curvature, we must determine the minimum or least value of the rac dius of curvature.

If the tangent at A is perpendicular to the axis, then to determine the right line BA, or the distance from the vertex A to the origin of the evolute, we must make s=o in the expression of the radius MC, and we shall have the value of BA. Lastly to find the equation of the evolute, draw CQ perpendicular to the axis, and call AB, a; BQ, 1; CQ, z; in the first place supposing i constant, we shall have MO=+*+yo, and = **+ýx

- y.

And then

- y

mr : YM :: MO : CÓ that is,

:: 4 :: *+9:CO=PQ=j (*+y). - y

- y Therefore AP+PQ-AB=i=r-at j (*++y"); values which together with the equation of the curve, enable us to find the equation of the evolute.

33. Hitherto we have supposed the ordinates to be parallel to one another. If they proceed from a fixed point or pole B, we may determine the radius MC in the following manner.

Conceive two ordinates BM, Bm indefinitely near each other, and CO, Co two perpendiculars to these ordinates ; then from B as a centre describe the arc Mr. This done, let BM=y, Mr=i, mr=j, Mm=; =V (**+y), MO=u. Because

B of the similar triangles Mrm, CMO, we have Mr : MÕ:: mr : CO :: Mm : MC

uy S: MC

c =

Taking the fluxion of this last equation (on the supposition that that is constant,) we have i= and the fluxion of CO


which is conco=-0Q= +uj.


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when y=.00,




y (z? +33) sø x-Y I Y

33 j—yz . This expression becomes merely


x y the ordinates are parallel, as we have already found. We shall now proceed to give a few examples.

The equation to the ellipse and hyperbela, when we place the origin of the abscissæ at the vertex, is expressed generally by yy= + prz


and it is evident that when a=00, we have yy=px; an 2a equation of the parabola, which consequently is merely an ellipse or hyperbola whose transverse axis is infinite. Hence the equation uy = px + per is general for all the conic sections. It will therefore serve to find their radius of curvature.

(y mal (26), if we call it n, the radius of curvature, supposing ¿ con. stant, will be expressed by ; and since in this example yy=

--y’y pe+ Pit, we have 2yj=pr+ paz; and again taking the fuxion of

2a this last equation we have aj ž+2y+=+?


and therefore y + x2 - y•j=> {=(p= +

n3 Hence

or the radius of curvature= -y' y

that is, in all the

App: conic sections the radius of curvature is equal to the cube of the nors mal divided by one-fourth of the parameler. Hence in the circle where n =

#p, the radius of curvature is always equal to the nora

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$ PP

p=a, BC

From the nature of evolutes AB + BC=MC. Therefore BC=


I p. But MN=V (pr+* pp) =V (1 pz +1 pp).

27 Therefore calling

a {(1+-14, which is

16 an expression for the length of any arc of the second cubical parabola of which the equation is z'=au.

36. Let the curve AMBа be a cycloid,

B and BODO' its generating circle, let also MOP be an ordinate perpendicular to BD. If we make BP=x,

A PM =y, BD= 2a,

la we shall have y= BO+v (2 ax-xx); now the fluxion of the

9 2





arc BOisa flur sin BO

cos BO

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Therefore y=(2


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(2 ar--r.)

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2 axor) cycloid. This premised, to find MC, the radius of curvature, suppose x

a 22 constant, and taking the fluxion we shall have y =

XV (2 ax-xxx
Consequently the radius MC--(+y)}

y 2 V2a(2a-1)=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 OĎ=AQ', and the arc OID, or the right line AN=arc ALQ. Now OD=CN, therefore CN=AQ, and consequently CQ=AN=the arc ALQ'; & property

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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.

38. Let the curve bethelogarithmic spiral ADM whose centre is A; we shall have cot MmA=


and Mr

B в taking the fluxion (supposing x constant) we shall have y =0, and the radius of evolution MC

becomes ? v (** + y). Therefore if we (ir+ y) - yr y 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 simi lar triangles Mrm and MAC give

Mm: Mr::MC: MA, that is or (z+ya) : ::MC: y Therefore MC=\ v (+ jo) as before.

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.

Hence also, if we draw AT perpendicular to AM, we shall have MTarc ADM. Consequently the logarithmic spiral and the cycloid are their own evolutes.

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

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