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PSLE

PSLE Foundation Mathematics

<p>A curve $C$ has parametric equations \[x = \sin 2\theta, \quad y = \cos^2 \theta \quad \text{for}\hspace{0.5em} 0 \le \theta \le \pi.\]</p>
<ol class="abc-list"><li data-list-item-id="imported-item-block-1-item-0" data-list-item-path="imported-item-block-1-item-0"><div class="list-item-paragraph">Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $\theta$, expressing your answer in a single trigonometric function.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="imported-item-block-1-item-1" data-list-item-path="imported-item-block-1-item-1"><div class="list-item-paragraph">Find the rate of decrease of the gradient of $C$ at $\theta = \frac{\pi}{12}$ if&nbsp;$\theta$ is increasing at a constant rate of $\frac{\pi}{10}\text{ rad/s}$. Leave your answer in an exact form.<span class="inline-marks">[3]</span></div></li></ol>

A curve $C$ has parametric equations \[x = \sin 2\theta, \quad y = \cos^2 \theta \quad \text{for } 0 \le \theta \le \pi.\]

Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $\theta$, expressing your answer in a single trigonometric function.

Find the rate of decrease of the gradient of $C$ at $\theta = \frac{\pi}{12}$ if $\theta$ is increasing at a constant rate of $\frac{\pi}{10}\text{ rad/s}$. Leave your answer in an exact form.

Connected Rates of Change

Differentiation of Parametric Equations

2026 TMJC WA2 Q3

Applications of Differentiation

Techniques of Differentiation

The air inside an airtight syringe is compressed so that its volume, $V\text{ cm}^3$, decreases. The air pressure, $P$ Newtons per $\text{cm}^2$, within the syringe is inversely proportional to the volume so that $PV=135$. At a given time, the volume of air is $20\text{ cm}^3$ and is decreasing at a rate of $8\text{ cm}^3$ per minute. Find the rate of increase in the air pressure at this time.

The air inside an airtight syringe is compressed so that its volume, $V\text{ cm}^3$, decreases.

The air pressure, $P$ Newtons per $\text{cm}^2$, within the syringe is inversely proportional to the volume so that $PV=135$.

At a given time, the volume of air is $20\text{ cm}^3$ and is decreasing at a rate of $8\text{ cm}^3$ per minute.

Find the rate of increase in the air pressure at this time.

IB AA HL May 2026 P2 TZA Q6

<p>Rectangle $OPQR$ has sides parallel to the coordinate axes, with $O$ at the origin, $P$ on the positive $x$-axis and $R$ on the positive $y$-axis. The vertex $Q(x,y)$ lies in the first quadrant on the curve $C$, whose equation is $x^2+2xy+4y^2=52$. The variables $x$ and $y$ are measured in units. Let&nbsp;$A$ be the area of the rectangle.</p>
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<ol class="i-list"><li data-list-item-id="implicit-rectangle-part-i" data-list-item-path="implicit-rectangle-part-i"><div class="list-item-paragraph">Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $x$ and $y$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="implicit-rectangle-part-ii" data-list-item-path="implicit-rectangle-part-ii"><div class="list-item-paragraph">Hence, show that&nbsp;$A$ has a stationary value when&nbsp;$x=2y$, and find the exact maximum value of $A$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="implicit-rectangle-part-iii" data-list-item-path="implicit-rectangle-part-iii"><div class="list-item-paragraph">At an instant, $Q=(6,1)$ and $x$ is increasing at a constant rate of $\frac{5}{2}$ units/s. Find the rate of change of $A$ at this instant.<span class="inline-marks">[3]</span></div></li></ol>

Rectangle $OPQR$ has sides parallel to the coordinate axes, with $O$ at the origin, $P$ on the positive $x$-axis and $R$ on the positive $y$-axis. The vertex $Q(x,y)$ lies in the first quadrant on the curve $C$, whose equation is $x^2+2xy+4y^2=52$. The variables $x$ and $y$ are measured in units. Let $A$ be the area of the rectangle.

