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PSLE

PSLE Foundation Mathematics

<p>A curve passing through the points $(1,2)$ and $(\frac{1}{4},-10)$ has a gradient which is inversely proportional to $x^2$. Find the equation of the curve.</p>

A curve passing through the points $(1,2)$ and $(\frac{1}{4},-10)$ has a gradient which is inversely proportional to $x^2$. Find the equation of the curve.

Solving Differential Equations by Direct Integration

Modelling with Differential Equations

Differential Equations

<p>The gradient of a curve at the point $(x, y)$ is proportional to $y^2$. <br>The curve passes through the points $(0, -2)$ and $(1, \frac{2}{3})$. <br>Find the equation of the curve.</p>

The gradient of a curve at the point $(x, y)$ is proportional to $y^2$. 
The curve passes through the points $(0, -2)$ and $(1, \frac{2}{3})$. 
Find the equation of the curve.

Solving Differential Equations by Separation of Variables

Word problems: recovering function

<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.

Sketching Solution Curves of 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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njg2Mzk5MzUxMDMteXhlNWc1bjY4bWwucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjczMTYyOSwiZXhwIjoxNzkyNzMxOTI5fQ.fULci_r7mIxb1w-MxWHDhE0K5jaWU58uWJSB-WsOipM" 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

Applications of Differentiation

<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:

Equilibrium Solutions and Long-Term Behaviour

Assumptions and Limitations of Differential Equation Models

2024 SAJC P1 Q10

<p>A spherical tank with negligible thickness and internal radius $a$cm contains water. At time $t$s, the water surface is at a height $x$cm above the lowest point of the tank and the volume of water in the tank, $V$cm$^{3}$, is given by $V=\frac{1}{3}\pi {{x}^{2}}\left( 3a-x \right)$. Water flows from the tank, through an outlet at its lowest point, at a rate $\pi k\sqrt{x}$cm$^{3}$ s$^{-1}$, where&nbsp;$k$ is a positive constant.</p>
<ol class="i-list"><li data-list-item-id="item-1761296198700-oygyrwpls" data-list-item-path="item-1761296198700-oygyrwpls"><div class="list-item-paragraph">Show that&nbsp;$\left( 2ax-{{x}^{2}} \right)\frac{\mathrm{d}x}{\mathrm{d}t}=-k\sqrt{x}$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1761296199187-a25b073m9" data-list-item-path="item-1761296199187-a25b073m9"><div class="list-item-paragraph">Find the general solution for&nbsp;$t$ in terms of $x$,$a$ and $k$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1761296200177-oxmgxsosr" data-list-item-path="item-1761296200177-oxmgxsosr"><div class="list-item-paragraph">Find the ratio ${{T}_{1}}:{{T}_{2}}$, where&nbsp;${{T}_{1}}$ is the time taken to empty the tank when initially it is completely full, and ${{T}_{2}}$ is the time taken to empty the tank when initially it is half full.<span class="inline-marks">[4]</span></div></li></ol>

A spherical tank with negligible thickness and internal radius $a$cm contains water. At time $t$s, the water surface is at a height $x$cm above the lowest point of the tank and the volume of water in the tank, $V$cm$^{3}$, is given by $V=\frac{1}{3}\pi {{x}^{2}}\left( 3a-x \right)$. Water flows from the tank, through an outlet at its lowest point, at a rate $\pi k\sqrt{x}$cm$^{3}$ s$^{-1}$, where $k$ is a positive constant.

2019 NYJC P1 Q7

<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.

Solving Differential Equations by Substitution

2015 NJC P1 Q10

<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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjA0MjcyODYwNTctMmJndTI4dWh3YXMucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjczMTYyOSwiZXhwIjoxNzkyNzMxOTI5fQ.l8TNplfAUstC-EtJ3hTZn8c-GHbbljuKk4TRpD8QPOQ" 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 $k=3$.

