1. 放物運動
水平方向には前章の第1節で学んだ等速運動、垂直方向には等加速度運動をする動点の動きを考えてみましょう。
例えば、x 軸方向に一定の速度2、y 軸方向に一定の加速度 -9.8 で動く動点 P の時刻 t における位置を (x(t),y(t)) と表すと
>with(DEtools):
>dsolve({diff(x(t), t) = 2, diff(y(t), t, t) = -9.8, x(0) = 0, y(0) = 0, (D(y))(0) = 9}, {x(t), y(t)}) >plot([rhs(op(1, %)), rhs(op(2, %)), t = 0 .. 2], scaling = constrained) >animate(pointplot, [[rhs(op(1, `%%`)), rhs(op(2, `%%`))]], t = 0 .. 2, scaling = constrained, symbol = solidcircle, symbolsize = 40) >display({%, %%}) |
>with(DEtools):
>de1:=[diff(x(t),t)=x(t)*(1+y(t)),diff(y(t),t)=y(t)*(1+x(t))] >ini1:=seq([x(0)=0.2*i,y(0)=0.9-0.2*i],i=1..4) >DEplot(de1,[x(t),y(t)],t=0..30,[ini1],x=0..4,y=0..4,stepsize=0.1) |
[正解例]
>de2:=[diff(x(t),t)=-10*(x(t)-y(t)),diff(y(t),t)=-x(t)*z(t)+28*x(t)-y(t),diff(z(t),t)=x(t)*y(t)-8/3*z(t)]
>DEplot3d(de2,{x(t),y(t),z(t)},t=0..50,[[x(0)=10,y(0)=10,z=(0)=10]], scene=[x(t),y(t),z(t)],x=-50..50,y=-50..50,z=0..50,stepsize=0.01) |
[正解例]