所滚过的轨迹

倪举鹏

用软件画这参数方程图也很难吧，这个系统里面没有不动的点，轨迹确实难求。我是将这看作立体靠展开那个立体的侧面的

倪举鹏

谁能化简上式么

倪举鹏

x := int(-sin(theta)*sin(-arccos((1/6)*sqrt(1-cos(theta)^2/(1+2*cos(theta))^2)*(2+cos(theta))*sqrt(6)/(sqrt(1/(1+2*cos(theta)))*(1+cos(theta))))+int((1/36)*sqrt(6)*(2*sqrt(6)*sqrt((9*cos(theta)^3+15*cos(theta)^2+10*cos(theta)+2)/(2*cos(theta)^2+3*cos(theta)+1))*sqrt(3*cos(theta)^2+4*cos(theta)+1)*cos(theta)+sqrt(6)*sqrt((9*cos(theta)^3+15*cos(theta)^2+10*cos(theta)+2)/(2*cos(theta)^2+3*cos(theta)+1))*sqrt(3*cos(theta)^2+4*cos(theta)+1)-sin(theta)*sqrt(-(18*cos(theta)^3+6*cos(theta)^2-12*cos(theta)-12)/(2*cos(theta)^2+3*cos(theta)+1)))/(sqrt(3*cos(theta)^2+4*cos(theta)+1)*sqrt(1/(1+2*cos(theta)))*(2*cos(theta)^2+3*cos(theta)+1)), theta))/(sqrt(3*cos(theta)^2+4*cos(theta)+1)*(1+2*cos(theta))), theta)
y := int(sin(theta)*cos(-arccos((1/6)*sqrt(1-cos(theta)^2/(1+2*cos(theta))^2)*(2+cos(theta))*sqrt(6)/(sqrt(1/(1+2*cos(theta)))*(1+cos(theta))))+int((1/36)*sqrt(6)*(2*sqrt(6)*sqrt((9*cos(theta)^3+15*cos(theta)^2+10*cos(theta)+2)/(2*cos(theta)^2+3*cos(theta)+1))*sqrt(3*cos(theta)^2+4*cos(theta)+1)*cos(theta)+sqrt(6)*sqrt((9*cos(theta)^3+15*cos(theta)^2+10*cos(theta)+2)/(2*cos(theta)^2+3*cos(theta)+1))*sqrt(3*cos(theta)^2+4*cos(theta)+1)-sin(theta)*sqrt(-(18*cos(theta)^3+6*cos(theta)^2-12*cos(theta)-12)/(2*cos(theta)^2+3*cos(theta)+1)))/(sqrt(3*cos(theta)^2+4*cos(theta)+1)*sqrt(1/(1+2*cos(theta)))*(2*cos(theta)^2+3*cos(theta)+1)), theta))/(sqrt(3*cos(theta)^2+4*cos(theta)+1)*(1+2*cos(theta))), theta)

想不到方程更简洁的表示了……

倪举鹏

这个困难问题怎么不火啊

yinhow

倪举鹏 发表于 2014-7-1 17:17
这个困难问题怎么不火啊

一是很少有人关注，二是这个问题真的很难

wayne

倪举鹏 发表于 2014-7-1 17:17
这个困难问题怎么不火啊

这个题目非常有趣，而且的确很难。
我反正是一时没有想法，而且几乎忘记了我曾经对这题感兴趣了

倪举鹏

s := dsolve({(1/36)*sqrt(6)*(2*sqrt(6)*sqrt((9*cos(theta)^3+15*cos(theta)^2+10*cos(theta)+2)/(2*cos(theta)^2+3*cos(theta)+1))*sqrt(3*cos(theta)^2+4*cos(theta)+1)*cos(theta)+sqrt(6)*sqrt((9*cos(theta)^3+15*cos(theta)^2+10*cos(theta)+2)/(2*cos(theta)^2+3*cos(theta)+1))*sqrt(3*cos(theta)^2+4*cos(theta)+1)-sin(theta)*sqrt(-(18*cos(theta)^3+6*cos(theta)^2-12*cos(theta)-12)/(2*cos(theta)^2+3*cos(theta)+1)))/(sqrt(3*cos(theta)^2+4*cos(theta)+1)*sqrt(1/(1+2*cos(theta)))*(2*cos(theta)^2+3*cos(theta)+1)) = (D(k))(theta), k(0) = 0, x(0) = 0, y(0) = 0, (D(x))(theta) = -sin(theta)*sin(-arccos((1/6)*sqrt(1-cos(theta)^2/(1+2*cos(theta))^2)*(2+cos(theta))*sqrt(6)/(sqrt(1/(1+2*cos(theta)))*(1+cos(theta))))+k(theta))/(sqrt(3*cos(theta)^2+4*cos(theta)+1)*(1+2*cos(theta))), (D(y))(theta) = sin(theta)*cos(-arccos((1/6)*sqrt(1-cos(theta)^2/(1+2*cos(theta))^2)*(2+cos(theta))*sqrt(6)/(sqrt(1/(1+2*cos(theta)))*(1+cos(theta))))+k(theta))/(sqrt(3*cos(theta)^2+4*cos(theta)+1)*(1+2*cos(theta)))}, {k(theta), x(theta), y(theta)}, numeric);
print(outputredirected...); # input placeholder
proc(x_rkf45) ... end;
s(1.9106);
print(outputredirected...); # input placeholder
[theta = 1.9106, k(theta) = HFloat(0.008396269066916085),

x(theta) = HFloat(1.0790492896250343),

y(theta) = HFloat(1.2949603345514877)]
s(1.9);
print(outputredirected...); # input placeholder
[theta = 1.9, k(theta) = HFloat(0.1459008353432432),

x(theta) = HFloat(0.6974249724255157),

y(theta) = HFloat(1.2147960021472228)]

得到一个三元微分方程组，可以解数值解，我不会用maple画微分方程组参数解x(theta)，y(theta)的图

mathe

题目看不懂，是如何滚动，哪个点的轨迹？

倪举鹏

mathe 发表于 2015-3-3 17:42
题目看不懂，是如何滚动，哪个点的轨迹？

以竖着硬币和平面的切点为原点，整体朝x轴正方向滚动，开始竖着硬币的轨迹方程是上面的参数方程x(theta)，y(theta)     可以QQ发maple你

neluzyy1

好帖啊！顶起来啊啊 啊啊啊啊啊啊啊啊