2023 DHS P1 Q10

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

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

Derivative as the Gradient of a Curve

2023 NJC P1 Q11

<ol class="abc-list" start="2"><li><div class="list-item-paragraph">At the instant when the radius of a sphere is $r$cm, its volume, $V$cm$^{3}$, is increasing at the rate of<br>$\pi \left( \lambda -{{r}^{3}} \right)$cm$^{3}$min$^{-1}$, where&nbsp;$\lambda $ is a constant. Express $\frac{\mathrm{d}V}{\mathrm{d}t}$ in terms of $\frac{\mathrm{d}r}{\mathrm{d}t}$ and show that&nbsp;$4{{r}^{2}}\frac{\mathrm{d}r}{\mathrm{d}t}=\lambda -{{r}^{3}}$.<span class="inline-marks">[3]</span></div><div class="list-item-paragraph">If the initial volume of the sphere is zero, find $t$ when the volume is $\frac{\pi \lambda }{6}$cm$^{3}$.<span class="inline-marks">[5]</span></div><div class="list-item-paragraph">[Volume of sphere$=\frac{4}{3}\pi {{r}^{3}}$]</div></li></ol>

2003 RJC P1 Q16

<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.

Modelling with Multiple Rates of Change

N2017 P1 Q11

<p>In a chemical reaction, two substances $A$ and $B$ are combined to form a new substance $C$. The initial masses of $A$, $B$and $C$ are $8$, $6$ and $0$ units respectively. After time $t$ seconds, the masses of $A$ and $B$ are each reduced by $x$ units, and the mass of $C$ increases by $2x$ units. The rate of change of the mass of $C$ with respect to $t$ is proportional to the product of the masses of $A$ and $B$ at any time $t$.</p>
<ol class="i-list"><li data-list-item-id="item-1760253665607-n8a4tunsy" data-list-item-path="item-1760253665607-n8a4tunsy"><div class="list-item-paragraph">Write down a differential equation relating $x$ and $t$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-1760253666071-wemd8og4y" data-list-item-path="item-1760253666071-wemd8og4y"><div class="list-item-paragraph">It is observed that when&nbsp;$t=5$, $x=5$. By solving the differential equation in part <strong>(i)</strong>, show that</div><div class="list-item-paragraph" style="text-align: center">$x=\frac{24{{\left( \frac{9}{4} \right)}^{-\frac{t}{5}}}-24}{3{{\left( \frac{9}{4} \right)}^{-\frac{t}{5}}}-4}$.<span class="inline-marks">[7]</span></div></li><li data-list-item-id="item-1760253666783-dnqxmnci9" data-list-item-path="item-1760253666783-dnqxmnci9"><div class="list-item-paragraph">Sketch the graph of $x$ against $t$.<span class="inline-marks">[2]</span></div></li></ol>
<p>In another experiment with the same initial masses of substances $A$ and $B$ as before, the increase in the mass of $C$ is modelled by the differential equation</p>
<p style="text-align: center">$\frac{{{\mathrm{d}}^{2}}x}{\mathrm{d}{{t}^{2}}}=-8{{\mathrm{e}}^{-2t}}-\frac{16}{9}{{\left( t+1 \right)}^{-\frac{7}{3}}}$</p>
<ol class="i-list" start="4"><li data-list-item-id="item-1760253725748-wscdubg73" data-list-item-path="item-1760253725748-wscdubg73"><div class="list-item-paragraph">Given that&nbsp;$x$ eventually stabilises at $6$ units after a long time, find $x$ in terms of $t$.<span class="inline-marks">[4]</span></div></li></ol>

In a chemical reaction, two substances $A$ and $B$ are combined to form a new substance $C$. The initial masses of $A$, $B$and $C$ are $8$, $6$ and $0$ units respectively. After time $t$ seconds, the masses of $A$ and $B$ are each reduced by $x$ units, and the mass of $C$ increases by $2x$ units. The rate of change of the mass of $C$ with respect to $t$ is proportional to the product of the masses of $A$ and $B$ at any time $t$.

