题目

Problem

The growth of duckweed on a pond is being studied.

The surface area of the pond covered by duckweed, $A\text{ m}^2$ , at a time $t$ days after the start of the study is modelled by the equation

A=pq^t

where $p$ and $q$ are positive constants.

Figure 1 shows the linear relationship between $\log_{10}A$ and $t$ .

The points $(0,0.32)$ and $(8,0.56)$ lie on the line as shown.

(a) Find, to 3 decimal places, the value of $p$ and the value of $q$ .

(4)

Using the model with the values of $p$ and $q$ found in part (a),

(b) find the rate of increase of the surface area of the pond covered by duckweed, in $\text{m}^2/\text{day}$ , exactly 6 days after the start of the study.

Give your answer to 2 decimal places.

(3)

题目中文翻译

现正研究池塘中浮萍的生长情况。

研究开始后 $t$ 天时，池塘中被浮萍覆盖的表面积 $A\text{ m}^2$ 由下式建模：

A=pq^t

其中 $p$ 和 $q$ 为正常数。

图 1 显示了 $\log_{10}A$ 与 $t$ 的线性关系。

如图所示，点 $(0,0.32)$ 和 $(8,0.56)$ 在这条直线上。

(a) 求 $p$ 和 $q$ 的值，答案精确到小数点后 3 位。

在(a)中求得的 $p$ 和 $q$ 的基础上，

(b) 求研究开始后恰好 6 天时，池塘中被浮萍覆盖表面积的增长率，单位为 $\text{m}^2/\text{day}$ 。

答案精确到小数点后 2 位。

解答

(a)

Since

A=pq^t

take logarithms base 10:

\log_{10}A=\log_{10}(pq^t)

Using log laws,

\log_{10}A=\log_{10}p+t\log_{10}q

The graph of $\log_{10}A$ against $t$ is a straight line.

It passes through

(0,0.32)

so the intercept is

\log_{10}p=0.32

Therefore

p=10^{0.32}=2.089\ldots

\boxed{p=2.089}

to 3 decimal places.

The gradient is

\frac{0.56-0.32}{8-0}=0.03

\log_{10}q=0.03

Therefore

q=10^{0.03}=1.0715\ldots

Hence

\boxed{q=1.072}

to 3 decimal places.

(b)

The model is

A=pq^t

Differentiate with respect to $t$ :

\frac{dA}{dt}=p\ln q\cdot q^t

Using

p=2.089,\qquad q=1.072

t=6

we get

\frac{dA}{dt}=2.089\ln(1.072)(1.072)^6

Using a calculator,

\frac{dA}{dt}=0.218\ldots

Therefore, to 2 decimal places,

\boxed{0.22\text{ m}^2/\text{day}}

IAL 2021 June Q5

题目

解答

(a)

(b)