Auxion of the area of this parallelogram with the fluxion of the area of the curve, they being homogeneous quantities; and the fluxion of the area of the parallelogram being ax, he gets the fluxion of the area of the curve. From what has been said above, when we reduce these matters to calculation, there appears to be no absolute necessity for this; but it is more scientific to make the comparison between homogeneous quantities, than between those which are not homogeneous, and therefore the former method is always to be preferred in cases where it can be applied, notwithstanding the conclusions which are otherwise de duced are perfectly true and satisfactory.

20. The ingenious and justly celebrated author of the Ana lyst has endeavoured to show, that the principles of fluxions, as delivered by its author, are not founded upon reasoning strictly logical and conclusive. He lays this down as a Lemma: "If you make any supposition, and in virtue thereof deduce any conse quence; if you destroy that supposition, every consequence before deduced must be destroyed and rejected, so as from thence forward to be no more supplied or applied in the demonstration.” This, he thinks, is so plain as to need no proof. It may perhaps be admitted to be true, when we want to deduce the absolute value of a quantity which is to be obtained in virtue of a supposition; but it is not true when we want to obtain the relative values of quantities. He seems not to have attended to the con. nection which there must necessarily be between the two terms which constitute a ratio, and the two terms which express the ratio to which they approach as their limit, when you diminish them sine limite, called the limit of the ratio. It is agreed, that by diminishing the increments you approach to the ratio of the velocities which they had at the points from whence they began to be generated, and that by making them become indefinitely small, you arrive at a ratio indefinitely near to that of the veloci ties at those points. Now the two terms expressing the limit of a ratio must depend upon the terms themselves of the ratio, and the terms are obtained upon the supposition of the existence of an increment; the limit therefore is obtained upon the supposition of the existence of an increment; but the limit is a certain determinate invariable ratio, totally independent of the magni tude of the terms of the ratio, and consequently of the increment, as appears by ¶ 8. When we therefore deduce the limit by making the increments vanish, the effect of the prior existence of the terms of the ratio still remains in the terms which express the limit of the ratio. If the existence of the terms which express the limit of the ratio, depended upon the existence of the terms themselves of the ratio, the supposition which makes the latter vanish would necessarily make the former also vanish, and then no conclusion could be deduced by making the terms of the ratio vanish ;

but as that is not the case, the limit, which is obtained by making the terms become equal to nothing, contains an effect, after the increments are actually vanished, which depends upon their having existed. The lemma therefore of the author, however true it may be under some circumstances, cannot be applied against the reasoning upon which the Principles of Fluxions are founded. The author admits the conclusions to be true. He says, "I have no controversy about your conclusions, but only about your logic; and it must be remembered that I am not concerned about the truth of your theorems, but only about the way of coming at them." The above observations show, not only that our conclusions are true, but that they are deduced by steps which are perfectly satisfactory, and strictly logical.

ARTICLE XXI.

NEW QUESTIONS to be solved in the next Number. I. QUEST. 62. By A. Rabbit, at Uncle Sam's, Downbelow. Edward Shepherd, from English Yorkshire, has in page 198 of his Columbian Accountant, recorded and answered the following question.

“A, B, and C are to share £100000 in the propor❝tion of, and respectively; but C's part being "lost by his death, it is required to divide the whole "sum properly between the other two." This eminent author makes A's share=57142313, and B's

4285747 Now I demand the operation, by which we ought to find the shares of A and B; and also by what curious process Shepherd created the fractional tails of his answers.

II. QUEST. 63. By G. Baron, New-York.

Thirty political sages meet every evening, at T. H. Kennedy's Porter-House in New-York. They consist of Royalists, Federalists, Clintonians, and Lewisites, in an increasing continual proportion; and their third difference is 2. I demand the number of each.

III. QUEST. 64. By Henry Smith, New-York. At what times of the day do the hour and minute hands of a watch form one continued straight line? IV. QUEST. 65. By William Lenhart, Baltimore. Prove that a+6 has an infinite number of algebraic divisors.

