A container in the shape of an inverted cone has height $12 c m$ and radius $6 c m$ at the top. If it is filled with water at the rate of $12 c m^{3} / s e c$ . what is the rate of change in the height of water level when the tank is filled $8 c m$ ?

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Detailed Solution

Let $r, h, v$ are radius, height, volume of cone at ‘ $t$ ’

$A C = h; C D = r$

$O B = 6; O A = 12$

$\frac{d v}{d t} = 12 {cm}^{3} / \sec$

$△ A O B \approx △ A C D$

$\frac{O B}{C D} = \frac{O A}{C A} \Rightarrow \frac{6}{r} = \frac{12}{h} \Rightarrow r = \frac{h}{2}$

Volume of cone

$V = \frac{1}{3} π r^{2} h = \frac{1}{3} π {(\frac{h}{2})}^{2} h = \frac{π h^{3}}{12}$

$\Rightarrow \frac{d v}{d t} = \frac{π}{12} (3 h^{2}) \frac{d h}{d t}$

$\Rightarrow 12 = \frac{π}{12} (3 h^{2}) \frac{d h}{d t} \Rightarrow \frac{d h}{d t} = \frac{48}{π h^{2}}$

${(\frac{d h}{d t})}_{h = 8} = \frac{48}{π (8)^{2}} = \frac{3}{4 π} cm / \sec$

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