CalcGospel 國際數學圖譜
返回

P4 2025 Jan Q4

A Level / Edexcel / P4

IAL 2025 Jan Paper · Question 4

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, VV, of a spherical balloon is increasing at a constant rate of 70π cm3 s−170\pi\text{ cm}^3\text{ s}^{-1}.

Find the rate of increase of the radius of the balloon, in cm s−1\text{cm s}^{-1}, at the instant when the radius of the balloon is 5 cm5\text{ cm}.

[The volume VV of a sphere of radius rr is given by the formula V=43πr3V=\frac43\pi r^3]

(4)

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

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

dhdt=kh3\frac{\mathrm{d}h}{\mathrm{d}t}=\frac{k}{h^3}

where kk is a constant and tt hours is the time after monitoring began.

Given that

  • initially the depth of water was 4 cm4\text{ cm}
  • 55 hours after monitoring began, the depth of water was 6 cm6\text{ cm}
  • TT hours after monitoring began, the depth of water was 10 cm10\text{ cm}

solve the differential equation to find the value of TT.

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 VV 以恒定速率 70π cm3 s−170\pi\text{ cm}^3\text{ s}^{-1} 增加。

求当气球半径为 5 cm5\text{ cm} 时，其半径增加的速率，单位为 cm s−1\text{cm s}^{-1}。

[半径为 rr 的球体体积 VV 的公式为 V=43πr3V=\frac43\pi r^3]

(4)

(ii) 正在监测洞穴中的水深。

洞穴中某一点处，水深 hh（单位：cm）的增长率由微分方程

dhdt=kh3\frac{\mathrm{d}h}{\mathrm{d}t}=\frac{k}{h^3}

建模，其中 kk 为常数，tt（单位：小时）为开始监测后的时间。

已知：

  • 初始时水深为 4 cm4\text{ cm}；
  • 开始监测 55 小时后，水深为 6 cm6\text{ cm}；
  • 开始监测 TT 小时后，水深为 10 cm10\text{ cm}。

解这个微分方程，求 TT 的值。

答案保留到 1 位小数。

(6)

解答

(i)

解法一

思路

展开

用球体体积对半径的导数和链式法则，连接体积增长率与半径增长率，再代入 r=5r=5。

答题过程

展开

The volume is increasing at the rate

dVdt=70π\frac{\mathrm{d}V}{\mathrm{d}t}=70\pi

For a sphere,

V=43πr3V=\frac43\pi r^3

求导得

dVdr=4πr2\frac{\mathrm{d}V}{\mathrm{d}r}=4\pi r^2

Using the chain rule,

dVdt=dVdr⋅drdt\frac{\mathrm{d}V}{\mathrm{d}t} =\frac{\mathrm{d}V}{\mathrm{d}r}\cdot\frac{\mathrm{d}r}{\mathrm{d}t}

At the instant when

r=5r=5

we have

70π=4π(5)2drdt70\pi=4\pi(5)^2\frac{\mathrm{d}r}{\mathrm{d}t}

So

70π=100πdrdt70\pi=100\pi\frac{\mathrm{d}r}{\mathrm{d}t}

Therefore

drdt=70100=710\frac{\mathrm{d}r}{\mathrm{d}t}=\frac{70}{100}=\frac{7}{10}

Hence the rate of increase of the radius is

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

也可先由恒定体积增长率写 V=70πt+CV=70\pi t+C，再将球体公式代入：

43πr3=70πt+C.\frac43\pi r^3=70\pi t+C.

对时间求导同样得到 4πr2 dr/dt=70π4\pi r^2\,\mathrm{d}r/\mathrm{d}t=70\pi。常数的导数为零，不需要假设初始体积为零。

(ii)

解法一

思路

展开

分离变量并积分，先用初始水深确定积分常数，再用五小时后的水深求 kk，最后代入 h=10h=10。

答题过程

展开

We are given

dhdt=kh3\frac{\mathrm{d}h}{\mathrm{d}t}=\frac{k}{h^3}

Separate variables:

h3 dh=k dth^3\,\mathrm{d}h=k\,\mathrm{d}t

Integrate both sides:

∫h3 dh=∫k dt\int h^3\,\mathrm{d}h=\int k\,\mathrm{d}t

So

14h4=kt+C\frac14h^4=kt+C

Initially,

t=0,h=4t=0,\qquad h=4

Substitute these values:

14(44)=C\frac14(4^4)=C

Thus

C=64C=64

So

14h4=kt+64\frac14h^4=kt+64

After 55 hours,

h=6h=6

so

14(64)=5k+64\frac14(6^4)=5k+64

Now

14(64)=324\frac14(6^4)=324

Therefore

324=5k+64324=5k+64

So

5k=2605k=260

and hence

k=52k=52

Now when t=Tt=T,

h=10h=10

So

14(104)=52T+64\frac14(10^4)=52T+64

Thus

2500=52T+642500=52T+64

Hence

52T=243652T=2436

and

T=243652=46.846…T=\frac{2436}{52}=46.846\ldots

Therefore

T=46.8\boxed{T=46.8}

to one decimal place.

解法二

思路

展开

将初始和五小时后的水深直接作为定积分上下限，先求 kk；再积分至 h=10h=10 求 TT。

答题过程

展开

Separate variables:

h3 dh=k dt.h^3\,\mathrm{d}h=k\,\mathrm{d}t.

Between t=0t=0 and t=5t=5, the depth increases from 44 to 66. Hence

∫46h3 dh=∫05k dt.\int_4^6h^3\,\mathrm{d}h=\int_0^5k\,\mathrm{d}t.

Evaluating each side,

[14h4]46= 64−444= 324−64= 260,[kt]05= 5k.\begin{align*} \Big[\tfrac14h^4\Big]_4^6 =&\,\frac{6^4-4^4}{4}\\ =&\,324-64\\ =&\,260,\\ [kt]_0^5=&\,5k. \end{align*}

Thus 5k=2605k=260, so k=52k=52.

Between t=0t=0 and t=Tt=T, the depth increases from 44 to 1010, so

∫410h3 dh=∫0T52 dt.\int_4^{10}h^3\,\mathrm{d}h=\int_0^T52\,\mathrm{d}t.

Therefore

104−444= 52T,2500−64= 52T,T= 243652= 46.846….\begin{align*} \frac{10^4-4^4}{4}=&\,52T,\\ 2500-64=&\,52T,\\ T=&\,\frac{2436}{52}\\ =&\,46.846\ldots. \end{align*}

Hence T=46.8T=46.8 hours to one decimal place.

若将 kk 移到左边分离变量，积分得到 h4/(4k)=t+C1h^4/(4k)=t+C_1。用同样两组条件仍得 k=52k=52，于是 h4/208=t+16/13h^4/208=t+16/13，与第一法方程等价。