Answer to Question #258017 in Calculus for Russell

Question

A tube is spewing sand at a rate of one cubic meter per second. It generates a cone-shaped pile of material. The radius of the circle at the base of the cone is equal to its height. When the sand pile reaches 2 meters in height, how fast does it rise?

Answer 1

The volume of cone is

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

If $h=r,$ then

$$V=\dfrac{1}{3}\pi h^3$$

Differentiate both sides with respect to $t$

$$\dfrac{dV}{dt}=\dfrac{d}{dt}(\dfrac{1}{3}\pi h^3)$$

Use the Chain Rule

$$\dfrac{dV}{dt}=\pi h^2\dfrac{dh}{dt}$$

Solve for

$$\dfrac{dh}{dt}=\dfrac{\dfrac{dV}{dt}}{\pi h^2}$$

Given $\dfrac{dV}{dt}=1m^3/s, h=2m.$

$$\dfrac{dh}{dt}=\dfrac{1m^3/s}{\pi (2m)^2}=\dfrac{1}{4\pi}\ m/s\approx0.080\ m/s$$

When the sand pile reaches 2 meters in height, it raises at speed of $\dfrac{1}{4\pi}\ m/s\approx0.080\ m/s.$