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PSLE

PSLE Foundation Mathematics

<p>The variables $x$ and $y$ satisfy the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x}=y^3-\frac{y}{x}$, where&nbsp;$x,y&gt;0$. It is given that&nbsp;$y=2$ when&nbsp;$x=1$.</p>
<p>Euler's method is a numerical method for estimating a solution of a differential equation. Over a small horizontal step of length $h$, it uses a straight line with the gradient given by the differential equation at the current estimated point.</p>
<p>The estimated change in $y$ is therefore&nbsp;$h\times\text{current gradient}$. Let&nbsp;$y_n$ be the estimate of $y$ at $x=x_n$, with $x_0=1$ and $y_0=2$. For this differential equation, the rule is</p>
<p>\[x_{n+1}=x_n+h,\qquad y_{n+1}=y_n+h\left(y_n^3-\frac{y_n}{x_n}\right).\]</p>
<p>At each step, calculate the gradient using the current values of $x$ and $y$, then add $h$ times this gradient to the current estimate of $y$. Recalculate the gradient at the new estimated point before repeating. Keep unrounded values in subsequent calculations.</p>
<ol class="abc-list"><li data-list-item-id="p2-tzc-q12-item01" data-list-item-path="p2-tzc-q12-item01"><div class="list-item-paragraph">Take $h=0.01$. First calculate the gradient at $(1,2)$. Use this gradient in the Euler formula to show that the first estimate, at $x_1=1.01$, is $y_1=2.06$.</div><div class="list-item-paragraph">Then calculate $y_2$, showing your substitution and giving its value to 5 decimal places. Continue the process to estimate $y$ when&nbsp;$x=1.05$, giving this estimate to 3 significant figures.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="p2-tzc-q12-item02" data-list-item-path="p2-tzc-q12-item02"><div class="list-item-paragraph">Using the substitution $u=xy$, show that the differential equation can be written as&nbsp;$\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{u^3}{x^2}$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="p2-tzc-q12-item03" data-list-item-path="p2-tzc-q12-item03"><div class="list-item-paragraph">Hence, by separating the variables and using the given initial condition, solve the differential equation, giving $y$ as a function of $x$.<span class="inline-marks">[5]</span></div></li><li data-list-item-id="p2-tzc-q12-item04" data-list-item-path="p2-tzc-q12-item04"><div class="list-item-paragraph">State the largest possible domain of this solution containing $x=1$, and its corresponding range.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="p2-tzc-q12-item05" data-list-item-path="p2-tzc-q12-item05"><div class="list-item-paragraph">By considering the shape of the solution curve for&nbsp;$1\leq x\leq1.05$ and the straight-line steps in Euler's method, state whether the approximation in part (a) is an overestimate or an underestimate. Give a reason.<span class="inline-marks">[1]</span></div></li></ol>

The variables $x$ and $y$ satisfy the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x}=y^3-\frac{y}{x}$, where $x,y>0$. It is given that $y=2$ when $x=1$.

Euler's method is a numerical method for estimating a solution of a differential equation. Over a small horizontal step of length $h$, it uses a straight line with the gradient given by the differential equation at the current estimated point.

The estimated change in $y$ is therefore $h\times\text{current gradient}$. Let $y_n$ be the estimate of $y$ at $x=x_n$, with $x_0=1$ and $y_0=2$. For this differential equation, the rule is

\[x_{n+1}=x_n+h,\qquad y_{n+1}=y_n+h\left(y_n^3-\frac{y_n}{x_n}\right).\]

At each step, calculate the gradient using the current values of $x$ and $y$, then add $h$ times this gradient to the current estimate of $y$. Recalculate the gradient at the new estimated point before repeating. Keep unrounded values in subsequent calculations.

Euler's Method

Concavity of Graphs

Solving Differential Equations by Separation of Variables

Function Notation, Domain and Range

Solving Differential Equations by Substitution

Range of a Function

Product Rule

Domain of a Function

Sketching Solution Curves of Differential Equations

IB AA HL May 2026 P2 TZC Q12 [modified]

Techniques of Differentiation

Functions

Differential Equations

Applications of Differentiation

<p>A freshly brewed cup of tea, initially at a temperature of ${85^\circ\text{C}}$, is left out to cool in a room with a room temperature of ${25^\circ\text{C}}$. After ${20}$ minutes, the temperature of the tea is ${55^\circ\text{C}}$.</p>
<p>It is known that the rate of change of the temperature of the tea is proportional to the difference between the temperature of the tea and the room temperature.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Given that&nbsp;$\theta$ is the temperature of the tea $t$ minutes after the tea is left out to cool, show that&nbsp;$\theta = 25 + 60\mathrm{e}^{-kt}$, where&nbsp;$k$ is a constant to be determined.<span class="inline-marks">[5]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Sketch the graph of $\theta$ against $t$.<span class="inline-marks">[2]</span></div></li></ol>

A freshly brewed cup of tea, initially at a temperature of ${85^\circ\text{C}}$, is left out to cool in a room with a room temperature of ${25^\circ\text{C}}$. After ${20}$ minutes, the temperature of the tea is ${55^\circ\text{C}}$.

