1.08k Separable differential equations: dy/dx = f(x)g(y)

2. Find the particular solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 y ^ { 2 } } { \sqrt { 4 x + 5 } } \quad x > - \frac { 5 } { 4 }$$ for which \(y = \frac { 1 } { 3 }\) at \(x = - \frac { 1 } { 4 }\) giving your answer in the form \(y = \mathrm { f } ( x )\) (6)

9. \begin{figure}[h] \end{figure} Figure 4 shows a cylindrical tank that contains some water. The tank has an internal diameter of 8 m and an internal height of 4.2 m .
Water is flowing into the tank at a constant rate of \(( 0.6 \pi ) \mathrm { m } ^ { 3 }\) per minute. There is a tap at point \(T\) at the bottom of the tank. At time \(t\) minutes after the tap has been opened,

• the depth of the water is \(h\) metres
• the water is leaving the tank at a rate of \(( 0.15 \pi h ) \mathrm { m } ^ { 3 }\) per minute
  1. Show that

$$\frac { \mathrm { d } h } { \mathrm {~d} t } = \frac { 12 - 3 h } { 320 }$$ Given that the depth of the water in the tank is 0.5 m when the tap is opened,

find the time taken for the depth of water in the tank to reach 3.5 m .

1. The number of goats on an island is being monitored.

When monitoring began there were 3000 goats on the island.
In a simple model, the number of goats, \(x\), in thousands, is modelled by the equation $$x = \frac { k ( 9 t + 5 ) } { 4 t + 3 }$$ where \(k\) is a constant and \(t\) is the number of years after monitoring began.

1. Show that \(k = 1.8\)
2. Hence calculate the long-term population of goats predicted by this model. In a second model, the number of goats, \(x\), in thousands, is modelled by the differential equation $$3 \frac { \mathrm {~d} x } { \mathrm {~d} t } = x ( 9 - 2 x )$$
3. Write \(\frac { 3 } { x ( 9 - 2 x ) }\) in partial fraction form.
4. Solve the differential equation with the initial condition to show that $$x = \frac { 9 } { 2 + \mathrm { e } ^ { - 3 t } }$$
5. Find the long-term population of goats predicted by this second model.

8. Water is being heated in a kettle. At time \(t\) seconds, the temperature of the water is \(\theta ^ { \circ } \mathrm { C }\). The rate of increase of the temperature of the water at time \(t\) is modelled by the differential equation $$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = \lambda ( 120 - \theta ) \quad \theta \leqslant 100$$ where \(\lambda\) is a positive constant.
Given that \(\theta = 20\) when \(t = 0\)

1. solve this differential equation to show that $$\theta = 120 - 100 \mathrm { e } ^ { - \lambda t }$$ When the temperature of the water reaches \(100 ^ { \circ } \mathrm { C }\), the kettle switches off.
2. Given that \(\lambda = 0.01\), find the time, to the nearest second, when the kettle switches off.
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8

1. Solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 - x } { y - 3 }$$ giving the particular solution that satisfies the condition \(y = 4\) when \(x = 5\).
2. Show that this particular solution can be expressed in the form $$( x - a ) ^ { 2 } + ( y - b ) ^ { 2 } = k$$ where the values of the constants \(a , b\) and \(k\) are to be stated.
3. Hence sketch the graph of the particular solution, indicating clearly its main features.

8 Water flows out of a tank through a hole in the bottom and, at time \(t\) minutes, the depth of water in the tank is \(x\) metres. At any instant, the rate at which the depth of water in the tank is decreasing is proportional to the square root of the depth of water in the tank.

1. Write down a differential equation which models this situation.
2. When \(t = 0 , x = 2\); when \(t = 5 , x = 1\). Find \(t\) when \(x = 0.5\), giving your answer correct to 1 decimal place.

8 The height, \(h\) metres, of a shrub \(t\) years after planting is given by the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = \frac { 6 - h } { 20 }$$ A shrub is planted when its height is 1 m .

1. Show by integration that \(t = 20 \ln \left( \frac { 5 } { 6 - h } \right)\).
2. How long after planting will the shrub reach a height of 2 m ?
3. Find the height of the shrub 10 years after planting.
4. State the maximum possible height of the shrub.

