Question

Sand is falling onto a cone-shaped pile at 10 cubic feet per minute. The diameter of the base of this cone is always 3 times the height of the cone. At what rate is the height of the sand pile increasing when the pile is 5 feet high?

Answer 1

`(dh)/(dt) = 0.056 ft/min`

Step-by-step explanation:

rate of falling of cone
`\frac {dv}{dt} = 10 ft^3/min`

height of pile is 5 feet

diameter is 3 times the height of cone

d = 3h

2r =3h

`r (3)/(2) h`

volume of cone is given as

`v = (1)/(3) \pi [(2)/(3) h]^(2) h`

`v =(3)/(4) \pi* h^(3)`

`(dv)/(dt) =(3)/(4)*\pi*3*h^(2)(dh)/(dt)`

`\frac {dv}{dt} =(9)/(4)*\pi h^(2)}(dh)/(dt)`

`(dh)/(dt) = (4)/(9) \pi*5^(2)*10}`

`(dh)/(dt) = (40)/(706.5)`

`(dh)/(dt) = 0.056 ft/min`