tiply the two numbers together, found by the above proportion, and the product will be the answer.* And, as 1 : 4:36-1: 140. Then, As 1 4: 56-1: 220. : 220×140=30800 square feet, the Answer. PROB. VI. Suppose I would set out an orchard of 600 trees, so that the length shall be to the breadth as 3 to 2, and the distance of each tree, one from the other, 7 yards: How many trees must it be in length, and how many in breadth? and, How many square yards of ground do they stand on ? To resolve any question of this nature, say, as the ratio in length : is to the ratio in breadth :: so is the number of trees to a fourth number, whose square root is the number in breadth. And as the ratio in breadth is to the ratio in length; so is the number of trees to a fourth, whose root is the number in length. : As 3: 2 :: 600: 400. And√400≈20=number in breadth. As 23: 600 900. And 900=30 number in length. As 17:30-1: 203. And as 17 :: 20-1: to 133. And 203×133 26999 square yards, the Answer. PROB. VII. Admit a leaden pipe inch diameter will fill a cistern in 3 hours; I demand the diameter of another pipe which will fill the same cistern in 1 hour. RULE. As the given time is to the square of the given diameter, so is the required time to the square of the required diameter † 275 and 75x755625. Then, as 3h.: ·5625 :: : 1h. 1-6875 inversely, and 1.6875 1.3 inches nearly, Ans. PROB VIII. If a pipe whose diameter is 1.5 inches, fill a cistern in 5 hours, in what time will a pipe whose diameter is 3.5 inches fill the same? : 1.5×1 5=2·25; and 3·5 × 3·5 12:25. Then, as 2-25 ; 5 :: 12·25 918+hour, inversely. 55 min. 5 sec. Answer. PROB. IX. If a pipe 6 inches hore, will be 4 hours in running off a certain quantity of water, In what time will 3 pipes, each four inches bore, be in discharging double the quantity? 6×6=36. 4x4=16, and 16x3=48. Then, as 36: 4h. :: 48: 3h. inversely, and as 1w. : 3h. :: 2w. : 6h. Answer. PROB X. Given the diameter of a circle, to make another circle which shall be 2, 3, 4, &c. times greater or less than the given circle. RULE. Square the given diameter, and if the required circle be greater, multiply the square of the diameter by the given proportion, and the square root of the product will be the required diameter. But if the required circle be less, divide the square of the diameter by the given proportion, and the root of the quotient will be the diameter required. *The above rule will be found useful in planting trees, having the distance of ground between each given. † For more water will run as the area of the pipe is greater, and the areas of circular pipes vary as the square of their diameters. There is a circle whose diameter is 4 inches; I demand the dr ameter of a circle 3 times as large? 4X4 16; and 16x3=48; and √486.928+ inches, Answer. PROB. XI. To find the diameter of a circle equal in area, to an ellipsis, (or oval) whose transverse and conjugate diameters are given.* KULE. Multiply the two diameters of the ellipsis together, and the square root of that product will be the diameter of a circle equal to the ellipsis. Let the transverse diameter of an ellipsis be 48, and the conjugate 36 What is the diameter of an equal circle? 48x36 1728, and 1728=41.569+the Answer. Note. The square of the hypothenuse, or the longest side of a right angled triangle, is equal to the sum of the squares of the other two sides; and consequently the difference of the squares of the hypothenuse and either of the other sides is the square of the remaining side. PROB. XII. A line 36 yards long will exactly reach from the top of a fort to the opposite bank of a river, known to be 24 yards broad. The height of the wall is required? 36×36=1296; and 24×24=576. Then, 1296-576-720, and 720=26 83+yards, the Answer. PROB. XIII. The height of a tree growing in the centre of a circular island 44 feet in diameter, is 75 feet, and a line stretched from the top of it over to the hither edge of the water, is 256 teet. What is the breadth of the stream, provided the land on each side of the water be level? 256X256=65536; and 75=75=5625: Then, 65536-5625= 59911 and √59911=244·76+ and 244·76-44-222-76 feet, Ans. PROB. XIV. Suppose a ladder 60 feet long be so planted as to reach a window 37 feet from the ground, on one side of the street, and without moving it at the foot, will reach a window 23 feet high on the other side; I demand the breadth of the street? 102 64 feet the Answer. PROB. XV. Two ships sail from the same port; one goes due north 45 leagues, and the other due west 76 leagues: How far are they asunder ?† 88-32 leagues, Answer. PROB. XVI. Given the sum of two numbers, and the difference of their squares, to find those numbers. RULE. Divide the difference of their squares by the sum of the numbers, and the quotient will be their difference. The two num * The transverse and conjugate are the longest and shortest diameters of an ellipsis; they pass through the centre, and cross each other at right angles, and the diameter of the equal circle is the square root of the product of the diameters of the ellipsis. The square root may in the same manner be applied to navigation; and, when deprived of other means of solving problems of that nature, the following proportion will serve to find the course. As the sum of the hypothenuse (or distance) and half the greater leg (whether difference of latitude or departure) is to the less leg; so is 86, to the angle opposite the less leg. bers may then be found, from their sum and difference, by Prob. 4, page 57. Ex. The sum of two numbers is 32, and the difference of their squares is 256, what are the numbers? Ans. The greater is 20. The less 12. PROB. XVII. Given the difference of two numbers, and the difference of their squares, to find the numbers. RULE. Divide the difference of the squares by the difference of the numbers, and the quotient will be their sum. ceed by Prob. 4, p. 57. Then pro Ex. The difference of two numbers is 20, and the difference of their squares is 2000; what are the numbers? Ans. 60 the greater. 