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

Question

[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 $\theta$.[1] Show that the volume of the cone, $V$, is given by $$V^2 = \frac{a^6}{576\pi^4} \theta^4 (4\pi^2 - \theta^2)$$[3] Using differentiation, find the exact value of $\theta$ that will maximise $V$.[4] 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 $W = \frac{2\pi y^3}{3}$.[2] Calculate the rate of change of $y$ when $y=2$.[2]

Accepted Answer

(a)$r=\frac{a}{2\pi}\theta$ (c)$\theta=2\pi\sqrt{\frac{2}{3}}$ (e)$-\frac{1}{120}$cm s$^{-1}$

2025 TJC P2 Q4

2025 TJC P2 Q4