Question

A water tank has the shape of an inverted right circular cone with its axis vertical and vertex lowermost. Its semi-vertical angle is tan inverse(0.5). Water is poured into it at a constant rate of 5 cubic metre per hour. Find the rate at which the level of water is rising at the instant when the depth of water in the tank is 4 m.

Answer 1

Dear Student,$$\begin{gathered} \mathrm{We}\mathrm{know}, \\ \tan \left(\mathrm{\alpha }\right)=\frac{\mathrm{r}}{\mathrm{h}}.......1 \\ \mathrm{Here}, \\ \mathrm{r}-\mathrm{radius} \\ \mathrm{h}-\mathrm{hieght} \\ \mathrm{\alpha }-\mathrm{semi}\mathrm{vertical}\mathrm{angle} \\ \mathrm{Also},\mathrm{\alpha }={\tan }^{-1}\left(0.5\right) \\ \mathrm{Thus},\mathrm{Tan}\left({\tan }^{-1}\left(0.5\right)\right)=\frac{\mathrm{r}}{\mathrm{h}}...........\mathrm{Using}1 \\ \Rightarrow 0.5=\frac{\mathrm{r}}{\mathrm{h}} \\ \mathrm{Thus},\mathrm{r}=\frac{\mathrm{h}}{2} \\ \mathrm{Also},\mathrm{Volume}\mathrm{of}\mathrm{Cone},\mathrm{V}=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}=\frac{1}{3}\mathrm{\pi }{\left(\frac{\mathrm{h}}{2}\right)}^2\left(\mathrm{h}\right)=\frac{1}{12}{\mathrm{\pi h}}^3 \\ \mathrm{Now},\frac{d\mathrm{V}}{d\mathrm{t}}=\frac{d\mathrm{V}}{d\mathrm{h}}\times \frac{d\mathrm{h}}{d\mathrm{t}}..............\mathrm{Using}\mathrm{Chain}\mathrm{Rule} \\ \frac{d\mathrm{V}}{d\mathrm{t}}=\frac{d\left(\frac{1}{12}{\mathrm{\pi h}}^3\right)}{d\mathrm{r}}\times \frac{d\mathrm{h}}{d\mathrm{t}}=\frac{1}{4}{\mathrm{\pi h}}^2\frac{d\mathrm{h}}{d\mathrm{t}} \\ \mathrm{Now},\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{Water}\mathrm{is}\mathrm{poured}\mathrm{into}\mathrm{it}\mathrm{at}\mathrm{a}\mathrm{constant}\mathrm{rate}\mathrm{of}5\mathrm{cubic}\mathrm{metre}\mathrm{per}\mathrm{hour} \\ \mathrm{Thus},\frac{d\mathrm{V}}{d\mathrm{t}}=5{\mathrm{m}}^3/\mathrm{hr} \\ \mathrm{Thus},\frac{1}{4}{\mathrm{\pi h}}^2\frac{d\mathrm{h}}{d\mathrm{t}}=5 \\ \Rightarrow \frac{d\mathrm{h}}{d\mathrm{t}}=\frac{20}{{\mathrm{\pi h}}^2} \\ \mathrm{Thus},\frac{d\mathrm{h}}{d\mathrm{t}}\mathrm{at}\mathrm{h}=4\mathrm{m} \\ \Rightarrow \frac{d\mathrm{h}}{d\mathrm{t}}=\frac{20}{\mathrm{\pi }{\left(4\right)}^2}=\frac{5}{4\mathrm{\pi }}=0.39\mathrm{m}/\mathrm{h} \\ \mathrm{Thus},\mathrm{At}\mathrm{the}\mathrm{instant}\mathrm{when}\mathrm{h}=4\mathrm{m},\mathrm{Water}\mathrm{level}\mathrm{is}\mathrm{rising}\mathrm{at}\mathrm{a}\mathrm{rate}\mathrm{of}0.39\mathrm{m}\mathrm{per}\mathrm{hour}. \end{gathered}$$

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