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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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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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Toolbox:
  • If $y=f(x)$,then $\large\frac{dy}{dx}$ measures the rate of change of $y$ w.r.t $x$.
  • $\big(\large\frac{dy}{dx}\big)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$
Step 1:
Let $r$ be the radius,$h$ be the height and $v$ be the volume of the sand-cone at any time $t$.
$v=\large\frac{1}{3}$$\pi r^2h$
It is given $r=6h$
Therefore $v=\large\frac{1}{3}$$\pi(36h^2)h$
$v=12\pi h^3$
Differentiating w.r.t $t$ on both sides we get,
$\large\frac{dv}{dt}$$=36\pi h^2\large\frac{dh}{dt}$
Step 2:
It is given $\large\frac{dv}{dt}$$=12cm^3/sec$
Therefore $12=36\pi h^2.\large\frac{dh}{dt}$
$\Rightarrow \large\frac{dh}{dt}=\frac{1}{3\pi h^2}$
Step 3:
When $h=4$
$\big(\large\frac{dh}{dt}\big)_{at\;h=4}=\frac{1}{3\pi(4)^2}$
$\Rightarrow \large\frac{1}{48\pi}$
Hence the height of the sand cone is increasing at the rate of $\large\frac{1}{48\pi}$$cm/sec$
answered Jul 8, 2013 by sreemathi.v
 

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