题目
Problem
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
The depth of water, D metres, in a harbour on a particular day is given by the equation
D=8+5sin(6πt+3)0⩽t<24
where t is the number of hours after midnight.
(a) Show that the depth of water in the harbour at 2 am is just over 4 metres.
(1)
(b) Find, to the nearest minute, the first time after midday when the depth of water in the harbour is exactly 6 metres.
(5)
题目中文翻译
本题必须展示计算过程的所有步骤。
完全依赖计算器技术的解答不予接受。
某一天,港口的水深 D 米由以下方程给出:
D=8+5sin(6πt+3)
其中 0⩽t<24,t 为午夜之后经过的小时数。
(a) 证明凌晨2点时港口的水深略大于4米。(1)
(b) 求中午之后港口水深第一次恰好为6米的时间,精确到最近的分钟。(5)
解答
(a)
解法一
思路
展开
凌晨2点对应 t=2;用弧度模式计算,保留足够中间精度。
答题过程
展开
At 2 am, t=2. Hence
D==≈8+5sin(3π+3)4.066006762…4.07 metres.
Therefore the depth is just over 4 metres.
(b)
解法一
思路
展开
先由水深求正弦值,再用参考角和第三、第四象限列出当天的候选时间。选择最早的午后解,并把小时小数部分换成分钟。
答题过程
展开
Set D=6 and let θ=6πt+3. Then
6=sinθ=8+5sinθ−52.
Let α=arcsin(0.4)=0.411516846…. The solutions are
θ=or θ=π+α+2nπ2π−α+2nπ,n∈Z.
Since 0⩽t<24,
3⩽θ<4π+3.
There are four candidates in this interval:
θπ+α2π−α3π+α4π−αt=π6(θ−3)1.056361331…5.484482766…13.056361331…17.484482766…
The first time after midday is therefore
t==π6(3π+α−3)13.056361331….
The minutes after 13:00 are
60(t−13)=3.381679878….
To the nearest minute, the time is 13:03 (1:03 pm).
也可利用模型的周期 12 小时:两个上午时刻各加 12 小时,得到 13.056361… 和 17.484483…,其中较早者就是答案。这里保留未舍入参考角计算;可写为 13:03 或 1:03 pm。