cube of silver, whose side is 4 inches, be worth £.50, I demand the side of a cube of the like silver, whose value shall be 4 times as much? 4×4×4=64, and 64×4-256. 3 256-6-349+inches, Ans. 6th. There is a cubical vessel, whose side is 2 feet; I demand the side of a vessel, which shall contain three times as much? 3 2X2X2=8, and 8x3=24. /24-2.884-2ft. 10 inches, Ans. 7th. The 7th.* The diameter of a bushel measure being 18 height 8 inches, I demand the side of a cubic box, tain that quantity? inches, and the which shall conAns 12.907+ inches. 8. Suppose a ship of 500 tons has 89 feet keel, 36 feet beam, and is 16 feet deep in the hold: What are the dimensions of a ship of 200 tons, of the same mould and shape? 3 89x89x89=704969-cubed keel. As 500: 200 :: 704969: 281987.6 cube of the required keel. √281987·6=65·57 feet the required keel. As 89: 65·57 :: 36 : 26 522–264 feet, beam, nearly. As 89: 65-57 :: 16: 117 feet, depth of the hold, nearly. 9. From the proof of any cable to find the strength of any other. RULE. The strength of cables, and consequently the weights of their anchors, are as the cubes of their peripheries. If a cable, 12 inches about, require an anchor of 18cwt. Of what weight must an anchor be, for a 15 inch cable? Cwt. Cwt. As 12x12x12: 18 :: 15x15x15: 35·15625 Ans. 10. If a 15 inch cable require an anchor 35-15625cwt.: What must the circumference of a cable be, for an anchor of 18cwt. ? 3 As 35-15625: 15x15x15 :: 18: 1728, and ✔ 1728=12 Ans. EXTRACTION OF THE BIQUADRATE ROOT. RULE. Extract the square root of the resolvend, and then the square root of that root, and you will have the biquadrate root. What is the biquadrate root of 20736? TWO METHODS OF EXTRACTING THE BIQUADRATE ROOT BY APPROXIMATION. RULE I. 1. Divide the resolvend by six times the square of the assumed root, and from the quotient subtract part of the square of the assumed root. 2. Extract Multiply the fquare of the diameter by 7854, and the product by the height; the cube root of the laft product is the answer. See Menfuration of Superficies and Solids. Art. 30. 2. Extract the square root of the remainder. 3. Add of the assumed root to the square root, and the sum will be the true root, or an approximation to it. 4. For every succeeding operation, either in this or the following method, proceed in the same manner, as in the first, each time using the last approximated root for the assumed root. The biquadrate root of 20736 is required. Subtract. 10x10÷18 = 5.5555 ✔ 29.0044=5.38 Add of 10 =6·66 Approximated root 12.04, to be made the assum ed root for the next operation. RULE II. Divide the resolvend by four times the cube of the assumed root: to the quotient add three fourths of the assumed root, and the sum will be the true root, or an approximation to it. Let the biquadrate of 20736 be required, as before? 10x10x10x4=4000)20736(5.184 Add 3 of 10=7.5 Approximated root 12-684, to be made the assumed root for the next operation. EXTRACTION OF THE SURSOLID ROOT BY APPROX. IMATION. A PARTICULAR RULE.* 1. Divide the resolvend by five times the assumed root, and to the quotient add one twentieth part of the fourth power of the same root. 2. From the square root of this sum subtract one fourth pare of the square of the assumed root. 3. To the square root of the remainder add one half of the assumed root, and the sum is the root required, or an approximation to it. Note. This rule will give the root true to five places, at the least, (and generally to eight or nine places) at the first process Required A GENERAL RULE FOR EXTRACTING THE ROOTS OF ALL POWERS. *1. Prepare the given number, for extraction, by pointing off from the unit's place, as the required root directs. 2. Find the first figure of the root by trial, or by inspection into the table of powers, and subtract its power from the left hand period. 3. To the remainder bring down the first figure in the next period, and call it the dividend. 4. Involve the root to the next inferiour power to that which is given, and multiply it by the number denoting the given power, for a divisor 5. Find how many times the divisor may be had in the dividend, and the quotient will be another figure of the root. 6. Involve the whole root to the given power, and subtract it from the given number, as before. 7. Bring down the first figure of the next period to the remainder for a new dividend, to which find a new divisor, as before, and, in like manner, proceed till the whole be finished. EXAMPLES. The extracting of roots of very high powers will, by this rule, be a tedious operation: The following method, when practicable, will be much more conven ient. When the index of the power, whofe root is to be extracted, is a compofite number, take any two or more indices, whose product is equal to the giver index, and extract out of the given number a root answering to another of the ind es, and fo on to the last. Thus, the fourth root-square root of the fquare root; the fixth root=square root of the cube root ;-the eighth root = fquare root of the fc: rth root;-the ninth root = the cube root of the cube root-the tenth root = Iquare root of the fifth root :-the twelfth root the cube root of the fourth, &c. |