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in any term of the equation. 16. A line which continually approaches a curve, without ever meeting it, is called an asymptote of the curve.

1. To find the equation of a circle. See Figure 11.

Let the Radius=r=AD, AB=x, and BD=y.

1. By Euclid 1. 47, r2 =x2 +y2

2. Transposing, r2 —x2 =y2

3. Extracting the square root, ±√r2 −x2 =y 4. Transposing in the 2d, r2 — y2 = x2

5. Extracting the square root, ±√72 —y2 =x; then if Radius=1, y=√1−x2, and x= √1—y2.

II. To find the equation of the curve called the CissOID of Diocles. See Figure 15.

1. By similar triangies, AP: PM::AR:RN

2. Substituting PB for AR, AP ;PM::PB; RN 3. Multiplying extremes and means, AP×RN=PB×PM 4. Dividing by AP, RN_PB×PM

5. Involving to the 20

AP

power, RN

2

6. By Euclid III. 3, AR×RB=RN

PB2 XPM2

AP2

2

7. Putting PB for its equal AR, and AP for RB; then PB

XAP=RN2

8. Making the 5th and 7th equal, PB×AP=

PM2XPB2
PB2

9. Clearing of fractions PBXAP3=PM2 ×PB2
10. Dividing by PB, AP3=PM2 × PB
11. Substituting small letters, a3=y3 (b-x)

To prove that APM and ARN, are similar triangles. By Euclid I. 28, APM=ARN, and by the same proposition, AMP=ANR, and the angle at A is common; therefore the triangles APM and ARN are similar.

III. To find the equation of the CONCHOID of Nicomedes. See Figure 16.

1. By similar triangles CFM and MPE, CF: FM::PM

: PE

2. Multiplying extremes and means, PE× CF=PM× FM 3. Dividing by CF, PE=

PMXFM

CF

4. Involving to the 2d power, PE PE

2

PM2XFM

CF2

5. By Euclid I. 47, PE2 = EM2 – PM2
6. Making the 4th and 5th equal, EM2 –PM2

PM2 × FM2
CF2

=

7. Substituting small letters,

=

b2-y-y2 × x2
—y2:
(a+y)2

8. Multiplying by (a+b)3, (a+y)2 ×(b2 —y2)=x2y2

To prove that the triangles CFM and MPE, are similar. From the equality of right angles, CAE-EPM. By Euclid I. 15, PEM AEC; wherefore, by Euclid I. 32, the remaining angle PME=the remaining angle ECA; therefore the triangle ACE is similar to the triangle PEM. By Euclid I. 28, CAE CFM, and by the same proposition, CEA= CMF, and the angle at C is common; therefore the triangle CAE is similar to the triangle CFM; and EPM is also similar to CAE, and figures that are similar to the same figures are similar to each other: therefore, EMP is similar to CFM. Q. E. D.

We now take leave of Algebra. In conforming to the order in which the different branches are studied, Applica. tion of Algebra to Geometry should have preceded Euclid. But as this whole treatise is comprised in a single volume, the arrangement will occasion the student no inconvenience. Playfair's Geometry will next be considered, where we shall be very brief, as nearly all the theorems are demonstrated.

Note. The figures as they are drawn in Playfair's Euclid, will, with a slight alteration, answer for the demonstrations in Geometry which follow. In one instance, however, the alterations were so essential, that a new figure has been supplied.

GEOMETRY.

SUPPLEMENT, BOOK I. PROP. V. COR. 3.

Hence also a polygon may be inscribed in a circle, such that the excess of the circumference of the circle above the perimeter of the polygon may be less than any given line.

