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 (in ft/min) is the height of the pile increasing when the pile is 11 ft high? (round your answer to two decimal places.)

Answer 1

0.21ft/ml I think so

Answer 2

0.21ft/min

Explanation:

`\sf Volume \ at \ given \ time = (dV)/(dt)= 20 \ cubic \ feet /min`

h = d

h = 2r

`\sf r = (h)/(2)`

Volume of heap = (1/3) πr²h

`\sf V =(1)/(3)\pi r^2h\\\\V =(1)/(3)\pi ((h)/(2))^2h\\\\V =(1)/(3)\pi (h^2)/(4)*h\\\\V=(1)/(12)\pi h^3`

Take derivative w.r.t time,

`\sf (dV)/(dt)=(1)/(12)\pi * 3h^2 (dh)/(dt)`

h = 11 ft

`\sf 20 = (1)/(12)\pi *3*11*11*(dh)/(dt)\\\\20=(121)/(4)\pi (dh)/(dt)\\\\`

`\sf (20*4)/(121*\pi )=(dh)/(dt)\\\\ (80)/(120*3.14)=(dh)/(dt)`

`\sf (dh)/(dt)=0.21`

The height is increasing at the rate of 0.21 ft/min