Differential equations

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

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

It is given that \(\theta = \tan^{-1}\left(\frac{3x}{2}\right)\).

1. By writing \(\theta = \tan^{-1}\left(\frac{3x}{2}\right)\) as \(2\tan\theta = 3x\), use implicit differentiation to show that $$\frac{d\theta}{dx} = \frac{k}{4 + 9x^2}$$, where \(k\) is an integer. [3 marks]
2. Hence solve the differential equation $$9y(4 + 9x^2)\frac{dy}{dx} = \cosec 3y$$ given that \(x = 0\) when \(y = \frac{\pi}{3}\). Give your answer in the form \(\mathbf{g}(y) = \mathbf{h}(x)\). [7 marks]

Find the general solution of the differential equation $$(2x^3 - 3x^2 - 11x + 6)\frac{dy}{dx} = y(20x - 35).$$ Give your answer in the form \(y = f(x)\). [9]

2. Given that \(x \geqslant 2\), find the general solution of the differential equation $$( 2 x - 3 ) ( x - 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } = ( 2 x - 1 ) y$$

The points \((x, y)\) on the curve \(C\) satisfy \((x + 1)(x + 2) \frac{dy}{dx} = xy\). The line with equation \(y = 2x + 5\) is the tangent to \(C\) at a point \(P\).

1. Find the coordinates of \(P\). [4]
2. Find the equation of \(C\), giving your answer in the form \(y = f(x)\). [8]

A population of meerkats is being studied. The population is modelled by the differential equation $$\frac{dP}{dt} = \frac{1}{22}P(11 - 2P), \quad t \geq 0, \quad 0 < P < 5.5$$ where \(P\), in thousands, is the population of meerkats and \(t\) is the time measured in years since the study began. Given that there were 1000 meerkats in the population when the study began, determine the time taken, in years, for this population of meerkats to double. [7]

13. A scientist is studying a population of insects. The number of insects, \(N\), in the population, \(t\) days after the start of the study is modelled by the equation $$N = \frac { 240 } { 1 + k \mathrm { e } ^ { - \frac { t } { 16 } } }$$ where \(k\) is a constant.
Given that there were 50 insects at the start of the study,

1. find the value of \(k\)
2. use the model to find the value of \(t\) when \(N = 100\)
3. Show that $$\frac { \mathrm { d } N } { \mathrm {~d} t } = \frac { 1 } { p } N - \frac { 1 } { q } N ^ { 2 }$$ where \(p\) and \(q\) are integers to be found.
   

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11. A team of conservationists is studying the population of meerkats on a nature reserve. The population is modelled by the differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 15 } P ( 5 - P ) , \quad t \geqslant 0$$ where \(P\), in thousands, is the population of meerkats and \(t\) is the time measured in years since the study began. Given that when \(t = 0 , P = 1\),

1. solve the differential equation, giving your answer in the form $$P = \frac { a } { b + c \mathrm { e } ^ { - \frac { 1 } { 3 } t } }$$ where \(a\), \(b\) and \(c\) are integers.
2. Hence show that the population cannot exceed 5000

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.

9. Bacteria are growing on the surface of a dish in a laboratory. The area of the dish, \(A \mathrm {~cm} ^ { 2 }\), covered by the bacteria, \(t\) days after the bacteria start to grow, is modelled by the differential equation $$\frac { \mathrm { d } A } { \mathrm {~d} t } = \frac { A ^ { \frac { 3 } { 2 } } } { 5 t ^ { 2 } } \quad t > 0$$ Given that \(A = 2.25\) when \(t = 3\)

1. show that $$A = \left( \frac { p t } { q t + r } \right) ^ { 2 }$$ where \(p , q\) and \(r\) are integers to be found. According to the model, there is a limit to the area that will be covered by the bacteria.
2. Find the value of this limit. \includegraphics[max width=\textwidth, alt={}, center]{79ac81c3-cd05-4f28-8840-3c8a6960e7b7-31_2255_50_314_34}
   

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Water is leaking from a container. After \(t\) seconds, the depth of water in the container is \(x\) cm, and the volume of water is \(V\) cm\(^3\), where \(V = \frac{1}{3}x^3\). The rate at which water is lost is proportional to \(x\), so that \(\frac{dV}{dt} = -kx\), where \(k\) is a constant.

1. Show that \(x \frac{dx}{dt} = -k\). [3]

Initially, the depth of water in the container is 10 cm.