In another experiment with the same initial masses of substances $A$ and $B$ as before, the increase in the mass of $C$ is modelled by the differential equation

$\frac{{{\mathrm{d}}^{2}}x}{\mathrm{d}{{t}^{2}}}=-8{{\mathrm{e}}^{-2t}}-\frac{16}{9}{{\left( t+1 \right)}^{-\frac{7}{3}}}$

Second-Order Linear Differential Equations

2022 NYJC P1 Q11

<p>Scientists are investigating how the temperature of water changes in various environments.</p>
<ol class="i-list"><li><div class="list-item-paragraph">The scientists begin by investigating how hot water cools.</div><div class="list-item-paragraph">The water is heated in a container and then placed in a room which is kept at a constant temperature of $16\,{}^\circ \mathrm{C}$. The temperature of the water $t$ minutes after it is placed in the room is $\theta \,{}^\circ \mathrm{C}\mathrm{.}$ This temperature decreases at a rate proportional to the difference between the temperature of the water and the temperature of the room. The temperature of the water falls from a value of $80\,{}^\circ \mathrm{C}$ to $32\,{}^\circ \mathrm{C}$ in the first $30$ minutes.</div><ol class="abc-list"><li><div class="list-item-paragraph">Write down the differential equation for this situation. Solve this differential equation to get $\theta $ as an exact function of $t$.</div><p style="text-align: right;">[6]</p></li><li><div class="list-item-paragraph">Find the temperature of the water $45$ minutes after it is placed in the room.</div><p style="text-align: right;">[1]</p></li></ol></li><li><div class="list-item-paragraph">The scientists then model the thickness of ice on a pond.</div><div class="list-item-paragraph">In winter the surface of the water in the pond freezes. Once the thickness of the ice reaches $3$ cm, it is safe to skate on the ice. The freezing of the water is modelled by a differential equation in which the rate of change of the thickness of ice is inversely proportional to its thickness. It is given that $T=0$ when $t=0.$ After $60$ minutes, the ice is $1$ cm thick.</div><div class="list-item-paragraph">Find the time from when freezing commences until the ice is first safe to skate on.</div><p style="text-align: right;">[6]</p></li></ol>

Scientists are investigating how the temperature of water changes in various environments.

Newton’s Law of Cooling and Warming

N2019 P1 Q11

<p>In a research laboratory, scientists carry out experiments to produce a new substance $C$. $C$ is produced when two substances $A$ and $B$ are reacted together in a chemical reaction. At time $t$ hours, the amount of $C$ is $x$ grams, and the amounts of $A$ and $B$ present are $\left( ax \right)$ grams and $\left( bx \right)$ grams respectively, where&nbsp;$a$ and $b$ are real constants. At the start of the experiment, there are no traces of $C$ found. At any instant, the rate at which the amount of $C$ increases is proportional to the product of the amount of $A$ and the amount of $B$ present at that instant.</p>
<ol class="i-list"><li data-list-item-id="item-1760182288890-cn4k3vjm7" data-list-item-path="item-1760182288890-cn4k3vjm7"><div class="list-item-paragraph">In an experiment, a scientist carries out the experiment with $b=a$.</div><ol class="abc-list"><li data-list-item-id="subitem-1760182291224-2v1sxyd4d" data-list-item-path="item-1760182288890-cn4k3vjm7/subitem-1760182291224-2v1sxyd4d"><div class="list-item-paragraph">Obtain a differential equation relating $x$ and $t$ solve for&nbsp;$x$ in terms of $t$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="subitem-1760182291906-wbsxp2r3t" data-list-item-path="item-1760182288890-cn4k3vjm7/subitem-1760182291906-wbsxp2r3t"><div class="list-item-paragraph">Sketch the graph of $x$ against $t$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="subitem-1760182293637-e9zitoztp" data-list-item-path="item-1760182288890-cn4k3vjm7/subitem-1760182293637-e9zitoztp"><div class="list-item-paragraph">Given that there is $\frac{1}{2}a$ grams of $C$ produced $1$ hour after the experiment started, find the time needed for&nbsp;$\frac{4}{5}a$ grams of $C$ to be produced.<span class="inline-marks">[2]</span></div></li></ol></li><li data-list-item-id="item-1760182294501-sq073evps" data-list-item-path="item-1760182294501-sq073evps"><div class="list-item-paragraph">Another scientist carries out the experiment with $b &lt; a$.</div><ol class="abc-list"><li data-list-item-id="subitem-1760182297020-kl2wwly3u" data-list-item-path="item-1760182294501-sq073evps/subitem-1760182297020-kl2wwly3u"><div class="list-item-paragraph">Obtain a differential equation relating $x$ and $t$ and solve for&nbsp;$x$ in terms of $t$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="subitem-1760182297868-g630a6ovb" data-list-item-path="item-1760182294501-sq073evps/subitem-1760182297868-g630a6ovb"><div class="list-item-paragraph"><p>Using your answer in part <strong>(ii)(a)</strong>¸ find the amount of $C$ produced after a long time.</p><span class="inline-marks">[2]</span></div></li></ol></li></ol>

