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PSLE

PSLE Foundation Mathematics

<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.

Equilibrium Solutions and Long-Term Behaviour

Solving Differential Equations by Separation of Variables

Sketching Solution Curves of Differential Equations

Modelling with Multiple Rates of Change

2025 SAJC P1 Q12

Differential Equations

<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

<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.

Partial Fractions

2025 HCI P1 Q10

Techniques of Integration

<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.

Sketching Exponential Functions and Solving Equations Graphically

Problems Involving Maximisation and Minimisation

Asymptotes and Discontinuities

Graphing Techniques

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>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>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>Liquid is poured into a container at a constant rate of $30$ cm$^{3}$s$^{-1}$ and leaks out at a rate which is proportional to the volume of liquid in the container. At time $t$ seconds, the volume of the water in the container is $V$cm$^{3}$. When the volume reaches $450$ cm$^{3}$, it decreases at a rate of $30$ cm$^{3}$s$^{-1}$ .</p>
<p>Show that&nbsp;$-15\frac{\mathrm{d}V}{\mathrm{d}t}=2V-450$.<span class="inline-marks">[2]</span></p>
<p>It is given that&nbsp;$V=1000$ cm$^{3}$ when&nbsp;$t=0$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Show that&nbsp;$V=A{{e}^{\alpha \,t}}+B$, where&nbsp;$\alpha $, $A$ and $B$ are constants to be determined.<span class="inline-marks">[4]</span></div></li><li><div class="list-item-paragraph">Sketch the graph of $V$ against $t$.<span class="inline-marks">[1]</span></div></li></ol>

Liquid is poured into a container at a constant rate of $30$ cm$^{3}$s$^{-1}$ and leaks out at a rate which is proportional to the volume of liquid in the container. At time $t$ seconds, the volume of the water in the container is $V$cm$^{3}$. When the volume reaches $450$ cm$^{3}$, it decreases at a rate of $30$ cm$^{3}$s$^{-1}$ .

Show that $-15\frac{\mathrm{d}V}{\mathrm{d}t}=2V-450$.

It is given that $V=1000$ cm$^{3}$ when $t=0$.

2003 HCI P2 Q2

<ol class="abc-list"><li><div class="list-item-paragraph">Show that the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x}-{{y}^{2}}=-\frac{2y}{x}$ can reduced to $\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{{{u}^{2}}}{{{x}^{2}}}$ using the substituion $y=\frac{u}{{{x}^{2}}}$. Hence, find $y$ in terms of $x$, given that&nbsp;$y=3$ when&nbsp;$x=1$.<span class="inline-marks">[6]</span></div></li><li><div class="list-item-paragraph">At time $t=0$, there are $1000$ students studying in Tenpin Junior College. At time $t$ years, the admission rate is equal to one tenth of the number of students $N$ studying in the college. There is a constant expulsion of $10$ students per year.</div><div class="list-item-paragraph">Assuming nobody transfers into or out of the college, show that&nbsp;$10\frac{\mathrm{d}N}{\mathrm{d}t}=N-100$.</div><div class="list-item-paragraph">Solve the differential equation to find $N$ in terms of $t$. Find the time taken for the number of students studying in Tenpin Junior College to increase to $1200$.<span class="inline-marks">[6]</span></div></li></ol>

