A container is in the shape of an inverted cone has height $8 m$ and radius 6m at the top. If it is filled with water at the rate of $2 c u b i c m e t r e / m m$ , how fast is the height of water chaning when the level is $4 m$ .

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Detailed Solution

Given $\frac{d V}{d t} = 2 c u b i c m e t r e / m m$

Now $\frac{r}{h} = \frac{6}{8} \Rightarrow r = \frac{3 h}{4}$

$V = \frac{1}{3} π r^{2} h = \frac{1}{3} π h {(\frac{3 h}{4})}^{2} = \frac{3 π h^{3}}{16}$

$\frac{d V}{d t} = \frac{3 π}{16} (3 h^{2} \frac{d h}{d t})$

$2 = \frac{3 π}{16} \times 3 \times 4 \times \frac{d h}{d t}; \frac{d h}{d t} = \frac{2}{9 π} m / \min$

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