Answer:
[tex]\frac{dh}{dt} = 0.056 ft/min[/tex]
Explanation:
rate of falling of cone[tex] \frac {dv}{dt} = 10 ft^3/min[/tex]
height of pile is 5 feet
diameter is 3 times the height of cone
d = 3h
2r =3h
[tex]r \frac{3}{2} h[/tex]
volume of cone is given as
[tex]v = \frac{1}{3} \pi [\frac{2}{3} h]^{2} h[/tex]
[tex]v =\frac{3}{4} \pi* h^{3}[/tex]
[tex]\frac{dv}{dt} =\frac{3}{4}*\pi*3*h^{2}\frac{dh}{dt}[/tex]
[tex]\frac {dv}{dt} =\frac{9}{4}*\pi h^{2}}\frac{dh}{dt}[/tex]
[tex]\frac{dh}{dt} = \frac{4}{9} \pi*5^{2}*10}[/tex]
[tex]\frac{dh}{dt} = \frac{40}{706.5}[/tex]
[tex]\frac{dh}{dt} = 0.056 ft/min[/tex]