Implicit Differentiation

Problems Involving Maximisation and Minimisation

Rectangle under an implicit curve

<p>Two rigid rods $OA$ and $OB$, of lengths $5\text{ m}$ and $8\text{ m}$ respectively, are joined at $O$. The angle $\angle AOB=\theta$ is increasing at a constant rate of $0.14\text{ rad s}^{-1}$. At time $t$ seconds, $AB=x\text{ m}$ and the area of triangle $AOB$ is $K\text{ m}^2$. At the instant when&nbsp;$\theta=\frac{\pi}{3}$, find</p>
<ol class="i-list"><li data-list-item-id="rods-q-i" data-list-item-path="rods-q-i"><div class="list-item-paragraph">the rate at which $x$ is increasing;<span class="inline-marks">[3]</span></div></li><li data-list-item-id="rods-q-ii" data-list-item-path="rods-q-ii"><div class="list-item-paragraph">the rate at which $K$ is increasing.<span class="inline-marks">[2]</span></div></li></ol>

Two rigid rods $OA$ and $OB$, of lengths $5\text{ m}$ and $8\text{ m}$ respectively, are joined at $O$. The angle $\angle AOB=\theta$ is increasing at a constant rate of $0.14\text{ rad s}^{-1}$. At time $t$ seconds, $AB=x\text{ m}$ and the area of triangle $AOB$ is $K\text{ m}^2$. At the instant when $\theta=\frac{\pi}{3}$, find

Changing separation and area of two hinged rods

<p>A solid cylindrical candle burns so that both its radius $r$ cm and height $h$ cm decrease over time, while remaining in the fixed proportion $h=4r$. The total surface area of the candle, including both circular ends, is decreasing at a constant rate of $6\pi\text{ cm}^2\text{ min}^{-1}$. Find the rate at which the volume of the candle is decreasing at the instant when&nbsp;$r=3$.<span class="inline-marks">[5]</span></p>

A solid cylindrical candle burns so that both its radius $r$ cm and height $h$ cm decrease over time, while remaining in the fixed proportion $h=4r$. The total surface area of the candle, including both circular ends, is decreasing at a constant rate of $6\pi\text{ cm}^2\text{ min}^{-1}$. Find the rate at which the volume of the candle is decreasing at the instant when $r=3$.

Volume change of a proportionally burning cylindrical candle

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<p>The diagram shows a hemispherical bowl, of inside radius $15\text{ cm}$, fixed so its circular rim is horizontal. When the depth of water in the bowl is $h\text{ cm}$, the volume, $V\text{ cm}^3$, of water in the bowl is given by $V=\dfrac{1}{3}\pi h^2(45-h).$</p>
<p>The bowl is initially empty, and water is poured into the bowl at the constant rate of $10\pi\text{ cm}^3$ per second.</p>
<ol class="abc-list"><li data-list-item-id="imported-item-block-4-item-0" data-list-item-path="imported-item-block-4-item-0"><div class="list-item-paragraph">Find the time taken to fill the bowl.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="imported-item-block-4-item-1" data-list-item-path="imported-item-block-4-item-1"><div class="list-item-paragraph">Find the exact rate at which the depth of water is increasing when the depth of water is $12\text{ cm}$.<span class="inline-marks">[4]</span></div></li></ol>
<p>The bowl is emptied, and water is poured into the bowl at the variable rate of $10\pi t\text{ cm}^3$ per second, where&nbsp;$t$ seconds is the time from when pouring started.</p>
<ol class="abc-list" start="3"><li data-list-item-id="item-1787122464331-8w1rq71ge" data-list-item-path="item-1787122464331-8w1rq71ge"><div class="list-item-paragraph"><p>Find the time taken to pour $972\pi\text{ cm}^3$ of water into the bowl.</p><span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1787122465108-w2kgz23s2" data-list-item-path="item-1787122465108-w2kgz23s2"><div class="list-item-paragraph"><p>Find the rate at which the depth of water is increasing when the volume of water in the bowl is $972\pi\text{ cm}^3$.</p><span class="inline-marks">[4]</span></div></li></ol>