1. Show by integration that \(x = \sqrt{100 - 2kt}\). [4]
2. Given that the container empties after 50 seconds, find \(k\). [2]

Once the container is empty, water is poured into it at a constant rate of 1 cm\(^3\) per second. The container continues to lose water as before.

1. Show that, \(t\) seconds after starting to pour the water in, \(\frac{dx}{dt} = \frac{1-x}{x^2}\). [2]
2. Show that \(\frac{1}{1-x} - x - 1 = \frac{x^2}{1-x}\). Hence solve the differential equation in part (iv) to show that $$t = \ln\left(\frac{1}{1-x}\right) - \frac{1}{2}x^2 - x.$$ [6]
3. Show that the depth cannot reach 1 cm. [1]

6 A hot drink is cooling. The temperature of the drink at time \(t\) minutes is \(T ^ { \circ } \mathrm { C }\).
The rate of decrease in temperature of the drink is proportional to \(( T - 20 )\).

1. Write down a differential equation to describe the temperature of the drink as a function of time.
2. When \(t = 0\), the temperature of the drink is \(90 ^ { \circ } \mathrm { C }\) and the temperature is decreasing at a rate of \(4.9 ^ { \circ } \mathrm { C }\) per minute. Determine how long it takes for the drink to cool from \(90 ^ { \circ } \mathrm { C }\) to \(40 ^ { \circ } \mathrm { C }\).

6. A hot piece of metal is dropped into a cool liquid. As the metal cools, its temperature \(T\) degrees Celsius, \(t\) minutes after it enters the liquid, is modelled by $$T = 300 \mathrm { e } ^ { - 0.04 t } + 20 , \quad t \geqslant 0$$

1. Find the temperature of the piece of metal as it enters the liquid.
2. Find the value of \(t\) for which \(T = 180\), giving your answer to 3 significant figures. (Solutions based entirely on graphical or numerical methods are not acceptable.)
3. Show, by differentiation, that the rate, in degrees Celsius per minute, at which the temperature of the metal is changing, is given by the expression $$\frac { 20 - T } { 25 }$$

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8. A bottle of water is put into a refrigerator. The temperature inside the refrigerator remains constant at \(3 ^ { \circ } \mathrm { C }\) and \(t\) minutes after the bottle is placed in the refrigerator the temperature of the water in the bottle is \(\theta ^ { \circ } \mathrm { C }\). The rate of change of the temperature of the water in the bottle is modelled by the differential equation, $$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = \frac { ( 3 - \theta ) } { 125 }$$

1. By solving the differential equation, show that, $$\theta = A \mathrm { e } ^ { - 0.008 t } + 3$$ where \(A\) is a constant. Given that the temperature of the water in the bottle when it was put in the refrigerator was \(16 ^ { \circ } \mathrm { C }\),
2. find the time taken for the temperature of the water in the bottle to fall to \(10 ^ { \circ } \mathrm { C }\), giving your answer to the nearest minute.

The variable \(y\) satisfies the differential equation $$2\frac{dy}{dx} = 5 - 2y, \quad \text{where } x \geqslant 0.$$ Given that \(y = 1\) when \(x = 0\), find an expression for \(y\) in terms of \(x\). [5]

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1. The number of fish in a lake is decreasing. After \(t\) years, there are \(x\) fish in the lake. The rate of decrease of the number of fish is proportional to the number of fish currently in the lake.
   1. Formulate a differential equation, in the variables \(x\) and \(t\) and a constant of proportionality \(k\), where \(k > 0\), to model the rate at which the number of fish in the lake is decreasing.
   2. At a certain time, there were 20000 fish in the lake and the rate of decrease was 500 fish per year. Find the value of \(k\).
2. The equation $$P = 2000 - A \mathrm { e } ^ { - 0.05 t }$$ is proposed as a model for the number of fish, \(P\), in another lake, where \(t\) is the time in years and \(A\) is a positive constant. On 1 January 2008, a biologist estimated that there were 700 fish in this lake.
   1. Taking 1 January 2008 as \(t = 0\), find the value of \(A\).
   2. Hence find the year during which, according to this model, the number of fish in this lake will first exceed 1900.

