Question

求大神解数学题

Answer 1

取A为坐标原点, AP为+x轴, AO为+y轴。 另设P(p, 0), 即AP = p
AC的弧长 = 半径AO*∠AOC的弧度θ = 1*θ = θ = (2/3)AP = 2p/3
令C(c, c'), 则c = OCsin∠AOC = sinθ = sin(2p/3)
c' = OA - OCcosθ = 1 - cos(2p/3)
CP的方程为: (y - 0)/[1 - cos(2p/3) - 0] = (x - p)/[sin(2p/3) - p]
令x = 0， 得M的纵坐标m = p[cos(2p/3) - 1]/[sin(2p/3) - p]
m和p均为时间t的函数， 两边对t求导:

dm/dt = {p(cos(2p/3) - 1]'[sin(2p/3) - p] - p[cos(2p/3) - 1][sin(2p/3) - p]'}/[sin(2p/3) - p]²
右边为一个分式和dp/dt的积，其余应当可以自己做。不清楚可以再问。

追问

一阶导数dm/dp有问题

首先求dm/dp

对不起！是我看错了，应该没问题。