It is known that the rate of change of the temperature of the tea is proportional to the difference between the temperature of the tea and the room temperature.

Newton’s Law of Cooling and Warming

2025 ACJC P1 Q7

<p>It is given that&nbsp;$2 \frac{\mathrm{d}y}{\mathrm{d}x} + \frac{2}{x} - \ln x = y$.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Use the substitution $y = \ln \left( \frac{u}{x} \right)$ to show that the differential equation can be reduced to $\frac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{f}(u)$, where the function $\mathrm{f}(u)$ is to be found.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Given that&nbsp;$y$ has a minimum value at $x = 3$, solve the differential equation $2 \frac{\mathrm{d}y}{\mathrm{d}x} + \frac{2}{x} - \ln x = y$, to find the particular solution for&nbsp;$y$ in terms of $x$.</span><span class="inline-marks">[5]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Sketch the graph of this particular solution.</span><span class="inline-marks">[2]</span></div></li></ol>

It is given that $2 \frac{\mathrm{d}y}{\mathrm{d}x} + \frac{2}{x} - \ln x = y$.

Use the substitution $y = \ln \left( \frac{u}{x} \right)$ to show that the differential equation can be reduced to $\frac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{f}(u)$, where the function $\mathrm{f}(u)$ is to be found.

Given that $y$ has a minimum value at $x = 3$, solve the differential equation $2 \frac{\mathrm{d}y}{\mathrm{d}x} + \frac{2}{x} - \ln x = y$, to find the particular solution for $y$ in terms of $x$.

Sketch the graph of this particular solution.

Derivatives of Natural Logarithmic Functions

Chain Rule

2025 VJC P1 Q11

<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

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

<p>To prepare for the General Election, a political party studies the effectiveness of spreading information on Instantglam, a social media platform, by uploading a particular video. The number of Instantglam users who have viewed the video increases at a rate that is directly proportional to the number of Instantglam users who have yet to view the video. The number of Instantglam users who have viewed the video $t$ hours after the video is uploaded is $x$. The total number of Instantglam users is $P$, assumed to be fixed.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Write down a differential equation and show that&nbsp;$x = P(1 - e^{-kt})$, where&nbsp;$k$ is a positive constant.<span class="inline-marks">[4]</span></div></li></ol>
<p>It is known that&nbsp;$12$ hours after the video is uploaded, half the total number of Instantglam users have viewed the video.</p>
<ol class="abc-list" start="2"><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Estimate to the nearest hour, the time needed for&nbsp;$80\%$ of the total number of Instantglam users to have viewed the video.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">Sketch the graph of $x$ against $t$.<span class="inline-marks">[2]</span></div></li></ol>

To prepare for the General Election, a political party studies the effectiveness of spreading information on Instantglam, a social media platform, by uploading a particular video. The number of Instantglam users who have viewed the video increases at a rate that is directly proportional to the number of Instantglam users who have yet to view the video. The number of Instantglam users who have viewed the video $t$ hours after the video is uploaded is $x$. The total number of Instantglam users is $P$, assumed to be fixed.

Write down a differential equation and show that $x = P(1 - e^{-kt})$, where $k$ is a positive constant.

It is known that $12$ hours after the video is uploaded, half the total number of Instantglam users have viewed the video.

Modelling with Differential Equations

2025 JPJC P2 Q1

<p>A Koi pond farmer introduces a circular patch of duckweed at the center of his circular pond to enhance water purification and provide supplemental nutrition for his Koi, as shown in the figure below.</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768639935103-yxe5g5n68ml.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njg2Mzk5MzUxMDMteXhlNWc1bjY4bWwucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjczMTQ0MywiZXhwIjoxNzkyNzMxNzQzfQ.PgO20ne64atABrL26OK1d75wlMZNkhMYDCo7W_VDKzk" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<p>The duckweed grows over time $t$ days, forming a continuous circular patch on the pond's surface. It is assumed that the duckweed forms a thin, floating layer with negligible thickness. The rate of increase of the area covered by duckweed is directly proportional to the area of the pond that remains uncovered by the duckweed.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Given that the radius of the pond is $10$ m, show that the radius, $r$ m of the circular patch of duckweed satisfies the differential equation $\frac{\mathrm{d}r}{\mathrm{d}t} = \frac{k(100 - r^2)}{2r}$, $k &gt; 0$.</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">The initial radius of the circular patch of duckweed is $1$ m. Solve the differential equation, expressing $r$ in the form $r = \sqrt{P - Qe^{-kt}}$, where&nbsp;$P$ and $Q$ are constants to be determined.</span><span class="inline-marks">[6]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Sketch the graph of $r$ against $t$.</span><span class="inline-marks">[3]</span></div></li></ol>

