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PSLE

PSLE Foundation Mathematics

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

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

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

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

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

Equilibrium Solutions and Long-Term Behaviour

Solving Differential Equations by Separation of Variables

Mixing and Concentration Models

Assumptions and Limitations of Differential Equation Models

Differential Equations

<p>A cylindrical container of height $h$ metres is originally completely filled with water. The water is flowing out through an opening at the base of the container into a pipe. The rate of flow of water through the opening at time $t$ seconds is proportional to the square root of the height of the water in the container, $y$ metres. Since the container is cylindrical, the amount of water in the container is also proportional to the height of water, so the differential equation relating $y$ and $t$ can be written as</p>
<p>\[\frac{\mathrm{d}y}{\mathrm{d}t} = -k\sqrt{y}\]</p>
<p>where $k$ is a positive constant.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Solve this differential equation to find $\sqrt{y}$ in terms of $t$, $h$ and $k$.</span><span class="inline-marks">[4]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Hence find the time taken for the container to empty, in terms of $h$ and $k$.</span><span class="inline-marks">[2]</span></div></li></ol>
<p>It is found that the water was draining too fast. To regulate the flow of water, a mechanical device is placed at the opening. This device slows the flow of water into the pipe by covering all or part of the opening, where the area covered varies with time. It is proposed to model this situation using the revised differential equation</p>
<p>\[\frac{\mathrm{d}y}{\mathrm{d}t} = -k\left( \frac{1 + \cos t}{2} \right)\sqrt{y},\]</p>
<p>where $k$ is the same constant as before.</p>
<ol class="abc-list" start="3"><li><div class="list-item-content"><span class="list-item-text">With reference to the possible values of $\cos t$, explain how the revised differential equation corresponds to a model with slower flow of water.</span><span class="inline-marks">[1]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Given again that the container is originally completely filled with water, solve the revised differential equation to find $\sqrt{y}$ in terms of $t$, $h$ and $k$.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">In the case where $h = 10$ and $k = 0.1$, find the ratio of the time taken in the revised model to the time taken in the original model for the container to empty. Give your answer to $3$ significant figures.</span><span class="inline-marks">[2]</span></div></li></ol>

A cylindrical container of height $h$ metres is originally completely filled with water. The water is flowing out through an opening at the base of the container into a pipe. The rate of flow of water through the opening at time $t$ seconds is proportional to the square root of the height of the water in the container, $y$ metres. Since the container is cylindrical, the amount of water in the container is also proportional to the height of water, so the differential equation relating $y$ and $t$ can be written as

\[\frac{\mathrm{d}y}{\mathrm{d}t} = -k\sqrt{y}\]

Solve this differential equation to find $\sqrt{y}$ in terms of $t$, $h$ and $k$.

Hence find the time taken for the container to empty, in terms of $h$ and $k$.

It is found that the water was draining too fast. To regulate the flow of water, a mechanical device is placed at the opening. This device slows the flow of water into the pipe by covering all or part of the opening, where the area covered varies with time. It is proposed to model this situation using the revised differential equation

\[\frac{\mathrm{d}y}{\mathrm{d}t} = -k\left( \frac{1 + \cos t}{2} \right)\sqrt{y},\]

where $k$ is the same constant as before.

With reference to the possible values of $\cos t$, explain how the revised differential equation corresponds to a model with slower flow of water.

Given again that the container is originally completely filled with water, solve the revised differential equation to find $\sqrt{y}$ in terms of $t$, $h$ and $k$.

In the case where $h = 10$ and $k = 0.1$, find the ratio of the time taken in the revised model to the time taken in the original model for the container to empty. Give your answer to $3$ significant figures.

2025 EJC P1 Q11

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

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

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

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

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

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

Sketching Exponential Functions and Solving Equations Graphically

Asymptotes and Discontinuities

Sketching Solution Curves of Differential Equations

Graphing Techniques

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

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

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

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

Modelling with Differential Equations

2024 SAJC P1 Q10

<p>A researcher is investigating the spread of a certain disease in a town with a population of $3000$ people. The researcher suggests that $I$, the number of people infected by the disease at time $t$ days satisfies the differential equation $\frac{\mathrm{d}I}{\mathrm{d}t}=kI\left( 3000-I \right)$, where $k$ is a positive constant.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Given that $I=30$ when $t=0$, show that $I=\frac{3000}{1+99{{e}^{-3000kt}}}$.</div><p style="text-align: right;">[6]</p></li><li><div class="list-item-paragraph">It is further observed that $I=240$ when $t=7$, find the time it takes for $90\%$ of the population to be infected by the disease.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">State, in the context of this question, one assumption needed to model the spread of the disease in the town by the given differential equation.</div><p style="text-align: right;">[1]</p></li></ol>

A researcher is investigating the spread of a certain disease in a town with a population of $3000$ people. The researcher suggests that $I$, the number of people infected by the disease at time $t$ days satisfies the differential equation $\frac{\mathrm{d}I}{\mathrm{d}t}=kI\left( 3000-I \right)$, where $k$ is a positive constant.

