Also Op=OR- Rp, since we go in the positive direction from O to R, and in the negative from R to p;

so that the polar equation of the curve, O being the pole and AD the initial line, will be

(r±b) cos 0 = a.

PROBLEM 156. To draw the Conchoid, the constants a and b being given (Fig. 158.)

= b.

Draw the line OD, and make OA on it = a, and AD, Ad each

Through A draw LAM perpendicular to OA; LM will be an asymptote of the curve. Draw any line OP through O meeting LM in R, and on it make RP = Rp = b.

By definition P and p will be points on the curve, and similarly any additional number of points may be determined. The curve is evidently symmetrical about OD.

If b is less than a, the form of the curve is that shewn by the dotted lines.

When ba the point O is a cusp on the curve.

To draw the normal at any point Q.

Let OQ meet LM in r; draw rG perpendicular to LM and OG perpendicular to OQ intersecting in G, which will be a point on the required normal; for the line OQ is moving so that it always passes through O while a fixed point on it is travelling along

LM; i.e. at the moment the line is moving along OQ (or turning about some point on OG), and also along LM (or turning about some point in rG), i.e. G is the centre of instantaneous rotation.

The Witch of Agnesi.

Let AB (fig. 159) be a diameter of a circle, NM a line perpendicular to AB meeting it in N and the circle in M. If P be taken on NM produced so that

the locus of the point P is the curve called the Witch.

If a be the radius of the circle we have from the above

which is the equation to the curve referred to rectangular axes with A as origin and AB axis of x.

PROBLEM 157. To describe the Witch of Agnesi corresponding to a circle of given diameter (Fig. 159).

Let AB be the given diameter, C′ its centre; draw the tangent at B, and through A draw any number of lines AE, AF,...&c., cutting the circle in E, F, &c., and the tangent at B in e, f,... &c. Lines drawn through E and e respectively parallel and perpendicular to the tangent will intersect in Q, a point on the curve; similarly lines through F and fintersect in R, and so any number of points can be determined.

The construction is obvious from the definition of the curve.

The curve is symmetrical about AB and cuts the diameter perpendicular to AB at distances from the centre equal to the diameter; the tangents at these points pass through B.

If CB be bisected in D and DK be drawn perpendicular to AB meeting the curve in K, K is a point of inflection on the curve. The tangent to the circle at A is an asymptote to the curve.

To draw the tangent at any point T.

Through T draw tTv parallel to AB meeting the tangent at B in t and the asymptote in v. Draw Aw perpendicular to At meet

ing the ordinate through C, the centre of the circle in w. The

tangent at T is parallel to vw.

THE CATENARY,

The curve in which a heavy inextensible string, freely suspended from two points, hangs under the action of gravity, is called the Catenary. If the mass of a unit length of the string is everywhere constant, i.e. if the string is of uniform density and thickness, the curve in which the string hangs is called the Common Catenary.

Investigation of the conditions of the statical equilibrium of the string gives for the curve of the common catenary the wellknown equation

the axis of y being a vertical line through the lowest point of the curve, and the axis of x a horizontal line in the plane of the string at a distance c below the lowest point. c is the length of string, the weight of which measures the tension at the lowest point, and e is the base of Napierian logarithms.

At a distance c from the origin measured along the axis of x, the corresponding value of y is

and so on;

and if we make c the unit of length the corresponding

C

The third column of the following table gives the value of at the corresponding points along the axis of x as shewn by the first column

e = 2·718281828...

log, 10 = 43429448...

PROBLEM 158. To draw the common catenary, the unit c being

given.

Example 1. (c=OA) fig. 160.

Draw the horizontal line Ox and the vertical line Oy. On Oy measure A C. A will be the lowest point of the curve. Set off from O along Ox lengths Oa = ab = bd = c, and draw the ordinates through a, b, d... parallel to Oy.

1

On the ordinate through a measure from a a length ap1 = (the number in third column of above table opposite x = c) x c, i.e. 1·54308 × c (e.g. if c is "it is only necessary to measure off on a diagonal scale of half inches a length 1·54). p, will be a point on the curve. Similarly on the ordinate through b measure bp,= (number in column 3 opposite x = 2c) x c, i.e. 3.76217 × c. P2 will × be a point on the curve. Similarly for ordinate through d.

P2

2

Points can of course be found between A and p1 by using the fractions of c given in the table.

Example 2. (c=0A) fig. 161.

The points P1, P2, P3, P1 on the ordinates through a, b, d, e,

where

4

are given by the table: the next point furnished by the table would be on the ordinate through ƒ, where ef = Oe. Points on

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