Question

sand falls from an overhead bin and accumulates in a conical pile with a radius that is always four times its height. suppose the height of the pile infcreases at a rate of 1cm/s when the pile is 12 cm hight. at what rate is the sand leaving the bin at that instant

Answer 1

Answer:

$\frac{dv}{dt} =7239.168 cm/sec$

Step-by-step explanation:

From the question we are told that:

Rate $\frac{dh}{dt}=1cm$

Height $h=12cm$

Radius $r=4h$

Generally the equation for Volume of Cone is mathematically given by

$V=\frac{1}{3}\pi r^2h$

$V=\frac{1}{3}\pi (4h)^2h$

Differentiating

$\frac{dv}{dt} =\frac{16}{3}\pi3h^2\frac{dh}{dt}$

$\frac{dv}{dt} =\frac{16}{3}*3.142*3*12^2*1$

$\frac{dv}{dt} =7239.168 cm/sec$