A container in the shape of an inverted cone of radius 3 metres and vertical height 4.5 metres is initially filled with liquid fertiliser. This fertiliser is released through a hole in the bottom of the container at a rate of $$ 0.01 \mathrm{~m}^{3} $$ per second. At time $$ t $$ seconds the fertiliser remaining in the container forms an inverted cone of height $$ h $$ metres. [The volume of a cone is $$ V=\frac{1}{3} \pi r^{2} h $$.] (i) Show that $$ h^{2} \frac{\mathrm{~d} h}{\mathrm{~d} t}=-\frac{9}{400 \pi} $$. (ii) Express $$ h $$ in terms of $$ t $$. (iii) Find the time it takes to empty the container, giving your answer to the nearest minute.

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Nunez Cruz · Accepted Answer

**Part (i):** $$ h^{2} \frac{\mathrm{d} h}{\mathrm{d} t} = -\frac{9}{400 \pi} $$ **Part (ii):** $$ h(t) = \left(91.125 - \frac{27}{400 \pi} t\right)^{1/3} $$ meters **Part (iii):** It takes approximately **71 minutes** to empty the container.

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