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Question

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 $$4 cm.$$


Solution

The volume of a cone $$(V)$$ with radius (r) and height $$(h)$$ is given by,
$$\displaystyle V=\frac{1}{3} \pi r^2 h$$

It is given that,
$$\Rightarrow \displaystyle h=\frac{1}{6}r$$

$$\Rightarrow r=6h$$

$$\therefore\displaystyle V=\frac{1}{3}\pi (6h)^2h=12 \pi h^3$$ 

The rate of change of volume with respect to time $$(t)$$ is given by, 
$$\displaystyle \frac{dV}{dt}=12x\frac{d}{dh}(h^3)\cdot \frac{dh}{dt}=12 \pi (3h^2)\frac{dh}{dt}=36\pi h^2\frac{dh}{dt}$$

It is also given that $$\cfrac{dV}{dt}=12 cm^3/s$$
Therefore, when $$h = 4$$ cm, we have: 
$$\displaystyle 12=36 \pi (4)^2 \frac{dh}{dt}$$
$$\Rightarrow \displaystyle \frac{dh}{dt}=\frac{12}{36\pi(16)} =\frac{1}{48 \pi}$$
Hence, when the height of the sand cone is $$4 cm$$, its height is increasing at the rate of $$\cfrac{1}{48 \pi}$$cm/s

Mathematics
RS Agarwal
Standard XII

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