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P2 2024 May R Q5

A Level / Edexcel / P2

IAL 2024 May (R) Paper · Question 5

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, DD metres, in a harbour on a particular day is given by the equation

D= 8+5sin⁡ ⁣(πt6+3)0⩽t<24\begin{align*} D=&\,8+5\sin\!\big(\frac{\pi t}{6}+3\big)\qquad0\leqslant t<24 \end{align*}

where tt 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 DD 米由以下方程给出:

D=8+5sin⁡(πt6+3)D=8+5\sin\bigg(\frac{\pi t}{6}+3\bigg)

其中 0⩽t<240\leqslant t<24,tt 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

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凌晨2点对应 t=2t=2;用弧度模式计算,保留足够中间精度。

答题过程

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At 2 am, t=2t=2. Hence

D= 8+5sin⁡(π3+3)= 4.066006762…≈ 4.07 metres.\begin{align*} D=&\,8+5\sin\bigg(\frac{\pi}{3}+3\bigg)\\ =&\,4.066006762\ldots\\ \approx&\,4.07\text{ metres}. \end{align*}

Therefore the depth is just over 44 metres.

(b)

解法一

思路

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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

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Set D=6D=6 and let θ=πt6+3\theta=\dfrac{\pi t}{6}+3. Then

6= 8+5sin⁡θsin⁡θ= −25.\begin{align*} 6=&\,8+5\sin\theta\\ \sin\theta=&\,-\frac25. \end{align*}

Let α=arcsin⁡(0.4)=0.411516846…\alpha=\arcsin(0.4)=0.411516846\ldots. The solutions are

θ= π+α+2nπor θ= 2π−α+2nπ,n∈Z.\begin{align*} \theta=&\,\pi+\alpha+2n\pi\\ \text{or }\theta=&\,2\pi-\alpha+2n\pi, \qquad n\in\mathbb Z. \end{align*}

Since 0⩽t<240\leqslant t<24,

3⩽θ<4π+3.3\leqslant\theta<4\pi+3.

There are four candidates in this interval:

θt=6π(θ−3)π+α1.056361331…2π−α5.484482766…3π+α13.056361331…4π−α17.484482766…\begin{array}{c|c} \theta&t=\frac6\pi(\theta-3)\\\hline \pi+\alpha&1.056361331\ldots\\ 2\pi-\alpha&5.484482766\ldots\\ 3\pi+\alpha&13.056361331\ldots\\ 4\pi-\alpha&17.484482766\ldots \end{array}

The first time after midday is therefore

t= 6π(3π+α−3)= 13.056361331….\begin{align*} t=&\,\frac6\pi(3\pi+\alpha-3)\\ =&\,13.056361331\ldots. \end{align*}

The minutes after 13:00 are

60(t−13)=3.381679878….60(t-13)=3.381679878\ldots.

To the nearest minute, the time is 13:03 (1:03 pm).

也可利用模型的周期 1212 小时:两个上午时刻各加 1212 小时,得到 13.056361…13.056361\ldots 和 17.484483…17.484483\ldots,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。