two values of AT and AQ give the position of the asymptotes the same as we found in article (49) conic sections. In the logarithmic curve, we have := A log y, and = Ay. y Therefore Y=A. Consequently its subtangent is always equal to y the modulus (art 69, page 429) 27. Let there be any curve BOC with another curve BMA, such, that if we prolong the ordinates OP of the first till they meet with O. á the second curve, the line MO) may be some function of arc BO; 91 MI it is proposed to draw through the 101T P given point M the tangent MT. Conceive the ordinate mp indefinitely near to MP, and Mr parallel to the tangent at the point 0; if we make the arc BO = 2, MO – u, we shall have mr = u, „=2u P M=0o=ž, and w: 23:4: OT="Now u being a function of 3, we shall have by taking the fluxions of the equation which expresses the given relation; and consequently TO, or the point T will be determined, whence it will be easy to draw the tangent MT. 6 Suppose, for example, that u= and OT=.=arc BO. If BOC is a circular arc, then AMB, is a cycloid, and this construction is the same as we have already given. In the quadratrix, if we compute the abscisse from the centre, we have (art 76 page 432) y=%.cote; thereforey = cot y 2, we shall have is= b a CI сxx a a a a sin 2 CX, a cx cot ard But — ty = OT, as may be pro , a ved by the two similar triangles MOT, and the fluxional triangle formed by drawing an ordinate indefinitely near to MP. (We employ -y, because y diminishes, as x augments.) Therefore OT cot me; and adding to each side sin crt са a When CM=CT, or at the point D, we have, as before explained, СМ: the base CD="a, and consequently CT = We must there CD fore take CT a third proportional to the base CD and the radius CM; this will give the point T, through which and through the point M if we draw the line MT, it will be the tangent required. 28. To draw tangents to spirals, we must resolve the following problem. Let there be described a circle with any radius CA, and let there be a curve CKM such, that drawing the radius CMN, the line CM may be any function of the arc ABŇ, it is required to draw through the given point M the tangent MT. Conceive the two radii CMN, Cmn indefinitely near to each other, and the little arc Mr described from the centre C, and with the radius CM; and then draw CT perpendicular to CM. This done, let CM=y, ABN = x, CA=a; we shall then have CA: CM::Nn: Mr or Q : y :: * : Mr = yet ay For example, let y = then CKM will be the spiral of Archi or ar a aa aa a a ay medes, and we shall have and CT= y T__axy=XY. y MQO'. In the hyperbolic spiral whose equation is xy=ab, we shall have ay+y*=0, y:= 'xyy -xy, and CT = b; as we before found it (art 85 page 435). 29. In the logarithmic spiral in which the angle CMT is constant, conceive the radii CM, Cm indefinitely near together, and from C as a centre and with any radius CN, describe a circle; make CM=%, CN= B a; and marking on the cir D cumference of the circle a fixed K point A, suppose the abscissa AN=r, which will give this T proportion, CN: Nn::CM : Mr Q : :: 2 : Mrs. M7 :72 N or a therefore log %> = + a constant quantity C; because the Auxion of the equation log za is the same as that of log == +C. Now the equation log x==+C, shews 1°, that this spiral makes at an infinite number of revolutions about its centre, as well in apa proaching towards it, as in receding from it; for in place of a we may successively substitute o +*, x + 2, x + 37, &c. + 2, -2*+*, &c. m being the circumference ANB. 119. That if we make C=log C, we shall have log de loge, or =că, and ==C'e- ; therefore at the point A where =o, we have CD=C. IIIdly. That if the abscissæ are taken in arithmetical progression as *, 22, 3x, &c. the ordinates will form the geometrical progression C'cae Cea, ce a &c. IVthly. That if t=c, we have =c', a property of the circle which, as we already know, cuts all its radii at right angles. These examples will enable the student to draw tangents to all sorts of curves whether geometrical or mechanical. Examples for Practice. 1. Let it be required to draw a tangent to the ellipse whose b equation is ye a 2. Required the expression for the sultangent to the cissoid of opos which the equation is y= 3. Required the subtangent of the conchoid whose equation is : b+y waa—yy). y 4. Required the subtangent of the parabolic spiral of which the equation is y=a-v px, (See art. 82 page 484) OF INVOLUTE AND EVOLUTE CURVES M 30. Suppose a thread ABC to be lapped closely upon any curve BC whose origin is at B, and to which AB is a M tangent at that point; if we gradually unlap this thread, keeping it always equally stretched, its extremity Awill B describe a curve AM, which A P will have the following properties. Ist. The tangent MC of the curve BC will always be perpendiular tothe curve AM; Ildly. The length of MC will be equal to the line AB + the arc BC; IIIdly. The indefinitely little arc Mm may be considered as a circular arc described from the centre C with the radius CM; IVly. The point C will be the point of concidence, or re-union of the two normals MN, mn, which are supposed to be indefinitely close to each other. 31. The curve BC is called the evolute of the curve AM: and reciprocally the curve AM, the involute of the curve BC; the line MC is the radius of the evolute, it is also called the radius of curvature, or of osculation, &c. This premised, we shall proceed to determine for any point M the radius MC of the evolute BC which we suppose known, Let MP, mp, be two perpendiculars to the axis AQ, indefinitely near to each other, and CO, TM two parallels to the same axis; if we call MO=u, AP=0, PM=y, Mm, or v (+2+y)=s, we shall have Mr : Mm :: MO: MC that is ic s : u : MC=us But while AP, PM, and MO vary, MC becoming mc, does not change; therefore the fluxion of the equation MC=" being taken, =” we shall have (us+su):=us x and since i=mr=ý, we shall find and consequently that MC = sy that u = S X y For the sake of simplifying this expression, let us suppose one of these fuxions to be constant, the element å of the curve, for . example, and we shall have. MC = sy – į v(x+y) if we had supposed j constant, we should have found § 5 == : 83 si _(2x +y") } whence s = which gives MC= (X)* y 2 But if, as is usually done, we suppose x to be constant, then ja _(x++y) } MC= -X y 32. As the curvature of circles varies in the inverse ratio of their radii, it follows that in two different points of any curve, the degrees of curvature are inversely as the radii of the evolute. Therefore in order to find in what points the curve has the greatest y x нь |