A Koi pond farmer introduces a circular patch of duckweed at the center of his circular pond to enhance water purification and provide supplemental nutrition for his Koi, as shown in the figure below.

The duckweed grows over time $t$ days, forming a continuous circular patch on the pond's surface. It is assumed that the duckweed forms a thin, floating layer with negligible thickness. The rate of increase of the area covered by duckweed is directly proportional to the area of the pond that remains uncovered by the duckweed.

Given that the radius of the pond is $10$ m, show that the radius, $r$ m of the circular patch of duckweed satisfies the differential equation $\frac{\mathrm{d}r}{\mathrm{d}t} = \frac{k(100 - r^2)}{2r}$, $k > 0$.

The initial radius of the circular patch of duckweed is $1$ m. Solve the differential equation, expressing $r$ in the form $r = \sqrt{P - Qe^{-kt}}$, where $P$ and $Q$ are constants to be determined.

Connected Rates of Change

2025 CJC P1 Q9

<p>The variables $w$, $x$ and $y$ are connected by the following differential equations:</p>
<p style="text-align: center">$\frac{\mathrm{d}w}{\mathrm{d}x} = -\frac{3}{2}w - 2 \quad\quad (A)$</p>
<p style="text-align: center">$\frac{\mathrm{d}y}{\mathrm{d}x} = w \qquad\qquad\quad\,\,\, (B)$</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Solve equation (A) to find $w$ in terms of $x$.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Hence find $y$ in terms of $x$.</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Different arbitrary constants found in part <strong>(ii)</strong> yields different solution curves. Some of the solution curves are straight lines. Write down the equation of one of these lines. On a single diagram, sketch your line together with a non-linear solution curve that has your line as an asymptote, indicating clearly any axes intercepts.<span class="inline-marks">[3]</span></div></li></ol>

The variables $w$, $x$ and $y$ are connected by the following differential equations:

$\frac{\mathrm{d}w}{\mathrm{d}x} = -\frac{3}{2}w - 2 \quad\quad (A)$

$\frac{\mathrm{d}y}{\mathrm{d}x} = w \qquad\qquad\quad\,\,\, (B)$

Solve equation (A) to find $w$ in terms of $x$.

Sketching Exponential Functions and Solving Equations Graphically

Solving Differential Equations by Direct Integration

Asymptotes and Discontinuities

Order, General and Particular Solutions of Differential Equations

2016 SAJC P1 Q8 [Modified]

Graphing Techniques

<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.

Mixing and Concentration Models

Assumptions and Limitations of Differential Equation Models

<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

<p>A freshly brewed cup of tea, initially at a temperature of $85$$^\circ C$, is left out to cool in a room with a room temperature of $25$$^\circ C$. After $20$ minutes, the temperature of the tea is $55$$^\circ C$. It is known that the rate of change of the temperature of the tea is proportional to the difference between the temperature of the tea and the room temperature.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Given that $\theta$ is the temperature of the tea $t$ minutes after the tea is left out to cool, show that $\theta = 25 + 60\mathrm{e}^{-kt}$, where $k$ is a constant to be determined.</span><span class="inline-marks">[5]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Sketch the graph of $\theta$ against $t$.</span><span class="inline-marks">[2]</span></div></li></ol>

A freshly brewed cup of tea, initially at a temperature of $85$$^\circ C$, is left out to cool in a room with a room temperature of $25$$^\circ C$. After $20$ minutes, the temperature of the tea is $55$$^\circ C$. It is known that the rate of change of the temperature of the tea is proportional to the difference between the temperature of the tea and the room temperature.

Given that $\theta$ is the temperature of the tea $t$ minutes after the tea is left out to cool, show that $\theta = 25 + 60\mathrm{e}^{-kt}$, where $k$ is a constant to be determined.

Sketch the graph of $\theta$ against $t$.