The diagram shows a hemispherical bowl, of inside radius $15\text{ cm}$, fixed so its circular rim is horizontal. When the depth of water in the bowl is $h\text{ cm}$, the volume, $V\text{ cm}^3$, of water in the bowl is given by $V=\dfrac{1}{3}\pi h^2(45-h).$

The bowl is initially empty, and water is poured into the bowl at the constant rate of $10\pi\text{ cm}^3$ per second.

Find the exact rate at which the depth of water is increasing when the depth of water is $12\text{ cm}$.

The bowl is emptied, and water is poured into the bowl at the variable rate of $10\pi t\text{ cm}^3$ per second, where $t$ seconds is the time from when pouring started.

N2025 P2 Q4

<p>The volume of a spherical balloon is increasing at a constant rate of $32\pi\text{ cm}^3\text{ s}^{-1}$. Find the rate of increase of its radius, $r$ cm, at the instant when&nbsp;$r=4$.</p>

The volume of a spherical balloon is increasing at a constant rate of $32\pi\text{ cm}^3\text{ s}^{-1}$. Find the rate of increase of its radius, $r$ cm, at the instant when $r=4$.

Connected rates of change

Differentiation

<ol class="i-list"><li data-list-item-id="imported-item-block-1-item-0" data-list-item-path="imported-item-block-1-item-0"><div class="list-item-paragraph">Sketch the graph with equation $x^2 + (y - r)^2 = r^2$, where&nbsp;$r &gt; 0$ and $y \le r$.<span class="inline-marks">[2]</span></div></li></ol>
<p>A hemispherical bowl of fixed radius $r\text{ cm}$ is filled with water. Water drains out from a hole at the bottom of the bowl at a constant rate.</p>
<p>Use your graph in part <strong>(i)</strong> to show that when the depth of water is $h\text{ cm}$ (where&nbsp;$h \le r$), the volume of water in the bowl is given by</p>
<p style="text-align: center">\[V = \frac{\pi h^2}{3}(3r - h),\]</p>
<p style="text-align: right">[4]</p>
<ol class="i-list" start="2"><li data-list-item-id="imported-item-block-6-item-0" data-list-item-path="imported-item-block-6-item-0"><div class="list-item-paragraph">Given that a full bowl of water would become empty in $24\text{ s}$, find the rate of decrease, in terms of $r$ and $h$, of the depth of water in the bowl at the instant when the depth of water is $h\text{ cm}$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="imported-item-block-6-item-1" data-list-item-path="imported-item-block-6-item-1"><div class="list-item-paragraph">Without any differentiation, determine, in terms of $r$, the slowest rate at which the depth of water is decreasing.<span class="inline-marks">[2]</span></div></li></ol>

Sketch the graph with equation $x^2 + (y - r)^2 = r^2$, where $r > 0$ and $y \le r$.

A hemispherical bowl of fixed radius $r\text{ cm}$ is filled with water. Water drains out from a hole at the bottom of the bowl at a constant rate.

Use your graph in part **(i)** to show that when the depth of water is $h\text{ cm}$ (where $h \le r$), the volume of water in the bowl is given by

Given that a full bowl of water would become empty in $24\text{ s}$, find the rate of decrease, in terms of $r$ and $h$, of the depth of water in the bowl at the instant when the depth of water is $h\text{ cm}$.

Without any differentiation, determine, in terms of $r$, the slowest rate at which the depth of water is decreasing.

