SOLUTIONS.

THE THEORY OF CURVES.

THE NATURE AND CONSTRUCTION OF CURVES.

1.

To describe the curve whose equation is x1 a2 x3

+ a2 y2 = 0, we have, by differentiating

(2x3 — 0....
a2x) dx + a3ydy = 0 .... (a)

And (6x2 — a2)dx2 + a2dy2 + a2yd'y ≈ 0....(b)

First let us determine the limits or maxima and minima of y;

whence by substituting in the given equation we get, y= 0 for a minimum value at the origin A, (See Fig. 1).

and y = ±

α
2

indicates two pairs of maximum values PM,

PM'; pm, pm' corresponding to the two values of x.
To find the limits of r we have

And substituting in the given equation there will result x=a, which indicates the points B, b.

Again, to find the multiple points (if any) we have

dy ax – 2x3 = 0, and it being found, by equation (6),

dx

=

a2y = 0

',

that x = 0, y = 0, are the only values deduced from the numerator and denominator, which satisfy the given equation, we have a double point at A.

Also, since when x and y each = 0,

=±1=± tan. the tangents at the

4

point A are inclined to Bb at an angle of 45 degrees. Again, to find the points of inflexion, we have

.. we have a point of inflexion at the origin A.

Since there are no infinite branches, the curve cannot have an asymptote.

This curve which has been called the lemniscata, is the locus of the extremity of that part of the ordinate of a circle whose radius is (a), which is constantly taken a sin. 6. cos. 0, 0 being the angle at = the centre subtended by that ordinate.

And AM'B and Am'b are equal to either of these.

.. the whole area A M B M' Ambm'A =

4a2
3

N. B. Unless the branches of an oval intersect, it is not perfectly 'quadrable. See Waring, Vince, &c.

(2) Let E F (Fig. 3) = y, AE = x

AD, the diameter of the generating

AHD, = a. Then AH

being joined, by the property of the cycloid, we have

y=2AH=2/AE. AD, by similar A,

= 2 √ ax.

y4ax, the equation to the curve, which is therefore a parabola, whose vertex is A, and focus D.

Now the area of a cycloid is known to be triple that of its gene

2

rating O, and the area of a parabola of its circumscribing

rectangle; hence

3

AFGD: ABCD ::

AD

AD x DG: 3A HD

3

But if the circumference of a circle whose diameter is 1,

(3) Let CB or CD or CR (Fig. 4) = r, Cb = a, PM =y, CP = x. Then since M bisects Rr in M, we have

y = PM =

RN+rn RN

a

=

++ ==

But RN: y::r: CM (= √ x2+y3)

.. RN

Hence y =

√ r2 — a2 — 4y2+4ay which is the equation to the

curve referred to rectangular co-ordinates.

To find the equation referred to polar co-ordinates, we have MCP = 0 and CM = e)

(putting the

if Cb be bisected in v, and vt be drawn parallel to AD, it will be an asymptote to the branch xM.

Let y = a

Then x = r2 — a2 = Fb, or the curve passes through F, the intersection of br with the circle.

Similar properties belong to each of the branches generated by the other quadrants.

Let the origin of abscissæ be at A, (Fig. 5), x being then = 0, and y = a = AC.

Again, let x = ± a = AB or Ab.

Then yco BN or BN, which are.. asymptotes to the branches CR, Cr respectively.

When is > a, the values of y are negative, and the corresponding branches are Ss, Tt; which, since when x is infinite,