In order to determine this sort of points, we may conceive the tangent at M as being at once a tangent to the two parts MA, MÕ; and upon this supposition we may imagine on both sides of the point À two elements Mm, Mm' in one right line ; from which it follows that the radius of evolution or curvature at the point M must then be infinite. But as these elements M may also be supposed to become gradually less and less, so as both ultimately to vanish, the radius of curvature must then be reduced to zero. 41. Therefore at the point of in P flerion the radius of curvature is always either infinite or nothing. . my Hence we must find the second fluxion of the equation of the curve, supposing & constant, and we shall have a finite value of - which must be put equal to either zero or infinity. By means of this equation and that of the curve, we shall be able to determine those values of x and y, which belong to the point or points of inflexion if there are several of them. 42. When the ordinates proceed from a fixed point, we shall have z? + y* — y © -=o oro Ex. I. Let there be given the first cubical parabola whose equation is y'=a 'x, we shall have y=rs a), y=8ixas, and y? aš3 32797 ma inflexion; therefore r=0. And consequently the point of inflexion is at the origin or vertex. Ex. II. Let the curve be the conchoid of Nicomedes, of which the equation is y=V (aa—xx). First, taking the fuxion we b+—, and again supposing instant, wo have y= -; have_y a+ 3a 'b x?_2a 46 30 at the point of inflexion. (az_*)(aa-xx) Consequently x +36 x'—2a +6=0, an equation which being resolved will give for x the value that belongs to the point of inflexion. Examples. Ex. 1. Required the point of inflexion in the curve of which the ara equation is y= a' + % Ex. 2. Find the point of inflexion in the curve whose equation is ay=av (ax +.r')? Ex. 3. Required the point of contrary flexure in the curve defined by the equation ay*=a*x +23? 43. If the ordinate MP of any curve BM mm is greater or less than K those ordinates which precede it, as pm, and than those which follow it, as p'm', it is B then called the maximum or minimum ordinate, that is, the PPP PPP greatest or least of the ordinates. The method which teaches us to determine these sorts of quantities is called the method of maxima and minima. 44. If CM is the radius of the osculating circle, or circle of equal curvature, it is evident that the ordinate MP must be greater or less than any other ordinate corresponding to any point of the arc KMD, described with the radius CM; consequently the ordinate MP (prolonged in the case of a minimum) passes through the centre of the osculating circle: therefore the tangent at M is parallel to the axis AP, and consequently the subtangent=.. Y. Henee y =0. y Now y may be considered as any function of the abscissa AP (*); therefore to determine in what cases a quantity y depending upon : may become a maximum or minimum, we must find the fluxion of the equation which expresses the relation that they bear to each other, and make the quantity Yequal to zero. The resulting equation combined with the original one, will give the values of and y, in which y is a maximum or minimum. 45. But to distinguish which case takes place, we must observe that at a maximum point the radius of curvature is positive, and that it is negative at a minimum point. Now the expression for the oscu (1+y) 1 lating radius is ; and as y Therefore if y is a maximum,y must be negative, and if it is a mi which determines this sort of ordinates. A Here MP may be considered both as a maximum and a minimum with respect to the two branches MB, MB. But this is a particular case included in the one of which we have spoken, and of which we shall proceed to give a few examples. 