题目

Problem

A scientist monitored the growth of bacteria on a dish over a 30-day period.

The area, $N\text{ mm}^2$ , of the dish covered by bacteria, $t$ days after monitoring began, is modelled by the equation

\log_{10}N=0.0646t+1.478\qquad 0\le t\le 30

(a) Show that this equation may be written in the form

N=ab^t

where $a$ and $b$ are constants to be found. Give the value of $a$ to the nearest integer and give the value of $b$ to 3 significant figures.

(4)

(b) Use the model to find the area of the dish covered by bacteria 30 days after monitoring began. Give your answer, in $\text{mm}^2$ , to 2 significant figures.

(2)

题目中文翻译

一位科学家在 30 天期间监测培养皿中细菌的生长。

监测开始后 $t$ 天时，被细菌覆盖的培养皿面积 $N\text{ mm}^2$ 由下式建模：

\log_{10}N=0.0646t+1.478\qquad 0\le t\le 30

(a) 证明该方程可写为

N=ab^t

的形式，其中 $a$ 和 $b$ 为待求常数。 $a$ 取最接近的整数， $b$ 保留 3 位有效数字。

(b) 利用该模型求监测开始 30 天后被细菌覆盖的培养皿面积，单位为 $\text{mm}^2$ ，答案保留 2 位有效数字。

解答

(a)

We are given

\log_{10}N=0.0646t+1.478

Raise 10 to the power of both sides:

N=10^{0.0646t+1.478}

Use the index law

10^{A+B}=10^A\cdot 10^B

N=10^{1.478}\cdot 10^{0.0646t}

Now

10^{0.0646t}=\left(10^{0.0646}\right)^t

Therefore

N=10^{1.478}\left(10^{0.0646}\right)^t

This is in the form

N=ab^t

where

a=10^{1.478}

and

b=10^{0.0646}

Using a calculator,

a=30.0607\ldots

so, to the nearest integer,

a=30

Also,

b=1.160379\ldots

so, to 3 significant figures,

b=1.16

Hence

\boxed{N=30(1.16)^t}

to the requested accuracy.

(b)

Using the model from part (a),

N=30(1.16)^t

When

t=30

we get

N=30(1.16)^{30}

Using a calculator,

N=2581.82\ldots

Using the unrounded original model gives

N=10^{0.0646(30)+1.478}=2606.15\ldots

So, to 2 significant figures,

\boxed{N=2600\text{ mm}^2}

IAL 2020 Oct Q2

题目

解答

(a)

(b)