Solving Differential Equations by Substitution

2007 TPJC CT Q3

<p>Sir Isaac Newton was a famous scientist renowned for his work on the laws of motion. One law states that, for an object falling vertically in a vacuum, the rate of change of velocity, $v$ ms$^{-1}$, with respect to time, $t$ seconds, is a constant, $c$.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text"></span></div><ol class="abc-list"><li><div class="list-item-paragraph">Write down a differential equation relating $v$, $t$ and $c$.</div><p style="text-align: right;">[1]</p></li><li><div class="list-item-paragraph">Initially the velocity of the object is $4$ ms$^{-1}$ and, after a further $2.5$ s, the velocity of the object is $29$ ms$^{-1}$. Find $v$ in terms of $t$ and state the value of $c$.</div><p style="text-align: right;">[3]</p></li></ol></li></ol>
<p>For an object falling vertically through the atmosphere, the rate of change of velocity is less than that for an object falling in a vacuum. The new rate of change of $v$ is modelled as the difference between the value of $c$ found in part <strong>(i)(b)</strong> and an amount proportional to the velocity $v$, with a constant of proportionality $k$.</p>
<ol class="i-list" start="2"><li><div class="list-item-paragraph">Given that in this case the initial velocity is zero, find $v$ in terms of $t$ and $k$.</div><p style="text-align: right;">[5]</p></li></ol>
<p>For an object falling through the atmosphere, the ‘terminal velocity’ is the value approached by the velocity after a long time.</p>
<ol class="i-list" start="3"><li><div class="list-item-paragraph">A falling object has initial velocity zero and terminal velocity $40$ ms$^{-1}$. Find how long it takes the object to reach $90\%$ of its terminal velocity.</div><p style="text-align: right;">[4]</p></li></ol>

Sir Isaac Newton was a famous scientist renowned for his work on the laws of motion. One law states that, for an object falling vertically in a vacuum, the rate of change of velocity, $v$ ms$^{-1}$, with respect to time, $t$ seconds, is a constant, $c$.

For an object falling vertically through the atmosphere, the rate of change of velocity is less than that for an object falling in a vacuum. The new rate of change of $v$ is modelled as the difference between the value of $c$ found in part **(i)(b)** and an amount proportional to the velocity $v$, with a constant of proportionality $k$.

For an object falling through the atmosphere, the ‘terminal velocity’ is the value approached by the velocity after a long time.

Solving Differential Equations by Direct Integration

Modelling with Differential Equations

N2017 P1 Q11

<p>A fish pond has a surface area of $100$m$^{2}$. Past observations on the growth of algae in similar ponds estimates that the area, $A$m$^{2}$, of algae present at time $t$ days is such that the rate at which $A$ is increasing is proportional to the product of the area of the pond covered by the algae and the area of the pond not covered by the algae.</p>
<p>The owner of the pond monitors the growth of algae in his pond. At $t=2$, he noted that the algae had grown to cover $11\%$ of the pond and was growing at a rate of $11$m$^{2}$ per day. He then decided to clear the algae at a constant rate of $11$m$^{2}$ each day, starting from $t=3$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Show that, from $t=3$, the differential equation relating $A$ and $t$ is</div><div class="list-item-paragraph" style="text-align: center">$\frac{\mathrm{d}A}{\mathrm{d}t}=-\frac{1}{89}\left( A-11 \right)\left( A-89 \right)$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Assuming that $A=20$ when $t=3$, show that $\frac{A-89}{A-11}=-\frac{23}{3}{{\mathrm{e}}^{q\left( 3-t \right)}}$, where $q$ is a positive constant to be determined.</div><p style="text-align: right;">[6]</p></li><li><div class="list-item-paragraph">To maintain the balance of the ecosystem in the pond, the owner wishes to ensure that algae always cover an area less than $80\%$ of the pond surface. Explain why this cannot be attained.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Find the value of $t$ when the maximum possible rate of change in the area of algae occurs.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">State one possible condition in which the model in part <strong>(i)</strong> is not valid.</div><p style="text-align: right;">[1]</p></li></ol>

A fish pond has a surface area of $100$m$^{2}$. Past observations on the growth of algae in similar ponds estimates that the area, $A$m$^{2}$, of algae present at time $t$ days is such that the rate at which $A$ is increasing is proportional to the product of the area of the pond covered by the algae and the area of the pond not covered by the algae.

The owner of the pond monitors the growth of algae in his pond. At $t=2$, he noted that the algae had grown to cover $11\%$ of the pond and was growing at a rate of $11$m$^{2}$ per day. He then decided to clear the algae at a constant rate of $11$m$^{2}$ each day, starting from $t=3$.