12
0
5 \end{array} \right) + s \left( \begin{array} { r } 1
- 4
- 2 \end{array} \right) .$$

1. Show that the lines intersect.
2. Find the angle between the lines.

\\

1. Show that, if \(y = \operatorname { cosec } x\), then \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) can be expressed as \(- \operatorname { cosec } x \cot x\).
2. Solve the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = - \sin x \tan x \cot t$$ given that \(x = \frac { 1 } { 6 } \pi\) when \(t = \frac { 1 } { 2 } \pi\).

8\\

1. Given that \(\frac { 2 t } { ( t + 1 ) ^ { 2 } }\) can be expressed in the form \(\frac { A } { t + 1 } + \frac { B } { ( t + 1 ) ^ { 2 } }\), find the values of the constants \(A\) and \(B\).
2. Show that the substitution \(t = \sqrt { 2 x - 1 }\) transforms \(\int \frac { 1 } { x + \sqrt { 2 x - 1 } } \mathrm {~d} x\) to \(\int \frac { 2 t } { ( t + 1 ) ^ { 2 } } \mathrm {~d} t\).
3. Hence find the exact value of \(\int _ { 1 } ^ { 5 } \frac { 1 } { x + \sqrt { 2 x - 1 } } \mathrm {~d} x\). 9 The parametric equations of a curve are $$x = 2 \theta + \sin 2 \theta , \quad y = 4 \sin \theta$$ and part of its graph is shown below. \includegraphics[max width=\textwidth, alt={}, center]{b8ba126f-c5fa-4828-9439-e5162a03ca5b-3_646_1150_1050_500}

1. Find the value of \(\theta\) at \(A\) and the value of \(\theta\) at \(B\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \sec \theta\).
3. At the point \(C\) on the curve, the gradient is 2 . Find the coordinates of \(C\), giving your answer in an exact form.

9 \includegraphics[max width=\textwidth, alt={}, center]{798da17d-0af5-4aa6-b731-564642dc28d5-4_572_917_294_607} A cylindrical container has a height of 200 cm . The container was initially full of a chemical but there is a leak from a hole in the base. When the leak is noticed, the container is half-full and the level of the chemical is dropping at a rate of 1 cm per minute. It is required to find for how many minutes the container has been leaking. To model the situation it is assumed that, when the depth of the chemical remaining is \(x \mathrm {~cm}\), the rate at which the level is dropping is proportional to \(\sqrt { } x\). Set up and solve an appropriate differential equation, and hence show that the container has been leaking for about 80 minutes.

6

1. Express \(\frac { 1 } { ( 2 x + 1 ) ( x + 1 ) }\) in partial fractions.
2. A curve passes through the point \(( 0,2 )\) and satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y } { ( 2 x + 1 ) ( x + 1 ) }$$ Show by integration that \(y = \frac { 4 x + 2 } { x + 1 }\). Section B (36 marks)

8 A curve has equation $$x ^ { 2 } + 4 y ^ { 2 } = k ^ { 2 } ,$$ where \(k\) is a positive constant.

1. Verify that $$x = k \cos \theta , \quad y = \frac { 1 } { 2 } k \sin \theta ,$$ are parametric equations for the curve.
2. Hence or otherwise show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { x } { 4 y }\).
3. Fig. 8 illustrates the curve for a particular value of \(k\). Write down this value of \(k\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{9a8332ec-2216-4e1f-9768-ef175c9e159b-4_657_1071_938_577} \captionsetup{labelformat=empty} \caption{Fig. 8}
   \end{figure}
4. Copy Fig. 8 and on the same axes sketch the curves for \(k = 1 , k = 3\) and \(k = 4\). On a map, the curves represent the contours of a mountain. A stream flows down the mountain. Its path on the map is always at right angles to the contour it is crossing.
5. Explain why the path of the stream is modelled by the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 y } { x } .$$
6. Solve this differential equation. Given that the path of the stream passes through the point \(( 2,1 )\), show that its equation is \(y = \frac { x ^ { 4 } } { 16 }\).