<ol class="i-list"><li><div class="list-item-paragraph">Use the substitution $w = x + y$, show that the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3 - x - y}{1 + x + y}$ can be reduced into $\frac{\mathrm{d}w}{\mathrm{d}x} = \frac{4}{1 + w}$. Solve this differential equation and deduce that the general solution can be written as $(x + y)^2 = 2(3x - y) + A$, where $A$ is an arbitrary constant.<span class="inline-marks">[5]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the particular solution for the above differential equation, given that the curve passes through $y$-intercept $(0, 5)$. Hence or otherwise, find the coordinates of the other $y$-intercept.</span><span class="inline-marks">[2]</span></div></li></ol>

Use the substitution $w = x + y$, show that the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3 - x - y}{1 + x + y}$ can be reduced into $\frac{\mathrm{d}w}{\mathrm{d}x} = \frac{4}{1 + w}$. Solve this differential equation and deduce that the general solution can be written as $(x + y)^2 = 2(3x - y) + A$, where $A$ is an arbitrary constant.

Find the particular solution for the above differential equation, given that the curve passes through $y$-intercept $(0, 5)$. Hence or otherwise, find the coordinates of the other $y$-intercept.

2024 MI P2 Q3

<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 population (in thousands) of fish present in a lake at time $t$ years is denoted by $x$. It is found that the growth rate of $x$ is proportional to $\left( 200-2t-x \right)$.</p>
<p>It is given that the initial population of the fish in the lake is $8000$ and the population grows at a rate of $16000$ per year initially. Show that the growth rate of $x$ at time $t$ years can be modelled by the differential equation</p>
<p style="text-align: center">$\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{200-2t-x}{12}$.</p><p style="text-align: right;">[2]</p>
<p>Find $x$ in terms of $t$ by using the substitution $u=2t+x$. Deduce, to the nearest number of years, the time taken for the fish to die out in this lake.</p><p style="text-align: right;">[6]</p>
<p>It is given that the solution curve that describes that population size of the fish at time $t$ years intersects the graph of $x=200-2t$ at the point $\left( {{t}_{1}},\mathrm{ }{{x}_{1}} \right)$. Describe, in context, what ${{t}_{1}}$ and ${{x}_{1}}$ represent.</p><p style="text-align: right;">[2]</p>

The population (in thousands) of fish present in a lake at time $t$ years is denoted by $x$. It is found that the growth rate of $x$ is proportional to $\left( 200-2t-x \right)$.

It is given that the initial population of the fish in the lake is $8000$ and the population grows at a rate of $16000$ per year initially. Show that the growth rate of $x$ at time $t$ years can be modelled by the differential equation

$\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{200-2t-x}{12}$.

Find $x$ in terms of $t$ by using the substitution $u=2t+x$. Deduce, to the nearest number of years, the time taken for the fish to die out in this lake.

It is given that the solution curve that describes that population size of the fish at time $t$ years intersects the graph of $x=200-2t$ at the point $\left( {{t}_{1}},\mathrm{ }{{x}_{1}} \right)$. Describe, in context, what ${{t}_{1}}$ and ${{x}_{1}}$ represent.

2015 NJC 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>Alice bakes a large pie and removes it from the oven at $1$pm. The temperature reading of the pie using a food thermometer is $200{}^\circ \mathrm{C}$ at that instant.</p>
<p>She left the pie on the kitchen countertop to cool down. The temperature of the kitchen is kept at a constant $32{}^\circ \mathrm{C}$. The temperature of the pie $t$ minutes after it is removed from the oven is $\theta {}^\circ \mathrm{C}$. This temperature decreases at a rate proportional to the difference between the temperature of the pie and the temperature of the kitchen.</p>
<p>At $1.15$pm, Alice checks the temperature of the pie again and it is $180{}^\circ \mathrm{C}$.</p>
<ol class="abc-list"><li data-list-item-id="item-1760697028857-cvyzfl505" data-list-item-path="item-1760697028857-cvyzfl505"><div class="list-item-paragraph">Find an exact expression for&nbsp;$\theta $ in terms of $t$, simplifying your answer.<span class="inline-marks">[5]</span></div></li><li data-list-item-id="item-1760697029352-skmh73k68" data-list-item-path="item-1760697029352-skmh73k68"><div class="list-item-paragraph">Sketch a graph of $\theta $ against $t$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1760697030747-btxgjonc1" data-list-item-path="item-1760697030747-btxgjonc1"><div class="list-item-paragraph">Food safety experts caution that bacteria grow fastest in food within the temperature range of $5{}^\circ \mathrm{C}$ to $60{}^\circ \mathrm{C}$, known as the temperature danger zone. Find the latest time to the nearest minute to safely store the pie to avoid this danger zone.<span class="inline-marks">[2]</span></div></li></ol>

Alice bakes a large pie and removes it from the oven at $1$pm. The temperature reading of the pie using a food thermometer is $200{}^\circ \mathrm{C}$ at that instant.

