题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

A population of fruit flies is being studied.

The number of fruit flies, $F$ , in the population, $t$ days after the start of the study, is modelled by the equation

F=\frac{350e^{kt}}{9+e^{kt}}

where $k$ is a constant.

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

(a) Find the number of fruit flies in the population at the start of the study.

(1)

Given that there are $200$ fruit flies in the population 15 days after the start of the study,

(b) show that $k=\dfrac1{15}\ln 12$

(3)

Given also that, when $t=T$ , the number of fruit flies in the population is increasing at a rate of $10$ per day,

(5)

题目中文翻译

本题中你必须写出解题过程的所有步骤。

不接受完全依赖计算器技术的解法。

某个果蝇种群正在被研究。

研究开始后 $t$ 天，该种群中的果蝇数量 $F$ 由方程

F=\frac{350e^{kt}}{9+e^{kt}}

建模，其中 $k$ 是常数。

利用该模型方程回答 (a)、(b) 和 (c)。

(a) 求研究开始时该种群中的果蝇数量。

已知研究开始 15 天后，种群中有 $200$ 只果蝇。

(b) 证明 $k=\dfrac1{15}\ln 12$ 。

又已知当 $t=T$ 时，果蝇数量的增长率为每天 $10$ 。

解答

(a)

解法一

思路

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研究开始时就是 $t=0$ 。代入模型后，利用 $e^0=1$ 即可。

答题过程

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At the start of the study,

t=0.

F=\frac{350e^{k\cdot0}}{9+e^{k\cdot0}}.

Since $e^0=1$ ,

F=\frac{350}{9+1}=35.

Therefore the number of fruit flies at the start of the study is

\boxed{35}.

(b)

解法一

思路

展开

把 $t=15$ 和 $F=200$ 代入模型，目标是解出 $e^{15k}$ 。得到 $e^{15k}=12$ 后，再两边取自然对数。

答题过程

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When $t=15$ ,

F=200.

Substitute these into the model:

200=\frac{350e^{15k}}{9+e^{15k}}.

Multiply both sides by $9+e^{15k}$ :

200(9+e^{15k})=350e^{15k}.

1800+200e^{15k}=350e^{15k}.

Hence

150e^{15k}=1800.

Therefore

e^{15k}=12.

Taking natural logarithms,

15k=\ln12.

\boxed{k=\frac1{15}\ln12}.

(c)

解法一

思路

展开

增长率是 $\frac{\mathrm{d}F}{\mathrm{d}t}$ ，所以先对模型求导。然后令导数等于 $10$ 。为了把方程变成二次方程，设 $u=e^{kT}$ ，解出两个正值，再用 $kT=\ln u$ 求 $T$ 。

答题过程

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The model is

F=\frac{350e^{kt}}{9+e^{kt}}.

Differentiate using the quotient rule:

\begin{align*} \frac{\mathrm{d}F}{\mathrm{d}t} =&\,\frac{350ke^{kt}(9+e^{kt}) -350e^{kt}\cdot ke^{kt}} {(9+e^{kt})^2} \\[2mm] =&\,\frac{3150ke^{kt}+350ke^{2kt}-350ke^{2kt}} {(9+e^{kt})^2} \\[2mm] =&\,\frac{3150ke^{kt}}{(9+e^{kt})^2}. \end{align*}

When $t=T$ , the population is increasing at a rate of $10$ per day, so

\frac{3150ke^{kT}}{(9+e^{kT})^2}=10.

Multiply by $(9+e^{kT})^2$ :

3150ke^{kT}=10(9+e^{kT})^2.

Divide by $10$ :

315ke^{kT}=(9+e^{kT})^2.

Expand the right-hand side:

315ke^{kT}=81+18e^{kT}+e^{2kT}.

e^{2kT}+(18-315k)e^{kT}+81=0.

From part (b),

k=\frac1{15}\ln12.

Let

u=e^{kT}.

Then

u^2+(18-315k)u+81=0.

Substitute $k=\frac1{15}\ln12$ :

u^2+\left(18-21\ln12\right)u+81=0.

Using the quadratic formula,

u=\frac{21\ln12-18 \pm\sqrt{(18-21\ln12)^2-324}}{2}.

This gives

u=2.5615\ldots \qquad\text{or}\qquad u=31.6214\ldots.

Since $u=e^{kT}$ ,

kT=\ln u.

T=\frac{\ln u}{k}.

For $u=2.5615\ldots$ ,

T=\frac{\ln(2.5615\ldots)}{\frac1{15}\ln12} =5.677\ldots.

For $u=31.6214\ldots$ ,

T=\frac{\ln(31.6214\ldots)}{\frac1{15}\ln12} =20.848\ldots.

Therefore, to one decimal place, the possible values of $T$ are

\boxed{T=5.7\quad\text{or}\quad T=20.8}.

IAL 2023 Jan Q10

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程