2016 VJC P1 Q9

<p>An upright vessel is in the shape of a right circular cone with vertical height $12\text{ cm}$ and base radius $5\text{ cm}$. The vessel is initially completely filled with water. Water then leaks from a small hole at the base at a constant rate of $20\pi\text{ cm}^3\text{/s}$. At time $t$ seconds, the water has depth $h\text{ cm}$ and its surface has radius $r\text{ cm}$.</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph"><p>Using similar triangles, show that the volume $V\text{ cm}^3$ of water remaining in the vessel is given by $V=25\pi \left(h- \frac{h^2}{12}+\frac{h^3}{432} \right)$.</p><span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Find the rate at which the depth of the water is decreasing at the instant when&nbsp;$h=6$.<span class="inline-marks">[3]</span></div></li></ol>

An upright vessel is in the shape of a right circular cone with vertical height $12\text{ cm}$ and base radius $5\text{ cm}$. The vessel is initially completely filled with water. Water then leaks from a small hole at the base at a constant rate of $20\pi\text{ cm}^3\text{/s}$. At time $t$ seconds, the water has depth $h\text{ cm}$ and its surface has radius $r\text{ cm}$.

Water Leaking from an Upright Conical Vessel

<p>Sand is poured onto a horizontal surface to form a right circular conical pile. The radius of the base of the pile is always $\frac{3}{4}$ of its vertical height. The volume of the pile is increasing at a constant rate of $10\text{ cm}^3\text{/s}$. Find the rate at which the curved surface area of the pile is increasing at the instant when its height is $4\text{ cm}$.<span class="inline-marks">[5]</span></p>

Sand is poured onto a horizontal surface to form a right circular conical pile. The radius of the base of the pile is always $\frac{3}{4}$ of its vertical height. The volume of the pile is increasing at a constant rate of $10\text{ cm}^3\text{/s}$. Find the rate at which the curved surface area of the pile is increasing at the instant when its height is $4\text{ cm}$.

Connected Rates for a Conical Sand Pile

<p>The diagram shows part of the graph $y = x^{\cos 2x}$, for&nbsp;$0 \le x \le 5$, which represents the path of a roller coaster. The horizontal distance travelled by the roller coaster is denoted by $x$ units and its vertical distance travelled is denoted by $y$ units.</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/1783573222728-q8fnjduy6xq.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODM1NzMyMjI3MjgtcThmbmpkdXk2eHEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzU1NjM3MSwiZXhwIjoxNzkzNTU2NjcxfQ.euiSMwMjBtjsKtzO4KO9weHafMw3spg2Y_daZzKn_8w" style="max-width: min(280px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Show that&nbsp;$\frac{\mathrm{d}y}{\mathrm{d}x} = x^{\cos 2x} \left[ \frac{\cos 2x}{x} - (2 \sin 2x) \ln x \right]$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">At the point on the graph where&nbsp;$x = \pi$, find the rate at which the roller coaster is moving vertically when it is moving horizontally at a rate of $8$ units per hour.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">Find the acute angle that the tangent to the graph where&nbsp;$x = \pi$ makes with the horizontal.<span class="inline-marks">[1]</span></div></li></ol>

The diagram shows part of the graph $y = x^{\cos 2x}$, for $0 \le x \le 5$, which represents the path of a roller coaster. The horizontal distance travelled by the roller coaster is denoted by $x$ units and its vertical distance travelled is denoted by $y$ units.

Derivatives of Trigonometric Functions

Logarithmic Differentiation

Derivative as the Gradient of a Curve

2024 ACJC P2 Q1

<p>The equation of a curve is $y = \frac{x}{e^{2x}}$.</p>
<ol class="abc-list"><li data-list-item-id="item-1" data-list-item-path="item-1"><div class="list-item-paragraph">Find an expression for&nbsp;$\frac{\mathrm{d}y}{\mathrm{d}x}$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-2" data-list-item-path="item-2"><div class="list-item-paragraph">Given that&nbsp;$x$ is changing at a constant rate of $0.18$ units per second, find the rate of change of $y$ when&nbsp;$x = 1$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-3" data-list-item-path="item-3"><div class="list-item-paragraph">Explain if the curve has more than one stationary point.<span class="inline-marks">[2]</span></div></li></ol>

The equation of a curve is $y = \frac{x}{e^{2x}}$.

