. Differential and integral calculus, an introductory course for colleges and engineering schools. DIFFERENTIAL CALCULUS §104 figure, can write down at once the formulae ds = Vdp2 + (pdd)2, -j- = coscf), -p = sin, ^- = tan0. These formulae do not need to be retained in memory, since theyare so readily derived from the figure. Examples. 1. To determine Dxs for the parabola y1 = 2mx. Differentiating the equation, yy = m, y = —; then 1 + V ~ y2 ~ 2x - DxS~ 7—~\~ 2x 2 Til 2. To determine Des for the parabola p = Differentiating, p = 1 — cos 02 m sin0 , , ,„ 8 m2 (1 — cos 0)2 (1 — cos 0)3 m 104. Ex
. Differential and integral calculus, an introductory course for colleges and engineering schools. DIFFERENTIAL CALCULUS §104 figure, can write down at once the formulae ds = Vdp2 + (pdd)2, -j- = coscf), -p = sin, ^- = tan0. These formulae do not need to be retained in memory, since theyare so readily derived from the figure. Examples. 1. To determine Dxs for the parabola y1 = 2mx. Differentiating the equation, yy = m, y = —; then 1 + V ~ y2 ~ 2x - DxS~ 7—~\~ 2x 2 Til 2. To determine Des for the parabola p = Differentiating, p = 1 — cos 02 m sin0 , , ,„ 8 m2 (1 — cos 0)2 (1 — cos 0)3 m 104. Exercises. 1. Find Dxs for the circle. 2. Find Z^s for the astroid, x* + 2/3 = a$. 3. Find Z)0s for the astroid, x = a cos3 6, y — a sin3 0. 4. Find D^s and Des for the cycloid, x = a(0 — sin 0), ?/ = a{\ — cos 0). 5. Find Des for the lemniscate, p2 = a2 cos 2 0. 6. Find D0s for the cardioid, p = 2 a(l — cos 0). n / - -\ 7. Find Dzs for the catenary, y = ^[ea -\- e °J. 8. Find D#s /or ^e epicycloid, and for the the equation at the foot of page CHAPTER XVSIMPLE FORMULA OF KINEMATICS 105. Resolution of Velocities and Accelerations. Let a mov-ing body have at P an (instantaneous) velocity vi along the linePM, while the line PM has at the same time an (instantaneous)velocity v2 in the direction PA and PB represent v\ andv2 in magnitude and now both motions become uni-form at P, the body will move in p,unit of time along PM to A, andin the same unit of time the segment PA will move to the positionBQ. Q will thus be the final position of the body. Since bothmotions are uniform, the body lies always upon the diagonal PQ,and has in reality reached Q by traversing the diagonal PQ in aunit of time. Instead, then, of regarding the body as having thetwo velocities Vi and v2 at P, we may regard it as having a singlevelocity v, represented in magnitude and direction by PQ. v ismade up of v\ and v2 and is termed their resultant, an
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