In a research laboratory, scientists carry out experiments to produce a new substance $C$. $C$ is produced when two substances $A$ and $B$ are reacted together in a chemical reaction. At time $t$ hours, the amount of $C$ is $x$ grams, and the amounts of $A$ and $B$ present are $\left( ax \right)$ grams and $\left( bx \right)$ grams respectively, where $a$ and $b$ are real constants. At the start of the experiment, there are no traces of $C$ found. At any instant, the rate at which the amount of $C$ increases is proportional to the product of the amount of $A$ and the amount of $B$ present at that instant.

2020 TJC P1 Q11

<p>A spherical container of radius $5$ m is formed by rotating the following circle $C$ about the $y$-axis.</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/1760169456333-9vxe3gvjjbg.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjAxNjk0NTYzMzMtOXZ4ZTNndmpqYmcucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjczMTYyOSwiZXhwIjoxNzkyNzMxOTI5fQ.we8Z82_K2z2J4s5heUjWCqAPQQ814Gxa2DQLdV4nFBQ" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<p>The container has negligible thickness, and the circle $C$ passes through the origin $O$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">State a cartesian equation of $C$.<span class="inline-marks">[1]</span></div></li></ol>
<p>Initially the spherical container is completely filled with water. Two engineers are calculating the time needed for the container to be completely drained from a small circular hole at the bottom. The volume of water in the container at time $t$ seconds is denoted by $V$ m$^{3}$.</p>
<p>[The volume of a sphere of radius $r$ is $\frac{4}{3}\pi {{r}^{3}}$.]</p>
<ol class="abc-list" start="2"><li><div class="list-item-paragraph">The first engineer proposes that the rate of change of $V$ with respect to $t$ is a constant $k$.<span class="inline-marks">[1]</span></div><ol class="i-list"><li><div class="list-item-paragraph">Write down a differential equation relating $V$, $t$ and $k$.<span class="inline-marks">[1]</span></div></li><li><div class="list-item-paragraph">Determine, with justification, the sign of $k$.<span class="inline-marks">[1]</span></div></li><li><div class="list-item-paragraph">Find $V$ in terms of $t$ and $k$, leaving your answer in exact form.<span class="inline-marks">[2]</span></div></li></ol></li><li><div class="list-item-paragraph">The second engineer argues that the rate at which water flows out from the hole will be at its greatest in the beginning, and decreases as the depth of water in the container decreases. He suggests using Torricelli’s law, which says that</div><div class="list-item-paragraph" style="text-align: center">$\frac{\mathrm{d}V}{\mathrm{d}t}=-\alpha \sqrt{20h}$,</div><div class="list-item-paragraph">where&nbsp;$\alpha $ m$^{2}$ is the area of the circular hole at the bottom of the container, and $h$m is the depth of water in the container at time $t$ seconds. The radius of the hole at the bottom of the container is found to be constant at $1$ cm.</div><ol class="i-list"><li><div class="list-item-paragraph">Show that&nbsp;$\left( 10{{h}^{\frac{1}{2}}}-{{h}^{\frac{3}{2}}} \right)\frac{\mathrm{d}h}{\mathrm{d}t}=-\frac{\sqrt{5}}{5000}$.<span class="inline-marks">[4]</span></div></li><li><div class="list-item-paragraph">Hence find the numerical value of $t$ when the container is completely drained.<span class="inline-marks">[4]</span></div></li></ol></li></ol>

A spherical container of radius $5$ m is formed by rotating the following circle $C$ about the $y$-axis.

