# A Course of Higher Mathematics. Volume II by V. I. Smirnov and A. J. Lohwater (Auth.)

By V. I. Smirnov and A. J. Lohwater (Auth.)

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Additional info for A Course of Higher Mathematics. Volume II

Example text

We remark that it is sometimes more con- 15] THE EQUATION y<*> = f(x) 47 venient to mark off the unit vector which we denoted by OP, as also the values of the functions g(z, y, z) and f(x, y, z), from some other point 0X of axis OY, instead of from the origin 0. This is done so as to avoid the lines PA0, PBQy . . > giving the directions of the step line, intersecting with the step line itself. Figure 18 illustrates the construction of the solution of equation (9), satisfying the initial conditions (10).

Voi V' lx-x. = t/6 > we obtain, on substituting x = x0, y = y0, y' = y' in (36) and (37): and the required solution becomes: dy -j *2/(y)dv + ir; Let a particle move along the x axis under the action of a force F(x) which depends only on the position of the point. The differential equation of motion is [13]: Let xQf vQ be the initial abscissa and initial velocity of the particle at t = 0: If-o ~ *°» da? dt = v0 If we multiply both sides of the equation by (dxjdt) dt and integrate, we get: x m mv T (l&f-li » = \$ F{x)6x x or m Y [wf-i F{x)dx== m T ^ m The first term on the left-hand side, m(d#/d£)2/2, consists of the kinetic energy, and the second term j — J F (x) dx\ of the potential energy of the moving particle; and it follows from (38) that the sum of the kinetic and potential energies remains constant throughout the motion.

The total loading is Jfc|d| = fc/2 2 The reactions at the supports O and L due to the elementary loading fc£d| are, in accordance with the usual law of levers: kHl- ■f) At A k & At —- d | and —=— d£. £_±,, , , - , - , . x-o = °; (34) i/Ix-r = °- The general solution is: -h Z2*3 ( x* , . n n \ Constants Gx and C2 are found from conditions (34): C2 = 0; C1 = 7 60 -^-^, whence finally: V = 3 " ^ (3* 5 - 10/2 ^3 + 7/%). To find the position and value of maximum deflection we put x = It, and re-write the above expression for y as: y = 36osr ( 3 < 5 - 1 0 t 3 + 7t) (°<«<1)- The derivative of the polynomial in brackets: 15*4 _ 30t* + 7 has only one zero in the interval (0, 1): *0 = y i - 2 ^ - j g - ~ 0 .