7 Data suggest that the number of cases of infection from a particular disease tends to oscillate between two values over a period of approximately 6 months.

1. Suppose that the number of cases, \(P\) thousand, after time \(t\) months is modelled by the equation \(P = \frac { 2 } { 2 - \sin t }\). Thus, when \(t = 0 , P = 1\).
   1. By considering the greatest and least values of \(\sin t\), write down the greatest and least values of \(P\) predicted by this model.
   2. Verify that \(P\) satisfies the differential equation \(\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } P ^ { 2 } \cos t\).
2. An alternative model is proposed, with differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } \left( 2 P ^ { 2 } - P \right) \cos t$$ As before, \(P = 1\) when \(t = 0\).
   1. Express \(\frac { 1 } { P ( 2 P - 1 ) }\) in partial fractions.
   2. Solve the differential equation (*) to show that $$\ln \left( \frac { 2 P - 1 } { P } \right) = \frac { 1 } { 2 } \sin t$$ This equation can be rearranged to give \(P = \frac { 1 } { 2 - \mathrm { e } ^ { \frac { 1 } { 2 } \sin t } }\).
   3. Find the greatest and least values of \(P\) predicted by this model. \begin{figure}[h]
      \includegraphics[alt={},max width=\textwidth]{9296c786-a42a-4aa5-b326-39adbb544cbc-05_609_622_301_719} \captionsetup{labelformat=empty} \caption{Fig. 8}
      \end{figure} In a theme park ride, a capsule C moves in a vertical plane (see Fig. 8). With respect to the axes shown, the path of C is modelled by the parametric equations $$x = 10 \cos \theta + 5 \cos 2 \theta , \quad y = 10 \sin \theta + 5 \sin 2 \theta , \quad ( 0 \leqslant \theta < 2 \pi )$$ where \(x\) and \(y\) are in metres.
      1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { \cos \theta + \cos 2 \theta } { \sin \theta + \sin 2 \theta }\). Verify that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(\theta = \frac { 1 } { 3 } \pi\). Hence find the exact coordinates of the highest point A on the path of C .
      2. Express \(x ^ { 2 } + y ^ { 2 }\) in terms of \(\theta\). Hence show that $$x ^ { 2 } + y ^ { 2 } = 125 + 100 \cos \theta$$
      3. Using this result, or otherwise, find the greatest and least distances of C from O . You are given that, at the point B on the path vertically above O , $$2 \cos ^ { 2 } \theta + 2 \cos \theta - 1 = 0$$
      4. Using this result, and the result in part (ii), find the distance OB. Give your answer to 3 significant figures. \section*{ADVANCED GCE UNIT MATHEMATICS (MEI)} Applications of Advanced Mathematics (C4) \section*{Paper B: Comprehension} \section*{THURSDAY 14 JUNE 2007} Afternoon
         Time: Up to 1 hour
         Additional materials:
         Rough paper
         MEI Examination Formulae and Tables (MF2) Candidate
         Name □
         Centre
         Number sufficient detail of the working to indicate that a correct method is being used. 1 This basic cycloid has parametric equations $$x = a \theta - a \sin \theta , \quad y = a - a \cos \theta$$
         \includegraphics[max width=\textwidth, alt={}]{9296c786-a42a-4aa5-b326-39adbb544cbc-10_307_1138_445_411}
         Find the coordinates of the points M and N , stating the value of \(\theta\) at each of them. Point M Point N 2 A sea wave has parametric equations (in suitable units) $$x = 7 \theta - 0.25 \sin \theta , \quad y = 0.25 \cos \theta$$ Find the wavelength and height of the wave.
         3 The graph below shows the profile of a wave.
         1. Assuming that it has parametric equations of the form given on line 68 , find the values of \(a\) and \(b\).
         2. Investigate whether the ratio of the trough length to the crest length is consistent with this shape. \includegraphics[max width=\textwidth, alt={}, center]{9296c786-a42a-4aa5-b326-39adbb544cbc-11_312_1141_623_415}
         3. \(\_\_\_\_\)
         4. \(\_\_\_\_\) 4 This diagram illustrates two wave shapes \(U\) and \(V\). They have the same wavelength and the same height. \includegraphics[max width=\textwidth, alt={}, center]{9296c786-a42a-4aa5-b326-39adbb544cbc-12_423_1552_356_205} One of the curves is a sine wave, the other is a curtate cycloid.
         5. State which is which, justifying your answer.
         6. \(\_\_\_\_\) The parametric equations for the curves are: $$x = a \theta , \quad y = b \cos \theta ,$$ and $$x = a \theta - b \sin \theta , \quad y = b \cos \theta .$$
         7. Show that the distance marked \(d\) on the diagram is equal to \(b\).
         8. Hence justify the statement in lines 109 to 111: "In such cases, the curtate cycloid and the sine curve with the same wavelength and height are very similar and so the sine curve is also a good model."
         9. \(\_\_\_\_\)
         10. \(\_\_\_\_\) 5 The diagram shows a curtate cycloid with scales given. Show that this curve could not be a scale drawing of the shape of a stable sea wave. \includegraphics[max width=\textwidth, alt={}, center]{9296c786-a42a-4aa5-b326-39adbb544cbc-13_289_1310_397_331}

