Sand is pouring from a pipe at the rate of 12 | vedclass

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(A) The volume of a cone $V$ with radius $r$ and height $h$ is given by $V = \frac{1}{3} \pi r^2 h$.Given that $h = \frac{1}{6} r$,we have $r = 6h$.Su

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$\frac{1}{48 \pi} 	ext{ cm/s}$

Vedclass · Answer

(A) The volume of a cone $V$ with radius $r$ and height $h$ is given by $V = \frac{1}{3} \pi r^2 h$.Given that $h = \frac{1}{6} r$,we have $r = 6h$.Substituting $r$ in the volume formula: $V = \frac{1}{3} \pi (6h)^2 h = \frac{1}{3} \pi (36h^2) h = 12 \pi h^3$.Differentiating with respect to time $t$: $\frac{dV}{dt} = 12 \pi (3h^2) \frac{dh}{dt} = 36 \pi h^2 \frac{dh}{dt}$.Given $\frac{dV}{dt} = 12 	ext{ cm}^3/	ext{s}$,we set $12 = 36 \pi h^2 \frac{dh}{dt}$.At $h = 4 	ext{ cm}$,$12 = 36 \pi (4)^2 \frac{dh}{dt} = 36 \pi (16) \frac{dh}{dt} = 576 \pi \frac{dh}{dt}$.Thus,$\frac{dh}{dt} = \frac{12}{576 \pi} = \frac{1}{48 \pi} 	ext{ cm/s}$.

Sand is pouring from a pipe at the rate of $12 \text{ cm}^3/\text{s}$. 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 $4 \text{ cm}$?

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