Product Rule

Derivatives of Exponential Functions

Chain Rule

Stationary Points and Their Nature

2024 GESS PRELIMS P2 Q8

<p>It is given that&nbsp;$y = \ln \left(\frac{2x}{5 - x} \right)$, where&nbsp;$0 &lt; x &lt; 5$.</p>
<ol class="abc-list"><li data-list-item-id="item-1" data-list-item-path="item-1"><div class="list-item-paragraph">If&nbsp;$y$ is decreasing at a rate of $10$ units/s, find the rate of change of $x$ when&nbsp;$y = \ln 8$.<span class="inline-marks">[5]</span></div></li><li data-list-item-id="item-2" data-list-item-path="item-2"><div class="list-item-paragraph">The equation of the normal to the curve at point $A$ is $8y = 8\ln 2 + 25 - 10x$. Find the $x$-coordinate of point $A$.<span class="inline-marks">[4]</span></div></li></ol>

It is given that $y = \ln \left(\frac{2x}{5 - x} \right)$, where $0 < x < 5$.

Derivatives of Natural Logarithmic Functions

Tangents and Normals at a Point

Equation of a Straight Line

2024 CGSS P1 Q9

Coordinate Geometry

<p>[The volume of a square-based pyramid is $\frac{1}{3} \times \text{base area} \times \text{height}$ .]</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/1779693069635-n5rr3e4b31.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Nzk2OTMwNjk2MzUtbjVycjNlNGIzMS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzNTU2MzcxLCJleHAiOjE3OTM1NTY2NzF9.OKjf0xTPKzFPyH6OBFlfhQEh0BAwC2x12x5cajeYtRQ" style="max-width: min(150px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<p>A manufacturer makes open containers in the shape of an inverted square pyramid with square base of side length $l$ cm, vertical height $h$ cm and fixed slant edge $k$ cm, as shown above. He decides that the volume of each container, $V\text{ cm}^3$, should be as large as possible.</p>
<ol class="abc-list"><li data-list-item-id="item-a"><div class="list-item-paragraph">Show that&nbsp;$V^2 = \frac{k^2l^4}{9} - \frac{l^6}{18}$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b"><div class="list-item-paragraph">Use differentiation to find, in terms of $k$, the exact maximum value of $V$, proving that it is a maximum.<span class="inline-marks">[6]</span></div></li><li data-list-item-id="item-c"><div class="list-item-paragraph">Water is poured into an empty container with the dimensions found in part <strong>(b)</strong> at a constant rate of $\frac{1}{3}\text{ cm}^3$ per second. It is given that at time $t$ seconds, the volume of water in the container is $W\text{ cm}^3$ and the water level is $p$ cm. Show that&nbsp;$W = \frac{4}{3}p^3$ and hence, find the rate of increase of $p$ at the instant when&nbsp;$p = \frac{k}{4}$.<span class="inline-marks">[4]</span></div></li></ol>

[The volume of a square-based pyramid is $\frac{1}{3} \times \text{base area} \times \text{height}$ .]

A manufacturer makes open containers in the shape of an inverted square pyramid with square base of side length $l$ cm, vertical height $h$ cm and fixed slant edge $k$ cm, as shown above. He decides that the volume of each container, $V\text{ cm}^3$, should be as large as possible.

