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IAL 2021 Oct Q9

A Level / Edexcel / P4

IAL 2021 Oct Paper · Question 9

题目

Problem

Figure 4 shows a cylindrical tank that contains some water.

The tank has an internal diameter of 88 m and an internal height of 4.24.2 m.

Water is flowing into the tank at a constant rate of (0.6π)(0.6\pi) m3^3 per minute.

There is a tap at point TT at the bottom of the tank.

At time tt minutes after the tap has been opened,

  • the depth of the water is hh metres
  • the water is leaving the tank at a rate of (0.15πh)(0.15\pi h) m3^3 per minute

(a) Show that

dhdt=12−3h320\frac{dh}{dt}=\frac{12-3h}{320}
(4)

Given that the depth of the water in the tank is 0.50.5 m when the tap is opened,

(b) find the time taken for the depth of water in the tank to reach 3.53.5 m.

(6)
题目中文翻译

图 4 显示了一个装有一些水的圆柱形水箱。

该水箱的内部直径为 88 m，内部高度为 4.24.2 m。

水正以恒定速率 (0.6π)(0.6\pi) m3^3 每分钟流入水箱。

水箱底部的点 TT 处有一个水龙头。

水龙头打开后 tt 分钟时，

  • 水深为 hh 米；
  • 水正以 (0.15πh)(0.15\pi h) m3^3 每分钟的速率流出水箱。

(a) 证明

dhdt=12−3h320\frac{dh}{dt}=\frac{12-3h}{320}

已知打开水龙头时水箱中的水深为 0.50.5 m，

(b) 求水深达到 3.53.5 m 所需的时间。

解答

(a)

解法一

思路

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圆柱水箱半径为 44 m，因此水的体积是 V=16πhV=16\pi h。水量的净变化率等于流入率减去流出率，再用链式法则 dVdt=dVdhdhdt\frac{\mathrm{d}V}{\mathrm{d}t}=\frac{\mathrm{d}V}{\mathrm{d}h}\frac{\mathrm{d}h}{\mathrm{d}t} 求出水深变化率。

答题过程

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The radius of the cylindrical tank is 44 m, so the volume of water is

V=π(42)h=16πh.V=\pi(4^2)h=16\pi h.

Hence,

dVdh=16π.\frac{\mathrm{d}V}{\mathrm{d}h}=16\pi.

The net rate of change of the volume is

dVdt=0.6π−0.15πh.\frac{\mathrm{d}V}{\mathrm{d}t} =0.6\pi-0.15\pi h.

Using the chain rule,

dVdt=dVdhdhdt,\frac{\mathrm{d}V}{\mathrm{d}t} =\frac{\mathrm{d}V}{\mathrm{d}h} \frac{\mathrm{d}h}{\mathrm{d}t},

so

0.6π−0.15πh=16πdhdt.0.6\pi-0.15\pi h =16\pi\frac{\mathrm{d}h}{\mathrm{d}t}.

Therefore,

dhdt= 0.6−0.15h16= 12−3h320,\begin{align*} \frac{\mathrm{d}h}{\mathrm{d}t} =&\,\frac{0.6-0.15h}{16}\\ =&\,\boxed{\frac{12-3h}{320}}, \end{align*}

as required.

(b)

解法一

思路

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承接 (a)，分离变量并积分，利用初始条件 t=0,h=0.5t=0,h=0.5 求积分常数，再代入 h=3.5h=3.5 求时间。对数相减可合并为 ln⁡7\ln7。

答题过程

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From part (a),

dhdt=12−3h320.\frac{\mathrm{d}h}{\mathrm{d}t} =\frac{12-3h}{320}.

Separating variables gives

∫112−3h dh=∫1320 dt.\int\frac{1}{12-3h}\,\mathrm{d}h =\int\frac{1}{320}\,\mathrm{d}t.

Therefore,

−13ln⁡(12−3h)=t320+C.-\frac13\ln(12-3h) =\frac{t}{320}+C.

When t=0t=0 and h=0.5h=0.5,

C=−13ln⁡(10.5).C=-\frac13\ln(10.5).

Thus,

−13ln⁡(12−3h)=t320−13ln⁡(10.5).-\frac13\ln(12-3h) =\frac{t}{320}-\frac13\ln(10.5).

When h=3.5h=3.5,

−13ln⁡(1.5)=t320−13ln⁡(10.5).-\frac13\ln(1.5) =\frac{t}{320}-\frac13\ln(10.5).

Hence,

t320= 13ln⁡(10.51.5)= 13ln⁡7,\begin{align*} \frac{t}{320} =&\,\frac13\ln\bigg(\frac{10.5}{1.5}\bigg)\\ =&\,\frac13\ln7, \end{align*}

so

t= 3203ln⁡7= 207.56…\begin{align*} t =&\,\frac{320}{3}\ln7\\ =&\,207.56\ldots \end{align*}

Therefore, the required time is

208 minutes\boxed{208\text{ minutes}}

to the nearest minute.

解法二

思路

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也可以在分离变量后直接把初始状态和目标状态作为定积分上下限，这样无需另求积分常数，并能直接得到所需时间。

答题过程

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Using the initial and final conditions directly,

∫0.53.532012−3h dh=∫0t1 dτ.\int_{0.5}^{3.5}\frac{320}{12-3h}\,\mathrm{d}h =\int_0^t 1\,\mathrm{d}\tau.

Hence,

t= [−3203ln⁡(12−3h)]0.53.5= −3203(ln⁡1.5−ln⁡10.5)= 3203ln⁡7= 207.56…\begin{align*} t =&\,\bigg[-\frac{320}{3} \ln(12-3h)\bigg]_{0.5}^{3.5}\\ =&\,-\frac{320}{3} \big(\ln1.5-\ln10.5\big)\\ =&\,\frac{320}{3}\ln7\\ =&\,207.56\ldots \end{align*}

Therefore,

t=208 minutes\boxed{t=208\text{ minutes}}

to the nearest minute.