Logistic Growth Models

2013 JJC P2 Q2

<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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjA0MjcyODYwNTctMmJndTI4dWh3YXMucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjczMTY5NywiZXhwIjoxNzkyNzMxOTk3fQ.n824CIK1aE8HrWWR6EXPj1-RTyTGAdWcVVPBMXdeYF0" 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 $k=3$.

Connected Rates of Change

2023 DHS P1 Q10

Applications of Differentiation

<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 $t=3$.

Problems Involving Maximisation and Minimisation

Modelling with Multiple Rates of Change

2021 DHS Promo Q12

<p>Scientists are interested in the population of a particular species. They attempt to model the population $P$ at time $t$ days using a differential equation. Initially the population is observed to be $50$ and after $10$ days the population is $100$.</p>
<p>The first model the scientists use assumes that the rate of change of the population is proportional to the population.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Write down a differential equation for this model and solve it for $P$ in terms of $t$.</div><p style="text-align: right;">[5]</p></li></ol>
<p>To allow for constraints on population growth, the model is refined to</p>
<p style="text-align: center">$\frac{\mathrm{d}P}{\mathrm{d}t}=\lambda P\left( 500-P \right)$</p>
<p>where $\lambda $ is a constant.</p>
<ol class="abc-list" start="2"><li><div class="list-item-paragraph">Solve this differential equation to find $P$ in terms of $t$.</div><p style="text-align: right;">[6]</p></li><li><div class="list-item-paragraph">Using the refined model, state the population of this species in the long term. Comment on how this value suggests the refined model is an improvement on the first model.</div><p style="text-align: right;">[2]</p></li></ol>

Scientists are interested in the population of a particular species. They attempt to model the population $P$ at time $t$ days using a differential equation. Initially the population is observed to be $50$ and after $10$ days the population is $100$.

The first model the scientists use assumes that the rate of change of the population is proportional to the population.

To allow for constraints on population growth, the model is refined to

$\frac{\mathrm{d}P}{\mathrm{d}t}=\lambda P\left( 500-P \right)$

Exponential Growth and Decay Models

N2022 P1 Q12

<ol class="abc-list" start="2"><li><div class="list-item-paragraph">A freshly brewed cup of coffee has an initial temperature $95{}^\circ C$. It is placed in a room where the temperature is a constant at $20{}^\circ C$. As the coffee cools down, the rate of decrease of its temperature $\theta {}^\circ C$ after time $t$ minutes is proportional to the temperature difference $\left( \theta -20 \right){}^\circ C$. It is cooling at a rate of $1{}^\circ C$ per minute when its temperature is $70{}^\circ C$.</div><ol class="i-list"><li><div class="list-item-paragraph">By setting up a differential equation, find the rate of cooling of the coffee when its temperature is $40{}^\circ C$ and show that $\theta =20+75{{e}^{-0.02t}}$ .</div></li><li><div class="list-item-paragraph">Comment on whether the model can be regarded as a good model of the situation in the real world.</div></li></ol></li></ol>

Newton’s Law of Cooling and Warming

2014 NJC P1 Q11

<p>A tank initially contains $2000$ litres of water and $20$ kg of dissolved salt. Brine with $C$ kg of salt per $1000$ litres is entering the tank at $5$ litres per minute. The amount of salt in the tank at time $t$ minutes is $x$ kg.</p>
<p>Assume that the solution is always uniformly mixed.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Show that&nbsp;$\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{1}{400}\left( 2C-x \right)$. Hence determine the value of $C$ if the amount of salt in the tank remains constant at $120$ kg after certain time has passed.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Find $x$ in terms of $t$.<span class="inline-marks">[5]</span></div></li><li><div class="list-item-paragraph">Sketch the graph of the particular solution, including the coordinates of the point(s) where the graph crosses the axes and the equations of any asymptotes. Find the time $t$ when the amount of salt in the tank is $60$ kg, giving your answer to the nearest minute.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">State one assumption for the above model to be valid.<span class="inline-marks">[1]</span></div></li></ol>

A tank initially contains $2000$ litres of water and $20$ kg of dissolved salt. Brine with $C$ kg of salt per $1000$ litres is entering the tank at $5$ litres per minute. The amount of salt in the tank at time $t$ minutes is $x$ kg.

