but, when we proceed to the limit, by making dy less than any assignable magnitude, the two quantities 277, 27 (y + dy), become equal to each other; hence, replacing indefinitely small differences by differentials, we see that

ds

2ty, dS = 27yds. ds

CHAPTER X.

CURVES REFERRED TO POLAR COORDINATES.

Tangency. 152. In this chapter we shall investigate formulæ, in relation to curves referred to polar coordinates, analogous to those which in the preceding chapters have been established in regard to curves referred to rectilinear coordinates. We shall begin with the investigation of the formulæ of tangency.

Let P and Q be any two neighbouring points of a curve AB. Let S be the pole, and SX an indefinite straight line through S. Join SP, SQ, and let SP = r, LPSX = 0, SQ = r + dr, QSX = 0 + 80. Through P and Q draw the

L 0 indefinite straight line RR and let TT" be the tangent at P, which will be the ultimate position of the line RR when the point Q approaches indefinitely near to P. From S draw SY, ST, at right angles to PT, PS, respectively: ST is called the subtangent at the point P. Let SY = P, ST = 0, L SPT = 0, LSPR = $, arc AP = s, arc PQ = ds, chord PQ = c.

or

2

Then from the geometry it is plain that

c sin d' = (r + dr) sin 80, ο δε

sin ᏧᎾ sin $' = (r + dr)

..(1); os 80

80 and

c cos d' + p = (r + dr) cos 80,

80

sin? cos

2

SA or cos $ + 2r

cos 80 ........

(2). os 80

809 4

80

2 Now ultimately, when 80 becomes less than any assignable quantity, es ds

sin
1,
$' = , p + dr = r,

1,
as Ꮎ

80

or

2

tan •

(5).

From (3) and (4), we have also

rde

dr Adding together the squares of (3) and (4), we get

p%d0% dr2 1

dsa ds'

+

Differential of an Area. 153. Let P and Q be any two neighbouring points in a curve AB. Join SA, SP, SQ. With S as a centre describe

or

two circular arcs, PP', QQ', cutting SQ and SP produced, respectively in P and Q. Let A, A + SA, denote the areas ASP, ASQ, respectively. Then it is evident that SA is intermediate in magnitude between the two circular sectors SPP, SQQ', that is,

*r*80 L SAL] (r + dr) so,

SA
122 L}(r + dr)?.

80 But ultimately, when 80, and therefore dr, becomes less than any assignable magnitude, the two quantities į ma, 1 (r + dr),

δΑ

dA assume a ratio of equality: hence, replacing by

80 do have

dA
de

dA = } rod0. Cor. If SA be taken as the axis of x, and a perpendicular to SA through S as the axis of y,

then
x = r cos 0, y = r sin ,
dx

dr cos 0 - p sin 0 do, dy - dr sin 0 + r cos 0 d0;

we

or

Diagram of Differentials. 154. From any point S, draw any line S, P, equal to r"; produce S,P, to a point R, such that P,R, = dr. From the point R, draw R, Q, at right angles to S, R, and equal to rdo.

1

1

1

1

1

1

1

1

Join Q,S, Q,P1, and from S, draw S, Y, S,T, to meet Q,P, produced, in Y,, T, the line S, Y, being at right angles to P, T, and the line S,T, being at right angles to S, P. Then will P,Q, - ds, ZS,PAT, = P, S,Y, = P, S, T, = v, area S,P,Q, = dA.

The truth of this proposition is manifest from the formulæ already established,

sin
ф

ds rdo, cos p. ds = dr,
p = r sind,

V = q tan , dA = } råde. We have only to remember this diagram in order to be able to call to mind all the polar formulæ of tangency.

Radius of Curvature in terms of r and p. 155. Let denote the angle between the tangent TP of a curve AB, and the line SX.