Deriving the differential equation

5 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 \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 75\).

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

10 A model for the height, \(h\) metres, of a certain type of tree at time \(t\) years after being planted assumes that, while the tree is growing, the rate of increase in height is proportional to \(( 9 - h ) ^ { \frac { 1 } { 3 } }\). It is given that, when \(t = 0 , h = 1\) and \(\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.2\).

1. Show that \(h\) and \(t\) satisfy the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.1 ( 9 - h ) ^ { \frac { 1 } { 3 } } .$$
2. Solve this differential equation, and obtain an expression for \(h\) in terms of \(t\).
3. Find the maximum height of the tree and the time taken to reach this height after planting.
4. Calculate the time taken to reach half the maximum height. \footnotetext{Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. }

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.

9 The population of a country at time \(t\) years is \(N\) millions. At any time, \(N\) is assumed to increase at a rate proportional to the product of \(N\) and \(( 1 - 0.01 N )\). When \(t = 0 , N = 20\) and \(\frac { \mathrm { d } N } { \mathrm {~d} t } = 0.32\).

1. Treating \(N\) and \(t\) as continuous variables, show that they satisfy the differential equation $$\frac { \mathrm { d } N } { \mathrm {~d} t } = 0.02 N ( 1 - 0.01 N )$$
2. Solve the differential equation, obtaining an expression for \(t\) in terms of \(N\).
3. Find the time at which the population will be double its value at \(t = 0\).

9 In an experiment to study the spread of a soil disease, an area of \(10 \mathrm {~m} ^ { 2 }\) of soil was exposed to infection. In a simple model, it is assumed that the infected area grows at a rate which is proportional to the product of the infected area and the uninfected area. Initially, \(5 \mathrm {~m} ^ { 2 }\) was infected and the rate of growth of the infected area was \(0.1 \mathrm {~m} ^ { 2 }\) per day. At time \(t\) days after the start of the experiment, an area \(a \mathrm {~m} ^ { 2 }\) is infected and an area \(( 10 - a ) \mathrm { m } ^ { 2 }\) is uninfected.

1. Show that \(\frac { \mathrm { d } a } { \mathrm {~d} t } = 0.004 a ( 10 - a )\).
2. By first expressing \(\frac { 1 } { a ( 10 - a ) }\) in partial fractions, solve this differential equation, obtaining an expression for \(t\) in terms of \(a\).
3. Find the time taken for \(90 \%\) of the soil area to become infected, according to this model.

10 A certain substance is formed in a chemical reaction. The mass of substance formed \(t\) seconds after the start of the reaction is \(x\) grams. At any time the rate of formation of the substance is proportional to \(( 20 - x )\). When \(t = 0 , x = 0\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 1\).

1. Show that \(x\) and \(t\) satisfy the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = 0.05 ( 20 - x ) .$$
2. Find, in any form, the solution of this differential equation.
3. Find \(x\) when \(t = 10\), giving your answer correct to 1 decimal place.
4. State what happens to the value of \(x\) as \(t\) becomes very large. \footnotetext{Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. }

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?

7 A curve is such that the gradient at a general point with coordinates \(( x , y )\) is proportional to \(\frac { y } { \sqrt { x + 1 } }\). The curve passes through the points with coordinates \(( 0,1 )\) and \(( 3 , e )\). By setting up and solving a differential equation, find the equation of the curve, expressing \(y\) in terms of \(x\).

11 In a field there are 300 plants of a certain species, all of which can be infected by a particular disease. At time \(t\) after the first plant is infected there are \(x\) infected plants. The rate of change of \(x\) is proportional to the product of the number of plants infected and the number of plants that are not yet infected. The variables \(x\) and \(t\) are treated as continuous, and it is given that \(\frac { \mathrm { dx } } { \mathrm { dt } } = 0.2\) and \(x = 1\) when \(t = 0\).

