LOGARITHMS.

TO FIND THE CONTENTS OF SOLIDS.

101. ANY solid which is formed by the revolution of a curve about its axis (see last fig.), may also be conceived to be generated by the motion of the plane of an expanding circle, moving perpendicularly along the axis. And therefore the area of that circle being drawn into the fluxion of the axis, will produce the fluxion of the solid. That is, AD X area of the circle BCF, whose radius is DE, or diameter BE, is the fluxion of the solid, by art. 9.

365

102. Hence, if AD-X, DE—y, c=3·1416; because cy2 is equal to the area of the circle BCF: therefore cy is the fluxion of the solid. Consequently if, from the given equation of the curve, the value of either y2 or x be found, and that value substituted for it in the expression cy2x, the fluent of the resulting quantity, being taken, will be the solidity of the figure proposed.

EXAMPLES.

EXAM. 1. To find the solidity of a sphere, or any segment. The equation to the generating circle being y2 = ax - x2, where a denotes the diameter, by substitution, the general fluxion of the solid cy2x, becomes caxx - cx2x, the fluent of which gives cax2-cx3, or cx2 (3a-2x), for the solid content of the spherical segment BAE, whose height AD is X.

3

When the segment becomes equal to the whole sphere, then xa, and the above expression for the solidity, becomes ¿ca3 for the solid content of the whole sphere.

And these deductions agree with the rules before given and demonstrated in the Mensuration of Solids.

EXAM. 2. To find the solidity of a spheroid.
EXAM. 3. To find the solidity of a paraboloid.
EXAM. 4. To find the solidity of an hyperboloid.

TO FIND LOGARITHMS.

108. Ir has been proved, art 23, that the fluxion of the byperbolic logarithm of a quantity, is equal to the fluxion of the quantity divided by the same quantity. Therefore, when any quantity is proposed, to find its logarithm; take the fluxion of that quantity, and divide it by the same quantity; then take the fluent of the quotient, either in a series or otherwise, and it will be the logarithm sought: when corrected as usual, if need be; that is, the hyperbolic logarithm.

104. But. for any other logarithm, multiply the hyperbolic logarithm, above found, by the modulus of the system, for the logarithm sought. Note.

FLUXIONS..

•

Note. The modulus of the hyperbolic logarithms, is 1; and the modulus of the common logarithms, is 43429448190 &c.; and, in general, the modulus of any system, is equal to the logarithm of 10 in that system divided by the number 2.3025850929940, &c. which is the hyp. log. of 10. Also, the hyp. log. of any number, is in proportion to the com. log. of the same number, as unity or 1 is to 43429 &c. or as the number 2.302585 &c. is to 1; and therefore, if the common log. of any number be multiplied by 2.302585 &c. it will give the hyp. log. of the same number; or if the hyp. log. be divided by 2.302585 &c. or multiplied by 43429 &c. it will give the common logarithm.

366

EXAM. 1. To find the log. of

Denoting any proposed number z, whose logarithm is required to be found, by the compound expression a+, the fluxion of the number z, is

XC and the fluxion of

>

a

log. of

and the log. of

Writing

Div. these numb. and

subtr. their logs, gives log.

a

a±x

Also, because

1÷

x for x, gives log.

Making

a± x..
a

α

xx

x

x2 x

a α2
a

A 3

a4

these terms give the logarithm of z or x2 X3 + 2α2 a

XI

&c.

a + x

a

a+x

+

is

the prod. gives log. a2x2

α2

X2

of a +x

a

a

X

a

a

+

3a3-4a4

X

X2

x3

α- X
a

x4

a 2a2 Sa3 4α4 a+x· 2x 2x3 2x5

+ + &c. X α 3a35a5

—0—log.

.

X x2
is +-+
a 2a2

2:2

a

x3x

or log.

2α2

+ &c.

α.

a + x

X3

3ast

x7

X4

+ -of- + &c.

a2 2a4 3a6

Now, for an example in numbers, suppose it were required to compute the common logarithm of the number 2. This will be best done by the series,

a+x

7a7

X4

404

&c.

&c.

&c.

XC

=2, gives a=3x; conseq., and

X

a

INFLECTIONS.

367

which is the constant factor for every succeeding term; also,
2m=2×·43429448190·868588964; therefore the calcula-
tion will be conveniently made, by first dividing this number
by 3 then the quotients successively by 9, and lastly these
quotients in order by the respective numbers 1, 3, 5, 7, 9, &c.
and after that, adding all the terms together, as follows:
3) ·868588964

9

•289529654

9

9

9

9

9

9

9

1) 3)

32169962

3574440 5

397160

27

44129 9

4903

11

545

13

61

15)

289529654
32169962

3574440

397160

44129

4903

545

61.

Sum of the terms gives log. 2·301029995

EXAM. 2. To find the log. of "+

a+x

b.