7 In a chemical reaction a compound \(X\) is formed from a compound \(Y\). The masses in grams of \(X\) and \(Y\) present at time \(t\) seconds after the start of the reaction are \(x\) and \(y\) respectively. The sum of the two masses is equal to 100 grams throughout the reaction. At any time, the rate of formation of \(X\) is proportional to the mass of \(Y\) at that time. When \(t = 0 , x = 5\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 1.9\).

1. Show that \(x\) satisfies the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = 0.02 ( 100 - x ) .$$
2. Solve this differential equation, obtaining an expression for \(x\) in terms of \(t\).
3. State what happens to the value of \(x\) as \(t\) becomes very large.

8 The equation of a curve is such that \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 2 x - 1\). Given that the curve has a minimum point at \(( 3 , - 10 )\), find the coordinates of the maximum point.

6 A curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 x ^ { \frac { 1 } { 2 } } - 6\) and the point \(( 9,2 )\) lies on the curve.

1. Find the equation of the curve.
2. Find the \(x\)-coordinate of the stationary point on the curve and determine the nature of the stationary point.

1. The variables \(x\) and \(y\) satisfy the differential equation

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 2 y ^ { 2 } - x - 1$$ where \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3\) and \(y = 0\) at \(x = 0\) Use the approximations $$\left( \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } \right) _ { n } \approx \frac { \left( y _ { n + 1 } - 2 y _ { n } + y _ { n - 1 } \right) } { h ^ { 2 } } \text { and } \left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) _ { n } \approx \frac { \left( y _ { n + 1 } - y _ { n - 1 } \right) } { 2 h }$$ with \(h = 0.1\) to find an estimate for the value of \(y\) at \(x = 0.2\)

7 When a stone is dropped into still water, ripples move outwards forming a circle of rippled water. At time \(t\) seconds after the stone hits the water the radius of the circle of ripples is increasing at a rate that is inversely proportional to the radius When the radius is 200 cm the rate of increase of the radius is 5 cm per second. Write down the differential equation that represents this situation.

7 The height of the tide in a certain harbour is \(h\) metres at time \(t\) hours. Successive high tides occur every 12 hours. The rate of change of the height of the tide can be modelled by a function of the form \(a \cos ( k t )\), where \(a\) and \(k\) are constants. The largest value of this rate of change is 1.3 metres per hour. Write down a differential equation in the variables \(h\) and \(t\). State the values of the constants \(a\) and \(k\).

\includegraphics{figure_9} A container in the shape of a cuboid has a square base of side \(x\) and a height of \((10 - x)\). It is given that \(x\) varies with time, \(t\), where \(t > 0\). The container decreases in volume at a rate which is inversely proportional to \(t\). When \(t = \frac{1}{10}\), \(x = \frac{1}{2}\) and the rate of decrease of \(x\) is \(\frac{20}{37}\).

1. Show that \(x\) and \(t\) satisfy the differential equation $$\frac{dx}{dt} = \frac{-1}{2t(20x - 3x^2)}$$ [5]
2. Solve the differential equation, obtaining an expression for \(t\) in terms of \(x\). [6]

The coordinates \((x, y)\) of a general point on a curve satisfy the differential equation $$x\frac{dy}{dx} = (2 - x^2)y.$$ The curve passes through the point \((1, 1)\). Find the equation of the curve, obtaining an expression for \(y\) in terms of \(x\). [7]

6 A certain curve is such that its gradient at a point \(( x , y )\) is proportional to \(x y\). At the point \(( 1,2 )\) the gradient is 4 .

1. By setting up and solving a differential equation, show that the equation of the curve is \(y = 2 \mathrm { e } ^ { x ^ { 2 } - 1 }\).
2. State the gradient of the curve at the point \(( - 1,2 )\) and sketch the curve.

11 A curve, \(C\), passes through the point with coordinates \(( 1,6 )\) The gradient of \(C\) is given by $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 6 } ( x y ) ^ { 2 }$$ Show that \(C\) intersects the coordinate axes at exactly one point and state the coordinates of this point. Fully justify your answer.

11. In a science experiment, a radio active particle, \(N\), decays over time, \(t\), measured in minutes. The rate of decay of a particle is proportional to the number of particles remaining. Write down a suitable equation for the rate of change of the number of particles, \(N\) in terms of \(t\).

11. In a science experiment, a radio active particle, \(N\), decays over time, \(t\), measured in minutes. The rate of decay of a particle is proportional to the number of particles remaining. Write down a suitable equation for the rate of change of the number of particles, \(N\) in terms of \(t\).

