Answer: (a) \(135\pi\). (b) \(18\pi\). (c) radius \(9\), height \(12\). (d)(i) \(r=\frac34h\). (d)(ii) \(\frac{3}{\pi}\approx0.955\text{ cm s}^{-1}\).

(a) The curved surface area of the cone is the area of the sector:

\(\frac12(15)^2\left(\frac{6\pi}{5}\right)=135\pi.\)

(b) The circumference of the circular top is the arc length of the sector:

\(15\left(\frac{6\pi}{5}\right)=18\pi.\)

(c) If \(R\) is the radius of the circular top, then

\(2\pi R=18\pi,\)

so \(R=9\).

The slant height of the cone is \(15\), so by Pythagoras,

\(h^2+9^2=15^2.\)

Thus

\(h=12.\)

(d)(i) By similarity,

\(\frac{r}{h}=\frac{9}{12}=\frac34.\)

Therefore

\(r=\frac34h.\)

(d)(ii) The volume of water is

\(V=\frac13\pi r^2h=\frac13\pi\left(\frac34h\right)^2h=\frac{3\pi}{16}h^3.\)

Differentiate with respect to \(h\):

\(\frac{\mathrm{d}V}{\mathrm{d}h}=\frac{9\pi}{16}h^2.\)

When \(h=4\),

\(\frac{\mathrm{d}V}{\mathrm{d}h}=9\pi.\)

Since \(\frac{\mathrm{d}V}{\mathrm{d}t}=27\),

\(\frac{\mathrm{d}h}{\mathrm{d}t}=\frac{\frac{\mathrm{d}V}{\mathrm{d}t}}{\frac{\mathrm{d}V}{\mathrm{d}h}}=\frac{27}{9\pi}=\frac{3}{\pi}=0.9549\ldots.\)