1.08l Interpret differential equation solutions: in context

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.

10 The number of birds of a certain species in a forested region is recorded over several years. At time \(t\) years, the number of birds is \(N\), where \(N\) is treated as a continuous variable. The variation in the number of birds is modelled by $$\frac { \mathrm { d } N } { \mathrm {~d} t } = \frac { N ( 1800 - N ) } { 3600 }$$ It is given that \(N = 300\) when \(t = 0\).

1. Find an expression for \(N\) in terms of \(t\).
2. According to the model, how many birds will there be after a long time?

9 In a chemical reaction, a compound \(X\) is formed from two compounds \(Y\) and \(Z\). The masses in grams of \(X , Y\) and \(Z\) present at time \(t\) seconds after the start of the reaction are \(x , 10 - x\) and \(20 - x\) respectively. At any time the rate of formation of \(X\) is proportional to the product of the masses of \(Y\) and \(Z\) present at the time. When \(t = 0 , x = 0\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 2\).

1. Show that \(x\) and \(t\) satisfy the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = 0.01 ( 10 - x ) ( 20 - x )$$
2. Solve this differential equation and obtain an expression for \(x\) in terms of \(t\).
3. State what happens to the value of \(x\) when \(t\) becomes large.

10 Naturalists are managing a wildlife reserve to increase the number of plants of a rare species. The number of plants at time \(t\) years is denoted by \(N\), where \(N\) is treated as a continuous variable.

1. It is given that the rate of increase of \(N\) with respect to \(t\) is proportional to ( \(N - 150\) ). Write down a differential equation relating \(N , t\) and a constant of proportionality.
2. Initially, when \(t = 0\), the number of plants was 650 . It was noted that, at a time when there were 900 plants, the number of plants was increasing at a rate of 60 per year. Express \(N\) in terms of \(t\).
3. The naturalists had a target of increasing the number of plants from 650 to 2000 within 15 years. Will this target be met?

10 A large field of area \(4 \mathrm {~km} ^ { 2 }\) is becoming infected with a soil disease. At time \(t\) years the area infected is \(x \mathrm {~km} ^ { 2 }\) and the rate of growth of the infected area is given by the differential equation \(\frac { \mathrm { d } x } { \mathrm {~d} t } = k x ( 4 - x )\), where \(k\) is a positive constant. It is given that when \(t = 0 , x = 0.4\) and that when \(t = 2 , x = 2\).

1. Solve the differential equation and show that \(k = \frac { 1 } { 4 } \ln 3\).
2. Find the value of \(t\) when \(90 \%\) of the area of the field is infected.

9. (a) Use the substitution \(u = 4 - \sqrt { } x\) to find $$\int \frac { \mathrm { d } x } { 4 - \sqrt { } x }$$ A team of scientists is studying a species of slow growing tree.
The rate of change in height of a tree in this species is modelled by the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = \frac { 4 - \sqrt { } h } { 20 }$$ where \(h\) is the height in metres and \(t\) is the time measured in years after the tree is planted.
(b) Find the range in values of \(h\) for which the height of a tree in this species is increasing.
(c) Given that one of these trees is 1 metre high when it is planted, calculate the time it would take to reach a height of 10 metres. Write your answer to 3 significant figures. \includegraphics[max width=\textwidth, alt={}, center]{5b698944-41ac-4072-b5e1-c580b7752c39-31_154_145_2599_1804}

5. (a) Obtain the general solution of the differential equation $$\frac { \mathrm { d } S } { \mathrm {~d} t } - 0.1 S = t$$ (b) The differential equation in part (a) is used to model the assets, \(\pounds S\) million, of a bank \(t\) years after it was set up. Given that the initial assets of the bank were \(\pounds 200\) million, use your answer to part (a) to estimate, to the nearest \(\pounds\) million, the assets of the bank 10 years after it was set up.

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 \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 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?

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.

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.

8. The diagram shows a hemispherical bowl of radius 5 cm . The bowl is filled with water but the water leaks from a hole at the base of the bowl. At time \(t\) minutes, the depth of water is \(h \mathrm {~cm}\) and the volume of water in the bowl is \(V \mathrm {~cm} ^ { 3 }\), where $$V = \frac { 1 } { 3 } \pi h ^ { 2 } ( 15 - h ) .$$ In a model it is assumed that the rate at which the volume of water in the bowl decreases is proportional to \(V\).