The container has negligible thickness, and the circle $C$ passes through the origin $O$.

Initially the spherical container is completely filled with water. Two engineers are calculating the time needed for the container to be completely drained from a small circular hole at the bottom. The volume of water in the container at time $t$ seconds is denoted by $V$ m$^{3}$.

[The volume of a sphere of radius $r$ is $\frac{4}{3}\pi {{r}^{3}}$.]

Volume of Revolution about the y-axis

Graphs of Conic Sections

2022 EJC P1 Q10

Applications of Integration

Graphing Techniques

<p>Two chemicals $X$ and $Y$ react to form a chemical $Z$ without any loss of mass. It is known that one part of $X$ combines with two parts of $Y$ to give three parts of $Z$. For example, $1.5$ g of $X$ combines with $3.0$ g of $Y$ to give $4.5$ g of $Z$. Let&nbsp;$z$ g be the mass of $Z$ formed $t$ minutes after the reaction started. The rate of change of $z$ with respect to $t$ is proportional to the product of the remaining masses of $X$ and $Y$ not reacted at any time $t$. Initially, there are $10$ g of $X$, $15$ g of $Y$ and $0$ g of $Z$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Show that</div><div class="list-item-paragraph" style="text-align: center">$\frac{\mathrm{d}z}{\mathrm{d}t}=k\left( 30-z \right)\left( 45-2z \right)$,</div><div class="list-item-paragraph">where&nbsp;$k$ is a positive constant.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">It is observed that the mass of $Z$ is $10$ g after $5$ min. Solve the differential equation in part <strong>(a)</strong> and find $z$ in terms of $t$.<span class="inline-marks">[8]</span></div></li><li><div class="list-item-paragraph">State, with justification, the mass of $Z$ that can be formed after a long time. Hence, or otherwise, find the corresponding remaining masses of $X$ and $Y$.<span class="inline-marks">[2]</span></div></li></ol>

Two chemicals $X$ and $Y$ react to form a chemical $Z$ without any loss of mass. It is known that one part of $X$ combines with two parts of $Y$ to give three parts of $Z$. For example, $1.5$ g of $X$ combines with $3.0$ g of $Y$ to give $4.5$ g of $Z$. Let $z$ g be the mass of $Z$ formed $t$ minutes after the reaction started. The rate of change of $z$ with respect to $t$ is proportional to the product of the remaining masses of $X$ and $Y$ not reacted at any time $t$. Initially, there are $10$ g of $X$, $15$ g of $Y$ and $0$ g of $Z$.

2024 DHS P2 Q5

<p>A blogger reviews a data plan with unlimited data. The download speed is $x$ megabytes per second, $t$ seconds after the download has started. This download speed decreases at a rate proportional to the difference between $x$ and the eventual limit of $40$ megabytes per second. The download speed falls from a value of $60$ megabytes per second to $50$ megabytes per second in the first $30$ minutes.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Write down a differential equation for this situation. Solve this differential equation to get&nbsp;$x$ as an exact function of $t$.<span class="inline-marks">[5]</span></div></li></ol>
<p>The total amount of data downloaded daily by time $t$ seconds is $y$ gigabytes. The fair usage policy enforces that the download speed will be reduced to a speed which cannot be used to stream videos if&nbsp;$y&gt;500$. You may assume that&nbsp;$1$ gigabyte is $1000$ megabytes.</p>
<ol class="i-list" start="2"><li><div class="list-item-paragraph">The blogger starts using the data plan on the morning of a particular day to stream videos. How long would he be able to stream videos continuously? Give your answer in hours to $1$ decimal place.<span class="inline-marks">[3]</span></div></li></ol>

A blogger reviews a data plan with unlimited data. The download speed is $x$ megabytes per second, $t$ seconds after the download has started. This download speed decreases at a rate proportional to the difference between $x$ and the eventual limit of $40$ megabytes per second. The download speed falls from a value of $60$ megabytes per second to $50$ megabytes per second in the first $30$ minutes.

