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

Answer 1

Answer:

0.25 feet per minute

Step-by-step explanation:

Gravel is being dumped from a conveyor belt at a rate of 20 ft3/min. Since we are told that the shape formed is a cone, the rate of change of the volume of the cone.

$\dfrac{dV}{dt}=20 ft^3/min$

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

Since base diameter = Height of the Cone

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: $\dfrac{3\pi h^2}{12}\dfrac{dh}{dt}=20$

We want to determine how fast is the height of the pile is increasing when the pile is 10 feet high.

$h=10 feet \\\\\dfrac{3\pi *10^2}{12}\dfrac{dh}{dt}=20\\25\pi \dfrac{dh}{dt}=20\\ \dfrac{dh}{dt}= \dfrac{20}{25\pi}\\ \dfrac{dh}{dt}=0.25 feet per minute (to two decimal places.)$

When the pile is 10 feet high, the height of the pile is increasing at a rate of 0.25 feet per minute