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IB AA HL Math

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N(A) E Math

PSLE

PSLE Foundation Mathematics

<p>The spread of an infection through a population of rabbits is modelled by a differential equation.</p>
<p>At time $t$ days, $N$ thousand rabbits are infected.</p>
<p>The model is $\frac{\mathrm{d}N}{\mathrm{d}t}=k(-3+4N-N^2)$, where&nbsp;$t\geq0$ and $k$ is a positive constant.</p>
<p>Initially, $1500$ rabbits are infected.</p>
<ol class="abc-list"><li data-list-item-id="h2-rabbits-6" data-list-item-path="h2-rabbits-6"><div class="list-item-paragraph">Express $\frac1{-3+4x-x^2}$ in the form $\frac{A}{x-1}+\frac{B}{3-x}$, where&nbsp;$A,B\in\mathbb{R}$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="h2-rabbits-8" data-list-item-path="h2-rabbits-8"><div class="list-item-paragraph">By separating the variables and using part (a), show that&nbsp;$\ln\left(\frac{N-1}{3-N}\right)=2kt-\ln p$, where&nbsp;$p$ is a positive constant to be determined. You may assume that&nbsp;$1&lt;N&lt;3$.<span class="inline-marks">[8]</span></div></li><li data-list-item-id="h2-rabbits-10" data-list-item-path="h2-rabbits-10"><div class="list-item-paragraph">Hence express $N$ explicitly in terms of $t$ and $k$, in the form $N=\frac{a+be^{-2kt}}{1+ce^{-2kt}}$, where&nbsp;$a$, $b$ and $c$ are integers to be determined.<span class="inline-marks">[5]</span></div></li><li data-list-item-id="h2-rabbits-12" data-list-item-path="h2-rabbits-12"><div class="list-item-paragraph">According to the model, the number of infected rabbits approaches a limiting value, $L$, as&nbsp;$t$ becomes large.</div><div class="list-item-paragraph">Find the value of $L$.<span class="inline-marks">[2]</span></div></li></ol>

The spread of an infection through a population of rabbits is modelled by a differential equation.

At time $t$ days, $N$ thousand rabbits are infected.

The model is $\frac{\mathrm{d}N}{\mathrm{d}t}=k(-3+4N-N^2)$, where $t\geq0$ and $k$ is a positive constant.

Initially, $1500$ rabbits are infected.

Express $\frac1{-3+4x-x^2}$ in the form $\frac{A}{x-1}+\frac{B}{3-x}$, where $A,B\in\mathbb{R}$.

Equilibrium Solutions and Long-Term Behaviour

Solving Differential Equations by Separation of Variables

Integration Using Partial Fractions

Partial Fractions

IB AA HL May 2026 P1 TZA Q12 (H2 adaptation)

Differential Equations

Techniques of Integration

<p>A tank contains $500$ litres of brine in which $75$ kg of salt is dissolved. Brine containing $3$ kg of salt per $10$ litres of water flows in at a rate of $20$ litres per minute, and the mixture flows out at the same rate.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Show that, for&nbsp;$t \ge 0, \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{150 - x}{25}$, where&nbsp;$x$ is the amount of dissolved salt in the tank at time $t \ge 0$.</span></div></li><li><div class="list-item-content"><span class="list-item-text">When will there be $125$ kg of dissolved salt in the tank?</span></div></li><li><div class="list-item-content"><span class="list-item-text">How much dissolved salt is in the tank after a long time?</span></div></li><li><div class="list-item-content"><span class="list-item-text">State an assumption used in the model.</span></div></li></ol>

A tank contains $500$ litres of brine in which $75$ kg of salt is dissolved. Brine containing $3$ kg of salt per $10$ litres of water flows in at a rate of $20$ litres per minute, and the mixture flows out at the same rate.

Show that, for $t \ge 0, \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{150 - x}{25}$, where $x$ is the amount of dissolved salt in the tank at time $t \ge 0$.

When will there be $125$ kg of dissolved salt in the tank?

How much dissolved salt is in the tank after a long time?

Mixing and Concentration Models

Assumptions and Limitations of Differential Equation Models

<p>As a tree grows, the rate of increase of its height, $h$ m, with respect to time, $t$ years after planting, is modelled by the differential equation</p>
<p style="text-align: center">\[\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{1}{10}\sqrt{\left(16 - \frac{1}{2}h\right)}\]</p>
<p>The tree is planted as a seedling of negligible height, so that&nbsp;$h=0$ when&nbsp;$t=0$.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">State the maximum height of the tree, according to this model.</span><span class="inline-marks">[1]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find an expression for&nbsp;$t$ in terms of $h$, and hence find the time the tree takes to reach half its maximum height.</span><span class="inline-marks">[5]</span></div></li></ol>

As a tree grows, the rate of increase of its height, $h$ m, with respect to time, $t$ years after planting, is modelled by the differential equation

\[\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{1}{10}\sqrt{\left(16 - \frac{1}{2}h\right)}\]

The tree is planted as a seedling of negligible height, so that $h=0$ when $t=0$.

State the maximum height of the tree, according to this model.

Find an expression for $t$ in terms of $h$, and hence find the time the tree takes to reach half its maximum height.

