题目

Problem

A population of a rare species of toad is being studied.

The number of toads, $N$ , in the population, $t$ years after the start of the study, is modelled by the equation

N=\frac{900e^{0.12t}}{2e^{0.12t}+1}\qquad t\ge 0,\ t\in\mathbb{R}

According to this model,

(a) calculate the number of toads in the population at the start of the study,

(1)

(b) find the value of $t$ when there are 420 toads in the population, giving your answer to 2 decimal places.

(4)

(1)

题目中文翻译

现正研究一种稀有蟾蜍的种群。

研究开始后 $t$ 年时，种群中的蟾蜍数量 $N$ 由下式建模：

N=\frac{900e^{0.12t}}{2e^{0.12t}+1}\qquad t\ge 0,\ t\in\mathbb{R}

根据该模型，

(a) 计算研究开始时种群中的蟾蜍数量；

(b) 当种群中有 420 只蟾蜍时，求 $t$ 的值，答案精确到小数点后 2 位；

解答

(a)

At the start of the study,

t=0

Substitute $t=0$ into the model:

N=\frac{900e^{0.12(0)}}{2e^{0.12(0)}+1}

Since

e^0=1

we get

N=\frac{900}{2+1}

Therefore

N=300

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

\boxed{300}

(b)

We are given

N=420

420=\frac{900e^{0.12t}}{2e^{0.12t}+1}

Let

u=e^{0.12t}

Then

420=\frac{900u}{2u+1}

Multiply by $2u+1$ :

420(2u+1)=900u

Expand:

840u+420=900u

420=60u

Hence

u=7

Therefore

e^{0.12t}=7

Take natural logarithms:

0.12t=\ln7

t=\frac{\ln7}{0.12}

Using a calculator,

t=16.2158\ldots

Therefore

\boxed{t=16.22}

to 2 decimal places.

(c)

Since

e^{0.12t}>0

let

u=e^{0.12t}

where $u>0$ . Then

N=\frac{900u}{2u+1}

If the population reached $500$ , then

500=\frac{900u}{2u+1}

500(2u+1)=900u

This gives

1000u+500=900u

and therefore

u=-5

But this is impossible because

u=e^{0.12t}>0

Therefore, according to this model, the population can never reach

\boxed{500}

IAL 2020 Jan Q1

题目

解答

(a)

(b)

(c)