Question

Pebbles are poured out of a tube at one cubic meter per second. It forms a pile which has the shape of a cone. The height of the cone is equal to the radius of circular base. How fast is the pile of pebbles rising when it is 2 meters high

Answer 1

Let V = volume of pebble in pile at a time t

$Given\ \frac{dv}{dt}=1m^3/s$

$Find \ \frac{dh}{dt}, when \ h=2m$

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

Diameter = height

2r = h

$r=\frac{h}{2}$

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

$V=\frac{1}{12}\pi h^3$

$\frac{dv}{dt}=\frac{1}{12\pi}(3h^2.\frac{dh}{dt})$

$\frac{dv}{dt}=\frac{3}{12}\pi h^2 \frac{dh}{dt}$

because $\frac{dv}{dt}=1$

$\frac{3}{12}\pi h^2 \frac{dh}{dt}=1$

$\frac{dh}{dt}=\frac{12}{3\pi h^2 }$

$\frac{dh}{dt}|_{h=2}=\frac{12}{3\pi (2)^2 }$

$=0.3183m/s$

The pile pebbles rise at $0.3183m/s$