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For example, to find the fluent of
Here, by dividing the numerator by the denominator, the proposed fluxion becomes x-2x+3x2x-5x3x+8x1 x − &c; then the fluents of all the terms being taken, give x-x2+x3-4x4+3x5-&c. for the fluent sought.
Again, to find the fluent of x1—2o.
Here, by extracting the root, or expanding the radical quantity 1x3, the given fluxion becomes x-1 x2x-1 x1xxx - &c. Then the fluents of all the terms, being taken, give x-3x3-_x5—7},x7 —&c. for the fluent sought.
FINDING OF FLUENTS.
To Correct the Fluent of any Given Fluxion.
46. The fluxion found from a given fluent, is always perfect and complete; but the fluent found from a given fluxion is not always so; as it often wants a correction, to make it contemporaneous with that required by the problem under consideration, &c.: for, the fluent of any given fluxion, as may be either x, which is found by the rule, or it may be x+c, or x-c, that is x plus or minus some constant quantity c; because both x and xc have the same fluxion x, and the finding of the constant quantity c, to be added or subtracted with the fluent as found by the foregoing rules, is called correcting the fluent.
Now this correction is to be determined from the nature of the problem in hand, by which we come to know the relation which the fluent quantities have to each other at some certain point or time. Reduce, therefore, the general fluential equation, supposed to be found by the foregoing rules, to that point or time; then if the equation be true, it is correct; but if not, it wants a correction; and the quantity of the correction, is the difference between the two general sides of the equation when reduced to that particular point. Hence the general rule for the correction is this:
Connect the constant, but indeterminate, quantity c, with one side of the fluential equation, as determined by the foregoing rules; then, in this equation, substitute for the variable quantities, such values as they are known to have at any particular state, place, or time; and then, from that particular state of the equation, find the value of c, the constant quantity of the correction.
47. Exam. 1. To find the correct fluent of 2=ax3x.
The general fluent is z=ax4, or z=ax2 + c, taking in the correction c.
Now, if it be known that z and x begin together, or that z is =0, when x = 0; then writing 0 for both x and 2, the general equation becomes 0=0+c, or=c; so that, the value of c being 0, the correct fluents are z=ɑxa.
FINDING OF FLUENTS.
But if z be =0, when x is = b, any known quantity; then substituting 0 for z, and b for x, in the general equation, it becomes 0=ab1+c, and hence we find c—— —aba; which being written for c in the general fluential equation, it becomes z= ax4-ab, for the correct fluents.
Or, if it be known that z is = some quantity d, when x is some other quantity as b; then substituting d for z, and b for x, in the general fluential equation z=ax4 + c, it becomes d=ab+c; and hence is deduced the value of the correction, namely, cd-ab4; consequently, writing this value for c in the general equation, it becomes zax1 — ab4d, for the correct equation of the fluents in this case.
48. And hence arises another easy and general way of correcting the fluents, which is this: In the general equation of the fluents write the particular values of the quantities which they are known to have at any certain time or position; then subtract the sides of the resulting particular equation from the corresponding sides of the general one, and the remainders will give the correct equation of the fluents sought.
So, the general equation being = ax*;
or z—ax1 —ɑba+d, the correct fluents as before.
EXAM. 2. To find the correct fluents of -0 O when x is =α.
EXAM. 3. To find the correct fluents of z and being=0 at the same time.
EXAM. 5. To find the correct fluents of z:
ing z and x to begin together.
2αx EXAM. 4. To find the correct fluent of z= ; supposing a+x z and x to begin to flow together, or to be each=0 at the same time.
= 5xx; z being
Z 3x √ α + x
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OF FLUXIONS AND FLUENTS.
ART. 49. In art 42, &c. is given a compendious table of various forms of fluxions and fluents, the truth of which it may be proper here in the first place to prove.
50. As to most of those forms indeed, they will be easily proved, by only taking the fluxions of the forms of fluents, in the last column, by means of the rules before given in art. 30 of the direct method; by which they will be found to produce the corresponding fluxions in the 2d column of the table. Thus, the 1st and 2d forms of fluents will be proved by the 1st of the said rules for fluxions; the 3d and 4th forms, of fluents by the 4th rule for fluxions; the 5th and 6th forms, by the 3d rule of fluxions; the 7th, 8th, 9th, 10th, 12th, 14th forms, by the 6th rule of fluxions: the 17th form, by the 7th rule of fluxions: the 18th form, by the 8th rule of fluxions. So that there remains only to prove the 11th, 13th, 15th, and 16th forms.
51. Now, as to the 16th form, that is proved by the 2d example in art. 98, where it appears that x (dx-x2) is the fluxion of the circular segment, whose diameter is d, and versed sine x. And the remaining three forms, viz. the 11th, 13th, and 15th, will be proved by means of the rectifications of circular arcs, in art. 96.
52. Thus, for the 11th form, it appears by that art. that the fluxion of the circular arc z, whose radius is r and tangent t,
p2i is == r2 + t then is i = nx 22-1
Now put t= x22, or t2= x
x, and r2 + 12 = a + x
radius 1 and tang. ✔
X arc to radius / ɑ and tang. xã
and the fluent is
X arc to
and a = 42
in n°. XI.
53. And, for the latter form of the fluent in the same no; because the coefficient of the former of these, viz. -is dou
ble of the coefficient of the latter, therefore the arc
in the latter case, must be double the arc in the former But the cosine of double an arc, to radius 1 and tangent t
FLUXIONS AND FLUENTS.
; and because t2
tuted for 13 in the cosine
as in the latter case of the 11th form.
√ (1 − Z ) = √
then these two
√(1 − x2)
54. Again, for the first case of the fluent in the 13th form. By art. 61, the fluxion of the circular arc z, to radius and sine y, is z
√ (r2 — y2)?
Now put y √. or y2
√(1—32) to the radius 1.
Now, put s✔
× √ (a —x"), and y = √ - ×jnx\"
by the former case, this substi
1 — t2
55. And, as the coefficient, in the latter case of the said
form, is the half of the coefficient in the former case, there
fore the arc in the latter case must be double of the arc in
56. Again, for the first case of fluent in the 15th form.
and secant s, is z=
to radius 1.
as in the table of forms, for the first case of
and therefore its fluent is, that is
; hence &√ (s2 — 1) =