Question

Sand pours from a chute and forms a conical pile whose height is always equal to its base diameter. The height of the pile increases at a rate of 5 feet/hour. Find the rate of change of the volume of the sand..?

in the conical pile, when the height of the pile is 4 feet.

Answer 1

`10pi`

Explanation:

Volume of a cone is:

`V = 1/3pir^2h`

height of pile increases at a rate of 5 feet per hr.

Let

`h=5t`

If height is always equal to diameter then diameter is increasing by 5 units per hr, which means radius in increasing by 2.5 units per hr.

Let `r=2.5t`

Then we have:

`V=1/3pi(2.5t)^2*5t`

`(dV)/dt(1/3pi(2.5t)^2*5t)=15.625pit^2=(125pit^2)/8`

When pile is 4 feet high. Using `h=5t` from above:

`4=5t=>t=4/5`

`(125pi(4/5)^2)/8=10pi`