Sand is pouring from a pipe at the rate of $12 {cm}^{3} / s$ . The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is $4 cm$ ?

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Solution

Let $r$ be the radius and $h$ be the height of the cone

Given:

$h = \frac{r}{6}$

Volume of cone,

$V = \frac{1}{3} π r^{2} h$

$\Rightarrow V = \frac{1}{3} π (6 h)^{2} h [∵ h = \frac{r}{6}]$

$\Rightarrow V = 12 π h^{3}$

Given,

$\frac{d V}{d t} = 12 {cm}^{3} / s$

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

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

$\Rightarrow \frac{d}{d t} (h^{3}) = \frac{1}{π}$

$\Rightarrow 3 h^{2} \frac{d h}{d t} = \frac{1}{π}$

$\Rightarrow \frac{d h}{d t} = \frac{1}{3 π h^{2}}$

When

$h = 4 cm$

$\frac{d h}{d t} = \frac{1}{3 π (4)^{2}}$

$∴ \frac{d h}{d t} = \frac{1}{48 π} cm/s$

Therefore, the height of the sand cone is increasing at the rate of

$\frac{1}{48 π} cm/sec$

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