Question

Gravel is being dumped from a conveyor belt at a rate of 15 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 12 ft high? (Round your answer to two decimal places.)

Answer 1

Answer:

0.13 ft/min

Step-by-step explanation:

We are given that

$\frac{dV}{dt}=15ft^3/min$

We have to find the increasing rate of change of height of pile  when the pile is 12 ft high.

Let d be the diameter of pile

Height of pile=h

d=h

Radius of pile,r=$\frac{d}{2}=\frac{h}{2}$

Volume of pile=$\frac{1}{3}\pi r^2 h=\frac{1}{12}\pi h^3$

$\frac{dV}{dt}=\frac{1}{4}\pi h^2\frac{dh}{dt}$

h=12 ft

Substitute the values

$15=\frac{1}{4}\pi(12)^2\frac{dh}{dt}$

$\frac{dh}{dt}=\frac{15\times 4}{\pi(12)^2}$

$\frac{dh}{dt}=0.13ft/min$