Question

求解美国微积分应用题！

the hand of a clock in some tower are approxiately 3m and 2.5m in length. How fast is the distance between the tips of the hands changing at 9:00? (Hint: use the law of cosine)

question: the rate is ____m/hr at 9:00

Answer 1

根据余弦定理，两表针尖端的距离L的平方为
L^2 = 3.0^2 + 2.5^2 - 2*3.0*2.5*cosθ = 15.25 - 15*cosθ
其中 θ = t*(2π - 2π/12)
两端对 t 求导
2 L dL/dt = 15*sinθ * (2π - 2π/12)
v = dL/dt ，将 t = 9.00 代入，得
v(9.00) = 11.0616 m/hr

Answer 2

n年不做这题目了，我来试试，做错的话就当是抛砖引玉，给你个思路
12.061 m/hr

追问

试了是错的啊老兄。。我是在网上的数学作业 ，要做到正确答案才能过。。