The total amount of data downloaded daily by time $t$ seconds is $y$ gigabytes. The fair usage policy enforces that the download speed will be reduced to a speed which cannot be used to stream videos if $y>500$. You may assume that $1$ gigabyte is $1000$ megabytes.

Definite Integral

2020 NJC P1 Q8

<p>The diagram shows a curved mirror such that any ray of light emanating from the source located at the origin $O$ of a Cartesian coordinate system is always reflected parallel to the $x$-axis.</p>
<figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1758545698789-flljqbypk3.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NTg1NDU2OTg3ODktZmxsanFieXBrMy5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNzMxNjI5LCJleHAiOjE3OTI3MzE5Mjl9.B-9a-WJqqGEzyM0jPiToXaXM6dPxLG5SeTHs7yTdRsc" width="400" class="aligncenter" /></figure>
<p>The ray emanating from $O$ strikes the curved mirror at $P$ with coordinates $\left( x,y \right)$. Let $\alpha $ be the acute angle which $OP$ makes with the tangent to the curved mirror at $P$ and let $\beta $ be the angle which $OP$ makes with the $x$-axis.</p>
<p>Assume that the law of reflection of light holds, in other words, the reflected ray also makes an angle of $\alpha $ with the tangent to the curved mirror at $P$.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Express $\beta $ in terms of $\alpha $.</span></div><p style="text-align: right;">[2]</p></li><li><div class="list-item-content"><span class="list-item-text">Write down expressions for $\tan \,\alpha $ and $\tan \beta $ in terms of $x$, $y$ and $\frac{\mathrm{d}y}{\mathrm{d}x}$, whichever is appropriate.</span></div><p style="text-align: right;">[2]</p></li><li><div class="list-item-content"><span class="list-item-text">Using the results in **(i)** and **(ii)**, prove that</span></div><div class="list-item-paragraph" style="text-align: center;">$\left[ 1-{{\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)}^{2}} \right]\left( \frac{y}{x} \right)=2\frac{\mathrm{d}y}{\mathrm{d}x}$. (1)</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-content"><span class="list-item-text">Use the substitution ${p = \frac{\mathrm{d}y}{\mathrm{d}x}}$ to show that the differential equation (1) can be reduced to the differential equation</span></div><div class="list-item-paragraph" style="text-align: center;">$y\frac{\mathrm{d}y}{\mathrm{d}x}+x=\pm \sqrt{{{x}^{2}}+{{y}^{2}}}$. (2)</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-content"><span class="list-item-text">Use the substitution ${z = x^2 + y^2}$ to find the general solution of the differential equation (2).</span></div><p style="text-align: right;">[5]</p></li></ol>

The diagram shows a curved mirror such that any ray of light emanating from the source located at the origin $O$ of a Cartesian coordinate system is always reflected parallel to the $x$-axis.

The ray emanating from $O$ strikes the curved mirror at $P$ with coordinates $\left( x,y \right)$. Let $\alpha $ be the acute angle which $OP$ makes with the tangent to the curved mirror at $P$ and let $\beta $ be the angle which $OP$ makes with the $x$-axis.

Assume that the law of reflection of light holds, in other words, the reflected ray also makes an angle of $\alpha $ with the tangent to the curved mirror at $P$.

Write down expressions for $\tan \,\alpha $ and $\tan \beta $ in terms of $x$, $y$ and $\frac{\mathrm{d}y}{\mathrm{d}x}$, whichever is appropriate.

Using the results in **(i)** and **(ii)**, prove that

Use the substitution ${p = \frac{\mathrm{d}y}{\mathrm{d}x}}$ to show that the differential equation (1) can be reduced to the differential equation

Use the substitution ${z = x^2 + y^2}$ to find the general solution of the differential equation (2).

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