From the nature of evolutes AB+BC=MC. Therefore BC= -p. But MN=√(px+3pp) =√ (} p2+{PP)·

MC-p:

MN3

9

PP

{(1+25)3 −1}, which is

4 a

an expression for the length of any arc of the second cubical para

bola of which the equation is z1—a u3.

36. Let the curve AMBa be a cycloid, and BODO' its generating circle, let also MOP be an ordinate perpendicular to BD. If we make BP=x, PM =y, BD= = 2a, we shall have y= BO+√(2 ax-xx); now the fluxion of the arc BO is flux sin BO

a

A

P

M

I

N

B

M

a

Therefore y

(2 a—x)

√(2 ax-rx)

cos BO

√(2ax-xx)

=3‚√(2 a—3) = ± √ (2 ax—xx), the fluxional equation of the

=*

cycloid.

This premised, to find MC, the radius of curvature, suppose constant, and taking the fluxion we shall have ÿ

22a (2a-x)-2 OD: Now MNC is parallel to OD, since (art 72 page 431) the tangent MT is parallel to OB; consequently OD-MN-NC.

Hence it follows 1st, that at the point A the radius of the evolute is zero, and consequently that the evolute passes through this point. 2ly. That the radius of the evolute at the point B is the line BE, double the line BD.

37. To determine the evolute ACE, complete the rectangle AE, and on the side AB'-DE-BD, as a diameter, describe a semicircle A'Q'B', draw AQ parallel to CM, and join C and Q'; this done, the angle NAQ NDO. Therefore OD-AQ, and the arc OID, or the right line AN-are ALQ. Now OD-CN, therefore CN-AQ, and consequently CQ=AN the arc ALQ'; & property

peculiar to the common cycloid. Hence the evolute ACE is a semicycloid equal to AMB, the one proposed, and differing from it only in position. We should have arrived at the same result, if we had directly investigated the equation of the evolute, by what was shewn (32).

The arc AC-MC-2 AQ'; therefore any arc of the cycloid is double the corresponding chord of the generating circle. Thus MB= 2 OB, AMB-2 BD, and the whole cycloid A Ba is quadruple the diameter BD.

Y (x2+y3) î

x (x2 + y2) — y x y

B

becomes √(x2 + y2). Therefore if we

draw AC perpendicular to MA, and MC perpendicular to the tangent at M,their point of concourse C will be in the evolute; for the similar triangles Mrm and MAC give

Mm Mr:: MC: MA,

39. The angle ACM-90°-AMC-AMT; from which it follows that the evolute ABC is the same logarithmic spiral ADM; it is merely disposed in a different manner. Hence the tangent MC is equal in length to the spiral ABC, though this last makes an infinite number of revolutions about the point A. also, if we draw AT perpendicular to AM, we shall have MT= arc ADM. Consequently the logarithmic spiral and the cycloid are their own evolutes.

Hence

OF POINTS OF INFLEXION IN CURVES, AND OF THE MAXIMA AND MINIMA OF FUNCTIONS OF ONE OR MORE VARIABLE QUANTITIES.

40. If a curve from being convex, becomes concave, the point M at which this change occurs, is called a point of inflexion.

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, MO; and upon this supposition we may imagine on both sides of the point M 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 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.

M

A

P

41. Therefore at the point of inflexion the radius of curvature is always either infinite or nothing.

Consequently supposing & constant, we shall always have (x2+y2)*

Hence we must find the second fluxion of the equation of the curve, supposing a constant, and we shall have a finite value of x

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

x2 + y2 — y Y—o or ∞

Ex. I. Let there be given the first cubical parabola whose equation is y3—a 2x, we shall have y=x a3, y=} x−1 x × a§, and

inflexion; therefore x-o.

is at the origin or vertex.

x2

& x 3/ aao, at the point of And consequently the point of inflexion

Ex. II. Let the curve be the conchoid of Nicomedes, of which the

√(aa-xx). First, taking the fuxion we

-x (aab+x3)

and again supposing a stant, we

have-y

a2 x3+3a b x2-2a b

o at the point of inflexion.

(α2 x3 —x3) √ (aa—xx)

Consequently +3b x2—2a 1b=o, 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

Ex. 2. Find the point of inflexion in the curve whose equation is ay=a√(ax+x2) ?

Ex. 3. Required the point of contrary flexure in the curve defined by the equation aya2x+x3?

thod 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. Henee0.

y

Now y may be considered as any function of the abscissa AP (*); therefore to determine in what cases a quantity y depending upon r 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 equal to zero. The resulting equa

tion combined with the original one, will give the values of x 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

which determines this sort of ordinates. 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 a-r, and we shall have ax-xx for the expression of the maximum or minimum. Let then y=ar—rr, and we shall have a—2x=o, whence x=a. Now in order to determine whether this solution gives a maximum or minimum, take the fluxion of the equation-a-2x, and we have

1=−2, a negative quantity; from which it follows that the value ==/ gives a maximum y=‡ a".