※ 引述《coolching (小欧)》之铭言：
: 就算看解答也不知怎么来的...
: 题目:
: https://i.imgur.com/jtuxnT4.jpg
: 答案:
: https://i.imgur.com/YtSg15n.jpg
: 拜托版上大大3Q
Area = (1/2)(3 + 2t)(4 + 2t)sin(Pi/2 - t)
= (1/2)[4t^2 + 14t + 12]sin(Pi/2 - t)
dArea/dt = (1/2)[8t + 14]sin(Pi/2 - t) - (1/2)[4t^2 + 14t + 2]cos(Pi/2 - t)
题目应该是问当角度为Pi/2的瞬间，t = 0时
dArea/dt值为多少
=> dArea/dt = (1/2) * 14 * 1
= 7