Assumptions and Limitations of Differential Equation Models

2021 DHS Promo Q12

<p>The cylindrical tank in a research laboratory has a cross-sectional area of $4$ m$^{2}$. To cool the tank, water is pumped in and out of the tank simultaneously. The volume and height of the water in the tank at any time $t$ minutes is given by $V$(litres) and $h$ (metres) respectively. Clean water is pumped into the tank at a rate that is proportional to ${{h}^{2}}$ and the water is pumped out from the tank at a rate that is proportional to $h$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Assume that the water does not overflow and that there is no change to the height of the water when $h$ is $10$, show that $\frac{\mathrm{d}h}{\mathrm{d}t}=\frac{kh(h-10)}{4}$ where $k$ is a real constant.</div><p style="text-align: right;">[4]</p></li></ol>
<p>The tank was initially filled with clean water to a height of $2$ metres. When the height of the water is $5$ metres, the volume of water is increasing at a rate of $5.5$ litres per minute.</p>
<ol class="i-list" start="2"><li><div class="list-item-paragraph">Find the exact value of $k$. Hence find $h$ in terms of $t$.</div><p style="text-align: right;">[5]</p></li><li><div class="list-item-paragraph">Sketch a graph of $h$ against $t$. Hence, write down the minimum height of the cylindrical tank that will not result in the overflow of the water.</div><p style="text-align: right;">[3]</p></li></ol>

The cylindrical tank in a research laboratory has a cross-sectional area of $4$ m$^{2}$. To cool the tank, water is pumped in and out of the tank simultaneously. The volume and height of the water in the tank at any time $t$ minutes is given by $V$(litres) and $h$ (metres) respectively. Clean water is pumped into the tank at a rate that is proportional to ${{h}^{2}}$ and the water is pumped out from the tank at a rate that is proportional to $h$.

The tank was initially filled with clean water to a height of $2$ metres. When the height of the water is $5$ metres, the volume of water is increasing at a rate of $5.5$ litres per minute.

2022 ASRJC P1 Q11

<ol class="abc-list"><li data-list-item-id="item-1760174551718-1o7huipab" data-list-item-path="item-1760174551718-1o7huipab"><div class="list-item-paragraph">At the beginning of January 2010, Mr Wong’s fish farm has&nbsp;$40000$ fish breeding in a large pond. In the period between January and December each year, the number of fish increases by $9\%$. At the end of December each year, $5400$ fish are harvested and sold at the fish markets for&nbsp;$\$6$ per fish. No fish died or are being poached while he is in business.</div><ol class="i-list"><li data-list-item-id="subitem-1760174557298-4zrvqvude" data-list-item-path="item-1760174551718-1o7huipab/subitem-1760174557298-4zrvqvude"><div class="list-item-paragraph">Find the number of fish just after the $n$ th harvest, giving your answer in the form $A-B{{\left( 1.09 \right)}^{n}}$, where&nbsp;$A$ and $B$ are constants to be determined.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="subitem-1760174560126-11k7jhi9s" data-list-item-path="item-1760174551718-1o7huipab/subitem-1760174560126-11k7jhi9s"><div class="list-item-paragraph">Find the earliest month and year in which there will be no more fish in the pond.<span class="inline-marks">[2]</span></div><div class="list-item-paragraph">Mr Wong has been thinking of retiring since January 2010. Every December just after the harvest, a nearby fish farm offers to buy over Mr Wong’s business by paying $\$8$ per fish for his remaining stock.</div></li><li data-list-item-id="subitem-1760174560629-43i2upvxu" data-list-item-path="item-1760174551718-1o7huipab/subitem-1760174560629-43i2upvxu"><div class="list-item-paragraph">If he wishes to retire by selling off his fish farm business after the $n$th harvest, find the value of $n$ where the total gains is the maximum.<span class="inline-marks">[3]</span></div></li></ol></li><li data-list-item-id="item-1760174610931-gmdkckgy6" data-list-item-path="item-1760174610931-gmdkckgy6"><div class="list-item-paragraph">White-tailed deer can be found in forests of Southern Canada. A wildlife biologist is investigating the change of a population of white-tailed deer of size $x$ (thousands) at time $t$ (years). He observed that on average, their birth rate is $4000$ per year and their death rate is proportional to the square of the population present.</div><ol class="i-list"><li data-list-item-id="subitem-1760174621979-pqkqkoj5h" data-list-item-path="item-1760174610931-gmdkckgy6/subitem-1760174621979-pqkqkoj5h"><div class="list-item-paragraph">When the population was $5000$, the rate of increase of the population at that instant was $1750$ per year. Assuming that&nbsp;$x$ and $t$ are continuous variables, form a differential equation relating $x$ and $t$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="subitem-1760174622826-eheovyzd1" data-list-item-path="item-1760174610931-gmdkckgy6/subitem-1760174622826-eheovyzd1"><div class="list-item-paragraph">There were $2000$ deer in the forest initially. Solve this differential equation to obtain $x$ as an exact function of $t$.<span class="inline-marks">[4]</span></div></li></ol></li></ol>