She left the pie on the kitchen countertop to cool down. The temperature of the kitchen is kept at a constant $32{}^\circ \mathrm{C}$. The temperature of the pie $t$ minutes after it is removed from the oven is $\theta {}^\circ \mathrm{C}$. This temperature decreases at a rate proportional to the difference between the temperature of the pie and the temperature of the kitchen.

At $1.15$pm, Alice checks the temperature of the pie again and it is $180{}^\circ \mathrm{C}$.

2023 RI P1 Q6

<p>The variables $x$ and $y$ are related by the differential equation $\pi \frac{\mathrm{d}y}{\mathrm{d}x}+y\left( 3-\pi \tan x \right)=0$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Using the substitution $y=z\sec x$, show that $\frac{\mathrm{d}z}{\mathrm{d}x}=\frac{-3x}{\pi }$. Hence, show that the particular solution for which $y=2$ when $x=\frac{\pi }{3}$ can be expressed in the form $y\cos x={{\mathrm{e}}^{a+bx}}$ where $a$ and $b$ are constants to be determined.</div><p style="text-align: right;">[6]</p></li><li><div class="list-item-paragraph">For the graph of the solution found in part <strong>(a)</strong>, find the equations of the two vertical asymptotes closest to the $y$-axis.</div><p style="text-align: right;">[2]</p></li></ol>

The variables $x$ and $y$ are related by the differential equation $\pi \frac{\mathrm{d}y}{\mathrm{d}x}+y\left( 3-\pi \tan x \right)=0$.

2023 EJC P2 Q3

<p>Jim places an $8$ m rod $XY$ against a vertical wall, as shown in the diagram below. The points $O$, $X$ and $Y$ are coplanar, with $O$ being the point where the vertical wall and the flat ground meet. At time $t$ seconds, the ends $X$ and $Y$ are $x$ and $y$ meters from $O$ respectively.</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1760427286057-2bgu28uhwas.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjA0MjcyODYwNTctMmJndTI4dWh3YXMucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjczMTQ0MywiZXhwIjoxNzkyNzMxNzQzfQ.K86OL97-o-LuOMQbslFiC8qO8CalrwAzu7HuTBkN8Bs" style="max-width: min(350px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<p>Jim makes a conjecture that as the rod slips, $y$ decreases at a rate proportional to $y$.</p>
<ol class="abc-list"><li data-list-item-id="item-1760427292902-u34i496sy" data-list-item-path="item-1760427292902-u34i496sy"><div class="list-item-paragraph">Based on Jim’s conjecture, show that&nbsp;$\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{k\left( 64-{{x}^{2}} \right)}{x}$, where&nbsp;$k$ is a positive constant.<span class="inline-marks">[4]</span></div></li></ol>
<p>It is given that&nbsp;$X$ is $4$ meters aways from $O$ initially, and that&nbsp;$k=3$.</p>
<ol class="abc-list" start="2"><li data-list-item-id="item-1760427305764-m84l7qu0x" data-list-item-path="item-1760427305764-m84l7qu0x"><div class="list-item-paragraph">Find an expression for&nbsp;$x$ in terms of $t$. Hence, find the time taken for&nbsp;$Y$ to be $3$ meters from $O$ from the instance the rod starts sliding from the initial position, giving your answer to $1$ decimal place.<span class="inline-marks">[5]</span></div></li><li data-list-item-id="item-1760427309304-9vmf18sfj" data-list-item-path="item-1760427309304-9vmf18sfj"><div class="list-item-paragraph">Sketch the graph of $x$ against $t$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1760427310234-s22oogmom" data-list-item-path="item-1760427310234-s22oogmom"><div class="list-item-paragraph">Comment, with justification, on whether Jim’s conjecture is appropriate for the behaviour of the rod.<span class="inline-marks">[1]</span></div></li></ol>

Jim places an $8$ m rod $XY$ against a vertical wall, as shown in the diagram below. The points $O$, $X$ and $Y$ are coplanar, with $O$ being the point where the vertical wall and the flat ground meet. At time $t$ seconds, the ends $X$ and $Y$ are $x$ and $y$ meters from $O$ respectively.

Jim makes a conjecture that as the rod slips, $y$ decreases at a rate proportional to $y$.

It is given that $X$ is $4$ meters aways from $O$ initially, and that $k=3$.

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