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)

解法一

思路

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研究开始时 $t=0$ 且 $A=30$ 。把这两个初始值代入模型并解出 $p$ 。

答题过程

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At the start of the study, $t=0$ and $A=30$ . Hence

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

Therefore

30p+120=80p

\boxed{p=2.4}

as required.

解法二

思路

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也可以直接把题目要求证明的 $p=2.4$ 代回模型；若 $t=0$ 时确实得到 $A=30$ ，便完成验证。这是官方评分资料接受的替代路线。

答题过程

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Using $p=2.4$ and $t=0$ ,

A=\frac{80(2.4)e^0}{2.4e^0+4} =\frac{192}{6.4} =30

This agrees with the initial surface area, so

\boxed{p=2.4}

as required.

(b)

解法一

思路

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把 $p=2.4$ 、 $A=50$ 和 $t=T$ 代入模型，先把指数项 $e^{0.15T}$ 单独留在一边，再取自然对数。

答题过程

展开

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

50=\frac{192e^{0.15T}}{2.4e^{0.15T}+4}

Therefore

\begin{align*} 120e^{0.15T}+200=&\,192e^{0.15T} \\ 200=&\,72e^{0.15T} \\ e^{0.15T}=&\,\frac{25}{9} \end{align*}

Taking natural logarithms,

T=\frac{\ln(25/9)}{0.15} =6.810\ldots

Hence, to one decimal place,

\boxed{T=6.8}

(c)

解法一

思路

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把分子和分母同时除以 $pe^{0.15t}$ ，便可直接读出 $t\to\infty$ 时的水平渐近值。

答题过程

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Rewrite the model as

A=\frac{80}{1+\dfrac{4}{pe^{0.15t}}}

As $t\to\infty$ ,

\frac{4}{pe^{0.15t}}\to0

so $A\to80$ . Therefore the maximum possible surface area is

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

IAL 2022 Jan Q4

题目

解答

(a)

解法一

思路

答题过程

解法二

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