1. Show that \(x\) and \(t\) satisfy the differential equation $$1495 \frac { \mathrm { dx } } { \mathrm { dt } } = x ( 300 - x )$$
2. Using partial fractions, solve the differential equation and obtain an expression for \(t\) in terms of a single logarithm involving \(x\).
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10 Ira ch mical reactio a cm \(\mathbf { p } \quad X\) is fo med rm twœ \(\mathbf { m } \mathbf { p } \quad Y\) ad \(Z\).
Tb masses in g ams \(\varnothing \quad X , Y\) ad \(Z \mathrm { p }\) esen at time \(t\) secd s after th start \(\varnothing\) th reactin are \(x , \mathbb { Q } - x\) ad \(0 - x\) resp ctie ly. At ay time th rate 6 fo matin \(6 X\) is p p tio l to to pd t \(\mathbf { 6 }\) th masses \(6 Y\) ad \(Z \mathrm { p }\) esen at th time. Wh \(\mathrm { n } t = \rho x = 0 \mathrm {~d} \frac { \mathrm { dx } } { \mathrm { dt } } = 2\).

1. Sh the t \(x\) ad \(t\) satisfyt b d fferen ial eq tin $$\frac { \mathrm { dx } } { \mathrm { dt } } = \left( \frac { 1 } { 1 } \quad x \quad x \right) \left( \begin{array} { l l } 1 & x \end{array} \right) .$$
2. Sb this d fferen ial eq tin \(\mathbf { C }\) aira ressift \(\mathbf { D }\) irt erms \(\boldsymbol { 6 } t\).
3. State wh th p в to b \& le \(6 x\) wd \(n t\) b cm es larg If B e th follw ig lin dpg to cm p ete th an wer(s) to ay q stin (s), th q stin \(\mathrm { m } \quad \mathbf { b } \quad \mathrm { r } ( \mathrm { s } )\) ms tb clearlys n n

15 An isolated island is populated by rabbits and foxes. At time \(t\) the number of rabbits is \(x\) and the number of foxes is \(y\). It is assumed that:

• The number of foxes increases at a rate proportional to the number of rabbits. When there are 200 rabbits the number of foxes is increasing at a rate of 20 foxes per unit period of time.
• If there were no foxes present, the number of rabbits would increase by \(120 \%\) in a unit period of time.
• When both foxes and rabbits are present the foxes kill rabbits at a rate that is equal to \(110 \%\) of the current number of foxes.
• At time \(t = 0\), the number of foxes is 20 and the number of rabbits is 80 .

15

1. 1. Construct a mathematical model for the number of rabbits.
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2. (ii) Use this model to show that the number of rabbits has doubled after approximately 0.7 units of time.
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3. Suggest one way in which the model that you have used for the number of rabbits could be refined.
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11 During an industrial process substance \(X\) is converted into substance \(Z\). Some of the substance \(X\) goes through an intermediate phase, and is converted to substance \(Y\), before being converted to substance \(Z\). The situation is modelled by $$\frac { \mathrm { d } y } { \mathrm {~d} t } = 0.3 x - 0.2 y \text { and } \frac { \mathrm { d } z } { \mathrm {~d} t } = 0.2 y + 0.1 x$$ where \(x , y\) and \(z\) are the amounts in kg of \(X , Y\) and \(Z\) at time \(t\) hours after the process starts. Initially there is 10 kg of substance \(X\) and nothing of substances \(Y\) and \(Z\). The amount of substance \(X\) decreases exponentially. The initial rate of decrease is 4 kg per hour.

1. Show that \(x = A \mathrm { e } ^ { - 0.4 t }\), stating the value of \(A\).
2. (a) Show that \(\frac { \mathrm { d } x } { \mathrm {~d} t } + \frac { \mathrm { d } y } { \mathrm {~d} t } + \frac { \mathrm { d } z } { \mathrm {~d} t } = 0\).
   (b) Comment on this result in the context of the industrial process.
3. Express \(y\) in terms of \(t\).
4. Determine the maximum amount of substance \(Y\) present during the process.
5. How long does it take to produce 9 kg of substance \(Z\) ? \section*{END OF QUESTION PAPER} {www.ocr.org.uk}) after the live examination series. If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity. For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1GE.
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17 Two similar species, X and Y , of a small mammal compete for food and habitat. A model of this competition assumes, in a particular area, the following.

