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CURVATURE AND EVOLUTES.
To find the radius R of curvature to the common cycloid, whose equation is
y = sin." x + ✓ 2rx – x2
(1 + 24 = 1)
which reduces to
R = 2 x 2 a
x 2a – x.
118. Required the chord of curvature parallel to the axis, of the common parabola, whose equation is yo = px. dır?
dy? X 1+
119. By the preceding problem we learn that
1 (4.x + p).
2 p Again, in the expressions 2 – 0 + (y B) dy = 0, dx
(See Appendix to dy?
d.ru Simpson's Fluxions or Lacroix).
a, 8 are the co-ordinates of the centre of curvature, and .. of the evolute of a curve; and if we can eliminate from them y the co-ordinates of the curve, the result will be the equation to the evolute.
But since y? = px, by substituting for dy, dạy hence obtained, we easily get
4y y – B = 4x + piva
• y, and .. x
120. By the question, the equation to the curve is
= r sin. y dy
= q cos. y. dy?
121. bola is
By problem 118 the radius of curvature of the para
(4 x + p), and the equation to the parabola
:. at the vertex of the parabola, where x = 0,
123. To find the evolute of the common cycloid, whose equation is
y = vers. * * + N 2rx
x ✓ 2rx- x"
B + 2 y 2r3
za = vers."x + N 2rx :: B = vers." (a
4r) ✓ 2r (« - 4r) equation to the evolute, which is therefore a cycloid equal to the curve itself, but having its bake I base of the given cycloid at either extremity.
124. To find the points of contrary flexure of a curve, whose equation is
x = (ly)', we have 3ly
(2– ly) = 0 or 6, by the rule, which gives
ly= 2, or x= (ly) = 8, and y = e', which are the co-ordinates of the point required.
125. If R be the radius of the wheel, r the distance of the generating point from its centre, x the abscissa of the trochoid, measured from the vertex or highest point of it; then the equation to the curve is
x + 2rx -- x? (1)
dy R + g -
(2rx — 2)
Rx (R + r) r = 0
or 2rx X2 and we :. have two points of contrary flexure, whose abscissæ are
(R + r) Ř and 2r,
vers. * 2r reR spectively, the latter point being a ceratoid, as we learn by substituting in equation 2.