8 The new price of a particular make of car is \(\pounds 10000\). When its age is \(t\) years, the list price is \(\pounds V\). When \(t = 5 , V = 5000\). Aloke, Ben and Charlie all run outlets for used cars. Each of them has a different model for the depreciation.

1. Aloke claims that the rate of depreciation is constant. Write this claim as a differential equation.
   Solve the differential equation and hence find the value of a car that is 7 years old according to this model.
   Explain why this model breaks down for large \(t\).
2. Ben believes that the rate of depreciation is inversely proportional to the square root of the age of the car. Express this claim as a differential equation and hence find the value of a car that is 7 years old according to this model.
   Does this model ever break down?
3. Charlie believes that a better model is given by the differential equation $$\frac { \mathrm { d } V } { \mathrm {~d} t } = k V$$ Solve this differential equation and find the value of the car after 7 years according to this model.
   Does this model ever break down?
4. Further investigation reveals that the average value of this particular type of car when 8 years old is \(\pounds 3000\). Find the value of \(V\) when \(t = 8\) for the three models above. Which of the three models best predicts the value of \(V\) at this time?

7 Solve the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x } { y }\) given that when \(x = 1 , y = 2\).

9 Two astronomers wish to model the path of motion of a particle in two dimensions.
Experimental results show that the position of the particle can be found using the parametric equations $$x = 2 \cos \theta - \sin \theta + 2 \quad y = \cos \theta + 2 \sin \theta - 1 \quad \left( 0 \leq \theta \leq 360 ^ { \circ } \right)$$ One astronomer uses trigonometry.

1. Express \(2 \cos \theta - \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R\) and \(\alpha\) are constants to be determined. Show also that, for the same values of \(R\) and \(\alpha\), $$\cos \theta + 2 \sin \theta = R \sin ( \theta + \alpha )$$
2. Hence, or otherwise, show that the path of particle may be written in the form $$( x - 2 ) ^ { 2 } + ( y + 1 ) ^ { 2 } = 5$$ Describe the path of the particle. The second astronomer sets up a first order differential equation with the condition that \(x = 4\) when \(y = 0\).
3. Verify that the point with parameter \(\theta = 0\) has coordinates \(( 4,0 )\).
4. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). Deduce that \(x\) and \(y\) satisfy the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { x - 2 } { y + 1 }$$
5. Solve this differential equation, using the condition that \(y = 0\) when \(x = 4\). Hence show that the two solutions give the same cartesian equation for the path of particle.

1. During a chemical reaction, a compound is being made from two other substances.

At time \(t\) hours after the start of the reaction, \(x \mathrm {~g}\) of the compound has been produced. Assuming that \(x = 0\) initially, and that $$\frac { \mathrm { d } x } { \mathrm {~d} t } = 2 ( x - 6 ) ( x - 3 )$$

1. show that it takes approximately 7 minutes to produce 2 g of the compound.
2. Explain why it is not possible to produce 3 g of the compound.
   