1. Show that $$\frac { \mathrm { d } h } { \mathrm {~d} t } = - \frac { k h ( 15 - h ) } { 3 ( 10 - h ) } ,$$ where \(k\) is a positive constant.
2. Express \(\frac { 3 ( 10 - h ) } { h ( 15 - h ) }\) in partial fractions. Given that when \(t = 0 , h = 5\),
3. show that $$h ^ { 2 } ( 15 - h ) = 250 \mathrm { e } ^ { - k t } .$$ Given also that when \(t = 2 , h = 4\),
4. find the value of \(k\) to 3 significant figures.

10 In a chemical reaction, a compound \(X\) is formed from two compounds \(Y\) and \(Z\).
The masses in grams of \(X , Y\) and \(Z\) present at time \(t\) seconds after the start of the reaction are \(x , 10 - x\) and \(20 - x\) respectively. At any time the rate of formation of \(X\) is proportional to the product of the masses of \(Y\) and \(Z\) present at the time. When \(t = 0 , x = 0\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 2\).

1. Show that \(x\) and \(t\) satisfy the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = 0.01 ( 10 - x ) ( 20 - x ) .$$
2. Solve this differential equation and obtain an expression for \(x\) in terms of \(t\).
3. State what happens to the value of \(x\) when \(t\) becomes large.

9 The temperature of a freezer is \(- 20 ^ { \circ } \mathrm { C }\). A container of a liquid is placed in the freezer. The rate at which the temperature, \(\theta ^ { \circ } \mathrm { C }\), of a liquid decreases is proportional to the difference in temperature between the liquid and its surroundings. The situation is modelled by the differential equation $$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = - k ( \theta + 20 ) ,$$ where time \(t\) is in minutes and \(k\) is a positive constant.

1. Express \(\theta\) in terms of \(t , k\) and an arbitrary constant. Initially the temperature of the liquid in the container is \(40 ^ { \circ } \mathrm { C }\) and, at this instant, the liquid is cooling at a rate of \(3 ^ { \circ } \mathrm { C }\) per minute. The liquid freezes at \(0 ^ { \circ } \mathrm { C }\).
2. Find the value of \(k\) and find also the time it takes (to the nearest minute) for the liquid to freeze. The procedure is repeated on another occasion with a different liquid. The initial temperature of this liquid is \(90 ^ { \circ } \mathrm { C }\). After 19 minutes its temperature is \(0 ^ { \circ } \mathrm { C }\).
3. Without any further calculation, explain what you can deduce about the value of \(k\) in this case.

9 A tank contains water which is heated by an electric water heater working under the action of a thermostat. The temperature of the water, \(\theta ^ { \circ } \mathrm { C }\), may be modelled as follows. When the water heater is first switched on, \(\theta = 40\). The heater causes the temperature to increase at a rate \(k _ { 1 } { } ^ { \circ } \mathrm { C }\) per second, where \(k _ { 1 }\) is a constant, until \(\theta = 60\). The heater then switches off.

1. Write down, in terms of \(k _ { 1 }\), how long it takes for the temperature to increase from \(40 ^ { \circ } \mathrm { C }\) to \(60 ^ { \circ } \mathrm { C }\). The temperature of the water then immediately starts to decrease at a variable rate \(k _ { 2 } ( \theta - 20 ) ^ { \circ } \mathrm { C }\) per second, where \(k _ { 2 }\) is a constant, until \(\theta = 40\).
2. Write down a differential equation to represent the situation as the temperature is decreasing.
3. Find the total length of time for the temperature to increase from \(40 ^ { \circ } \mathrm { C }\) to \(60 ^ { \circ } \mathrm { C }\) and then decrease to \(40 ^ { \circ } \mathrm { C }\). Give your answer in terms of \(k _ { 1 }\) and \(k _ { 2 }\). 4

12 A cake is cooling so that, \(t\) minutes after it is removed from an oven, its temperature is \(\theta ^ { \circ } \mathrm { C }\). When the cake is removed from the oven, its temperature is \(160 ^ { \circ } \mathrm { C }\). After 10 minutes its temperature has fallen to \(125 ^ { \circ } \mathrm { C }\).

1. In a simple model, the rate of decrease of the temperature of the cake is assumed to be constant.
   1. Write down a differential equation for this model.
   2. Solve this differential equation to find \(\theta\) in terms of \(t\).
   3. State one limitation of this model.
2. In a revised model, the rate of decrease of the temperature of the cake is proportional to the difference between the temperature of the cake and the temperature of the room. The temperature of the room is a constant \(20 ^ { \circ } \mathrm { C }\).
   1. Write down a differential equation for this revised model.
   2. Solve this differential equation to find \(\theta\) in terms of \(t\).
3. The cake can be decorated when its temperature is \(25 ^ { \circ } \mathrm { C }\). Find the difference in time between when the two models would predict that the cake can be decorated, giving your answer correct to the nearest minute. \section*{END OF QUESTION PAPER}

1. (a) Express \(\frac { 3 } { ( 2 x - 1 ) ( x + 1 ) }\) in partial fractions.

