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

Answer:

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

Step-by-step 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 $\frac{x}{2}$.

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

$V=\frac{1}{3}\pi r^2h$

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

$V=\frac{1}{3}\pi (\frac{x}{2})^2(x)$

$V=\frac{1}{3}\pi\frac{x^2}{4}(x)$

$V=\frac{1}{12}\pi x^3$

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

$\frac{dV}{dt}=\frac{1}{12}\pi\cdot 3x^2\cdot \frac{dx}{dt}$

$\frac{dV}{dt}=\frac{1}{4}\pi\cdot x^2\cdot \frac{dx}{dt}$

Upon substituting $x=10$ and $\frac{dx}{dt}=5$, we will get:

$\frac{dV}{dt}=\frac{1}{4}\pi\cdot (10)^2\cdot 5$

$\frac{dV}{dt}=\frac{1}{4}\pi\cdot 100\cdot 5$

$\frac{dV}{dt}=\pi\cdot 25\cdot 5$

$\frac{dV}{dt}=125\pi$

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