• In the absence of the other species, each species would increase at a rate proportional to the number present with the same constant of proportionality in each case.
• The competition reduces the rate of increase of each species by an amount proportional to the number of the other species present.

So if the numbers of species X and Y present at time \(t\) years are \(x\) and \(y\) respectively, the model gives the differential equations
\(\frac { d x } { d t } = k x - a y\) and \(\frac { d y } { d t } = k y - b x\),
where \(k , a\) and \(b\) are positive constants.

1. 1. Show that the general solution for \(x\) is \(x = A e ^ { ( k + n ) t } + B e ^ { ( k - n ) t }\), where \(n = \sqrt { a b }\) and \(A\) and \(B\) are arbitrary constants.
   2. Hence find the general solution for \(y\) in terms of \(A , B , k , n , a\) and \(t\). Observations suggest that suitable values for the model are \(k = 0.015 , a = 0.04\) and \(b = 0.01\). You should use these values in the rest of this question.
2. When \(t = 0\), the numbers present of species X and Y in this area are \(x _ { 0 }\) and \(y _ { 0 }\) respectively.
   1. Show that \(\mathrm { x } = \frac { 1 } { 2 } \left( \mathrm { x } _ { 0 } - 2 \mathrm { y } _ { 0 } \right) \mathrm { e } ^ { 0.035 \mathrm { t } } + \frac { 1 } { 2 } \left( \mathrm { x } _ { 0 } + 2 \mathrm { y } _ { 0 } \right) \mathrm { e } ^ { - 0.005 \mathrm { t } }\).
   2. Hence show that \(y = \frac { 1 } { 4 } \left( x _ { 0 } + 2 y _ { 0 } \right) e ^ { - 0.005 t } - \frac { 1 } { 4 } \left( x _ { 0 } - 2 y _ { 0 } \right) e ^ { 0.035 t }\).
3. Use initial values \(x _ { 0 } = 500\) and \(y _ { 0 } = 300\) with the results in part (b) to determine what the model predicts for each of the following questions.
   1. What numbers of each species will be present after 25 years?
   2. In this question you must show detailed reasoning. When will the numbers of the two species be equal?
   3. Does either species ever disappear from the area? Justify your answer.
4. Different initial values will apply in other areas where the two species compete, but previous studies indicate that one species or the other will eventually dominate in any given area.
   1. Identify a relationship between \(x _ { 0 }\) and \(y _ { 0 }\) where the model does not predict this outcome.
   2. Explain what the model predicts in the long term for this exceptional case.

10 In a colony of seabirds, there are \(y\) birds at time \(t\) years. 10

1. The rate of reduction in the number of birds due to birds dying or leaving the colony is proportional to the number of birds. In one year the reduction in the number of birds due to birds dying or leaving the colony is equal to \(16 \%\) of the number of birds at the start of the year. If no birds are born or join the colony, find the constant \(k\) such that $$\frac { \mathrm { d } y } { \mathrm {~d} t } = - k y$$ Give your answer to three significant figures.
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2. A wildlife protection group takes measures to support the colony.
   The rate of reduction in the number of birds due to birds dying or leaving the colony is the same as in part (a), but in addition:
   • The rate of increase in the number of birds due to births is \(20 t\) per year.
   • The wildlife protection group brings 45 birds into the colony each year.
   Write down a first-order differential equation for \(y\) and \(t\)
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3. The initial number of birds is 340 Solve your differential equation from part (b) to find \(y\) in terms of \(t\)
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4. Describe two limitations of the model you have used. Limitation 1 \(\_\_\_\_\)
   Limitation 2 \(\_\_\_\_\)