题目

Problem

The total mass of gold, $G$ tonnes, extracted from a mine is modelled by the equation

G=40-30e^{1-0.05t}\qquad t\ge k,\ G\ge 0

where $t$ is the number of years after 1st January 1800.

Figure 2 shows a sketch of $G$ against $t$ .

Use the equation of the model to answer parts (a), (b) and (c).

(a) (i) Find the value of $k$ .

(ii) Hence find the year and month in which gold started being extracted from the mine.

(3)

(b) Find the total mass of gold extracted from the mine up to 1st January 1870.

(2)

There is a limit to the mass of gold that can be extracted from the mine.

(1)

题目中文翻译

从某矿山中已开采出的黄金总质量 $G$ （单位：吨）由方程

G=40-30e^{1-0.05t}\qquad t\ge k,\ G\ge 0

建模，其中 $t$ 是自 1800 年 1 月 1 日起经过的年数。

图 2 给出了 $G$ 关于 $t$ 的示意图。

利用该模型方程回答 (a)、(b) 和 (c)。

(a) (i) 求 $k$ 的值。

(ii) 由此求开始从矿山开采黄金的年份和月份。

(b) 求到 1870 年 1 月 1 日为止，从矿山中开采出的黄金总质量。

可开采出的黄金总质量有一个上限。

解答

(a)(i)

Gold starts being extracted when

G=0

Using the model,

0=40-30e^{1-0.05t}

30e^{1-0.05t}=40

Hence

e^{1-0.05t}=\frac43

Taking natural logarithms,

1-0.05t=\ln\frac43

Therefore

0.05t=1-\ln\frac43

t=20\left(1-\ln\frac43\right)

Thus

\boxed{k=20\left(1-\ln\frac43\right)=14.2\text{ to 3 s.f.}}

(a)(ii)

The value $k=14.2\ldots$ means gold started being extracted about $14.2$ years after 1st January 1800.

This is in the year $1814$ .

The decimal part is

0.246\ldots

of a year, which is about

0.246\ldots\times 12=2.95\ldots

months after January.

So gold started being extracted in

\boxed{\text{March 1814}}

(b)

1st January 1870 is $70$ years after 1st January 1800, so use $t=70$ .

G=40-30e^{1-0.05(70)}

Therefore

G=40-30e^{-2.5}

G=37.537\ldots

Hence the total mass extracted is

\boxed{37.5\text{ tonnes}}

to 3 significant figures.

(c)

As $t$ becomes very large,

e^{1-0.05t}\to 0

Therefore

G=40-30e^{1-0.05t}\to 40

So the limiting mass is

\boxed{40\text{ tonnes}}

IAL 2021 Oct Q3

题目

解答

(a)(i)

(a)(ii)

(b)

(c)