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.

B

P

R

X

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, L PSX = 0, SQ = = r + dr, ≤ QSX = 0 + 80. Through P and Q draw the 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 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 = v, ≤ SPT = 4, ▲ SPR = 6', arc AP =s, arc PQ = ds, chord PQ = C.

Now ultimately, when d0 becomes less than any assignable

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

(5).

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

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,

But ultimately, when 80, and therefore dr, becomes less than any assignable magnitude, the two quantities ¦ r2, ¦ (r + dr)2, assume a ratio of equality: hence, replacing by

have

SA dA

we

COR. If SA be taken as the axis of x, and a perpendicular to SA through S as the axis of y, then

154. From any point S, draw any line SP, equal to r; produce SP, to a point R, such that PR, dr. From the point R, draw R,Q, at right angles to SR1, and equal to rde.

=

Join QS, QP1, and from S draw SY, ST1, to meet QP ̧ produced, in Y,, T1, the line S, Y, being at right angles to P,T, and the line S, T, being at right angles to SP1. Then will

1 1

1 1

PQ = ds, LS,P,T= 4, SY, = p, ST, = v, area SP ̧Q1 = dA. The truth of this proposition is manifest from the formulæ already established,

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.

Then

p

= r sin &,

dp = dr sin p + r cos o do ;

or, multiplying by ds and observing that sin p ds = rd0, and

ds = dr,

dp ds = rd0dr + rdødr = rdr (d0 + dp) = rdrd↓.

But we know that, p denoting the radius of curvature at P,

Chord of Curvature through the Pole.

156. Let the radius vector PS, produced if necessary, meet the osculating circle at P in a point p; and let Pp = q. Then it is plain that

9 = 2p sin p

* This method of investigating the expression for is given in the Cambridge Mathematical Journal for November, 1840.