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 14 ft high? (Round your answer to two decimal places.)

Answer 1

Answer:

The height of the pile is increasing $\frac{20}{49\pi}$ a minute when the pile is 14ft high.

Step-by-step explanation:

The volume of a cone is given by the following formula:

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

We have that the diameter and the height are equal, so $r = \frac{h}{2}$

So

$V = \frac{\pi h^{3}}{12}$

Let's derivate this equation, using implicit derivatives.

$\frac{dV}{dt} = \frac{\pi h^{2}}{4}\frac{dh}{dt}$

In this problem, we have to:

Find $\frac{dh}{dt}$, when $\frac{dV}{dt} = 20, h = 14$. So

$\frac{dV}{dt} = \frac{\pi h^{2}}{4}\frac{dh}{dt}$

$20 = \frac{196\pi}{4}\frac{dh}{dt}$

$\frac{dh}{dt} = \frac{20}{49\pi}$

The height of the pile is increasing $\frac{20}{49\pi}$ a minute when the pile is 14ft high.