题目

Problem

The growth of a weed on the surface of a pond is being studied.

The surface area of the pond covered by the weed, $A$ m $^2$ , is modelled by the equation

A=\frac{80pe^{0.15t}}{pe^{0.15t}+4}

where $p$ is a positive constant and $t$ is the number of days after the start of the study.

Given that

• $30$ m $^2$ of the surface of the pond was covered by the weed at the start of the study

• $50$ m $^2$ of the surface of the pond was covered by the weed $T$ days after the start of the study

(a) show that $p=2.4$

(2)

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

Solutions relying entirely on graphical or numerical methods are not acceptable.

(4)

The weed grows until it covers the surface of the pond.

(1)

题目中文翻译

正在研究池塘表面一种杂草的生长。

池塘表面被杂草覆盖的面积 $A$ （单位：m $^2$ ）由方程

A=\frac{80pe^{0.15t}}{pe^{0.15t}+4}

建模，其中 $p$ 是正常数， $t$ 是研究开始后的天数。

已知

• 研究开始时，池塘表面有 $30$ m $^2$ 被杂草覆盖

• 研究开始后 $T$ 天，池塘表面有 $50$ m $^2$ 被杂草覆盖

(a) 证明 $p=2.4$ 。

(b) 求 $T$ 的值，答案精确到小数点后 1 位。

不接受完全依赖图像法或数值法的解法。

杂草会继续生长，直到覆盖整个池塘表面。

解答

(a)

At the start of the study,

t=0

and

A=30

Substitute these into the model:

30=\frac{80pe^{0}}{pe^{0}+4}

Since $e^0=1$ ,

30=\frac{80p}{p+4}

30(p+4)=80p

Therefore

30p+120=80p

Hence

120=50p

and so

\boxed{p=2.4}

as required.

(b)

Using $p=2.4$ and $A=50$ ,

50=\frac{80(2.4)e^{0.15T}}{2.4e^{0.15T}+4}

50(2.4e^{0.15T}+4)=192e^{0.15T}

Expand:

120e^{0.15T}+200=192e^{0.15T}

Therefore

200=72e^{0.15T}

e^{0.15T}=\frac{200}{72}

Taking natural logarithms,

0.15T=\ln\left(\frac{200}{72}\right)

Hence

T=\frac{\ln\left(\frac{200}{72}\right)}{0.15}

T=6.810\ldots

Therefore

\boxed{T=6.8}

to one decimal place.

(c)

As $t$ becomes very large,

e^{0.15t}\to\infty

A=\frac{80pe^{0.15t}}{pe^{0.15t}+4}

the term $4$ becomes negligible compared with $pe^{0.15t}$ .

A\to 80

Therefore the maximum possible surface area of the pond is

\boxed{80\text{ m}^2}

IAL 2022 Jan Q4

题目

解答

(a)

(b)

(c)