47. Ex. 1. Let it be required to divide a right line a into two parts such that their rectangle may be either a maximum or a minimum. Call one of the parts x, the other will be at, and we shall have ates for the expression of the maximum or minimum. Let then y=arər, and we shall have y=-2x=0, whence x=ļa. Now in order to determine whether this solution gives a maximum or minimum, take the fluxion of the equation=-2x, and we have =-2, a negative quantity; from which it follows that the value -= gives a maximum y=ta'. =m, 2 ab a Ex. 2. To find which pair of conjugate diameters of an ellipse make the least angle with each other. Let m, n, be these diameters, P the angle which they form with each other. Then by (Art. 59, Conic Sec.) we have mn sin P=ab, and ab flux sin P m+n*—a+b'. Therefore sin P= and n (a+b-n) -ab (a* +6_-2 n) =0; consequently — ab (a* +62—2 n')=0. na (a* +6–12)} Now since ab cannot be equal to zero, we must have the factor (a*+6). a' +6_2 n=0, whence n = 2 Hence the equal conjugate diameters of an ellipse are those which by their intersection form the least angle as required. The sine of that angle is. a' +60 6 2 tan U Let 2 tan U -tan U, we shall have sin P 1+tan C seca U 2 sin U cos U=sin 2 U; therefore the angle P is equal to that formed by the two lines drawn from the two extremities of the conjugate axis to either extremity of the transverse axis. Ex. 3. Of all triangles constructed on the M V same base AB, and having the same perimeter, which has the greatest surface. Let the semi-perimeter=9, the base AB=a, the side AM = x, MB will = 29 Therefore calling y the surface, we shall have y=N9 A and therefore 2 log y=log 9+log (9-a) +log (9-x)+log (a +r-9); consequently + : :) ato-9 9 Hence since y cannot=0, we must have the second factor 2 equal to zero, and therefore a +1-939-*, or 2 q-a-=MB=t, and consequently the triangle required is isosceles. From this it follows that among all isoperimetrical triangles, or triangles having the same perimeter, the one which has the greatest surface is equilateral. For if AMB is the triangle required, it is evident that it must have a greater surface than any other isoperimetrical triangle AMB, constructed on the same base AB; therefore AM -- MB. In the same manner it may be proved that AM-AB. 49. Hitherto we have only considered the maximum or minimum of the function of a single variable r. To find in what cases any function Y of two variables x and y becomes a maximum or a minimum, we may employ the following method. 2 y. 2 1 2 =0 Let us suppose that y has already the value adapted to render the function Y a maximum or minimum ; we shall then only have to find the proper value of x, that is to say we must take the fuxion of the function Y supposing z only to vary, and equate the co-efficient of i to zero Pursuing a similar mode of reasoning, we shall find that to obtain y we must take the fluxion of the function Y making only y to vary, and then equate the co-efficient of y to zero. Hence it follows that if ï is represented generally by Pi+Qy, we must have P=0, and Q=0, two equations which will give the values of I and y proper to render the function Y a marimum or a minimum. It is easy to see that this same reasoning applies, whatever be the number of variable quantities of which Y may represent a function. Hence in general to determine those values of the variable quantilies which will render the function Y a maximum or minimum, we must take the entire flucion of Y, and equate to zero the co-efficient of the fluxion of each variable, which will give as many equations as there are unknown quantities. Ex. I. Let it be required to divide a given number a in three parts, whose product may be a maximum. Calling x and y two of these parts, the third will be expressed by t-y, and we shall have for the product xy (amr-y). The fluxion of this expression is =(a–2—y) yz + (a—2y—5) zý. Equating separately to zero the co-efficient of ., and that of y, we shall have a—2x - Y=o=a_2y—. Hence y=x=fa. Conse. quently the given number must be divided into three equal parts. Ex. 2. Let it now be required to determine among all isoperime. trical triangles, that which has the greatest surface. We have before Let x, y, be two of its sides, 29 the perimeter, then 2q---- y will be the other side, and the surface'will be v {9(9~r) (9-y) (x+y-9)}. This must be a maximum ; if we call it Y, we shall have 2 log Ylog o=log (7–) + log (9-y) + log (x+y-9). Therefore Y= Y: 1 Yy 1 -) + equating to zero 2 , ' *+-99--* 2 *+y-99 the co-efficient of 2 and also that of y, we have x+y=9=9-y=q-«; hence ==y=29=29–8–y. Consequently the triangle is equi. 3 lateral, as we before found. Examples for Practice. 1. Of all the squares which may be inscribed in a given square, which is the least? 2. Of all fractions, which is the one that exceeds its mth power by the greatest possible quantity ? |