N2015 P2 Q1

<p>Welding fumes contain a dangerous amount of particulates. To protect workers in a workshop, an extraction device remove particulates in the air continuously. Let&nbsp;$V$ mg represent the mass of particulates in the air in the workshop at time $t$ min after the extraction device is activated.</p>
<p>Particulates are produced at a constant rate of $0.32$ mg/min. The rate at which particulates are removed is proportional to the mass of particulates in the air. When the mass of particulates in the air in the workshop is $18.75$ mg, the mass of particulates in the air increases at a rate of $0.12$ mg/min.</p>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">Show that&nbsp;$\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{4}{375}(30-V)$.</span><span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">Solve the differential equation, given that the workshop is initially free of particulates.</span><span class="inline-marks">[5]</span></div></li><li data-list-item-id="item-iii"><div class="list-item-content"><span class="list-item-text">Sketch the graph of $V$ against $t$.</span><span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-iv"><div class="list-item-content"><span class="list-item-text">It is recommended that the mass of particulates in this workshop should be kept below $32$ mg. Comment on whether the extraction device is effective enough to meet the recommendation.</span><span class="inline-marks">[1]</span></div></li></ol>
<p>The extraction device becomes less efficient and now removes particulates at half its original rate.</p>
<ol class="i-list" start="5"><li data-list-item-id="item-v"><div class="list-item-content"><span class="list-item-text">Write down a new differential equation to model this.</span><span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-vi"><div class="list-item-content"><span class="list-item-text">The mass of particulates in the air in the workshop is said to reach a steady-state when it remains constant. Find the value of $V$ at steady-state.</span><span class="inline-marks">[1]</span></div></li></ol>

Welding fumes contain a dangerous amount of particulates. To protect workers in a workshop, an extraction device remove particulates in the air continuously. Let $V$ mg represent the mass of particulates in the air in the workshop at time $t$ min after the extraction device is activated.

Particulates are produced at a constant rate of $0.32$ mg/min. The rate at which particulates are removed is proportional to the mass of particulates in the air. When the mass of particulates in the air in the workshop is $18.75$ mg, the mass of particulates in the air increases at a rate of $0.12$ mg/min.

Show that $\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{4}{375}(30-V)$.

Solve the differential equation, given that the workshop is initially free of particulates.

It is recommended that the mass of particulates in this workshop should be kept below $32$ mg. Comment on whether the extraction device is effective enough to meet the recommendation.

The extraction device becomes less efficient and now removes particulates at half its original rate.

Write down a new differential equation to model this.

The mass of particulates in the air in the workshop is said to reach a steady-state when it remains constant. Find the value of $V$ at steady-state.

Sketching Solution Curves of Differential Equations

Modelling with Multiple Rates of Change

2025 SAJC P1 Q12

<p>A cafe roasts its own coffee beans and packages them to be sold. The amount of roasted coffee beans remaining in the cafe at time $t$ days is denoted by $x$ kg. The cafe produces roasted coffee beans at a fixed rate of $10$ kg/day, and sells the roasted coffee beans at a rate proportional to $x^2$ kg/day.</p>
<ol class="abc-list"><li data-list-item-id="item-cafe-a" data-list-item-path="item-cafe-a"><div class="list-item-paragraph">Given that there is no change in the amount of roasted coffee beans remaining in the cafe when there is $5$ kg of beans remaining, show that&nbsp;$\frac{\mathrm{d}x}{\mathrm{d}t} = 10 - \frac{2}{5}x^2$. Find the general solution of this differential equation. (You do not have to make $x$ the subject.)<span class="inline-marks">[3]</span></div></li></ol>
<p>Due to improvement in roasting capabilities of the cafe, it is now able to produce roasted coffee beans at the rate of $3x$ kg/day, and sells the roasted coffee beans at a rate proportional to $x^2$ kg/day.</p>
<ol class="abc-list" start="2"><li data-list-item-id="item-cafe-b" data-list-item-path="item-cafe-b"><div class="list-item-paragraph">Given that amount of roasted coffee beans is increasing at the rate of $6$ kg/day when the amount of roasted coffee beans remaining in the cafe is $6$ kg, write down the modified differential equation relating $\frac{\mathrm{d}x}{\mathrm{d}t}$ and $x$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-cafe-c" data-list-item-path="item-cafe-c"><div class="list-item-paragraph"><p>Given that the initial amount of roasted coffee beans in the cafe is $13.5$ kg, solve the differential equation in part <strong>(b)</strong>, expressing $x$ in terms of $t$. Deduce the amount of roasted coffee beans remaining in the cafe in the long run, and sketch the curve of $x$ against $t$.</p><span class="inline-marks">[6]</span></div></li><li data-list-item-id="item-cafe-d" data-list-item-path="item-cafe-d"><div class="list-item-paragraph"><p>Based on the curve in part <strong>(c)</strong>, explain in context why this might be favourable for the cafe owner.</p><span class="inline-marks">[1]</span></div></li></ol>

A cafe roasts its own coffee beans and packages them to be sold. The amount of roasted coffee beans remaining in the cafe at time $t$ days is denoted by $x$ kg. The cafe produces roasted coffee beans at a fixed rate of $10$ kg/day, and sells the roasted coffee beans at a rate proportional to $x^2$ kg/day.

