Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

THE USE OF THE CUBE ROOT.

Case I.

To find the side of a cube that shall be equal in solidity to any given solid, as a giobe, cylinder, prism, cone, &c. Rule.

Extract the cube root of the solid content of the given body, which root will be the side of the cube required.

Examples.

1. There is a stone of a cubic form, which contains 21952 solid feet; what is the superficial content of one of its sides? Answ. 28 feet.

Case II.

Having the dimension of any solid body, to find the dimensions of another similar solid, that shall be any number of times greater or less than the solid given.

Rule.

Multiply the cube of each side by the difference between the solid given and that required, if greater (or divide by the difference if less) than the solid given; then extract the cube root of each product or quotient, which will give the dimensions of the solid required.

Examples.

2. Suppose the length of a ship's keel to be 125 feet, the breadth of the midship beam 25 feet, and the depthof the hold 15 feet; I demand the dimensions of another ship of the same form, that will carry three times the burthen?

Length of the keel, 180.28 feet.

Answ. Breadth of the beam, 36.05

[blocks in formation]

3. Again, I demand the dimensions of another ship of the same form, that shall be only half the burthen of that whose dimensions are given as above?

Answ.

Length of the keel, 99.21 feet.

Breadth of the beam, 19.84
Depth of the hold,

11.9

Case

Case III.

Having the dimension and capacity of a solid, to find the dimensions of a similar solid of a different capacity.

Rule.

Like solids are in triplicate proportion to their homologous sides, therefore it will be as the cube of a dimension is to its given weight so is the cube: of any like. dimension to the weight sought.

Examples.

4. If a ship of 300 tons burthen be 75 feet long in the keel, I demand the burthen of another ship, whose keel is 100 feet long? Answ. 711.111 tons.

5. Suppose a ball of 4 inches diameter weighs 18ib. I demand the diameter of another that weighs 114lb. ì Answ. 7.4 inches.

Case IV.

To find two mean proportionals between two given numbers.

Rule.

Divide the greater extreme by the less, and the cube root of the quotient, multiplied by the less extreme, gives the lesser mean; multiply the said cube root by the lesser mean, and the product will be the greater mean proportional.

Examples.

6. What are the two mean proportionals between 7 and 189 ? Answ. 21 and 68.

7. Find two mean proportionals between 4 and 256 ? Answ. 16 and 64.

A

CHAP. III,

ARITHMETICAL PROPORTION.

NY rank of numbers encreasing one above another

ceed each other by 1; or 2, 4 6, 8, 10, &c. 1, 3, 5, 7, 9, &c. whose common Excess is 2; or contrariwise decreasing by a common Difference, as 5, 4, 3, 2, 1; 10, 8, 6, 4, 2, are said to be in Arithmetical Progression.

The

The numbers which form an Arithmetical Progression are called Terms of the Progression, and the number whereby the latter term exceeds or is deficient of the former is called the Common Difference.

Bs 2, 4, 6, 8, 10, these numbers are the terms, and 2 the common difference.

From the infinity of number, it is easy to conceive that an increasing Arithmetical progression may be infinitely continued; but a decreasing progression cannot be continued further than till the last term becomes less than the common difference..

Numbers in Arithmetical progression have sundry peculiar properties, of which are the following.

Proposition 1.

In any increasing series, if the first term be added to the product of the common difference multiplied by the number of terms less 1, the sum will be the last term: And in a decreasing series if the said product be subtracted from the first term, the remainder will be the last term.

Proposition 2.

If three numbers are in Arithmetical progression, the sum of the two extremes will be double the mean.

Example.

If 2, 3, 4, 2 + 46 and 3 + 36
4, 6, 8, 4+ 8 = 12 and 6 +.6 = 12
4, 3, 2, 4 + 2 = 6, &c,

Proposition 3.

If four numbers be in Arithmetical progression, the sum of the means is equal to the sum of the extremes.

Example.

2, 4, 6, 8

{8, 6, 4, 22+8=10 and 4+6=10

Problem I.

Having two or more numbers in Arithmetical progression, none less than the common difference, to continue the progression upwards and downwards.

Subtract the less from the greater, and thereby find the common difference, which add to the greater and subtract from the less, and so will two extreme terms be found, of which extremes the less being diminished and the greater encreased by the same common difference, we get two

other

other extreme terms, and thus we may continue the progression upwards as far as we please, and downwards till we find a number less than the common difference, which is the first term of an encreasing progression.

Example. Let 8 and 10 be given, and let it be required to continue the progression both ways. By deducting 8 from 10 we find the common difference 2, whence we get this progression:

0, 2, 4, 6, [8, 10,] 13, 14, 16, 18.

Proposition 4.

In any Arithmetical progression whose number of terms is odd, the mean or middle term being doubled, is equal to the sum of any two terms equally distant therefrom: and if the number of terms be even, the sum of the means or two middle terms is equal to the sum of any two terms equally distant therefrom.

Example. In 1, 3, 5, 7, 9, 11, 13. 7+714 and 59; 3+11; 1+13 each 5+

14.

Again, 1, 3, 5, 7, 9, 11, 13, 15, 7+916 anl 5+11; 3+13; 1+15 each 16.

Cor. Hence if any two numbers be added, and their sum halved, that half is an Arithmetical mean between the said two numbers.

In an Arithmetical progression these five things are more especially to be noted; (1) the first term; (2) the last term; (3) the common difference; (4) the number of terms; (5) the sum of the series; any three of which being given the rest may be found, which admits of 21 problems, but the following 3 seeming principally useful in Arithmetic, for brevity sake we confine ourselves thereto.

[ocr errors]

Problem II.

Having the first term, the common difference and num ber of terms, to find the last term.

Rule.

the sun

Multiply the number of terms less 1 by the common difference, and to that product add the first term, is the last term required.

Example.

Let 1 be the first term, 2 the common difference, and 11 the last term; 2 × 10(=20)+121 last term.

1

Cor. If the first term be common difference, then the first term multiplied by the number of terms will produce the last.

Problem

Problem III.

Having the first term, the last term, and the number of terms, to find the sum of the series,

Rule.

Add the first and last terms together, and multiply half the sum by the number of terms, or the whole sum by half the number of terms, and the product is the sum of the

series.

Example.

Let the first term be 1; the last 21, and number of terms 11; then

1+21 22 whose 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 2}, half is 11 X 11 =

121 sum of the series, or 1+ 1920 × 5 half the num ber of terms) 100 the sum of the series of 10 terms.

Problem IV.

Having given the first term, common difference and number of terms, to find the sum.

Rule.

First find the last term by Prob. 2. and then the sum by Prob. 3.

Examples.

1. How many strokes doth a regular clock strike in a natural day, or 24 hours? Answ. 156.

For at one o'clock it strikes 1, at two, 2, &c. so it is required to find the sum of an Arithmetical progression 1, 2, 3, &c. up to 12, where we have 1 the first term, 1 cominon difference; the number of terms 12, whence the last term is found to be 12. So then 1+12= 136 half the number of terms produceth 78 the numr ber of strokes in the first 12 hours, which being doubled gives 156 in 24 hours.

2. A man buys 17 yards of kersey: for the first yard he gave 2 shillings; for the last 10s. the price of each yard encreasing in an Arithmetical progression; how much dist the whole amount to?

Answ 51. 2s.

3. How many strokes doth the clocks of Venice (which go on to be 2 o'clock) strike in the compass of a natural day? Answ. 300.

[ocr errors]
« ΠροηγούμενηΣυνέχεια »