Question

Sand pouring from a chute forms a conical pile whose height is always equal to the diameter. If the height increases at a constant rate of 5 ft/min, at what rate is sand pouring from the chute when the pile is 10 ft high?

Answer 1

`125\pi \frac{\text{ft}^3}{\text{min}}`.

Explanation:

Let x represent height of the cone.

We have been given that Sand pouring from a chute forms a conical pile whose height is always equal to the diameter.

We know that radius is half the diameter, so radius of cone would be
`(x)/(2)`.

We will use volume of cone formula to solve our given problem.

`V=(1)/(3)\pi r^2h`

Upon substituting the value of height and radius in terms of x, we will get:

`V=(1)/(3)\pi ((x)/(2))^2(x)`

`V=(1)/(3)\pi(x^2)/(4)(x)`

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

Now, we will take the derivative of volume with respect to time as:

`(dV)/(dt)=(1)/(12)\pi\cdot 3x^2\cdot (dx)/(dt)`

`(dV)/(dt)=(1)/(4)\pi\cdot x^2\cdot (dx)/(dt)`

Upon substituting
`x=10` and
`(dx)/(dt)=5`, we will get:

`(dV)/(dt)=(1)/(4)\pi\cdot (10)^2\cdot 5`

`(dV)/(dt)=(1)/(4)\pi\cdot 100\cdot 5`

`(dV)/(dt)=\pi\cdot 25\cdot 5`

`(dV)/(dt)=125\pi`

Therefore, the sand is pouring from the chute at a rate of
`125\pi \frac{\text{ft}^3}{\text{min}}`.