Given that there is no change in the amount of roasted coffee beans remaining in the cafe when there is $5$ kg of beans remaining, show that $\frac{\mathrm{d}x}{\mathrm{d}t} = 10 - \frac{2}{5}x^2$. Find the general solution of this differential equation. (You do not have to make $x$ the subject.)

Due to improvement in roasting capabilities of the cafe, it is now able to produce roasted coffee beans at the rate of $3x$ kg/day, and sells the roasted coffee beans at a rate proportional to $x^2$ kg/day.

Given that amount of roasted coffee beans is increasing at the rate of $6$ kg/day when the amount of roasted coffee beans remaining in the cafe is $6$ kg, write down the modified differential equation relating $\frac{\mathrm{d}x}{\mathrm{d}t}$ and $x$.

Logistic Growth Models

2025 RVHS P1 Q13

<p>Scientists are studying the growth of a newly discovered biomass that entered Earth's atmosphere together with an asteroid. A sample of $100$ grams of the biomass was collected initially. It was found that on the next day, the sample grew to $110$ grams. The scientists observed that the rate of growth of biomass was proportional to the amount of biomass present. The amount of biomass at time $t$ days is denoted by $B$ grams.</p>
<ol class="abc-list"><li data-list-item-id="item-a-4" data-list-item-path="item-a-4"><div class="list-item-paragraph">Write down a differential equation relating $B$ and $t$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-b-4" data-list-item-path="item-b-4"><div class="list-item-paragraph">Solve the differential equation obtained in part <strong>(a)</strong>, expressing $B$ in terms of $t$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-c-4" data-list-item-path="item-c-4"><div class="list-item-paragraph">Sketch the graph of $B$ against $t$, and explain what would happen if the biomass was left on its own.<span class="inline-marks">[2]</span></div></li></ol>
<p>To limit the growth of the biomass, the scientist collected a second sample of $100$ grams of the biomass and introduced a nutrient depletion serum to the sample the moment it was collected. The rate of growth of the biomass for this second sample is modelled by the differential equation</p>
<p style="text-align: center">\[\frac{\mathrm{d}B}{\mathrm{d}t} = \frac{mB}{3 + 2t^2}\]</p>
<p>where&nbsp;$m$ is a constant. After $1$ day, the second sample grew to $101$ grams.</p>
<ol class="abc-list" start="4"><li data-list-item-id="item-d-4" data-list-item-path="item-d-4"><div class="list-item-paragraph">For this second sample, find $B$ in terms of $t$.<span class="inline-marks">[5]</span></div></li><li data-list-item-id="item-e-4" data-list-item-path="item-e-4"><div class="list-item-paragraph">Determine the amount of the second biomass sample after a long time, leaving your answer to $5$ decimal places.<span class="inline-marks">[1]</span></div></li></ol>

Scientists are studying the growth of a newly discovered biomass that entered Earth's atmosphere together with an asteroid. A sample of $100$ grams of the biomass was collected initially. It was found that on the next day, the sample grew to $110$ grams. The scientists observed that the rate of growth of biomass was proportional to the amount of biomass present. The amount of biomass at time $t$ days is denoted by $B$ grams.

Write down a differential equation relating $B$ and $t$.

Sketch the graph of $B$ against $t$, and explain what would happen if the biomass was left on its own.

To limit the growth of the biomass, the scientist collected a second sample of $100$ grams of the biomass and introduced a nutrient depletion serum to the sample the moment it was collected. The rate of growth of the biomass for this second sample is modelled by the differential equation

\[\frac{\mathrm{d}B}{\mathrm{d}t} = \frac{mB}{3 + 2t^2}\]

where $m$ is a constant. After $1$ day, the second sample grew to $101$ grams.

For this second sample, find $B$ in terms of $t$.

Determine the amount of the second biomass sample after a long time, leaving your answer to $5$ decimal places.

Exponential Growth and Decay Models

2025 NYJC P1 Q12

<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Express $\frac{1}{Q(500 - Q)}$ in partial fractions.<span class="inline-marks">[2]</span></div></li></ol>
<ol class="abc-list" start="2"><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph"><p>A group of scientists is monitoring the population of a particular species of birds on an island. At time $t$ years after the start of the monitoring, the number of birds on the island is $Q$. The scientists observe that the birth rate is proportional to the bird's population and the birds are dying at a rate proportional to the square of the bird's population. There were $1000$ birds on the island when the scientists first started monitoring the population. They discover that the population remains unchanged when there are $500$ birds.</p></div><ol class="i-list"><li data-list-item-id="item-b-i" data-list-item-path="item-b/item-b-i"><div class="list-item-paragraph">Show that the differential equation relating $Q$ and $t$ is given by\[\frac{\mathrm{d}Q}{\mathrm{d}t} = kQ(500 - Q)\]where&nbsp;$k$ is a positive constant.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b-ii" data-list-item-path="item-b/item-b-ii"><div class="list-item-paragraph">Hence, solve the differential equation, expressing $Q$ in terms of $k$ and $t$.<span class="inline-marks">[5]</span></div></li></ol></li></ol>

Express $\frac{1}{Q(500 - Q)}$ in partial fractions.

