Question

Grain pouring from a chute at the rate of 12 ft^3/min forms a conical pile whose height is always 4 times its radius. How fast is the height of the pile increasing at the instant the pile is 3 ft. high?

Answer 1

Given:

The rate of grain pouring from chute is given as

$\frac{dV}{dt}=12\text{ }\frac{ft^3}{\min }$

The relation between height and radius is given

$h=4r$

It is also given that height is h=3 ft.

Explanation:

The volume of cone is

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

Substitute the value of h=4r.

$V=\frac{1}{3}\pi(\frac{h}{4})^2h=\frac{\pi}{48}h^3$

Now take the derivative of the volume with respect to t,

$\frac{dV}{d\text{ t}}=\frac{\pi}{48}\times3h^2\times\frac{dh}{\text{ dt}}$

Substitute the value of h and dh/dt,

$12=\frac{\pi}{48}\times3\times9\times\frac{dh}{dt}$$\frac{12\times48}{\pi\times3\times9}=\frac{dh}{d\text{ t}}$$\frac{dh}{dt}=\frac{4\times16}{\pi\times3}=\frac{64}{3\pi}\frac{ft}{\min }$

Answer:

Hence the height increasing at the rate of

$\frac{64}{3\pi}\frac{ft}{\min }$