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CURVATURE AND EVOLUTES.

117.

To find the radius R of curvature to the common cycloid, whose equation is

y = sin." x + 2rx – x2
we have
d.x

dy?
Х 1+
- d’y

dir?
X N 2ax x

R =

(

(1 + 24 = 1)

a

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+

dx2

R =

-doy

(

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But

dy
dx

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R=

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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?

døy
1+ + (y -B)
dx

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

dx dx2

4y y – B = 4x + piva

р

y, and .. x

р P.

and X,

2

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2.1

2

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120. By the question, the equation to the curve is
X = r. vers. y = rar cos. Y,

dr

= r sin. y dy

dox and

= q cos. y. dy?

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121. bola is

By problem 118 the radius of curvature of the para

R =

2NP

(4 x + p), and the equation to the parabola

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:. at the vertex of the parabola, where x = 0,

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123. To find the evolute of the common cycloid, whose equation is

y = vers. * * + N 2rx
Here
dy

2r
dx
doy
dix

x ✓ 2rx- x"

.

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4r): the

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.

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CONTRARY FLEXURE.

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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

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dy?

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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

R

x + 2rx -- x? (1)

vers,

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dy R + g -
dir

2rx 22
d'y Rx-(R+r)r
...
d.x2

(2rx — 2)
which being put = 0 or, will give

Rx (R + r) r = 0

or 2rx X2 and we :. have two points of contrary flexure, whose abscissæ are

(R + r) Ř and 2r,
and ordinates
R

R
Ř VRʻ

vers. * 2r reR spectively, the latter point being a ceratoid, as we learn by substituting in equation 2.

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