2025 HCI P1 Q10

<p>A swimming pool contains $375000$ litres of pure water. Water containing $s$ milligrams of free chlorine per litre flows into the pool at a rate of $10$ litres per minute. The pool is also draining at a rate of $10$ litres per minute, such that the volume of the water in the pool remains constant.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Show that the rate at which the mass of the free chlorine, $x$ grams, is changing in the pool over time, $t$ minutes, can be modelled by the differential equation<br>\[\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{375s - x}{37500},\]State a necessary assumption for the above model to be valid.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Given that no free chlorine is present in the pool initially, find $x$ in terms of $t$ and $s$.</span><span class="inline-marks">[4]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find, in terms of $s$, the mass of free chlorine present in the pool after one hour.</span><span class="inline-marks">[1]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Sketch the graph of $x$ against $t$ which is relevant to the context, labelling the point(s) where the curve crosses the axes and the equation(s) of the asymptote(s). Hence determine the mass of free chlorine present in the pool after a long period in terms of $s$.</span><span class="inline-marks">[3]</span></div></li></ol>
<p>To prevent health complications, the recommended safe level of free chlorine to be used is between $375$ grams and $1125$ grams in a pool which contains $375000$ litres of water. Find the range of values of $s$ that should be used.<span class="inline-marks">[2]</span></p>

A swimming pool contains $375000$ litres of pure water. Water containing $s$ milligrams of free chlorine per litre flows into the pool at a rate of $10$ litres per minute. The pool is also draining at a rate of $10$ litres per minute, such that the volume of the water in the pool remains constant.

Given that no free chlorine is present in the pool initially, find $x$ in terms of $t$ and $s$.

Find, in terms of $s$, the mass of free chlorine present in the pool after one hour.

Sketch the graph of $x$ against $t$ which is relevant to the context, labelling the point(s) where the curve crosses the axes and the equation(s) of the asymptote(s). Hence determine the mass of free chlorine present in the pool after a long period in terms of $s$.

To prevent health complications, the recommended safe level of free chlorine to be used is between $375$ grams and $1125$ grams in a pool which contains $375000$ litres of water. Find the range of values of $s$ that should be used.

Sketching Exponential Functions and Solving Equations Graphically

Asymptotes and Discontinuities

Graphing Techniques

<p>The mass of a person depends both on daily rate of energy intake and on a daily rate of energy expenditure.</p>
<p>In this question, mass is in kg, time is in days, and energy intake and energy expenditure are measured in Calories per day.</p>
<p>The rate of change of a person's mass with respect to time is proportional to the difference between energy intake and energy expenditure.</p>
<p>Andrew has a mass of $M$ kg and his energy intake is fixed at $C$ Calories per day. For every kg of his mass, he expends $30$ Calories per day.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Show that $\frac{\mathrm{d}M}{\mathrm{d}t} = k(C - 30M)$, where $t$ is time and $k$ is a constant.<span class="inline-marks">[1]</span></div></li></ol>
<p>Andrew's initial mass is $110$ kg.</p>
<ol class="abc-list" start="2"><li><div class="list-item-content"><span class="list-item-text">Find the energy intake such that he maintains his mass at $110$ kg.</span><span class="inline-marks">[1]</span></div></li></ol>
<p>As part of a new health plan, Andrew fixes his energy intake at $80$% of the value found in part <strong>(b)</strong>.</p>
<ol class="abc-list" start="3"><li><div class="list-item-paragraph">By solving the differential equation in part <strong>(a)</strong>, show that Andrew's mass while he is on the plan satisfies the equation $M = 88 + 22\mathrm{e}^{-30kt}$.<span class="inline-marks">[4]</span></div></li></ol>
<p>Andrew's mass after $75$ days on the plan is $100$ kg.</p>
<ol class="abc-list" start="4"><li><div class="list-item-paragraph">Find the number of additional days required on the plan for Andrew's mass to fall below $96$ kg.<span class="inline-marks">[4]</span></div></li><li><div class="list-item-paragraph"></div><ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Sketch a graph of Andrew's mass while on this plan. Explain why Andrew cannot achieve a mass of $80$ kg using this plan.</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">State the range of possible values of energy intake for which Andrew could achieve a mass of $80$ kg.</span><span class="inline-marks">[1]</span></div></li></ol></li></ol>

The mass of a person depends both on daily rate of energy intake and on a daily rate of energy expenditure.

In this question, mass is in kg, time is in days, and energy intake and energy expenditure are measured in Calories per day.

The rate of change of a person's mass with respect to time is proportional to the difference between energy intake and energy expenditure.

Andrew has a mass of $M$ kg and his energy intake is fixed at $C$ Calories per day. For every kg of his mass, he expends $30$ Calories per day.

Find the energy intake such that he maintains his mass at $110$ kg.

As part of a new health plan, Andrew fixes his energy intake at $80$% of the value found in part **(b)**.

Andrew's mass after $75$ days on the plan is $100$ kg.

Find the number of additional days required on the plan for Andrew's mass to fall below $96$ kg.

