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 exactness.

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=1333, &c. that is, ·583+1·333= 1.916 approximated root.

Now assume 1916 for the root. Then, by the second process,

the root is

7

3x1.916

2+3 × 1·916=1.9126, &c.

2d. What is the cube root of 9? Let 2 be the assumed root as before. Then,+2=2·08 the approximated root. Now assume 2.08. Then,

9

3×2.08

[2+3×2·08=2·08008, &c.

3d. What is the cube root of 282? Let 6 be the assumed root.Then, 6x6x3=108)282(2 611, &c. and 2-611+ of 6-6 611 approximated root. Now assume 6 611, and it will be 6·611×6·611 X3=131-116)282(2 1507, &c. and 2 1507 +3 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+3 of 12·426=12014 a farther approximated root, and so on.

APPLICATION AND USE OF THE CUBE ROOT.

1. To find two mean proportionals between any two given num、 bers.

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?

3

750÷÷6=125, and 125=5. Then, 5x6=30=least, and 30x 5-10 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 32; What will a bullet of the same metal weigh, whose diameter is 3 inches?

6×6×6=216. 3x3x3=27. As 216: 321 :: 27: 4, 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?

Ans. £2880. 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.t 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?

3

4x4x4=64, and 64x4=256. √2566349+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? Ans. 2ft. 103 inches. 7th. The diameter of a bushel measure being 18 inches, and the height 8 inches, I demand the side of a cubic box, which shall contain that quantity. Ans. 12 907+inches.

As two mean proportionals are required to two given numbers, there will be four terms in the proportion, in which the first is to the second, as the second to the third, and the third to the fourth. The numbers therefore belong to a geometrical progression of four terms. The first part of the rule is explained in Prob. VIII. of Geometrical Progression, and the second and third parts of the Fule are evident from the proof of Prob. I. of Geometrical Progression.

The solid, called a cube, has its length and breadth and height all equal. As the number of solid feet, inches, &c. in a cube are found by multiplying the height and length and breadth together, that is, by multiplying one side into itself twice, the third power of a number is called the cube of that number.

Multiply the square of the diameter by 7354, and the product by the height; the cube root of the last product is the answer. See Mensuration of Superficies and Solids, Art. 30.

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

89×89×89=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=26 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.

What

As 12x12x12: 18 :: 15x15×15: 35·15625 Ans. 10. If a 15 inch cable require an anchor 35-15625cwt. : must the circumference of a cable be, for an anchor of 18cwt? 12 inches, Answer.

EXTRACTION OF THE BIQUADRATE ROOT.

RULE.

Extract the square root of the resolvend, and then the root of that root, and you will have the biquadrate root. What is the biquadrate root of 20736?

square

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 the square root of the remainder.

3. Add of the assumed root to the square root, and the sum will 33 be the true root, or an approximation to it.

4. For every succeeding operation, either in this or the following method, proceea 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.

Here 10 is the assumed root.

10x10x 6=600)20736(34.56

Subtract 10x10÷18 = 5.5555.

29.0044 5.38

Add of 10 =6.66

Approximated root 12.04, to be made the as

sumed 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?
The assumed root is 10

10x10x10x4=4000)20736(5.184

Add of 107.5

Approximated root 12-684, to be made the assumed

root for the next operation.

EXTRACTION OF THE SURSOLID ROOT BY APPROXΙΜΑ ΓΙΟΝ.

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 part 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 the sureolid root of 281950621875?

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 bad 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 remain der for a new dividend, to which find a new divisor, as before, and, in like manner, proceed till the whole be finished.

23

EXAMPLES.

1st. What is the cube root of 20346417 ?

20346417(273

2×2×2=8 root of the 1st. period, or 1st. Subtrahend.

8 = 1st. Subtrah.2×2=4(=next inferiour power,) and, 4x3 (the index of the given pow.)= 12 1st. Divisor.

22x3=12)123=Dividend 27x27x27=19683=2d. Subtrahend.

273

=

27X27=729 (next inferior power) and, 19683=2d. Subt.729×3 (=index of the given pow.) = 2187 2d. Divisor. 272x3=2187)6634-2d.Di. 273x273×273-27346417-3d. Subtra.

2733 = 20346417=3d. Subtrahend.

*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 convenient.

When the index of the power, whose root is to be extracted, is a composite number, take any two or more indices, whose product is equal to the given index and extract out of the given number a root answering to another of the indices, and so on to the last.

Thus, the fourth root square root of the square root; the sixth root square root of the cube root; the eighth root square root of the fourth root; the ninth root the cube root of the cube root; the tenth root square root of the fifth root; the twelfth root the cube root of the fourth, &c.

The general rule for extracting the roots of all powers, may be illustrated in the same way, as those for the square and cube roots. Any student may at once see the truth of the rule, in exhausting the several products of the case illustrating the rule for the cube root. And the same will be evident by raising the number to any higher power.