1. A bath is filled with hot water which is allowed to cool. The temperature of the water is \(\theta ^ { \circ } \mathrm { C }\) after cooling for \(t\) minutes and the temperature of the room is assumed to remain constant at \(20 ^ { \circ } \mathrm { C }\).

Given that the rate at which the temperature of the water decreases is proportional to the difference in temperature between the water and the room,

1. write down a differential equation connecting \(\theta\) and \(t\). Given also that the temperature of the water is initially \(37 ^ { \circ } \mathrm { C }\) and that it is \(36 ^ { \circ } \mathrm { C }\) after cooling for four minutes,
2. find, to 3 significant figures, the temperature of the water after ten minutes. Advice suggests that the temperature of the water should be allowed to cool to \(33 ^ { \circ } \mathrm { C }\) before a child gets in.
3. Find, to the nearest second, how long a child should wait before getting into the bath.
   

5. Given that \(y = - 2\) when \(x = 1\), solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = y ^ { 2 } \sqrt { x }$$ giving your answer in the form \(y = \mathrm { f } ( x )\).

8. When a plague of locusts attacks a wheat crop, the proportion of the crop destroyed after \(t\) hours is denoted by \(x\). In a model, it is assumed that the rate at which the crop is destroyed is proportional to \(x ( 1 - x )\). A plague of locusts is discovered in a wheat crop when one quarter of the crop has been destroyed. Given that the rate of destruction at this instant is such that if it remained constant, the crop would be completely destroyed in a further six hours,

1. show that \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 2 } { 3 } x ( 1 - x )\),
2. find the percentage of the crop destroyed three hours after the plague of locusts is first discovered.

8. A small town had a population of 9000 in the year 2001. In a model, it is assumed that the population of the town, \(P\), at time \(t\) years after 2001 satisfies the differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } = 0.05 P \mathrm { e } ^ { - 0.05 t }$$

1. Show that, according to the model, the population of the town in 2011 will be 13300 to 3 significant figures.
2. Find the value which the population of the town will approach in the long term, according to the model.

7. A mathematician is selling goods at a car boot sale. She believes that the rate at which she makes sales depends on the length of time since the start of the sale, \(t\) hours, and the total value of sales she has made up to that time, \(\pounds x\). She uses the model $$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { k ( 5 - t ) } { x }$$ where \(k\) is a constant.
Given that after two hours she has made sales of \(\pounds 96\) in total,

1. solve the differential equation and show that she made \(\pounds 72\) in the first hour of the sale. The mathematician believes that is it not worth staying at the sale once she is making sales at a rate of less than \(\pounds 10\) per hour.
2. Verify that at 3 hours and 5 minutes after the start of the sale, she should have already left.

1. The gradient at any point \(( x , y )\) on a curve is proportional to \(\sqrt { y }\).

Given that the curve passes through the point with coordinates \(( 0,4 )\),

1. show that the equation of the curve can be written in the form $$2 \sqrt { y } = k x + 4$$ where \(k\) is a positive constant. Given also that the curve passes through the point with coordinates ( 2,9 ),
2. find the equation of the curve in the form \(y = \mathrm { f } ( x )\).
   

6. The number of people, \(n\), in a queue at a Post Office \(t\) minutes after it opens is modelled by the differential equation $$\frac { \mathrm { d } n } { \mathrm {~d} t } = \mathrm { e } ^ { 0.5 t } - 5 , \quad t \geq 0$$

1. Find, to the nearest second, the time when the model predicts that there will be the least number of people in the queue.
2. Given that there are 20 people in the queue when the Post Office opens, solve the differential equation.
3. Explain why this model would not be appropriate for large values of \(t\).

7

1. Express \(\frac { 3 } { ( y - 2 ) ( y + 1 ) }\) in partial fractions.
   [0pt] [3]
2. Hence, given that \(x\) and \(y\) satisfy the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = x ^ { 2 } ( y - 2 ) ( y + 1 )$$ show that \(\frac { y - 2 } { y + 1 } = A \mathrm { e } ^ { x ^ { 3 } }\), where \(A\) is a constant.