3. Required that number x of which the xth root is a maximum? 4. I am desirous of constructing a cylindrical measure of a given capacity, and of which the internal surface may be a minimum. What must be the ratio between the height of the measure and the diameter of its base ? 5. Among all the cylinders which may be inscribed in the same sphere, which is that whose convex surface is a marimum ? 6. Which among these cylinders has the greatest solid content ? 7. What must be the dimensions of the greatest cylinder that can be inscribed in a given cone ? 8. Of all triangles standing on the same base, and inscribed in the same circle, which is the greatest ? 9. Which, on the contrary, will be the least of all the triangles that may be circumscribed about the same circle? OF VANISHING FRACTIONS. 50. We sometimes meet with algebraic expressions in the form of fractions, which on certain suppositions become equal to . Such 22-a? for example, is the quantity when x=a. Now though in appearance indeterminate, these results are nevertheless susceptible of determinate values ; and the following is a method for ascertaining them. P Let be a fraction whose numerator and denominator are func Q tions of a, both of which become zero when x=a. To find the value of this fraction, substitute r+r in place of x in P and Q, and we shall have. ; then making x=a in this latter fraction, Q+Q P it will become. ; and this will be the value of the proposed frac Q tion, on the supposition of x=a+i, or of r=a, provided however P+P that the terms of the fraction do not again destroy one another Q upon making r=a. Ex. 1. Required the value of x?_a* when r=a? Here Per 2x x -a', and Q=r-a; therefore 2r=2a. Q Ex. 2. Let there be the geometrical progression ?, ??, ??,...." of 20+1_ which the sum is: ; required the value of this sum when x=1 Taking the fluxions as directed, we shall find=(n+1)x"—1= Q n, as is evident. Ex. 3. Let there be proposed the quantity V (2a os-x4)—apa's 4-V å x3 which becomes when r=a. Taking the fluxions and proceeding a 16 Va's as before, we shall have „(2a *r—24) 3.3 3 4 a ga, the value of the proposed quantity. P 51. But if it happens that on substituting a instead of x in. this ... 2 fraction also becomes; we must treat it in the same manner as the first, and so on, till we arrive at a value of which one term at least is finite. Ex. If we take the fluxion of the identical equation x+8 toy n+1 after dividing both sides by, we shall have it 1-3 27 which becomes (1-3) nesc n+1 R 1-2" (x+1)+n(n+2)x when r=l. Therefore -; but this Q -2 (1-1) new expression gives also... on substituting 1 for x; we must therefore take the fluxions of its numerator and denominator separately, and we shall have n.2%-1 (n+1)*+ n(n+1) (n+2) x" which 2 on making x=1, gives. n (n+1) for the sum of the arithmetical 2 progression 1, 2, 3,.... Ex. 2. In the quadratrix y="** tan Ce; and this expression ber comes Co., when «=a. Therefore y= a flux cot C2 when x=1 Examples for Practice. ar? + ac --2 acx 1. Required the value of the fraction bx*_2 bcx + bc 2. What is the value of the fraction a'b* when I=0. Note. By these principles we may in each naels indeterminate values of 0 x 0, and of We also reduce to t a since co = OF THE INVERSE METHOD OF FLUXIONS. In what has hitherto been delivered of the theory of Fluxions, we have supposed the relation of the variable quantities or fluents to be known, and have investigated that of their Auxions. But in the inverse method of fluxions, we propose to ascertain the relation of the fluents from that of their fluxions. 52. To indicate a fluent, or, as it is also frequently termed, an integral, we shall employ the leters; for example, fax, is the general expression of all the quantities which, when thrown into fluxions, produce ar ; and as ax may equally by derived from a simply, or or from ax +a constant quantity, to each integral we add a constant term C, which must be determined afterwards by the conditions of the problem. The quantity ax being in a measure the sum of all its elements ai, we call the expression / ax, the sum of ax; and to integrale, to sum, or to find the fluent, are synonimous expressions. If every fluxional expression were derived immediately from some algebraic function, every fluxion would have its integral or fluent ; but as we call any quantity affected with x, y, &c. a fluxion, there are many which are not susceptible of integration, because they cannot have been produced by taking the fluxion of any quantity: yx, for example, is of this number. Many others we have not hitherto been able to integrate but by approximation. Such are the fluxions of logarithms, of circular arcs, and in general of all those quantities which are called transcendental. We shall first treat of those which admit of exact, or algebraic integrals. OF QUANTITIES SUSCEPTIBLE OF AN EXACT INTEGRATION 53. Since the fluxion of this nx-1, it is evident that the fuent of n zu-1 must reciprocally be ?"; therefore Szmiandmaking m+1 m+1 1-1-m we shall have sem +C; a formula which m+1 m+1 or gives for the integration of a fluxional expression of one rerm only, a rule the reverse of that given for finding the fuxion of any algebraic quantity consisting of a single term. 54. Therefore to integrate a flucional expression of one term, increase by unity the exponent of the variable quantity, and then divide by the exponent thus increased, and by the fluxion of the variable. 55. This rule, however, is subject to an exception in the case of m=-1; for then the fluent becomes ++C, that is, it assumes an infinite form. But the fluxion zmi is in this case reduced to 5, which expression we have shewn in article 16 to be the fluxion of the hyperbolic logarithm of r, hence its integral is lx.Therefore /*; lx +C, and consequently all fluxional expressions consisting of one term, and containing only one variable quantity, may either be integrated exactly, or at least approximately, by means of logarithms. Several other fluxions may be integrated in the same manner. 56. Suppose y=r(a+bx+cro+&c.)=ax+bxë+cx ** + &c. and we shall have y=C+ax++ bx? c.r3 +&c. 3 Let y=ač (6+x)"; if we make 6+x=2, we shall have <=2, and m+1 j=az"ż; consequently y= (6+2)+1+C. When m+1 m+1 2 az a ar bta mth it m=-1, we shall have y=57 -=a log (6+x)+C=log c (6+2) by making C=log c. 57. Let now į = axhië (6 + **)", we shall havey=C+; n (m+1) (6+2")m+1; and in general, if we have y=x" + (a+bam)", we shall find, on developing this expression, y=a* x** + kak-1 box k (k-1) ak–2 62 22m+ 3+ &c. of which the integral is y=C+ 2 af 21+1 k k (k-1) ak-62 zem+n+1 n+1 m+n+1 *2 (2m+n+1) k (k-1) (k-2) ats 63 2-3m+1+1 +&c. + 2. 3 (3m+n+1) This integral will always be finite, when k is a positive whole number. But we must observe, that if after having expanded the binomial, there occur terms of the form, we must integrate them by logarithms. The method is the same for të (a + b +cr+&c.)! |