题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

(i) The volume, $V$ , of a spherical balloon is increasing at a constant rate of $70\pi\text{ cm}^3\text{ s}^{-1}$ .

Find the rate of increase of the radius of the balloon, in $\text{cm s}^{-1}$ , at the instant when the radius of the balloon is $5\text{ cm}$ .

\text{[The volume }V\text{ of a sphere of radius }r\text{ is given by the formula }V=\frac{4}{3}\pi r^3\text{.]}

(4)

(ii) The depth of water in a cave is being monitored.

The rate of increase in the depth of water, $h$ cm, at a particular point in the cave is modelled by the differential equation

\frac{dh}{dt}=\frac{k}{h^3}

where $k$ is a constant and $t$ hours is the time after monitoring began.

Given that

initially the depth of water was $4\text{ cm}$
$5$ hours after monitoring began, the depth of water was $6\text{ cm}$
$T$ hours after monitoring began, the depth of water was $10\text{ cm}$

solve the differential equation to find the value of $T$ .

Give your answer to one decimal place.

(6)

题目中文翻译

本题必须写出全部解题步骤。

不接受完全依赖计算器技术的解法。

(i) 一个球形气球的体积 $V$ 以恒定速率 $70\pi\text{ cm}^3\text{ s}^{-1}$ 增加。

求当气球半径为 $5\text{ cm}$ 时，其半径增加的速率，单位为 $\text{cm s}^{-1}$ 。

\text{[半径为 }r\text{ 的球体体积 }V\text{ 的公式为 }V=\frac{4}{3}\pi r^3\text{。]}

(ii) 正在监测洞穴中的水深。

洞穴中某一点处，水深 $h$ （单位：cm）的增长率由微分方程

\frac{dh}{dt}=\frac{k}{h^3}

建模，其中 $k$ 为常数， $t$ （单位：小时）为开始监测后的时间。

已知：

初始时水深为 $4\text{ cm}$ ；
开始监测 $5$ 小时后，水深为 $6\text{ cm}$ ；
开始监测 $T$ 小时后，水深为 $10\text{ cm}$ 。

解这个微分方程，求 $T$ 的值。

答案保留到 1 位小数。

解答

(i)

The volume is increasing at the rate

\frac{dV}{dt}=70\pi

For a sphere,

V=\frac43\pi r^3

求导得

\frac{dV}{dr}=4\pi r^2

Using the chain rule,

\frac{dV}{dt} =\frac{dV}{dr}\cdot\frac{dr}{dt}

At the instant when

r=5

we have

70\pi=4\pi(5)^2\frac{dr}{dt}

70\pi=100\pi\frac{dr}{dt}

Therefore

\frac{dr}{dt}=\frac{70}{100}=\frac{7}{10}

Hence the rate of increase of the radius is

\boxed{0.7\text{ cm s}^{-1}}

(ii)

We are given

\frac{dh}{dt}=\frac{k}{h^3}

Separate variables:

h^3\,dh=k\,dt

Integrate both sides:

\int h^3\,dh=\int k\,dt

\frac14h^4=kt+C

Initially,

t=0,\qquad h=4

Substitute these values:

\frac14(4^4)=C

Thus

C=64

\frac14h^4=kt+64

After $5$ hours,

h=6

\frac14(6^4)=5k+64

Now

\frac14(6^4)=324

Therefore

324=5k+64

5k=260

and hence

k=52

Now when $t=T$ ,

h=10

\frac14(10^4)=52T+64

Thus

2500=52T+64

Hence

52T=2436

and

T=\frac{2436}{52}=46.846\ldots

Therefore

\boxed{T=46.8}

to one decimal place.

IAL 2025 Jan Q4

题目

解答

(i)

(ii)