N2023 P1 Q10

<p>A particular species reproduce at a rate of $100k\%$, where $k$ is a positive constant. Due to natural causes, members of the population die at a rate that is proportional to the square of its population size with constant of proportionality $\frac{k}{n}$, where $n$ is a positive constant. It may be assumed that there are no other external factors that affect the growth of the population, whose size is represented by $x$.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Write, in terms of $k$ and $n$, a differential equation for the population growth rate.</span></div></li><li><div class="list-item-paragraph">Based on your differential equation in part <strong>(i)</strong>, find, in terms of $k$ and/or $n$, the size of the population when its growth is the fastest, and the growth rate at this instant.</div></li><li><div class="list-item-paragraph">Solve the differential equation in part <strong>(i)</strong>, given that the population is growing the fastest initially. Express $x$ in terms of $t$.</div></li><li><div class="list-item-content"><span class="list-item-text">Sketch the graph of $x$ against $t$, stating the equation of the asymptote in terms of $n$.</span></div></li><li><div class="list-item-content"><span class="list-item-text">With reference to your graph, explain the significance of $n$ in context.</span></div></li></ol>

A particular species reproduce at a rate of $100k\%$, where $k$ is a positive constant. Due to natural causes, members of the population die at a rate that is proportional to the square of its population size with constant of proportionality $\frac{k}{n}$, where $n$ is a positive constant. It may be assumed that there are no other external factors that affect the growth of the population, whose size is represented by $x$.

Write, in terms of $k$ and $n$, a differential equation for the population growth rate.

Sketch the graph of $x$ against $t$, stating the equation of the asymptote in terms of $n$.

With reference to your graph, explain the significance of $n$ in context.

Problems Involving Maximisation and Minimisation

Applications of Differentiation

<p>A company proposed to build a water detention tank to address the flooding problem in a village. To test the feasibility of the proposal, the company created a model of a water detention tank, with a capacity of $300$ cm$ {}^3 $. </p>
<p>Water is pumped into the model of the tank at a constant rate of $50$ cm$ {}^3 $/min and pumped out at a rate proportional to the volume of water in the tank. At time $t$ minutes, the volume of water in the model of the tank is $V$ cm$ {}^3 $.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Write down a differential equation for this situation.</span><span class="inline-marks">[1]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Solve this differential equation to obtain the general solution for&nbsp;$V$ in terms of $t$.</span><span class="inline-marks">[4]</span></div></li></ol>
<p>Initially, the volume of water in the tank is $100$ cm$ {}^3 $. After a minute, the volume increased to $130$ cm$ {}^3 $.</p>
<ol class="i-list" start="3"><li><div class="list-item-content"><span class="list-item-text">Hence, show that the particular solution for&nbsp;$V$ in terms of $t$ is $V = 289 - 189e^{-0.173t}$, correct to $3$ significant figures.</span><span class="inline-marks">[4]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find how long it will take for the tank to reach $95$% of its maximum capacity.</span><span class="inline-marks">[1]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Sketch the graph of $V$ against $t$.</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Explain, with justification, what will happen to the volume of water in the tank in the long run.</span><span class="inline-marks">[1]</span></div></li></ol>

A company proposed to build a water detention tank to address the flooding problem in a village. To test the feasibility of the proposal, the company created a model of a water detention tank, with a capacity of $300$ cm$ {}^3 $. 

Water is pumped into the model of the tank at a constant rate of $50$ cm$ {}^3 $/min and pumped out at a rate proportional to the volume of water in the tank. At time $t$ minutes, the volume of water in the model of the tank is $V$ cm$ {}^3 $.

Write down a differential equation for this situation.

Solve this differential equation to obtain the general solution for $V$ in terms of $t$.

Initially, the volume of water in the tank is $100$ cm$ {}^3 $. After a minute, the volume increased to $130$ cm$ {}^3 $.

Hence, show that the particular solution for $V$ in terms of $t$ is $V = 289 - 189e^{-0.173t}$, correct to $3$ significant figures.

Find how long it will take for the tank to reach $95$% of its maximum capacity.

Explain, with justification, what will happen to the volume of water in the tank in the long run.

 2024 TMJC P1 Q10

<p>Scientists are investigating the growth in length of a particular species of fish.</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">The scientists used the von Bertalanffy growth model to predict the length $L$ cm of the fish at a particular time, $t$ years after birth. It is said that the rate of growth in length of the fish is directly proportional to the difference between its theoretical maximum length of the fish, $60$ cm and its length at the time $t$ years. The constant of proportionality, $k$, is known as the growth coefficient which is always positive. At birth, it is known that the rate of growth when the fish is $10$ cm long is $5$ cm per year.</div><ol class="abc-list"><li data-list-item-id="item-i-a" data-list-item-path="item-i/item-i-a"><div class="list-item-paragraph">Write down a differential equation for this situation. Solve this differential equation to get&nbsp;$L$ as an exact function of $t$.<span class="inline-marks">[7]</span></div></li><li data-list-item-id="item-i-b" data-list-item-path="item-i/item-i-b"><div class="list-item-paragraph">Find the length of the fish $5$ years after it is born.<span class="inline-marks">[1]</span></div></li></ol></li></ol>
<ol class="i-list" start="2"><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Based on the data collected, one of the scientists proposed a second model to predict the length $L$ cm of the fish at a particular time, $t$ years after birth. The proposed model is as follows:</div><div class="list-item-paragraph">\[\frac{\mathrm{d}L}{\mathrm{d}t} = \frac{50}{L}\]</div><ol class="abc-list"><li data-list-item-id="item-ii-a" data-list-item-path="item-ii/item-ii-a"><div class="list-item-paragraph">Given that the length of the fish is $10$ cm long at birth, find $L$ in terms of $t$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-ii-b" data-list-item-path="item-ii/item-ii-b"><div class="list-item-paragraph"><p>Comment on the validity of the second model in modelling the length of fish, $L$ cm, at a particular time, $t$ years after birth as compared to the model in <strong>(i)</strong>.</p><span class="inline-marks">[1]</span></div></li></ol></li></ol>

Scientists are investigating the growth in length of a particular species of fish.

