Exponential growth/decay - approach to limit (dN/dt = k(N - N₀))

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 In a certain chemical reaction, a compound \(A\) is formed from a compound \(B\). The masses of \(A\) and \(B\) at time \(t\) after the start of the reaction are \(x\) and \(y\) respectively and the sum of the masses is equal to 50 throughout the reaction. At any time the rate of increase of the mass of \(A\) is proportional to the mass of \(B\) at that time.

1. Explain why \(\frac { \mathrm { d } x } { \mathrm {~d} t } = k ( 50 - x )\), where \(k\) is a constant.
   It is given that \(x = 0\) when \(t = 0\), and \(x = 25\) when \(t = 10\).
2. Solve the differential equation in part (i) and express \(x\) in terms of \(t\).

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.

7 A skydiver drops from a helicopter. Before she opens her parachute, her speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) after time \(t\) seconds is modelled by the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 \mathrm { e } ^ { - \frac { 1 } { 2 } t }$$ When \(t = 0 , v = 0\).

1. Find \(v\) in terms of \(t\).
2. According to this model, what is the speed of the skydiver in the long term? She opens her parachute when her speed is \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Her speed \(t\) seconds after this is \(w \mathrm {~m} \mathrm {~s} ^ { - 1 }\), and is modelled by the differential equation $$\frac { \mathrm { d } w } { \mathrm {~d} t } = - \frac { 1 } { 2 } ( w - 4 ) ( w + 5 )$$
3. Express \(\frac { 1 } { ( w - 4 ) ( w + 5 ) }\) in partial fractions.
4. Using this result, show that \(\frac { w - 4 } { w + 5 } = 0.4 \mathrm { e } ^ { - 4.5 t }\).
5. According to this model, what is the speed of the skydiver in the long term?

15 A model for the motion of a small object falling through a thick fluid can be expressed using the differential equation \(\frac { \mathrm { d } v } { \mathrm {~d} t } = 9.8 - k v\),
where \(v \mathrm {~ms} ^ { - 1 }\) is the velocity after \(t \mathrm {~s}\) and \(k\) is a positive constant.

1. Given that \(v = 0\) when \(t = 0\), solve the differential equation to find \(v\) in terms of \(t\) and \(k\).
2. Sketch the graph of \(v\) against \(t\). Experiments show that for large values of \(t\), the velocity tends to \(7 \mathrm {~ms} ^ { - 1 }\).
3. Find the value of \(k\).
4. Find the value of \(t\) for which \(v = 3.5\).

6. Liquid is poured into a container at a constant rate of \(30 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\). At time \(t\) seconds liquid is leaking from the container at a rate of \(\frac { 2 } { 15 } V \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\), where \(V \mathrm {~cm} ^ { 3 }\) is the volume of liquid in the container at that time.

1. Show that $$- 15 \frac { \mathrm {~d} V } { \mathrm {~d} t } = 2 V - 450$$ Given that \(V = 1000\) when \(t = 0\),
2. find the solution of the differential equation, in the form \(V = \mathrm { f } ( t )\).
3. Find the limiting value of \(V\) as \(t \rightarrow \infty\).

8

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.

13 Solve the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - k ( y - 10 )\), where \(k\) is a constant, given that \(y = 70\) when \(x = 0\) and \(y = 40\) when \(x = 1\). Express your answer in the form \(y = a + b \left( \frac { 1 } { 2 } \right) ^ { x }\) where \(a\) and \(b\) are constants to be found.
[0pt] [10]

In a certain industrial process, a substance is being produced in a container. The mass of the substance in the container \(t\) minutes after the start of the process is \(x\) grams. At any time, the rate of formation of the substance is proportional to its mass. Also, throughout the process, the substance is removed from the container at a constant rate of 25 grams per minute. When \(t = 0\), \(x = 1000\) and \(\frac{dx}{dt} = 75\).

1. Show that \(x\) and \(t\) satisfy the differential equation $$\frac{dx}{dt} = 0.1(x - 250).$$ [2]
2. Solve this differential equation, obtaining an expression for \(x\) in terms of \(t\). [6]

Liquid is pouring into a container at a constant rate of \(20\text{ cm}^3\text{s}^{-1}\) and is leaking out at a rate proportional to the volume of the liquid already in the container.

1. Explain why, at time \(t\) seconds, the volume, \(V\text{ cm}^3\), of liquid in the container satisfies the differential equation $$\frac{dV}{dt} = 20 - kV,$$ where \(k\) is a positive constant. [2]

The container is initially empty.

1. By solving the differential equation, show that $$V = A + Be^{-kt},$$ giving the values of \(A\) and \(B\) in terms of \(k\). [6]

Given also that \(\frac{dV}{dt} = 10\) when \(t = 5\),

1. find the volume of liquid in the container at 10 s after the start. [5]

Liquid is poured into a container at a constant rate of 30 cm\(^3\) s\(^{-1}\). At time \(t\) seconds liquid is leaking from the container at a rate of \(\frac{1}{5}V\) cm\(^3\) s\(^{-1}\), where \(V\) cm\(^3\) is the volume of liquid in the container at that time.

1. Show that $$-15 \frac{dV}{dt} = 2V - 450.$$ [3]

Given that \(V = 1000\) when \(t = 0\),

1. find the solution of the differential equation, in the form \(V = f(t)\). [7]
2. Find the limiting value of \(V\) as \(t \to \infty\). [1]

The rate of change of a certain population \(P\) at time \(t\) is modelled by the equation \(\frac{dP}{dt} = (100 - P)\). Initially \(P = 2000\).

1. Determine an expression for \(P\) in terms of \(t\). [7]
2. Describe how the population changes over time. [2]

A gardener stores rainwater in a cylindrical container. The container has a height of 130 centimetres. The gardener empties the water from the container through a hose. The hose is attached 5 centimetres from the bottom of the container. At time \(t\) minutes after the hose is switched on, the depth of water, \(h\) centimetres, in the container decreases at a rate which is proportional to \(h - 5\) Initially the container of water is full, and the depth of water is decreasing at a rate of 1.5 centimetres per minute.

1. Show that $$\frac{dh}{dt} = -0.012(h - 5)$$ [3 marks]
2. Solve the differential equation $$\frac{dh}{dt} = -0.012(h - 5)$$ to find an expression for \(h\) in terms of \(t\) [5 marks]
3. Find the time taken for the container to be half empty. Give your answer to the nearest minute. [2 marks]

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]