Assume that the solution is always uniformly mixed.

2019 YIJC P1 Q10

<ol class="abc-list" start="2"><li data-list-item-id="item-1784540554023-8rf4xper7" data-list-item-path="item-1784540554023-8rf4xper7"><div class="list-item-paragraph"><p>A water tank contains $500$ litres of a solution with $15$ kilograms of dissolved salt. Pure water enters the tank at a rate of $20$ litres per minute. The solution is drained from the tank at the same rate. Let $x$ kg be the amount of salt in the tank at time $t$ minutes.</p></div><ol class="i-list"><li data-list-item-id="subitem-1784540575751-btlxqj63r" data-list-item-path="item-1784540554023-8rf4xper7/subitem-1784540575751-btlxqj63r"><div class="list-item-paragraph"><p>Show that&nbsp;$\frac{\mathrm{d}x}{\mathrm{d}t}=-\frac{x}{25}$</p><span class="inline-marks">[1]</span></div></li><li data-list-item-id="subitem-1784540576263-es5euoy1d" data-list-item-path="item-1784540554023-8rf4xper7/subitem-1784540576263-es5euoy1d"><div class="list-item-paragraph"><p>State an assumption required for this model to be more appropriate.</p><span class="inline-marks">[1]</span></div></li></ol><div class="list-item-after-content"><p>Find the amount salt in the tank after $30$ minutes.</p><span class="inline-marks">[4]</span></div></li></ol>

2013 NJC P1 Q9

<p>The rate of temperature loss of an animal corpse can be estimated using Newton’s Law of Cooling, which states that the rate of change of temperature $\theta {}^\circ \mathrm{C}$, $t$ hours after death of an animal is proportional to the difference between its body temperature $\theta {}^\circ \mathrm{C}$ and the constant surrounding temperature ${{\theta }_{0}}{}^\circ \mathrm{C}$, where&nbsp;$\theta &gt;{{\theta }_{0}}$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Write down a differential equation for this situation. Solve this differential equation and show that the general solution of the above differential equation is given by $\theta ={{\theta }_{0}}+A{{e}^{-kt}}$, where&nbsp;$A$ and $k$ are positive constants.<span class="inline-marks">[3]</span></div></li></ol>
<p>It is given that&nbsp;${{\theta }_{0}}=24$, the initial value of $\theta $ is $36$ and the initial rate of temperature loss is $2.5{}^\circ \mathrm{C}$ per hour.</p>
<ol class="i-list" start="2"><li><div class="list-item-paragraph">Calculate the exact values of $A$ and $k$.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Hence, sketch the graph of $\theta $ against the $t$.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Explain why the rate of change of temperature of an animal corpse cannot be modelled by a constant rate of decrease of $1.5{}^\circ \mathrm{C}$ per hour.<span class="inline-marks">[1]</span></div></li></ol>

The rate of temperature loss of an animal corpse can be estimated using Newton’s Law of Cooling, which states that the rate of change of temperature $\theta {}^\circ \mathrm{C}$, $t$ hours after death of an animal is proportional to the difference between its body temperature $\theta {}^\circ \mathrm{C}$ and the constant surrounding temperature ${{\theta }_{0}}{}^\circ \mathrm{C}$, where $\theta >{{\theta }_{0}}$.

It is given that ${{\theta }_{0}}=24$, the initial value of $\theta $ is $36$ and the initial rate of temperature loss is $2.5{}^\circ \mathrm{C}$ per hour.