2024 VJC Promo Q12

<p>A curve $C$ has parametric equations</p>
<p style="text-align: center">\[x = \frac{\lambda}{1 + \lambda^3} \quad \text{and} \quad y = \frac{\lambda^2}{1 + \lambda^3}, \quad \text{for}\hspace{0.5em} \lambda &gt; 0.\]</p>
<p>The point $P$ is a variable point on $C$.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">With reference to the origin $O$, $OP$ forms the diagonal of the rectangle $OQPR$, where vertices $Q$ and $R$ lie on the $x$- and $y$-axis respectively. Using differentiation, find the value of $\lambda$ which maximises the area of rectangle $OQPR$. You need to show that your answer gives a maximum.<span class="inline-marks">[5]</span></div></li></ol>
<p>When the area of rectangle $OQPR$ is at its maximum, the rate of change of the $x$-coordinate of the point $P$ is $1$ unit per second.</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">Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ and hence determine the rate of change of the $y$-coordinate of the point $P$ at this instant.<span class="inline-marks">[4]</span></div></li></ol>

\[x = \frac{\lambda}{1 + \lambda^3} \quad \text{and} \quad y = \frac{\lambda^2}{1 + \lambda^3}, \quad \text{for } \lambda > 0.\]

The point $P$ is a variable point on $C$.

When the area of rectangle $OQPR$ is at its maximum, the rate of change of the $x$-coordinate of the point $P$ is $1$ unit per second.

Quotient Rule

2024 RI Promo Q4

<p>A coffee filter is of the shape of a right circular cone. The cone has a base radius $3\text{ cm}$ and height $12\text{ cm}$. The coffee powder, as seen in the cross section of the filter below, has a depth of $4\text{ cm}$. Let&nbsp;$h$ be the depth of the water in the filter at time $t$ seconds. Assume that the amount of water in the coffee powder is negligible.</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/1779353928397-4i5s7uwe5d.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzkzNTM5MjgzOTctNGk1czd1d2U1ZC5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzNTU2MzcxLCJleHAiOjE3OTM1NTY2NzF9.2lSRcaheXJB2l908ldRsYv7vp-w8LTMsg6-uwF-7Aw4" style="max-width: min(300px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<p>Initially, the filter contains only coffee powder. A machine dispenses hot water into the filter at a constant rate of $5\text{ cm}^3$ per second and the hot water drips out of the filter at a constant rate of $2\text{ cm}^3$ per second.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Find the exact rate of change of the depth of the water when the depth is $5\text{ cm}$.<span class="inline-marks">[5]</span></div></li><li><div class="list-item-paragraph">Given that&nbsp;$200\text{ cm}^3$ of hot water will be dispensed into the filter, determine if the hot water will overflow.<span class="inline-marks">[2]</span></div></li></ol>
<p>[The volume of a cone of base radius $r$ and height $h$ is given by $V = \frac{1}{3}\pi r^2 h$.]</p>

A coffee filter is of the shape of a right circular cone. The cone has a base radius $3\text{ cm}$ and height $12\text{ cm}$. The coffee powder, as seen in the cross section of the filter below, has a depth of $4\text{ cm}$. Let $h$ be the depth of the water in the filter at time $t$ seconds. Assume that the amount of water in the coffee powder is negligible.

Initially, the filter contains only coffee powder. A machine dispenses hot water into the filter at a constant rate of $5\text{ cm}^3$ per second and the hot water drips out of the filter at a constant rate of $2\text{ cm}^3$ per second.

[The volume of a cone of base radius $r$ and height $h$ is given by $V = \frac{1}{3}\pi r^2 h$.]

2019 TJC C2BT2 P2 Q1

<div class="graph-block" 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" data-width="400" data-height="300" style="width:400px;height:300px;max-width:100%;display:block;margin:0 auto;"></div>
<p>In optics, the focal length of a thin lens (negligible thickness) is the distance between the centre of the lens and the point at which all rays will meet after passing through the lens. The focal length is a fixed quantity for each piece of lens. The distances from the object $A$ to the lens and from the lens to the image $B$, for a lens of negligible thickness, in air, are related by the thin lens formula $\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$, where&nbsp;$u$ is the distance of an object from the lens, $v$ is the distance of the image from the lens and $f$ is the focal length of the lens. Suppose $A$ can vary with time $t$ (the object can move closer to or further away from the lens).</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Show that the further $A$ is from the lens, the nearer $B$ is to the lens.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">If&nbsp;$A$ is moving away from the lens at a rate of $2\text{ cm/s}$ when it is $45\text{ cm}$ from the lens, find the rate of change of distance of $B$ in terms of $f$.<span class="inline-marks">[4]</span></div></li></ol>

