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. 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.. α xx x x2 x a α2 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 x4 a 2a2 Sa3 4α4 a+x· 2x 2x3 2x5 + + &c. X α 3a35a5 —0—log. . X x2 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, 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 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 56737 4903 446 42 4 TO FIND THE POINTS OF INFLEXION, OR OF CON- A 105. THE Point of ] 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- 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; 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 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 yê xy ) ; X X x2 Y a2 ay 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 TO FIND THE RADIUS OF CURVATURE OF CURVES, 108. THE Curvature of a Circle is constant, or the same 109. Let Are be any curve, con- Ꮐ Now, by sim. triangles, the three lines Ed, de, Ee, B ture at the point E, their abscisses 48 365 or x, y, z, GE, GC, CE x + Gc. x =GE. y + GEÿ y; ❤ BG. v=GE ÿ +GE. y. are respectively as the three and the flux. of this eq. is GC .. x - x But since the two curves AE and BE have the same curva GE ·y; |