•289529654
10723321
714888

56737

4903

446

42

4

TO FIND THE POINTS OF INFLEXION, OR OF CON-
TRARY FLEXURE IN CURVES.

A

105. THE Point of
Inflexion in a curve, is
that point of it which
separates the concave
from the convex part,
lying between the two;
A
or where the curve
D
D
changes from concave to convex, or from convex to concave,
on the same side of the curve. Such as the point E in the
annexed figures, where the former of the two is concave
towards the axis AD, from A to E, but covex from E to F; and
on the contrary, the latter figure is convex from A to E, and
concave from E to F.

]

106. From the nature of curvature, as has been remarked before at art. 28, it is evident, that when a curve is concave towards an axis, then the fluxion of the ordinate decreases,

368

FLUXIONS.

or is in a decreasing ratio, with regard to fluxion of the ab-
sciss; but, on the contrary, that it increases, or is in an in-
creasing ratio to the fluxion of the absciss, when the curve is
convex towards the axis; and consequently those two fluxions
are in a constant ratio at the point of inflexion, where the
curve is neither convex nor concave; that is, is to y in a

1

X

constant ratio, or or is a constant quantity. But constant

20

Y

quantities have no fluxion, or their fluxion is equal to nothing;
so that in this case, the fluxion of? or of is equal to no-

XC

x

y

thing. And hence we have this general rule:

107. Put the given equation of the curve into fluxions ;,

X

from which find either

Then take the fluxion of this

or

X

y

ratio, or fraction, and put it equal to O or nothing; and from
this last equation find also the value of the same

X

Y XC

ory. Then put this latter value equal to the former, which will form an equation; from which, and the first given equation of the curve, x and y will be determined, being the absciss and ordinate answering to the point of inflexion in the curve, as required.

EXAMPLES.

EXAM. 1. To find the point of inflexion in the curve whose equation is ax2 =α2y + x2y,

ax2

Then the fluxion of this quantity

This equation in fluxions is 2axx = a3y + 2xyx + x2ÿ
X. а 3 + x 2·
which gives
y 2ax-2xy
made=0, gives 2xx (ax — xy) — (a2 +·x2) × (αx −
a2 + x2
x
and this again gives ==

yê xy ) ;

X

X x2

Y

a2

ay
Lastly, this value of being put equal the former, gives
a2 + x2 X ã2 + x2 1
2x a—y;

Y

; and hence 2x2=a2

α2

x2 a-y

or 3x3a2, and xa, the absciss.

x2,

Hence also, from the original equation, y =

ax2

a2 + x2

=1a, the ordinate of the point of inflexion sought.

EXAM. 2. To find the point of inflexion in a curve defined by the equation aya√ ax + x2.

EXAM. 3. To find the point of inflexion in a curve defined by the equation ay2 a2x + x3.

EXAM. 4.

Fas

ૐao

FLUXIONS.

EXAM. 4. To find the point of inflexion in the Conchoid of Nicomedes, which is gene, rated or constructed in this manner: From

a fixed point P, which is called the pole of FG

"

the conchoid, draw any number of right lines

PA, PB, PC, PE, &c. cutting the given line FD
in the points F, G, H, I, &c. : then make the
distances, FA, GB, HC, IE, &c. equal to each other, and equal
to a given line; then the curve line ABCE &c. will be the cons
choid; a curve so called by its inventor Nicomedes.

TO FIND THE RADIUS OF CURVATURE OF CURVES,

108. THE Curvature of a Circle is constant, or the same
in every point of it, and its radius is the radius of curvature.
But the case is different in other curves, every one of which
has its curvature continually varying, either increasing or
decreasing, and every point having a degree of curvature pe-
culiar to itself; and the radius of a circle which has the same
curvature with the curve at any given point, is the radius of
curvature at that point; which radius it is the business of this
chapter to find.

109. Let Are be any curve, con-
cave towards its axis AD; draw an or-
dinate De to the point E, where the
curvature is to be found: and sup-
pose Ec perpendicular to the curve,
and equal to the radius of curvature
sought, or equal to the radius of a
circle having the same curvature
there, and with that radius describe
the said equally curved circle BEе;
lastly, draw Ed parallel to AD, and de
parallel and indefinitely near to DE: thereby making Ed the
fluxion or increment of the absciss ad, also de the fluxion of
the ordinate DE, and Ee that of the curve AE.
Then put x-
AD, Y➡DE, Z—AE, and r≈CE the radius of curvature; then is
Ed=x, de=y, and Eez,

Ꮐ

Now, by sim. triangles, the three lines Ed, de, Ee,

B

ture at the point E, their abscisses
VOL. II.

48

365

or x, y, z,

GE, GC, CE
GC x =GE

x + Gc. x =GE. y + GEÿ

y;

❤

BG. v=GE ÿ +GE. y.

are respectively as the three
therefore

and the flux. of this eq. is GC
or, because GC —— BG, it is GC

..

x - x

But since the two curves AE and BE have the same curva

GE ·y;