In optics, the focal length of a thin lens (negligible thickness) is the distance between the centre of the lens and the point at which all rays will meet after passing through the lens. The focal length is a fixed quantity for each piece of lens. The distances from the object $A$ to the lens and from the lens to the image $B$, for a lens of negligible thickness, in air, are related by the thin lens formula $\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$, where $u$ is the distance of an object from the lens, $v$ is the distance of the image from the lens and $f$ is the focal length of the lens. Suppose $A$ can vary with time $t$ (the object can move closer to or further away from the lens).

Increasing and Decreasing Functions

<p>A conical water tank with vertex down has a radius of $10\text{ m}$ at the top and is $24\text{ m}$ high. If water flows into the tank at a rate $20\text{ m}^3\text{s}^{-1}$, how fast is the depth of the water increasing when the water is $16\text{ m}$ high?<span class="inline-marks">[4]</span></p>

A conical water tank with vertex down has a radius of $10\text{ m}$ at the top and is $24\text{ m}$ high. If water flows into the tank at a rate $20\text{ m}^3\text{s}^{-1}$, how fast is the depth of the water increasing when the water is $16\text{ m}$ high?

<p>A circular patch of water of negligible thickness is evaporating. The radius of the patch, $r$ cm is decreasing at a constant rate of $0.2$ cm/s. Find the rate of change of the area of this patch, $A$ $\text{cm}^2$ when the area of the patch is $225\pi$ $\text{cm}^2$.<span class="inline-marks">[5]</span></p>

A circular patch of water of negligible thickness is evaporating. The radius of the patch, $r$ cm is decreasing at a constant rate of 0.2 cm/s. Find the rate of change of the area of this patch, $A$ cm$^2$ when the area of the patch is $225\pi$ cm$^2$.

2025 CGSS PRELIMS P1 Q13

<p>Oil is dripping into an empty inverted right circular cone at the rate of $16\ \mathrm{cm}^3$ per second.</p>
<p>At each instant, the oil forms a smaller similar cone. Its height, $h$ cm, of the oil cone is three-quarters its radius, $r$ cm.</p>
<p>[Volume of a right circular cone = $\frac{1}{3} \times \pi \times r^2 \times h$]</p>
<ol class="abc-list"><li data-list-item-id="scgs-p1-q1-q-3-content-items-0" data-list-item-path="scgs-p1-q1-q-3-content-items-0"><div class="list-item-paragraph">Show that the volume, $V\ \mathrm{cm}^3$, of oil is $V = \frac{1}{4}\pi r^3$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="scgs-p1-q1-q-3-content-items-1" data-list-item-path="scgs-p1-q1-q-3-content-items-1"><div class="list-item-paragraph">Calculate, at the instant when the radius is 8 cm, the rate of change of the radius of the oil surface.<span class="inline-marks">[3]</span></div></li></ol>

Oil is dripping into an empty inverted right circular cone at the rate of $16\ \mathrm{cm}^3$ per second.

At each instant, the oil forms a smaller similar cone. Its height, $h$ cm, of the oil cone is three-quarters its radius, $r$ cm.

[Volume of a right circular cone = $\frac{1}{3} \times \pi \times r^2 \times h$]

Show that the volume, $V\ \mathrm{cm}^3$, of oil is $V = \frac{1}{4}\pi r^3$.

Calculate, at the instant when the radius is 8 cm, the rate of change of the radius of the oil surface.

Math Question Bank