V.QUEST. 66. By Thomas Whittaker, York-Town, Penn It is required to divide the arc of a semicircle into two arcs, so that the radius of the semicircle may be a mean proportional between the chords of those arcs. VI. QUEST. 67. By Peter Spangler, York-Town, Penn. If a trapezium be described about a circle, the sum of any two of its opposite sides is equal to the sum of the other two sides; required a demonstration.

VII. QUEST. 68. By Joseph Cowing, Alexandria. The base of a plane rectilineal triangle being 14, its area equal to double that of its inscribed square, and one of the angles at the base 53° 03'; quere the other two sides of the triangle?

VIII. QUEST. 69. By William Lenhart, Baltimore. Given the area of a plane rectilineal triangle 840; the length of a straight line drawn from the vertical angle to the middle of the base-39, and the vertical angle=61° 56': to determine the triangle.,

IX QUEST. 70. By John D. Craig, Philadelphia.

Describe a circle that shall touch the arc of the quadrant of a given circle, one of its bounding radii, and a semicircle described on the other bounding radius.

X. QUEST. 71. By David Sanford, Newtown, Conn.

Given the base and vertical angle of a plane recti-` lineal triangle, and the distance between the centres of its circumscribing and inscribing circles; to determine the triangle.

XI. PRIZE QUEST. 72. By Joseph Cowing, Alexandria.

Let a point be assumed in the circumference of a given circle; it is required to find, geometrically, another point in a straight line, passing through the assumed point and the centre of the circle, from which if a tangent be drawn, and a straight line drawn from the assumed point to the point of contact, this last shall obtain a given ratio to the straight line joining the assumed point and that which is required.

ARTICLE XXI.

SOLUTIONS of the QUESTIONS proposed in ARTICLE XXI.

I. QUEST. 62. Solved by A. Rabbit, at Uncle Sam's, Downbelow.

Because C's part was lost by his death, it is plain that::: 4 : 3 :: the share of A: the share of

B.

Hence 7: 4 :: £100000 : £571429=A's share;

and

7:3 £100000 £42857 B's share.

In answering this question, Shepherd first divided the given sum between A, B, and the ghost of C, in the ratios of, and ; and afterwards divided the share of the ghost between A and B, in the ratio of to. This is the curious process, by which Shepherd created the fractional tails of his answers.

II. QUEST. 63. Solved by Enoch Lewis, Westown, Pennsylvania.

Let the required number of Royalists, Federalists, Clintonians, and Lewisites, be respectively represented by x, xy, xy2 and xy3; then by the question and the nature of differentials, we have

(y3 + y2+y+1)Xx = 30, and (3-32+3y-1)xx

two equations we have

2. From these

y3323-1: y3+y2+y+1:1:15; or 7y323y2+22y8; which resolved, gives y=2. Hence is easily found x=2, xy=4, xy2=8, and xy3=16: consequently our 30 political sages consist of 2 Royalists, 4 Federalists, 8 Clintonians, and 16 Lewisites.

III. QUEST. 64. Solved by the proposer, Henry Smith, New-York.

It is plain that the hour and minute hands of a watch form one continued straight line at the hour of six. And, since the lengths of the two hands may be considered as equal; if the length of the arc described by the hour hand in one hour, be put =1; the length of that described by the minute hand, in the same time, will be = 12; and the difference of these two arcs will be 11. Now, when this difference becomes 12, the hands will again form a continued straight line: therefore by proportion 11: 12: 1 hour: 1 hour 5 min. Further, since the angular velocities of the two hands are always invariable, the required times are evidently in arithmetic progression; and their common difference is 1 hour, 5 min.

h h m

h

h

m

h

and 4 54, the hour and minute hands of a watch are diametrically opposite to each other, or form onę continued straight line.

IV. QUEST. 65. Solved by Enoch Lewis.

a+b; assume m=2, 3, 4, &c. and the expressions are still algebraical; whence, as the assumed value of m may be infinitely varied, the number of alge. braic divisors of a+b, is infinite.