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IAL 2024 Oct Q9

A Level / Edexcel / P4

IAL 2024 Oct Paper · Question 9

题目

Problem

(a) Express

1x(2x−1)\frac{1}{x(2x-1)}

in partial fractions.

(2)

The height above ground, hh metres, of a carriage on a fairground ride is modelled by the differential equation

dhdt=150h(2h−1)cos⁡(t10)\frac{dh}{dt}=\frac{1}{50}h(2h-1)\cos\left(\frac{t}{10}\right)

where tt seconds is the time after the start of the ride.

Given that, at the start of the ride, the carriage is 2.52.5 m above ground,

(b) solve the differential equation to show that, according to the model,

h=510−8eksin⁡(t10)h=\frac{5}{10-8e^{k\sin\left(\frac{t}{10}\right)}}

where kk is a constant to be found.

(6)

(c) Hence find, according to the model, the time taken for the carriage to reach its maximum height above ground for the 3rd time.

Give your answer to the nearest second.

(Solutions relying entirely on calculator technology are not acceptable.)

(2)
题目中文翻译

(a) 将

1x(2x−1)\frac{1}{x(2x-1)}

分解为部分分式。

游乐场设施中某个吊舱离地高度为 hh 米，其变化由微分方程

dhdt=150h(2h−1)cos⁡(t10)\frac{dh}{dt}=\frac{1}{50}h(2h-1)\cos\left(\frac{t}{10}\right)

建模，其中 tt 秒为开始乘坐后的时间。

已知开始时吊舱离地 2.52.5 m，

(b) 解该微分方程并证明，根据模型，

h=510−8eksin⁡(t10)h=\frac{5}{10-8e^{k\sin\left(\frac{t}{10}\right)}}

其中 kk 是待求常数。

(c) 进而根据该模型，求吊舱第 3 次达到最大离地高度所用的时间。

答案取最接近的整秒。

（不接受完全依赖计算器技术的解法。）

解答

(a)

Write

1x(2x−1)=Ax+B2x−1\frac{1}{x(2x-1)} =\frac{A}{x}+\frac{B}{2x-1}

Then

1=A(2x−1)+Bx1=A(2x-1)+Bx

Let x=0x=0:

1=−A1=-A

so

A=−1A=-1

Let x=12x=\dfrac12:

1=12B1=\frac12B

so

B=2B=2

Therefore

1x(2x−1)=−1x+22x−1\boxed{ \frac{1}{x(2x-1)} =-\frac1x+\frac{2}{2x-1} }

(b)

We are given

dhdt=150h(2h−1)cos⁡(t10)\frac{dh}{dt} =\frac{1}{50}h(2h-1)\cos\left(\frac{t}{10}\right)

Separate variables:

1h(2h−1) dh=150cos⁡(t10) dt\frac{1}{h(2h-1)}\,dh =\frac{1}{50}\cos\left(\frac{t}{10}\right)\,dt

Using the result from part (a),

1h(2h−1)=−1h+22h−1\frac{1}{h(2h-1)} =-\frac1h+\frac{2}{2h-1}

So

∫(−1h+22h−1) dh=∫150cos⁡(t10) dt\int\left(-\frac1h+\frac{2}{2h-1}\right)\,\mathrm{d}h =\int \frac{1}{50}\cos\left(\frac{t}{10}\right)\,\mathrm{d}t

Integrating gives

−ln⁡h+ln⁡(2h−1)=15sin⁡(t10)+C-\ln h+\ln(2h-1) =\frac15\sin\left(\frac{t}{10}\right)+C

Hence

ln⁡(2h−1h)=15sin⁡(t10)+C\ln\left(\frac{2h-1}{h}\right) =\frac15\sin\left(\frac{t}{10}\right)+C

Exponentiate both sides:

2h−1h=Ae15sin⁡(t10)\frac{2h-1}{h} =Ae^{\frac15\sin\left(\frac{t}{10}\right)}

where AA is a constant.

At the start of the ride,

t=0,h=2.5=52t=0,\qquad h=2.5=\frac52

Substitute these values:

2(52)−152=Ae0\frac{2\left(\frac52\right)-1}{\frac52} =Ae^0

So

452=A\frac{4}{\frac52}=A

and therefore

A=85A=\frac85

Thus

2h−1h=85e15sin⁡(t10)\frac{2h-1}{h} =\frac85 e^{\frac15\sin\left(\frac{t}{10}\right)}

Now rearrange to make hh the subject:

2−1h=85e15sin⁡(t10)2-\frac1h =\frac85 e^{\frac15\sin\left(\frac{t}{10}\right)}

So

1h=2−85e15sin⁡(t10)\frac1h =2-\frac85 e^{\frac15\sin\left(\frac{t}{10}\right)}

Therefore

h=12−85e15sin⁡(t10)h =\frac{1}{2-\frac85 e^{\frac15\sin\left(\frac{t}{10}\right)}}

Multiplying numerator and denominator by 55,

h=510−8e15sin⁡(t10)h =\frac{5}{10-8e^{\frac15\sin\left(\frac{t}{10}\right)}}

This is in the required form, so

k=15\boxed{k=\frac15}

(c)

From part (b),

h=510−8e15sin⁡(t10)h =\frac{5}{10-8e^{\frac15\sin\left(\frac{t}{10}\right)}}

To maximise hh, the denominator must be minimised.

Since

e15sin⁡(t10)e^{\frac15\sin\left(\frac{t}{10}\right)}

is largest when

sin⁡(t10)=1\sin\left(\frac{t}{10}\right)=1

the maximum heights occur when

t10=π2, 5π2, 9π2,…\frac{t}{10}=\frac{\pi}{2},\ \frac{5\pi}{2},\ \frac{9\pi}{2},\ldots

The 3rd time is therefore

t10=9π2\frac{t}{10}=\frac{9\pi}{2}

So

t=45πt=45\pi

Now

45π=141.37…45\pi=141.37\ldots

Hence the time taken is

141 seconds\boxed{141\text{ seconds}}

to the nearest second.