. Differential and integral calculus, an introductory course for colleges and engineering schools. s 0, ds ± = smd, (b) dx = cos 8 ds, dy = sin 9 ds. These equations may be derived directly from the figure if weremember that the sides of triangle PBT are dx, dy, ds. Problem. Derive III directly from the triangle PBQ. In Art. 84 it was shown that Dxy = -=p = tan 0. Then by II of the present articleDts Dxs = whence IV (a) D,x Vl + tan2 0 = sec 0, Dtx = cos 6 Dts, Dty = tan 6 Dtx = sin 0 Dts. rw\ dx (b) It nds COS 6 -r,, dt dy . Qdsdt dt Clearing IV (b) of fractions, we have a proof that III (b)
. Differential and integral calculus, an introductory course for colleges and engineering schools. s 0, ds ± = smd, (b) dx = cos 8 ds, dy = sin 9 ds. These equations may be derived directly from the figure if weremember that the sides of triangle PBT are dx, dy, ds. Problem. Derive III directly from the triangle PBQ. In Art. 84 it was shown that Dxy = -=p = tan 0. Then by II of the present articleDts Dxs = whence IV (a) D,x Vl + tan2 0 = sec 0, Dtx = cos 6 Dts, Dty = tan 6 Dtx = sin 0 Dts. rw\ dx (b) It nds COS 6 -r,, dt dy . Qdsdt dt Clearing IV (b) of fractions, we have a proof that III (b) holdtrue whatever the independent variable may be. 102. The Derivative ofthe Arc: Polar Coordinates. Let the coordinates of P and ^ Q be 6, p and 6 + A0, p+Ap. We seek to express Dds in terms of p and 6. As in the preceding article, we have As = (1 +e)c. We now draw PB perpendicular to OQ, and with 0 as center and p as radius we strike the circular arc PR. Then c2 = PB2 + BQ*.Now PB = psinA0 and BQ = OQ - OB = p + Ap - pcos A0 = Ap + p(l -cosA0). Hence c2 = p2 sin2 A0 + [Ap + p(l - cos A0)]2,. 144 DIFFERENTIAL CALCULUS §103 and As n±u/ ,/sinAfly [AP 1 -cosAfl]2 Taking limits and making use of I and II of Art. 11, we have V (a) n6s = Vf + (2>eP)2. From this formula and from Art. 97 (a) and from the figure follow without much difficulty (b) cos$ =^=Dsp, sinty =^~s= ?DSQ; The corresponding differential formula? are (b) dp = cos ()> ds, p dQ = sin (j> ds; p = P2~r- ,pds= p2 dQ. CIS Problem. Prove V(a) by differentiating as to d the equationsx = p cos 6, y = p sin 6and substituting the results in II of Art. 101. 103. The Method of Infinitesimals. Problems like those ofthe preceding article may be solved by the aid of the followingtheorem of infinitesimals: Theorem. In taking the limit of the ratio of two infinitesimals,each or either may be replaced by an infinitesimal whose ratio to theone replaced has the limit unity. In symbolic language this theorem runs Lim- = lim
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