8. A population growth is modelled by the differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } = k P ,$$ where \(P\) is the population, \(t\) is the time measured in days and \(k\) is a positive constant.
Given that the initial population is \(P _ { 0 }\),

1. solve the differential equation, giving \(P\) in terms of \(P _ { 0 } , k\) and \(t\). Given also that \(k = 2.5\),
2. find the time taken, to the nearest minute, for the population to reach \(2 P _ { 0 }\). In an improved model the differential equation is given as $$\frac { \mathrm { d } P } { \mathrm {~d} t } = \lambda P \cos \lambda t$$ where \(P\) is the population, \(t\) is the time measured in days and \(\lambda\) is a positive constant.
   Given, again, that the initial population is \(P _ { 0 }\) and that time is measured in days,
3. solve the second differential equation, giving \(P\) in terms of \(P _ { 0 } , \lambda\) and \(t\). Given also that \(\lambda = 2.5\),
4. find the time taken, to the nearest minute, for the population to reach \(2 P _ { 0 }\) for the first time, using the improved model.

7 A tank is shaped as a cuboid. The base has dimensions 10 cm by 10 cm . Initially the tank is empty. Water flows into the tank at \(25 \mathrm {~cm} ^ { 3 }\) per second. Water also leaks out of the tank at \(4 h ^ { 2 } \mathrm {~cm} ^ { 3 }\) per second, where \(h \mathrm {~cm}\) is the depth of the water after \(t\) seconds. Find the time taken for the water to reach a depth of 2 cm .

10 A water tank is in the shape of a cuboid with base area \(40000 \mathrm {~cm} ^ { 2 }\). At time \(t\) minutes the depth of water in the tank is \(h \mathrm {~cm}\). Water is pumped into the tank at a rate of \(50000 \mathrm {~cm} ^ { 3 }\) per minute. Water is leaking out of the tank through a hole in the bottom at a rate of \(600 \mathrm {~cm} ^ { 3 }\) per minute.

1. Show that \(200 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 250 - 3 h\).
   
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2. It is given that when \(t = 0 , h = 50\). Find the time taken for the depth of water in the tank to reach 80 cm . Give your answer correct to 2 significant figures.

7 A tank is shaped as a cuboid. The base has dimensions 10 cm by 10 cm . Initially the tank is empty. Water flows into the tank at \(25 \mathrm {~cm} ^ { 3 }\) per second. Water also leaks out of the tank at \(4 h ^ { 2 } \mathrm {~cm} ^ { 3 }\) per second, where \(h \mathrm {~cm}\) is the depth of the water after \(t\) seconds. Find the time taken for the water to reach a depth of 2 cm .

A particle moves so that its acceleration, \(a\text{ ms}^{-2}\), at time \(t\) seconds may be modelled in terms of its velocity, \(v\text{ ms}^{-1}\), as $$a = -0.1v^2$$ The initial velocity of the particle is \(4\text{ ms}^{-1}\)

1. By first forming a suitable differential equation, show that $$v = \frac{20}{5 + 2t}$$ [6 marks]
2. Find the acceleration of the particle when \(t = 5.5\) [2 marks]

8 Vicky has mass 65 kg and is skydiving. She steps out of a helicopter and falls vertically. She then waits a short period of time before opening her parachute. The parachute opens at time \(t = 0\) when her speed is \(19.6 \mathrm {~ms} ^ { - 1 }\), and she then experiences an air resistance force of magnitude \(260 v\) newtons, where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is her speed at time \(t\) seconds.