2022 SAJC P1 Q7

<p>A tank has a capacity of 100 litres. Initially, the tank contains 10 litres of water thoroughly mixed with 300 grams of salt. Salt water with a concentration of 5 g/litre is poured into the tank at a constant rate of 2 litres per minute, while the mixture flows out at a constant rate of 1 litre per minute. </p>
<p>Let&nbsp;$S$ denote the amount of dissolved salt in the tank (in grams) at time $t$ minutes after salt water is poured into the tank.</p>
<ol class="i-list"><li data-list-item-id="item-1759145628578-sxiuu5hpw" data-list-item-path="item-1759145628578-sxiuu5hpw"><div class="list-item-paragraph">Show that&nbsp;$\frac{\mathrm{d}S}{\mathrm{d}t}=10-\frac{S}{t+10}$, stating your assumption(s) clearly.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1759145629261-n2k09ecec" data-list-item-path="item-1759145629261-n2k09ecec"><div class="list-item-paragraph">By substituting $Q=\left( t+10 \right)S$, solve the differential equation in (i), and find $S$ in terms of $t$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-1759145630889-5xy7sgk7c" data-list-item-path="item-1759145630889-5xy7sgk7c"><div class="list-item-paragraph">Hence, find the concentration of salt in the tank at the point just before it overflows.<span class="inline-marks">[2]</span></div></li></ol>
<p>Once the volume of salt solution in the tank reaches 100 litres, the pouring stops, and the tank is allowed to drain off. The salt solution drains from the tank at a rate proportional to the volume of solution in the tank. Let&nbsp;$V$ denote the volume of solution in the tank at time $T$ after the tank starts to drain off.</p>
<ol class="i-list" start="4"><li data-list-item-id="item-1759145654830-s1565fpw8" data-list-item-path="item-1759145654830-s1565fpw8"><div class="list-item-paragraph">If the tanks takes 10 minutes to drain half of its contents, find $V$ in terms of  $T$.<span class="inline-marks">[4]</span></div></li></ol>

A tank has a capacity of 100 litres. Initially, the tank contains 10 litres of water thoroughly mixed with 300 grams of salt. Salt water with a concentration of 5 g/litre is poured into the tank at a constant rate of 2 litres per minute, while the mixture flows out at a constant rate of 1 litre per minute. 

Let $S$ denote the amount of dissolved salt in the tank (in grams) at time $t$ minutes after salt water is poured into the tank.

Show that $\frac{\mathrm{d}S}{\mathrm{d}t}=10-\frac{S}{t+10}$, stating your assumption(s) clearly.

By substituting $Q=\left( t+10 \right)S$, solve the differential equation in (i), and find $S$ in terms of $t$.

Hence, find the concentration of salt in the tank at the point just before it overflows.

Once the volume of salt solution in the tank reaches 100 litres, the pouring stops, and the tank is allowed to drain off. The salt solution drains from the tank at a rate proportional to the volume of solution in the tank. Let $V$ denote the volume of solution in the tank at time $T$ after the tank starts to drain off.

If the tanks takes 10 minutes to drain half of its contents, find $V$ in terms of  $T$.

Solving Differential Equations by Substitution

Solving Differential Equations by Direct Integration

2018 EJC JC2 MYE P2 Q4

<p>It has been suggested that the outbreak of H1N1 in Mexico in 2009 can be modelled by a special type of differential equation called Logistic Equation. In this model, $x$ denotes the number of infected individuals in the population of a town at any time $t$ days.</p>
<p>The differential equation is given by $\frac{\mathrm{d}x}{\mathrm{d}t}=0.2x\left( 1-\frac{x}{N} \right)$, where $0.2$ indicates the initial growth rate of the virus and $N$ can be taken to be the population. It was known that a small town has a population of $1000$ and initially $20$ people were infected with H1N1.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Show that $x=\frac{1000}{P+Q{{e}^{-0.2t}}}$, where $P$ and $Q$ are positive integers to be determined.</span></div><p style="text-align: right;">[5]</p></li><li><div class="list-item-content"><span class="list-item-text">Hence sketch the solution curve, indicating clearly asymptotes and axial intercepts if any.</span></div><p style="text-align: right;">[2]</p></li><li><div class="list-item-content"><span class="list-item-text">Deduce the number of days when $75\%$ of the population will be infected with H1N1. State one assumption you have made about the population in your calculation.</span></div><p style="text-align: right;">[3]</p></li></ol>

It has been suggested that the outbreak of H1N1 in Mexico in 2009 can be modelled by a special type of differential equation called Logistic Equation. In this model, $x$ denotes the number of infected individuals in the population of a town at any time $t$ days.

The differential equation is given by $\frac{\mathrm{d}x}{\mathrm{d}t}=0.2x\left( 1-\frac{x}{N} \right)$, where $0.2$ indicates the initial growth rate of the virus and $N$ can be taken to be the population. It was known that a small town has a population of $1000$ and initially $20$ people were infected with H1N1.

Show that $x=\frac{1000}{P+Q{{e}^{-0.2t}}}$, where $P$ and $Q$ are positive integers to be determined.

Hence sketch the solution curve, indicating clearly asymptotes and axial intercepts if any.

Deduce the number of days when $75\%$ of the population will be infected with H1N1. State one assumption you have made about the population in your calculation.

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