题目

Problem

A scientist is studying a population of fish in a lake. The number of fish, $N$ , in the population, $t$ years after the start of the study, is modelled by the equation

N=\frac{600e^{0.3t}}{2+e^{0.3t}}\qquad t\ge 0

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

(a) Find the number of fish in the lake at the start of the study.

(1)

(b) Find the upper limit to the number of fish in the lake.

(1)

(c) Find the time, after the start of the study, when there are predicted to be 500 fish in the lake. Give your answer in years and months to the nearest month.

(4)

(d) Show that

\frac{dN}{dt}=\frac{Ae^{0.3t}}{(2+e^{0.3t})^2}

where $A$ is a constant to be found.

(3)

Given that when $t=T$ , $\dfrac{dN}{dt}=8$

(e) find the value of $T$ to one decimal place.

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

(4)

题目中文翻译

一位科学家正在研究湖中鱼类种群。研究开始后 $t$ 年时，鱼群数量 $N$ 由下式建模：

N=\frac{600e^{0.3t}}{2+e^{0.3t}}\qquad t\ge 0

利用该模型方程回答(a)、(b)、(c)、(d) 和 (e)。

(a) 求研究开始时湖中的鱼数。

(b) 求湖中鱼数的上限。

(d) 证明

\frac{dN}{dt}=\frac{Ae^{0.3t}}{(2+e^{0.3t})^2}

其中 $A$ 为待求常数。

已知当 $t=T$ 时， $\dfrac{dN}{dt}=8$ 。

(e) 求 $T$ 的值，答案精确到小数点后 1 位。

（不接受完全依赖计算器技术的解法。）

解答

(a)

At the start of the study,

t=0

Substitute $t=0$ :

N=\frac{600e^0}{2+e^0}

Since

e^0=1

we get

N=\frac{600}{3}=200

So the number of fish at the start of the study is

\boxed{200}

(b)

As $t$ becomes very large,

e^{0.3t}

also becomes very large.

N=\frac{600e^{0.3t}}{2+e^{0.3t}}

the constant $2$ becomes negligible compared with $e^{0.3t}$ . So

N\to \frac{600e^{0.3t}}{e^{0.3t}}=600

Therefore the upper limit is

\boxed{600}

(c)

We need

N=500

500=\frac{600e^{0.3t}}{2+e^{0.3t}}

Multiply both sides by the denominator:

500(2+e^{0.3t})=600e^{0.3t}

Expand:

1000+500e^{0.3t}=600e^{0.3t}

Therefore

1000=100e^{0.3t}

e^{0.3t}=10

Take natural logarithms:

0.3t=\ln10

Hence

t=\frac{\ln10}{0.3}=7.675\ldots

This is 7 years plus

0.675\ldots\times 12=8.10\ldots

months.

Therefore, to the nearest month, the time is

\boxed{7\text{ years }8\text{ months}}

(d)

We have

N=\frac{600e^{0.3t}}{2+e^{0.3t}}

Use the quotient rule.

Let

u=600e^{0.3t},\qquad v=2+e^{0.3t}

Then

u'=180e^{0.3t}

and

v'=0.3e^{0.3t}

\frac{dN}{dt} =\frac{v u'-u v'}{v^2}

Therefore

\frac{dN}{dt} =\frac{(2+e^{0.3t})(180e^{0.3t})-(600e^{0.3t})(0.3e^{0.3t})}{(2+e^{0.3t})^2}

Simplify the numerator:

\begin{aligned} &(2+e^{0.3t})(180e^{0.3t})-(600e^{0.3t})(0.3e^{0.3t})\\ &=360e^{0.3t}+180e^{0.6t}-180e^{0.6t}\\ &=360e^{0.3t} \end{aligned}

Hence

\frac{dN}{dt}=\frac{360e^{0.3t}}{(2+e^{0.3t})^2}

\boxed{A=360}

(e)

We are told that

\frac{dN}{dt}=8

Using part (d),

8=\frac{360e^{0.3T}}{(2+e^{0.3T})^2}

Let

u=e^{0.3T}

Then

8=\frac{360u}{(2+u)^2}

8(2+u)^2=360u

Expand:

8(u^2+4u+4)=360u

Therefore

8u^2+32u+32=360u

8u^2-328u+32=0

Divide by $8$ :

u^2-41u+4=0

Use the quadratic formula:

u=\frac{41\pm\sqrt{41^2-4(1)(4)}}{2}

Thus

u=\frac{41\pm\sqrt{1665}}{2}

Since $T\geq 0$ , we need

u=e^{0.3T}\geq 1

so we take

u=\frac{41+\sqrt{1665}}{2}

Therefore

e^{0.3T}=\frac{41+\sqrt{1665}}{2}

Take natural logarithms:

0.3T=\ln\left(\frac{41+\sqrt{1665}}{2}\right)

T=\frac{1}{0.3}\ln\left(\frac{41+\sqrt{1665}}{2}\right)

Using a calculator,

T=12.3706\ldots

Therefore

\boxed{T=12.4}

to one decimal place.

IAL 2021 June Q8

题目

解答

(a)

(b)

(c)

(d)

(e)