Question

g Gravel is being dumped from a conveyor belt at a rate of 10 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 6 ft high

Answer 1

Answer:

0.3537 feet per minute.

Step-by-step explanation:

Gravel is being dumped from a conveyor belt at a rate of 10 ft3/min. Since we are told that the shape formed is a cone, the rate of change of the volume of the cone.

$\dfrac{dV}{dt}=10 ft^3/min$

$\text{Volume of a cone}=\dfrac{1}{3}\pi r^2 h$

If the Base Diameter = Height of the Cone

The radius of the Cone = h/2

Therefore,

$\text{Volume of the cone}=\dfrac{\pi h}{3} (\dfrac{h}{2}) ^2 \\V=\dfrac{\pi h^3}{12}$

$\text{Rate of Change of the Volume}, \dfrac{dV}{dt}=\dfrac{3\pi h^2}{12}\dfrac{dh}{dt}$

Therefore: $\dfrac{3\pi h^2}{12}\dfrac{dh}{dt}=10$

We want to determine how fast is the height of the pile is increasing when the pile is 6 feet high.

$When h=6 feet \\\dfrac{3\pi *6^2}{12}\dfrac{dh}{dt}=10\\9\pi \dfrac{dh}{dt}=10\\ \dfrac{dh}{dt}= \dfrac{10}{9\pi}\\ \dfrac{dh}{dt}=0.3537 feet per minute$

When the pile is 6 feet high, the height of the pile is increasing at a rate of 0.3537 feet per minute.