Problems involving inequalities

Finding Explicit Terms of Linear Recurrences

2020 ASRJC P1 Q11

Sequence and Series

<p>Glucose in the blood stream is reduced at a rate proportional to the amount of glucose present in the blood stream.</p>
<p>Let $G$ milligrams (mg) be the amount of glucose in $1$ decilitre (dL) of blood stream at time $t$ minutes and let $\lambda $ denote the positive constant of proportionality.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Write down a differential equation relating $G$, $\lambda $ and $t$. Solve this differential equation to find an expression of $G$ in terms of $\lambda $ and $t$.</div><p style="text-align: right;">[3]</p></li></ol>
<p>Through extensive research, doctors recommended that a healthy glucose level in the blood stream of an adult should be between $70$ mg/dL to $100$ mg/dL.</p>
<p>An adult patient, Neo, has $80$ mg/dL of glucose in his blood stream at time $t=0$ minutes. Due to a medical condition, he is not able to extract glucose from the food he eats. As he is not able to replenish the glucose in his blood stream, he is thus losing his glucose in the blood stream at a rate (in mg/dL per minute) corresponding to $\lambda =0.005$ per minute.</p>
<ol class="i-list" start="2"><li><div class="list-item-paragraph">Find the approximate time it would take for Neo’s glucose level in the blood stream to fall below the healthy range if there is no intake of glucose.</div><p style="text-align: right;">[2]</p></li></ol>
<p>In order to maintain Neo’s glucose level in the blood stream in the healthy range, glucose is injected into his blood stream intravenously at a constant rate of $\mu $ mg/dL per minute.</p>
<ol class="i-list" start="3"><li><div class="list-item-paragraph">Write down a differential equation relating $G$, $\mu $ and $t$ to model Neo’s situation. Solve this differential equation to find an expression of $G$ in terms of $\mu $ and $t$.</div><p style="text-align: right;">[5]</p></li><li><div class="list-item-paragraph">Explain whether it is recommended to keep injecting glucose at a rate of $0.7$ mg/dL per minute into Neo’s blood stream.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Find the maximum range of values of $\mu $ so that Neo’s glucose level in the blood stream will always be in the healthy range.</div><p style="text-align: right;">[1]</p></li></ol>

Glucose in the blood stream is reduced at a rate proportional to the amount of glucose present in the blood stream.

Let $G$ milligrams (mg) be the amount of glucose in $1$ decilitre (dL) of blood stream at time $t$ minutes and let $\lambda $ denote the positive constant of proportionality.

Through extensive research, doctors recommended that a healthy glucose level in the blood stream of an adult should be between $70$ mg/dL to $100$ mg/dL.