The scientists used the von Bertalanffy growth model to predict the length $L$ cm of the fish at a particular time, $t$ years after birth. It is said that the rate of growth in length of the fish is directly proportional to the difference between its theoretical maximum length of the fish, $60$ cm and its length at the time $t$ years. The constant of proportionality, $k$, is known as the growth coefficient which is always positive. At birth, it is known that the rate of growth when the fish is $10$ cm long is $5$ cm per year.

Based on the data collected, one of the scientists proposed a second model to predict the length $L$ cm of the fish at a particular time, $t$ years after birth. The proposed model is as follows:

Modelling with Differential Equations

2024 SAJC P1 Q10

<p>The height $x$, of a certain plant, at time $t$ is proportional to the amount of nutrients it absorbs. The plant loses nutrients at a rate proportional to the height of the plant and absorbs nutrients at a rate inversely proportional to the surface area of the stem of the plant. Taking the cross-section of the stem to be circular with radius $r=kx$, where $k$ is a constant, and the surface area of the stem to be $2\pi rx$,</p>
<ol class="i-list"><li><div class="list-item-paragraph">Show that $x$ satisfies the differential equation
</div><div class="list-item-paragraph" style="text-align: center">$\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{p}{x^{2}}-qx$,
</div><div class="list-item-paragraph">where $p$ and $q$ are positive constants.</div><p style="text-align: right;">[2]</p><div class="list-item-paragraph">For the case ${p = 2q}$, it is known that ${x = 0}$ and ${t = 0}$.</div></li><li><div class="list-item-paragraph">Solve this differential equation, obtaining an expression for $x$ in terms of $t$ and $q$.</div><p style="text-align: right;">[5]</p></li><li><div class="list-item-paragraph">Deduce the value of $x$ as $t$ becomes very large.</div><p style="text-align: right;">[1]</p></li></ol>

The height $x$, of a certain plant, at time $t$ is proportional to the amount of nutrients it absorbs. The plant loses nutrients at a rate proportional to the height of the plant and absorbs nutrients at a rate inversely proportional to the surface area of the stem of the plant. Taking the cross-section of the stem to be circular with radius $r=kx$, where $k$ is a constant, and the surface area of the stem to be $2\pi rx$,

DHS P2 Q3

<p>Water is being pumped from reservoir into a filtration device at a constant rate of $8$ gallons per hour. The filtration device processes the water and discharges it at a rate proportional to the volume of water currently in it. At time $t$ hours, the volume of water in the device is $v$ gallons. The filtration device is initially empty. After $10$ hours, the volume of water in the device is $5$ gallons.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Find an expression for $v$ in terms of $t$.</div><p style="text-align: right;">[7]</p></li><li><div class="list-item-paragraph">Hence find the volume of water in the filtration device in the long run.</div><p style="text-align: right;">[1]</p></li></ol>
<p>A fault in the filtration device is discovered when there are $4.5$ gallons of water in the device. The device now discharges water at a rate that is <strong>equal</strong> to the square of the volume of water currently in it.</p>
<ol class="abc-list" start="3"><li><div class="list-item-paragraph">Write down a differential equation to model the new situation.</div><p style="text-align: right;">[1]</p></li><li><div class="list-item-paragraph">Find the time (to the nearest minute) that elapsed from the moment the fault is discovered to the moment when the water in the device drops to $3$ gallons.</div><p style="text-align: right;">[4]</p></li></ol>

Water is being pumped from reservoir into a filtration device at a constant rate of $8$ gallons per hour. The filtration device processes the water and discharges it at a rate proportional to the volume of water currently in it. At time $t$ hours, the volume of water in the device is $v$ gallons. The filtration device is initially empty. After $10$ hours, the volume of water in the device is $5$ gallons.

A fault in the filtration device is discovered when there are $4.5$ gallons of water in the device. The device now discharges water at a rate that is **equal** to the square of the volume of water currently in it.

2023 TJC P1 Q10

<p>A tank contains $2$ m $^{3}$of water initially. Water flows into the tank at a constant rate of $4$ m$^{3}$s$^{-1}$and flows out at a rate which is proportional to the amount of water $V$ m$^{3}$ in the tank. The rate of change of volume of water in the tank is equal to $1$ m$^{3}$sx at the instant when&nbsp;$V=6$. By setting up and solving a differential equation, show that&nbsp;$V=8-6{{\mathrm{e}}^{-0.5t}}$.<span class="inline-marks">[7]</span></p>
<ol class="i-list"><li data-list-item-id="item-1760697249049-inzvnp10q" data-list-item-path="item-1760697249049-inzvnp10q"><div class="list-item-paragraph">What will happen to the volume of water in the tank eventually?<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-1760697249477-20qar9vtq" data-list-item-path="item-1760697249477-20qar9vtq"><div class="list-item-paragraph">Sketch a graph of $V$ against $t$.<span class="inline-marks">[2]</span></div></li></ol>

A tank contains $2$ m $^{3}$of water initially. Water flows into the tank at a constant rate of $4$ m$^{3}$s$^{-1}$and flows out at a rate which is proportional to the amount of water $V$ m$^{3}$ in the tank. The rate of change of volume of water in the tank is equal to $1$ m$^{3}$sx at the instant when $V=6$. By setting up and solving a differential equation, show that $V=8-6{{\mathrm{e}}^{-0.5t}}$.