Let NO be the given line. Take in NO, the part NP less than its half, and less also than AD, and let a polygon be inscribed in the circle ABC, so that the excess of the circumference of the circle above the perimeter of the po. lygon may be less than the square of NP. (Sup. I; 4. Cor. 1.) Let a side of this polygon be AB. Now, because in the triangle ABD, DM is drawn perpendicular to the base, the triangle ABD is equal to the rectangle contained by DM and half of AB, which is less than the rectangle contained by DG, or DA and half of AB; therefore the polygon of which one side is AB, is less than the rectangle contained by DA and AK, half the perimeter of the polygon. Suppose the point to be taken still further towards C, until we have the rectangle ADXAX equal to the inscribed polygon; the circle exceeds the inscribed polygon by the rectangle ADXHX: Therefore the rectangle ADXHX, is less than the square of NP; hence, since AD is greater that NP, HX is less than NP, and twice HX is less than twice NP: much more, then, is HK less than NP, and twice HK less than NO. But HK is the difference between half the perimeter of the inscribed polygon, whose side is AB, and half the circumference of the circle ABC. Therefore, twice HL is the difference between the whole perimeter of the inscribed polygon, and the whole circumfer ence of the circle, (Euc. V. 5.) Therefore the excess of the circumference of the circle above the perimeter of the inscribed polygon, is less than the given line NO.

Q. E. D. SUPPLEMENT, BOOK I. PROP. Q. In the same manner it may be demonstrated that Q is less than any polygon described about the circle GHL.

Describe (Sup. I. 3.) about the circle ABD, a polygon similar to the polygon ABCDEF, of which RS shall be the

side homologous to AB, and OT the perpendicular from the centre to RS, and OU the perpendicular from the center to AB; also about the circle GHL describe a polygon similar to the polygon GHKLMN, of which WV shall be the side homologous to GH, and PX the perpendicular from the center to WV, and PY the perpendicular from the center to GH. Then, because the polygons ABCDEF, and GHKLM have the same number of sides, the angle AOB is equal to the angle GPH, and their halves AOU and GPY are equal : Therefore (VI. 4.) RT : TO :: WX : XP. (V. Def. 14.)RT ': WX :: TO; XP, (V. 15,) RS; WV::TO; XP, and RS: WV:: AO GP, (V. 15.) RS: WV::AO: GP2 (VI. 4 and V. 15.) AB2 : GH2::AO2 : GP2, & (Sup. I. 3, 1 Cor.) Polygon ABCDEF: polygon GHKLMN::AB3 :GH*. Ex aquali, polygon ABCDEF : polygon GHKLMN::A02; GP2. (Sup. I. 3, 1 Cor.) polygon circumscribed about ABD polygon circumscribed about GHK::RS: WW2. Ex æquali, polygon circumscribed about ABD polygon circumscribed about GHK::polygon ABCDEF: polygon GHKLMN. But these polygons are as the squares of AD and GL: Therefore the polygon described about the circle ABD is to the polygon described about the circle GHK as AD is to GL2. But AD2 : GL::the circle ABD Q; therefore, the polygon described about ABD the polygon described about GHL::ABD : Q. Now the circle ABD<polygon described about ABD; therefore, the circle Q< polygon described about the circle GHL: Therefore (Sup. I. 4, 2 Cor.) Q is equal to the circle GHD. Q. E. D.

2

SUPPLEMENT, BOOK III. PROP. XVII. To prove that the cylinder ABCD is not greater than the parallelopiped EF.

Secondly, let the cylinder ABCD be greater than the parallelopiped EF. Take the parallelopiped EV, equal to the cylinder ABCD. About the circle ABG, describe (Sup. I. 4,) a polygon which shall differ from the circle by a врасе less than any given space, and cut off from the parallelogram EX, a part EZ equal to the circumscribed polygon : The point Z will fall between H and X. On the circum

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scribed polygon let an upright prism be constituted of the same altitude with the cylinder, which will therefore be greater than the cylinder, because it is without it; and, if through the point Y, a plane YZ be made to pass, it will cut off the parallelopiped OY, equal to the prism, (Sup. III. 8, 2 Cor.) because its base is equal to the base of the prism, and its altitude is the same. But the prism is greater than the cylinder ABCD, and the cylinder is by hypothesis equal to EV; therefore the prism circumscribed about the cylinder ABCD is greater than EV: it is also less, which is impossible. Therefore the cylinder ABCD is not greater than the parallelopiped ÈF, and it has been proved that it is not less; therefore they are equal.

LOGARITHMS.

HISTORY OF LOGARITHMS.

The history of Logarithms affords the more pleasure, because their invention is of so recent origin, that whatever is

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