Question

Gravel is being dumped from a conveyor belt at a rate of 20 ft3 /min and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 15 ft high

Answer 1

The height of the pile is increasing at the rate of
`\mathbf{ (20)/(56.25 \pi) \ \ \ \ \ ft/min}`

Explanation:

Given that :

Gravel is being dumped from a conveyor belt at a rate of 20 ft³ /min

i.e
`(dV)/(dt)= 20 \ ft^3/min`

we know that radius r is always twice the diameter d

i.e d = 2r

Given that :

the shape of a cone whose base diameter and height are always equal.

then d = h = 2r

h = 2r

r = h/2

The volume of a cone can be given by the formula:

`V = (\pi r^2 h)/(3)`

`V = (\pi (h/2)^2 h)/(3)`

`V = (1)/(12) \pi h^3`

`V = ( \pi h^3)/(12)`

Taking the differentiation of volume V with respect to time t; we have:

`(dV)/(dt )= ((d)/(dh)((\pi h^3)/(12))) * (dh)/(dt)`

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

we know that:

`(dV)/(dt)= 20 \ ft^3/min`

So;we have:

`20= ((\pi (15)^2)/(4) ) * (dh)/(dt)`

`20= 56.25 \pi * (dh)/(dt)`

`\mathbf{(dh)/(dt)= (20)/(56.25 \pi) \ \ \ \ \ ft/min}`

The height of the pile is increasing at the rate of
`\mathbf{ (20)/(56.25 \pi) \ \ \ \ \ ft/min}`