2013 MJC P1 Q7

<p>In military exercises, parachutists jump from rest and their motion is tracked by sensors tagged to their bodies. One parachutist jumps vertically from a helicopter. At time $t$ seconds, the parachutist is $x$ metres below the helicopter with velocity $v$ms$^{-1}$. It experiences a downward gravitational acceleration of $\mathrm{9}\mathrm{.8}$ms$^{-1}$ and an upward acceleration $R$ due to air resistance, where $R<9.8$. $R$ is directly proportional to $v$, with a constant of proportionality $k>0$.</p>
<p>The rate of change of $v$ for the parachutist is modelled as the difference between the downward gravitational acceleration and $R$. Write down a differential equation relating $v$, $t$ and $k$, and solve this differential equation to find an expression for $v$ in terms of $t$ and $k$.</p><p style="text-align: right;">[6]</p>
<p>The terminal velocity of a falling object is the value of its velocity after a long time. State the terminal velocity of the parachutist in terms of $k$.</p><p style="text-align: right;">[1]</p>

In military exercises, parachutists jump from rest and their motion is tracked by sensors tagged to their bodies. One parachutist jumps vertically from a helicopter. At time $t$ seconds, the parachutist is $x$ metres below the helicopter with velocity $v$ms$^{-1}$. It experiences a downward gravitational acceleration of $\mathrm{9}\mathrm{.8}$ms$^{-1}$ and an upward acceleration $R$ due to air resistance, where $R<9.8$. $R$ is directly proportional to $v$, with a constant of proportionality $k>0$.

The rate of change of $v$ for the parachutist is modelled as the difference between the downward gravitational acceleration and $R$. Write down a differential equation relating $v$, $t$ and $k$, and solve this differential equation to find an expression for $v$ in terms of $t$ and $k$.

The terminal velocity of a falling object is the value of its velocity after a long time. State the terminal velocity of the parachutist in terms of $k$.

2023 JPJC P2 Q2

<p>Baby Peter is pulling a toy wagon on a flat ground which is modelled by the $xy$-coordinate plane. Peter is initially at the origin $O$ and the toy wagon is located on the $y$-axis at ${{W}_{0}}\left( 0,6 \right)$. The toy wagon is tethered to him with a rope of length $6$ units. As he moves in the direction of the positive $x$-axis (given by point $P$), he pulls the toy wagon (given by the point ${{W}_{1}}$) using the rope ${{W}_{1}}P$ which is kept taut throughout the motion. The path of the toy wagon is given by curve $C$. It is known that the rope is always tangent to $C$.</p>
<div class="graph-block" data-graph-config="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" data-width="294" data-height="192" data-proportional style="width:294px;height:192px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:294px;--graph-aspect:294 / 192;"></div>
<ol class="abc-list"><li data-list-item-id="item-1760424713332-d3edcnc48" data-list-item-path="item-1760424713332-d3edcnc48"><div class="list-item-paragraph">Show that a differential equation for&nbsp;$C$ can be expressed as</div><div class="list-item-paragraph" style="text-align: center">$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\alpha y}{\sqrt{36-{{y}^{2}}}}$,</div><div class="list-item-paragraph">where&nbsp;$\alpha $ is a constant to be determined.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1760424714015-hwwmcikt1" data-list-item-path="item-1760424714015-hwwmcikt1"><div class="list-item-paragraph">Using the substitution ${{w}^{2}}=36-{{y}^{2}}$ where&nbsp;$w&gt;0$, show that the differential equation in <strong>(a)</strong> can be reduced to</div><div class="list-item-paragraph" style="text-align: center">$\frac{\mathrm{d}w}{\mathrm{d}x}=\frac{36-{{w}^{2}}}{{{w}^{2}}}$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1760424716340-uv4vdevsh" data-list-item-path="item-1760424716340-uv4vdevsh"><div class="list-item-paragraph">Find $x$ in terms of $y$.<span class="inline-marks">[6]</span></div></li><li data-list-item-id="item-1760424716925-s3jfoa81v" data-list-item-path="item-1760424716925-s3jfoa81v"><div class="list-item-paragraph">Determine the behaviour of $x$ as&nbsp;$y$ approaches $0$.<span class="inline-marks">[2]</span></div></li></ol>

Baby Peter is pulling a toy wagon on a flat ground which is modelled by the $xy$-coordinate plane. Peter is initially at the origin $O$ and the toy wagon is located on the $y$-axis at ${{W}_{0}}\left( 0,6 \right)$. The toy wagon is tethered to him with a rope of length $6$ units. As he moves in the direction of the positive $x$-axis (given by point $P$), he pulls the toy wagon (given by the point ${{W}_{1}}$) using the rope ${{W}_{1}}P$ which is kept taut throughout the motion. The path of the toy wagon is given by curve $C$. It is known that the rope is always tangent to $C$.

