Sand is pouring from a pipe at the rate of $12 c m^{3} / s$ . 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 $3 c m$ ?

Question

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Answer 1

Let $r = r a d i u s; h = h e i g h t; V =$ Volume of sand cone, $t = t i m e$

Given $h = 4 c m; \frac{d V}{d t} = 12 c m^{3} / s$ and $h = \frac{1}{6} r$ or
$r = 6 h$

$∴ V = \frac{1}{3} π r^{2} h$

$= \frac{1}{3} π (6 h)^{2} h = \frac{3 6 h ^{3} π}{3} = 12 π h^{3}$ (diff w.r.t. $t$ )

$∴ \frac{d V}{d t} = 12 π . 3 h^{2} \frac{d h}{d t}$

$12 = 12 π . 3 (4)^{2} \frac{d h}{d t}$

$∴ \frac{d h}{d t} = \frac{1}{4 8 π} c m / s$ .

Answer 2

Let

r = r a d i u s; h = h e i g h t; V =

Volume of sand cone,

t = t i m e

Given

h = 4 c m; \frac{d V}{d t} = 12 c m^{3} / s

and

h = \frac{1}{6} r

or

r = 6 h

∴ V = \frac{1}{3} π r^{2} h

= \frac{1}{3} π (6 h)^{2} h = \frac{3 6 h ^{3} π}{3} = 12 π h^{3}

(diff w.r.t.

t

)

∴ \frac{d V}{d t} = 12 π . 3 h^{2} \frac{d h}{d t}

12 = 12 π . 3 (4)^{2} \frac{d h}{d t}

∴ \frac{d h}{d t} = \frac{1}{4 8 π} c m / s

.