Question

At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 6 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude. At what rate is the height of the pile changing when the pile is 2 feet high?

Answer 1

Answer:

$0.2122\text{ ft per min}$

Step-by-step explanation:

Let r be the radius ( in feet ) of the cone,

So, the diameter of the cone = 2 × radius = 2r feet,

∵ The diameter of the base of the cone is three times the height,

If h represents the height of the cone,

⇒ 2r = 3 × h,

⇒ $r=\frac{3h}{2}$ feet,

∵ Volume of a cone,

$V=\frac{1}{3}\pi R^2 H$

Where,

R = radius, H = height,

Thus, the volume of the given cone is,

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

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

$=\frac{3}{4}\pi h^3$

Differentiating with respect to t ( time )

$\frac{dV}{dt}=\frac{9}{4}\pi h^2\frac{dh}{dt}$

We have,

$\frac{dV}{dt}=6\text{ cubic ft per min},h=2\text{ feet}$

$6=\frac{9}{4}\pi (2)^2 \frac{dh}{dt}$

$6=9\pi \frac{dh}{dt}$

$\implies \frac{dh}{dt}=\frac{6}{9\pi }\approx 0.2122\text{ ft per min}$