Derivative as the Gradient of a Curve

Solving Differential Equations by Substitution

2023 NJC P1 Q11

<p>Glucose is a type of sugar in our blood and our body uses it for energy. Having low sugar in the blood for long periods of time can cause health problems if it is not treated.</p>
<p>At 10:00 am, a patient is given glucose via an intravenous drip at a constant rate of $r$ units of glucose per hour. It is known that the rate at which the glucose in the blood stream is absorbed by the body, is proportional to the amount of glucose present in the blood stream. At the start of the treatment, the patient has ${{x}_{0}}$units of glucose in his blood stream.</p>
<p>It is given that $x$ denotes the amount of glucose present in the blood stream of the patient at time $t$ hours after he starts the treatment.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Form a differential equation in $x$ and show that $x=\left( {{x}_{0}}-\frac{r}{k} \right){{\mathrm{e}}^{-kt}}+\frac{r}{k}$ where $k$ is a positive constant.</div><p style="text-align: right;">[4]</p></li></ol>
<p>It is now given that $\frac{r}{k}=2{{x}_{0}}$.</p>
<ol class="abc-list" start="2"><li><div class="list-item-paragraph">State the theoretical limiting value of $x$ if the patient is given the treatment over a long period.</div><p style="text-align: right;">[1]</p></li><li><div class="list-item-paragraph">Sketch the graph of $x$ for $t\ge 0$.</div><p style="text-align: right;">[2]</p></li></ol>
<p>The patient has $10\%$ more glucose in his blood stream after $\frac{1}{2}$ hour of treatment.</p>
<ol class="abc-list" start="4"><li><div class="list-item-paragraph">At what time (to the nearest minute), would the glucose in his blood stream first exceed $90\%$ of the value found in part <strong>(b)</strong>?</div><p style="text-align: right;">[5]</p></li></ol>

Glucose is a type of sugar in our blood and our body uses it for energy. Having low sugar in the blood for long periods of time can cause health problems if it is not treated.

At 10:00 am, a patient is given glucose via an intravenous drip at a constant rate of $r$ units of glucose per hour. It is known that the rate at which the glucose in the blood stream is absorbed by the body, is proportional to the amount of glucose present in the blood stream. At the start of the treatment, the patient has ${{x}_{0}}$units of glucose in his blood stream.

It is given that $x$ denotes the amount of glucose present in the blood stream of the patient at time $t$ hours after he starts the treatment.

It is now given that $\frac{r}{k}=2{{x}_{0}}$.

The patient has $10\%$ more glucose in his blood stream after $\frac{1}{2}$ hour of treatment.

2023 ASRJC P1 Q10

<p>Two scientists are investigating the change of a certain population of size $n$ thousand at time $t$ years.</p>
<ol class="i-list"><li><div class="list-item-paragraph">One scientist suggests that $n$ and $t$ are related by the differential equation $\frac{{{\mathrm{d}}^{2}}n}{\mathrm{d}{{t}^{2}}}=10-6t$. Find the general solution of this differential equation. Sketch three members of the family of solution curves, given that $n=100$ when $t=0$.<span class="inline-marks">[5]</span></div></li><li><div class="list-item-paragraph">The other scientist suggests that $n$ and $t$ are related by the differential equation $\frac{\mathrm{d}n}{\mathrm{d}t}=3-0.02n$. Find $n$ in terms of $t$, given again that $n=100$ when $t=0$. Explain in simple terms what will eventually happen to the population using this model.<span class="inline-marks">[7]</span></div></li></ol>

Two scientists are investigating the change of a certain population of size $n$ thousand at time $t$ years.

Second-Order Linear Differential Equations

Solving Differential Equations by Direct Integration

Order, General and Particular Solutions of Differential Equations

N2009 P2 Q4

<p>In a chemical plant, a tank initially contains $5000$ litres of water dissolved with a certain amount of substance $X$. Solution containing $0.05$kg of substance $X$ per litre of water flows into the tank at a constant rate of $10$ litres per minute. The mixture is thoroughly stirred and the resulting solution flows out of the tank at the same rate such that the volume of liquid in the tank is kept constant. If&nbsp;$x$(kg) is the amount of the substance $X$ in the tank at time, $t$ (min), show that the rate of change of the amount of substance $X$ in the tank can be modelled by the differential equation $\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{250-x}{500}$.<span class="inline-marks">[2]</span></p>
<ol class="i-list"><li><div class="list-item-paragraph">Solve the differential equation, leaving your answer in the form $x=B-A{{\mathrm{e}}^{-\,Ct}}$ where&nbsp;$A$ is an arbitrary constant, and $B$ and $C$ are positive constants to be determined.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Deduce the long-term steady state amount of substance $X$ in the tank?<span class="inline-marks">[1]</span></div></li></ol>

In a chemical plant, a tank initially contains $5000$ litres of water dissolved with a certain amount of substance $X$. Solution containing $0.05$kg of substance $X$ per litre of water flows into the tank at a constant rate of $10$ litres per minute. The mixture is thoroughly stirred and the resulting solution flows out of the tank at the same rate such that the volume of liquid in the tank is kept constant. If $x$(kg) is the amount of the substance $X$ in the tank at time, $t$ (min), show that the rate of change of the amount of substance $X$ in the tank can be modelled by the differential equation $\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{250-x}{500}$.

Math Question Bank