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, metres, in a harbour on a particular day is given by the equation
D=8+5\sin\!\left(\frac{\pi t}{6}+3\right)\qquad0\leqslant t<24 \end{align*}$$ where $t$ is the number of hours after **midnight**. (a) Show that the depth of water in the harbour at 2 am is just over 4 metres. <div style="text-align: right;">(1)</div> (b) Find, to the nearest minute, the first time after **midday** when the depth of water in the harbour is exactly 6 metres. <div style="text-align: right;">(5)</div>解答
(a)
解法一
思路
展开
2 am 意味着 ，直接代入公式计算。
答题过程
展开
At 2 am, , so
D=8+5\sin\!\left(\frac{2\pi}{6}+3\right)=8+5\sin\!\left(\frac{\pi}{3}+3\right). \end{align*}$$ Using a calculator (in radians), $$\begin{align*} \sin\!\left(\frac{\pi}{3}+3\right)\approx\sin(4.047)\approx-0.789, \end{align*}$$ $$\begin{align*} D\approx8+5\times(-0.789)=8-3.947\approx4.05, \end{align*}$$ which is just over 4 metres. </details> ## (b) ### 解法一 #### 思路 <details> <summary>展开</summary> 设 $D=6$，解三角方程。Midday 是 $t=12$，所以需要找 $t>12$ 的第一个解。注意角度是以弧度为单位。 </details> #### 答题过程 <details> <summary>展开</summary> Setting $D=6$, $$\begin{align*} 6=8+5\sin\!\left(\frac{\pi t}{6}+3\right) \end{align*}$$ $$\begin{align*} \sin\!\left(\frac{\pi t}{6}+3\right)=-\frac{2}{5}=-0.4. \end{align*}$$ The principal value is $$\begin{align*} \frac{\pi t}{6}+3=\arcsin(-0.4)\approx-0.4115. \end{align*}$$ So $$\begin{align*} \frac{\pi t}{6}=-0.4115-3=-3.4115 \end{align*}$$ $$\begin{align*} t=\frac{-3.4115\times6}{\pi}\approx-6.517. \end{align*}$$ This is negative, so not in the range $0\leqslant t<24$. Using the general solution, $$\begin{align*} \frac{\pi t}{6}+3=n\pi+(-1)^n\times(-0.4115),\quad n\in\mathbb{Z}. \end{align*}$$ For $n=1$: $$\begin{align*} \frac{\pi t}{6}+3=\pi+0.4115=3.5531 \end{align*}$$ $$\begin{align*} \frac{\pi t}{6}=0.5531 \end{align*}$$ $$\begin{align*} t=\frac{0.5531\times6}{\pi}\approx1.059. \end{align*}$$ This is before midday. For $n=2$: $$\begin{align*} \frac{\pi t}{6}+3=2\pi-0.4115=5.8717 \end{align*}$$ $$\begin{align*} \frac{\pi t}{6}=2.8717 \end{align*}$$ $$\begin{align*} t=\frac{2.8717\times6}{\pi}\approx5.486. \end{align*}$$ Still before midday. For $n=3$: $$\begin{align*} \frac{\pi t}{6}+3=3\pi+0.4115=9.8347 \end{align*}$$ $$\begin{align*} \frac{\pi t}{6}=6.8347 \end{align*}$$ $$\begin{align*} t=\frac{6.8347\times6}{\pi}\approx13.064. \end{align*}$$ This is after midday ($t=12$). Converting $0.064$ hours to minutes: $0.064\times60\approx3.8$ minutes. So $t\approx13$ hours and 4 minutes after midnight, which is **1:04 pm**. </details> ### 解法二 #### 思路 <details> <summary>展开</summary> 设 $\theta=\frac{\pi t}{6}+3$，先解 $\sin\theta=-0.4$，找出所有 $\theta$ 值，再转回 $t$。 </details> #### 答题过程 <details> <summary>展开</summary> Let $\theta=\dfrac{\pi t}{6}+3$. Then $\sin\theta=-0.4$. The solutions in one cycle are $$\begin{align*} \theta_1=\arcsin(-0.4)\approx-0.4115,\qquad\theta_2=\pi-\arcsin(-0.4)\approx3.5531. \end{align*}$$ The general solutions are $$\begin{align*} \theta=2n\pi+(-0.4115)\quad\text{or}\quad\theta=2n\pi+3.5531,\quad n\in\mathbb{Z}. \end{align*}$$ For $0\leqslant t<24$, we need $3\leqslant\theta<4\pi+3$, i.e.\ $3\leqslant\theta<15.566$. From $\theta=2n\pi+3.5531$: - $n=1$: $\theta=9.8363$, $t=\dfrac{(9.8363-3)\times6}{\pi}\approx13.064$ - $n=2$: $\theta=16.119$ (too large) From $\theta=2n\pi-0.4115$: - $n=1$: $\theta=5.8717$, $t=\dfrac{(5.8717-3)\times6}{\pi}\approx5.486$ (before midday) - $n=2$: $\theta=12.155$, $t=\dfrac{(12.155-3)\times6}{\pi}\approx17.49$ The first solution after midday is $t\approx13.064$ hours, i.e.\ **1:04 pm**. </details>