题目

Problem

The population of a town was monitored.

Exactly $5$ years after monitoring began, the population was $58\,000$ .

Exactly $10$ years after monitoring began, the population was $65\,000$ .

Given that the population of the town, $P$ thousand, $t$ years after monitoring began can be modelled by the equation

\begin{align*} P^2=a+bt^3 \end{align*}

where $a$ and $b$ are constants,

(a) find the value of $a$ and the value of $b$ .

(3)

According to the model, exactly $T$ years after monitoring began, the population was $85\,000$ .

Making your method clear,

(b) find the value of $T$ , giving your answer to one decimal place.

(Solutions relying entirely on calculator technology are not acceptable.)

(2)

解答

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注意 $P$ 的单位是 thousand，所以 $58\,000$ 对应 $P=58$ ， $65\,000$ 对应 $P=65$ 。代入模型得到两个关于 $a,b$ 的方程。

展开

When $t=5$ , $P=58$ , so

\begin{align*} 58^2=&\,a+b(5)^3\\ 3364=&\,a+125b. \end{align*}

When $t=10$ , $P=65$ , so

\begin{align*} 65^2=&\,a+b(10)^3\\ 4225=&\,a+1000b. \end{align*}

Subtract:

\begin{align*} 4225-3364=&\,(a+1000b)-(a+125b)\\ 861=&\,875b\\ b=&\,\frac{861}{875}\\ =&\,0.984. \end{align*}

Then

\begin{align*} 3364=&\,a+125(0.984)\\ 3364=&\,a+123\\ a=&\,3241. \end{align*}

Therefore

\begin{align*} a=3241,\qquad b=0.984. \end{align*}

展开

$85\,000$ 对应 $P=85$ 。把 $P=85$ 、(a) 中的 $a,b$ 代入模型，解出 $T^3$ ，再开三次方。

展开

When the population is $85\,000$ ,

\begin{align*} P=85. \end{align*}

Use the model:

\begin{align*} 85^2=&\,3241+0.984T^3\\ 7225=&\,3241+0.984T^3\\ 3984=&\,0.984T^3\\ T^3=&\,\frac{3984}{0.984}\\ =&\,4048.780\ldots \end{align*}

Therefore

\begin{align*} T=&\,\sqrt[3]{4048.780\ldots}\\ =&\,15.9\ldots \end{align*}

\begin{align*} T=15.9. \end{align*}