题目

Problem

The percentage, $P$ , of the population of a small country who have access to the internet, is modelled by the equation

P=ab^t

where $a$ and $b$ are constants and $t$ is the number of years after the start of 2005.

Using the data for the years between the start of 2005 and the start of 2010, a graph is plotted of $\log_{10}P$ against $t$ .

The points are found to lie approximately on a straight line with gradient $0.09$ and intercept $0.68$ on the $\log_{10}P$ axis.

(a) Find, according to the model, the value of $a$ and the value of $b$ , giving your answers to 2 decimal places.

(4)

(b) In the context of the model, give a practical interpretation of the constant $a$ .

(1)

(2)

题目中文翻译

某小国人口中能够接入互联网的百分比 $P$ 由下式建模：

P=ab^t

其中 $a,b$ 为常数， $t$ 表示自 2005 年年初起经过的年数。

利用 2005 年年初到 2010 年年初之间的数据，作出 $\log_{10}P$ 关于 $t$ 的图像。

这些点大致落在一条斜率为 $0.09$ 、在 $\log_{10}P$ 轴上的截距为 $0.68$ 的直线上。

(a) 根据该模型求 $a$ 与 $b$ 的值，答案保留 2 位小数。

(b) 结合模型实际意义，说明常数 $a$ 的实际解释。

解答

The model is

P=ab^t

Take logarithms base 10:

\log_{10}P=\log_{10}(ab^t)

Using log laws,

\log_{10}P=\log_{10}a+t\log_{10}b

This is a straight-line equation in the form

Y=c+mt

where

Y=\log_{10}P

The intercept is $0.68$ , so

\log_{10}a=0.68

Therefore

a=10^{0.68}=4.7863\ldots

\boxed{a=4.79}

to 2 decimal places.

The gradient is $0.09$ , so

\log_{10}b=0.09

Therefore

b=10^{0.09}=1.2302\ldots

\boxed{b=1.23}

to 2 decimal places.

When

t=0

the model gives

P=ab^0=a

Since $t=0$ is the start of 2005, $a$ represents the percentage of the population who had access to the internet at the start of 2005.

\boxed{a\text{ is the initial percentage with internet access at the start of 2005}}

At the start of 2015, the number of years after the start of 2005 is

t=10

Using

P=4.79(1.23)^t

we get

P=4.79(1.23)^{10}

Using a calculator,

P=37.93\ldots

Therefore the estimate is

\boxed{38\%}