An adult patient, Neo, has $80$ mg/dL of glucose in his blood stream at time $t=0$ minutes. Due to a medical condition, he is not able to extract glucose from the food he eats. As he is not able to replenish the glucose in his blood stream, he is thus losing his glucose in the blood stream at a rate (in mg/dL per minute) corresponding to $\lambda =0.005$ per minute.

In order to maintain Neo’s glucose level in the blood stream in the healthy range, glucose is injected into his blood stream intravenously at a constant rate of $\mu $ mg/dL per minute.

Exponential Growth and Decay Models

2022 R1 P1 Q10

<p>Statistics show that the population of rodents in a small area follows a density-dependent population growth model. In this model, as the population approaches its carrying capacity, the growth rate slows down due to increased competition for limited resources, such as food, space, and mates.</p>
<p>In a particular small farm, a virus spread has a significant impact on the population of rodents. The population $t$ days after the start of the virus spread is denoted by $N$(in hundreds). The density-dependent population growth model states that the population of rodents increases at a rate inversely proportional to the square of the population. At the same time rodents die due to the virus spread at a rate proportional to the population. When&nbsp;$N=3$, the number of rodents remains constant.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Show that&nbsp;$\frac{\mathrm{d}N}{\mathrm{d}t}=\frac{k\left( 27-{{N}^{3}} \right)}{{{N}^{2}}}$, where&nbsp;$k$ is a positive constant.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">It is further given that when the virus spread started, there were $500$ rodents. Show that&nbsp;${{N}^{3}}=27+C{{\mathrm{e}}^{-3kt}}$, where&nbsp;$C$ is a positive constant to be determined.<span class="inline-marks">[6]</span></div></li><li><div class="list-item-paragraph">Sketch the graph of $N$ against $t$ and explain what will happen to the population of rodents in the farm in the long run.<span class="inline-marks">[3]</span></div></li></ol>

Statistics show that the population of rodents in a small area follows a density-dependent population growth model. In this model, as the population approaches its carrying capacity, the growth rate slows down due to increased competition for limited resources, such as food, space, and mates.

In a particular small farm, a virus spread has a significant impact on the population of rodents. The population $t$ days after the start of the virus spread is denoted by $N$(in hundreds). The density-dependent population growth model states that the population of rodents increases at a rate inversely proportional to the square of the population. At the same time rodents die due to the virus spread at a rate proportional to the population. When $N=3$, the number of rodents remains constant.

2023 VJC P1 Q11

<p>Scientists are investigating the effect of a disease on the number of sheep on a small island. They discover that every year the death rate of the sheep is greater than the birth rate of the sheep. The difference every year between the death rate and the birth rate for the population of sheep on the island is $3%$. The number of sheep on the island is $P$ at a time $t$ years after the scientists begin observations.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Write down a differential equation relating $P$ and $t$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Solve this differential equation to find an expression for $P$ in terms of $t$. Explain what happens to the number of sheep if this situation continues over many years.</div><p style="text-align: right;">[4]</p></li></ol>
<p>The scientists import sheep at a constant uniform rate of $n$ sheep per year. (The difference every year between the death rate and the birth rate remains at $3%$.)</p>
<ol class="i-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;">[2]</p></li><li><div class="list-item-paragraph">Solve the differential equation to find an expression for $P$ in terms of $t$ and $n$.</div><p style="text-align: right;">[4]</p></li><li><div class="list-item-paragraph">Given that the number of sheep settles down to $500$ after many years, find $n$.</div><p style="text-align: right;">[2]</p></li></ol>

Scientists are investigating the effect of a disease on the number of sheep on a small island. They discover that every year the death rate of the sheep is greater than the birth rate of the sheep. The difference every year between the death rate and the birth rate for the population of sheep on the island is $3%$. The number of sheep on the island is $P$ at a time $t$ years after the scientists begin observations.

The scientists import sheep at a constant uniform rate of $n$ sheep per year. (The difference every year between the death rate and the birth rate remains at $3%$.)

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