When chemical \(A\) and chemical \(B\) are mixed, oxygen is produced.
A scientist mixed these two chemicals and measured the total volume of oxygen produced over a period of time. The total volume of oxygen produced, \(V \mathrm {~m} ^ { 3 } , t\) hours after the chemicals were mixed, is modelled by the differential equation $$\frac { \mathrm { d } V } { \mathrm {~d} t } = \frac { 3 V } { ( 2 t - 1 ) ( t + 1 ) } \quad V \geqslant 0 \quad t \geqslant k$$ where \(k\) is a constant.
Given that exactly 2 hours after the chemicals were mixed, a total volume of \(3 \mathrm {~m} ^ { 3 }\) of oxygen had been produced,
(b) solve the differential equation to show that $$V = \frac { 3 ( 2 t - 1 ) } { ( t + 1 ) }$$ The scientist noticed that

• there was a time delay between the chemicals being mixed and oxygen being produced
• there was a limit to the total volume of oxygen produced

Deduce from the model
(c) (i) the time delay giving your answer in minutes,
(ii) the limit giving your answer in \(\mathrm { m } ^ { 3 }\)

7

1. 1. Solve the differential equation \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \sqrt { x } \sin \left( \frac { t } { 2 } \right)\) to find \(x\) in terms of \(t\).
   2. Given that \(x = 1\) when \(t = 0\), show that the solution can be written as $$x = ( a - \cos b t ) ^ { 2 }$$ where \(a\) and \(b\) are constants to be found.
2. The height, \(x\) metres, above the ground of a car in a fairground ride at time \(t\) seconds is modelled by the differential equation \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \sqrt { x } \sin \left( \frac { t } { 2 } \right)\). The car is 1 metre above the ground when \(t = 0\).
   1. Find the greatest height above the ground reached by the car during the ride.
   2. Find the value of \(t\) when the car is first 5 metres above the ground, giving your answer to one decimal place.

7 A biologist is investigating the growth of a population of a species of rodent. The biologist proposes the model $$N = \frac { 500 } { 1 + 9 \mathrm { e } ^ { - \frac { t } { 8 } } }$$ for the number of rodents, \(N\), in the population \(t\) weeks after the start of the investigation. Use this model to answer the following questions.

1. 1. Find the size of the population at the start of the investigation.
   2. Find the size of the population 24 weeks after the start of the investigation. your answer to the nearest whole number.
   3. Find the number of weeks that it will take the population to reach 400 . Give your answer in the form \(t = r \ln s\), where \(r\) and \(s\) are integers.
2. 1. Show that the rate of growth, \(\frac { \mathrm { d } N } { \mathrm {~d} t }\), is given by $$\frac { \mathrm { d } N } { \mathrm {~d} t } = \frac { N } { 4000 } ( 500 - N )$$
   2. The maximum rate of growth occurs after \(T\) weeks. Find the value of \(T\).

7 A giant snowball is melting. The snowball can be modelled as a sphere whose surface area is decreasing at a constant rate with respect to time. The surface area of the sphere is \(A \mathrm {~cm} ^ { 2 }\) at time \(t\) days after it begins to melt.

1. Write down a differential equation in terms of the variables \(A\) and \(t\) and a constant \(k\), where \(k > 0\), to model the melting snowball.
2. 1. Initially, the radius of the snowball is 60 cm , and 9 days later, the radius has halved. Show that \(A = 1200 \pi ( 12 - t )\).
      (You may assume that the surface area of a sphere is given by \(A = 4 \pi r ^ { 2 }\), where \(r\) is the radius.)
   2. Use this model to find the number of days that it takes the snowball to melt completely.

8

1. A water tank has a height of 2 metres. The depth of the water in the tank is \(h\) metres at time \(t\) minutes after water begins to enter the tank. The rate at which the depth of the water in the tank increases is proportional to the difference between the height of the tank and the depth of the water. Write down a differential equation in the variables \(h\) and \(t\) and a positive constant \(k\).
   (You are not required to solve your differential equation.)
2. 1. Another water tank is filling in such a way that \(t\) minutes after the water is turned on, the depth of the water, \(x\) metres, increases according to the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 1 } { 15 x \sqrt { 2 x - 1 } }$$ The depth of the water is 1 metre when the water is first turned on.
      Solve this differential equation to find \(t\) as a function of \(x\).
   2. Calculate the time taken for the depth of the water in the tank to reach 2 metres, giving your answer to the nearest 0.1 of a minute.
      (l mark)