<p>[The arc length \(L\) of a sector of radius \(a\) and angle \(\theta\) is given by \(L=a\theta\);</p>
<p>The volume \(V\) of a cone of base radius \(r\) and height \(h\) is given by \(V=\frac{1}{3}\pi r^2h\)]</p>
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<p>A metal sheet is shaped as a circular sector with angle \(\theta\) and fixed radius \(a\), as shown in Figure \(1\). This sector is then formed into a right circular cone with slant height \(a\) by joining the two radii \(OA\) and \(OB\) together as shown in Figure \(2\).</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">If&nbsp;\(\theta\) is measured in radians, state the base radius of the cone, \(r\), in term of \(a\) and \(\theta\).</span><span class="inline-marks">[1]</span></div></li><li><div class="list-item-paragraph">Show that the volume of the cone, \(V\), is given by \[V^2 = \frac{a^6}{576\pi^4} \theta^4 (4\pi^2 - \theta^2)\]<span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Using differentiation, find the exact value of \(\theta\) that will maximise \(V\).</span><span class="inline-marks">[4]</span></div></li></ol>
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<p>Two identical cones, as described in part (c), are joined together tip-to-tip to form a new object, as shown in Figure \(3\). The upper cone is initially filled with water, which leaks out into the lower cone at a rate of \(\frac{\pi}{15}\) cm\(^3\) s\(^{-1}\). At time \(t\) seconds, the height of the water in the upper cone is \(y\).</p>
<ol class="abc-list" start="4"><li><div class="list-item-content"><span class="list-item-text">Show that the volume of water in the cone, \(W\) cm\(^3\), is given by \(W = \frac{2\pi y^3}{3}\).</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Calculate the rate of change of \(y\) when&nbsp;\(y=2\).</span><span class="inline-marks">[2]</span></div></li></ol>

[The arc length \(L\) of a sector of radius \(a\) and angle \(\theta\) is given by \(L=a\theta\);

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

A metal sheet is shaped as a circular sector with angle \(\theta\) and fixed radius \(a\), as shown in Figure \(1\). This sector is then formed into a right circular cone with slant height \(a\) by joining the two radii \(OA\) and \(OB\) together as shown in Figure \(2\).

If \(\theta\) is measured in radians, state the base radius of the cone, \(r\), in term of \(a\) and \(\theta\).

Show that the volume of the cone, \(V\), is given by

Show that the volume of the cone, \(V\), is given by \[V^2 = \frac{a^6}{576\pi^4} \theta^4 (4\pi^2 - \theta^2)\]

Using differentiation, find the exact value of \(\theta\) that will maximise \(V\).

Two identical cones, as described in part (c), are joined together tip-to-tip to form a new object, as shown in Figure \(3\). The upper cone is initially filled with water, which leaks out into the lower cone at a rate of \(\frac{\pi}{15}\) cm\(^3\) s\(^{-1}\). At time \(t\) seconds, the height of the water in the upper cone is \(y\).

Show that the volume of water in the cone, \(W\) cm\(^3\), is given by \(W = \frac{2\pi y^3}{3}\).

2025 TJC P2 Q4

2025 TJC P2 Q4