To find the point of contrary flexure, we have

17. To find that curve whose normal has the same relation to the distance between the origin of abscissæ, and the intersection of the normal with the axis, that the ordinate of a parabola has to its corresponding abscissa, we have

y'px' in the parabola, and

. (Normal) p. (abscissa + subnormal) in the curve required.

:. (y. 1 + dy 2

dx2

2

Now if r be the radius of a O, and the origin of abscissæ be distant from its circumference by the interval a, we have

y1 = (2r+a) a + 2. (x + a) x, — x2.

The curve is therefore a circle.

THE NATURE AND

To trace the curve whose equation is (x2 + y3)*

Then y = 0, or the curve passes through the origin of abscissæ A, (see Fig. 1.)

Let x1 = AB or Ab,

Then y =0, or the curve passes through B and b, and y is imaginary for every greater value of x.

Also since the pairs of values of y, (±y) are the same for the positive as for the equal negative values of x, the curve will consist of two equal and similar ovals AMBA, AmbA, intersecting, (the curve is therefore quadrable) in A.

dy
Again = √√2.
dx

2x

x 8x2+1

√8x2+1×(√8x2+1 − 2x2 − 1)§

tan 0 (0 being the inclination of the tangent at any point to the line of abscissæ.)

Hence when the curve is parallel to the axis, tan. 0 = tan. 00, or

and it will be found, by the usual methods, that this point is double; that there is at A for each branch a point of single inflexion, and that the curve there cuts the axis at an angle of 45o.

To find at what angle the curve cuts the axis when x = a maximum, or when x1, we have

= those of the elastic curve, whose equation is y =

the abscissa x corresponding to p.

This curve is called the Lemniscata of James Bernoulli.

It is the locus of the intersection of the perpendicular from the centre upon the tangent of an equilateral hyperbola, whose semiaxes are unity.

For in the equilateral hyperbola

20. The subcontrary section of an oblique cone is a circle. Let a m b (Fig. 14,) be the section made by a plane passing through the cone parallel to the circular base AB, AVB being the triangular section of the cone made by a plane passing through the vertex and centre of the base. Also let A'm B' be the subcontrary section made by a plane plane AVB, and inclined to BV by the angle A'B'VA, and intersecting amb in mp.

Join Vm, Vp and produce them to M and P respectively; join PM.

Then since the planes amb, A'm B', are plane AVB, their intersection mp is ab. Also because PM is parallel to pm, (being the intersections of a plane with two parallel planes,) it is plane AVB, and .. 1 AB.

Again, from similar triangles,

AP: ap: VP: vp :: MP: mp

and BP bp :: VP: vp :: MP: mp

:

.. AP. BP ap. bp :: MP2 : mp2
But AP. BP = MP2 (by prop." of O)

Also since

:. ap. bp = mp3

pB'b = 2 A = a, and the vertical are

equal, the A A'ap, B'bp are similar,

. A'p ap: pb : B'p

and the subcontrary section A'm B' is a circle.

21. To find the curve by whose revolution round its line

of abscissæ, a solid is generated =

linder.

of its circumscribing cy

Let x and y be its co-ordinates, x beginning at the vertex.

..y3 cx is the equation to the curve, which is, therefore, the cubic parabola.

22.

To construct the spiral whose arc is the measure of the ratio between the ordinates which intercept it.

If CAP = 0, AP = and CP = x,

AC being

1 (Fig. 15)

But dz = √dp2 + p2 dė3, as we readily learn from the figure.

which being analogous to that of the involute of a circle whose radius is unit, we will attempt the construction by the aid of that

curve.