40 the less. Examples for the two preceding problems. 1. A and B played at marbles, having 14 apiece at first; B having lost some, would play no longer, and the difference of the squares of the numbers which each then had, was 336; pray how many did B lose? Ans. B lost 6. 2. Said Harry to Charles, my father gave me 12 apples more than he gave brother Jack, and the difference of the squares of our separate parcels was 288; Now, tell me how many he gave us, and you shall have half of mine. Ans. Harry's share 12. 18 EXTRACTION OF THE CUBE ROOT. A cube is any number multiplied by its square. To extract the cube root, is to find a number which, being multiplied into its square, shall produce the given number. FIRST METHOD. RULE. *1. Separate the given number into periods of three figures each, by putting a point over the unit figure, and every third figure beyond the place of units. 2. Find the greatest cube in the left hand period, and put its root in the quotient. 3. Subtract the cube, thus found, from the said period, and to the remainder bring down the next period, and call this the divi· dend. 4. Multiply the square of the quotient by 300, calling it the triple square, and the quotient by 30, calling it the triple quotient, and the sum of these call the divisor. 5. Seek how often the divisor may be had in the dividend, and place the result in the quotient. * The reason of pointing the given number, as directed in the rule, is obvious from Coroll. 2, to the Lemma made use of in demonstrating the square root. The process for extracting the Cube Root may be illustrated in the same manner as that for the Square Root. Take the same number 37, and multiply 6. Multiply the triple square by the last quotient figure, and write the product under the dividena; multiply the square of the last quotient figure by the triple quotient, and place this product under the last; under all, set the cube of the last quotient figure and call their sum the subtrahend. as before, collecting the twice 21 into one sum, as they belong to the same place and the operation will be simplified, 37 3=50653. 4410 3X30X7 3X32) 189=3×32 X7 3×302\18900=3×302 XT 3×3 As 27 or 27000 is the greatest cube, its root is 3 or 30, and that part of the cube is exhausted by this extraction. Collect those terms which belong to the same places, and we have 32 x7=63, and 2×32 ×7=126, and 63+126=3×32 x7=189; and 2×3×73-294 and 3×72147, and 294-+-147=441=3x3x7%, for a dividend, which divided by the divisor, formed according to the rule, the quotient is 7, for the next figure in the root. And it is evident on inspecting the work, that that part of the cube not exhausted is composed of the several products which form the subtrahend, according to the rule. The same may be shown in any other case, and the universality of the rule hence inferred. The other method of illustration, employed on the square root, is equally applicable in this case. 37-304-7, and 30+72=302-1-2×30×74-72 30-4-7 the multiplier. 30342×302X7+30×72 302X7+2X30X724-73 373-50653-3034-3×302×7+3×30×72÷73(30+7=37 Divisor 3x30243×30 303 )3X302X7+3x30x724-73 dividend. 3×302X7+3x30x72-1-73 subtrahend. It is evident that 303 is the greatest cube. When its root is extracted, the next three terms constitute the dividend; and, the several products formed by means of the quotient or second figure in the root, are precisely equal to the remaining parts of the power, whose root was to be found. The arithmetical demonstrations of the Rules for extracting either the square or cube root, are not only more consistent with the plan of an Arithmetick than demonstrations on the figure, called a square, and the solid, called a cube, but they are much more readily understood by those unaccustomed to the mathematical consideration of solid bodies. 7. 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 till the whole be finished. Note. The same rule must be observed for continuing the operation, and pointing for decimals, as in the square root. = 78875=1st. Subtrahend. 2.Div. 1689750)14161824-2d. Divid. 75X45X306= 75X30 125 78875-1st. Subtra. 1687500=2d. Trip. sq. 2030=2d. do. quo. To find the true denominator, to be placed under the remainder, after the operation is finished. In the extraction of the cube root, the quotient is said to be squared and tripled for a new divisor; but is not really so, till the triple number of the quotient be added to it; therefore when the operation is finished, it is but squaring the quotient, or root, then multiplying it by 3, and to that number adding the triple number of the root, when it will become the divisor, or true denominator to its own fraction, which fraction must be annexed to the quotient, to complete the root. |