1. When \(t > 0\) :
   1. show that the resultant downward force acting on Vicky is 65(9.8-4v) newtons
   2. show that \(\frac { \mathrm { d } v } { \mathrm {~d} t } = - 4 ( v - 2.45 )\).
2. By showing that \(\int \frac { 1 } { v - 2.45 } \mathrm {~d} v = - \int 4 \mathrm {~d} t\), find \(v\) in terms of \(t\).
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1. Show that the motion of the particle can be modelled by the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} t } + \frac { 1 } { 2 } v = \frac { 1 } { 4 } t$$ The particle is at rest when \(t = 0\).
2. Find \(v\) in terms of \(t\).
3. Find the velocity of the particle when \(t = 2\). When \(t = 2\) the force acting in the positive \(x\)-direction is replaced by a constant force of magnitude \(\frac { 1 } { 2 } \mathrm {~N}\) in the same direction.
4. Refine the differential equation given in part (a) to model the motion for \(t \geqslant 2\).
5. Use the refined model from part (d) to find an exact expression for \(v\) in terms of \(t\) for \(t \geqslant 2\). \(6 \quad A\) is a fixed point on a smooth horizontal surface. A particle \(P\) is initially held at \(A\) and released from rest. It subsequently performs simple harmonic motion in a straight line on the surface. After its release it is next at rest after 0.2 seconds at point \(B\) whose displacement is 0.2 m from \(A\). The point \(M\) is halfway between \(A\) and \(B\). The displacement of \(P\) from \(M\) at time \(t\) seconds after release is denoted by \(x \mathrm {~m}\).
   1. On the axes provided in the Printed Answer Booklet, sketch a graph of \(x\) against \(t\) for \(0 \leqslant t \leqslant 0.4\).
   2. Find the displacement of \(P\) from \(M\) at 0.75 seconds after release.

Sam goes on a diet. He assumes that his mass, \(m\) kg after \(t\) days, decreases at a rate that is inversely proportional to the cube root of his mass.

1. Construct a differential equation involving \(m\), \(t\) and a positive constant \(k\) to model this situation. [3 marks]
2. Explain why Sam's assumption may not be appropriate. [1 mark]

1. A spherical mint of radius 5 mm is placed in the mouth and sucked. Four minutes later, the radius of the mint is 3 mm .

In a simple model, the rate of decrease of the radius of the mint is inversely proportional to the square of the radius. Using this model and all the information given,

1. find an equation linking the radius of the mint and the time.
   (You should define the variables that you use.)
2. Hence find the total time taken for the mint to completely dissolve. Give your answer in minutes and seconds to the nearest second.
3. Suggest a limitation of the model.

4. 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.
   4. continued

7 Solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = y ^ { 2 } x \sin 3 x$$ given that \(y = 1\) when \(x = \frac { \pi } { 6 }\). Give your answer in the form \(y = \frac { 9 } { \mathrm { f } ( x ) }\).

A balloon in the shape of a sphere has volume \(V\) and radius \(r\). Air is pumped into the balloon at a constant rate of \(40\pi\) starting when time \(t = 0\) and \(r = 0\). At the same time, air begins to flow out of the balloon at a rate of \(0.8\pi r\). The balloon remains a sphere at all times.

1. Show that \(r\) and \(t\) satisfy the differential equation $$\frac{dr}{dt} = \frac{50 - r}{5r^2}.$$ [3]
2. Find the quotient and remainder when \(5r^2\) is divided by \(50 - r\). [3]
3. Solve the differential equation in part (a), obtaining an expression for \(t\) in terms of \(r\). [6]
4. Find the value of \(t\) when the radius of the balloon is 12. [1]

10. The amount of an antibiotic in the bloodstream, from a given dose, is modelled by the formula $$x = D \mathrm { e } ^ { - 0.2 t }$$ where \(x\) is the amount of the antibiotic in the bloodstream in milligrams, \(D\) is the dose given in milligrams and \(t\) is the time in hours after the antibiotic has been given. A first dose of 15 mg of the antibiotic is given.

1. Use the model to find the amount of the antibiotic in the bloodstream 4 hours after the dose is given. Give your answer in mg to 3 decimal places. A second dose of 15 mg is given 5 hours after the first dose has been given. Using the same model for the second dose,
2. show that the total amount of the antibiotic in the bloodstream 2 hours after the second dose is given is 13.754 mg to 3 decimal places. No more doses of the antibiotic are given. At time \(T\) hours after the second dose is given, the total amount of the antibiotic in the bloodstream is 7.5 mg .
3. Show that \(T = a \ln \left( b + \frac { b } { \mathrm { e } } \right)\), where \(a\) and \(b\) are integers to be determined.
   

10 A container in the shape of an inverted cone of radius 3 metres and vertical height 4.5 metres is initially filled with liquid fertiliser. This fertiliser is released through a hole in the bottom of the container at a rate of \(0.01 \mathrm {~m} ^ { 3 }\) per second. At time \(t\) seconds the fertiliser remaining in the container forms an inverted cone of height \(h\) metres.
[0pt] [The volume of a cone is \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\).]

1. Show that \(h ^ { 2 } \frac { \mathrm {~d} h } { \mathrm {~d} t } = - \frac { 9 } { 400 \pi }\).
2. Express \(h\) in terms of \(t\).
3. Find the time it takes to empty the container, giving your answer to the nearest minute.

1. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

A population of fruit flies is being studied.
The number of fruit flies, \(F\), in the population, \(t\) days after the start of the study, is modelled by the equation $$F = \frac { 350 \mathrm { e } ^ { k t } } { 9 + \mathrm { e } ^ { k t } }$$ where \(k\) is a constant.
Use the equation of the model to answer parts (a), (b) and (c).

1. Find the number of fruit flies in the population at the start of the study. Given that there are 200 fruit flies in the population 15 days after the start of the study,
2. show that \(k = \frac { 1 } { 15 } \ln 12\) Given also that, when \(t = T\), the number of fruit flies in the population is increasing at a rate of 10 per day,
3. find the possible values of \(T\), giving your answers to one decimal place.

1 The variables \(x\) and \(y\) are such that \(y = - 1\) when \(x = 0\) and $$\left( x + \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 3 } = y ^ { 2 } + x$$

1. Find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) when \(x = 0\).
2. Find also the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) when \(x = 0\).

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 In a chemical process, the mass \(M\) grams of a chemical at time \(t\) minutes is modelled by the differential equation $$\frac { \mathrm { d } M } { \mathrm {~d} t } = \frac { M } { t \left( 1 + t ^ { 2 } \right) }$$

1. Find \(\int \frac { t } { 1 + t ^ { 2 } } \mathrm {~d} t\).
2. Find constants \(A , B\) and \(C\) such that $$\frac { 1 } { t \left( 1 + t ^ { 2 } \right) } = \frac { A } { t } + \frac { B t + C } { 1 + t ^ { 2 } } .$$
3. Use integration, together with your results in parts (i) and (ii), to show that $$M = \frac { K t } { \sqrt { 1 + t ^ { 2 } } } ,$$ where \(K\) is a constant.
4. When \(t = 1 , M = 25\). Calculate \(K\). What is the mass of the chemical in the long term?

\includegraphics{figure_3} Fig. 9 shows a hemispherical bowl, of radius 10 cm, filled with water to a depth of \(x\) cm. It can be shown that the volume of water, \(V\) cm\(^3\), is given by $$V = \pi(10x^2 - \frac{1}{3}x^3).$$ Water is poured into a leaking hemispherical bowl of radius 10 cm. Initially, the bowl is empty. After \(t\) seconds, the volume of water is changing at a rate, in cm\(^3\) s\(^{-1}\), given by the equation $$\frac{dV}{dt} = k(20 - x),$$ where \(k\) is a constant.

1. Find \(\frac{dV}{dx}\), and hence show that \(\pi x \frac{dx}{dt} = k\). [4]
2. Solve this differential equation, and hence show that the bowl fills completely after \(T\) seconds, where \(T = \frac{50\pi}{k}\). [5]

Once the bowl is full, the supply of water to the bowl is switched off, and water then leaks out at a rate of \(kx\) cm\(^3\) s\(^{-1}\).

1. Show that, \(t\) seconds later, \(\pi(20 - x) \frac{dx}{dt} = -k\). [3]
2. Solve this differential equation. Hence show that the bowl empties in 37 seconds. [6]

4 The motion of a particle is modelled by the differential equation $$v \frac { \mathrm {~d} v } { \mathrm {~d} x } + 4 x = 0$$ where \(x\) is its displacement from a fixed point, and \(v\) is its velocity. Initially \(x = 1\) and \(v = 4\).

1. Solve the differential equation to show that \(v ^ { 2 } = 20 - 4 x ^ { 2 }\). Now consider motion for which \(x = \cos 2 t + 2 \sin 2 t\), where \(x\) is the displacement from a fixed point at time \(t\).
2. Verify that, when \(t = 0 , x = 1\). Use the fact that \(v = \frac { \mathrm { d } x } { \mathrm {~d} t }\) to verify that when \(t = 0 , v = 4\).
3. Express \(x\) in the form \(R \cos ( 2 t - \alpha )\), where \(R\) and \(\alpha\) are constants to be determined, and obtain the corresponding expression for \(v\). Hence or otherwise verify that, for this motion too, \(v ^ { 2 } = 20 - 4 x ^ { 2 }\).
4. Use your answers to part (iii) to find the maximum value of \(x\), and the earliest time at which \(x\) reaches this maximum value.