# related rates problem

Sand is falling into a conical pile at the rate of $200$ cubic inches per minute.The radius of the circular base of the pile is increasing at $0.2$ inches per minute. Find the rate of change of the height of the pile when its height is $20$ inches and the diameter of the base is $30$ inches.

The answer in the book is:

$$\frac{dv}{dt} = \frac{\pi}{3} 2r h \cdot \frac{dr}{dt} + \frac{\pi}{3} r^2 \frac{dh}{dt}$$

my question is why he added $\dfrac{\pi}{3} r^2 \dfrac{dh}{dt}$ to the equation ?

This is the product rule, combined with the chain rule. What you have is the derivative of the volume function $$V=\frac{\pi}{3}r^2h.$$
EDIT: You have two functions that are varying with respect to time, $\frac{\pi}{3}r^2$ and $h$. The deriviative of the first: $$\frac{d}{dt}\frac{\pi}{3}r^2=\frac{\pi}{3}(2r)\frac{dr}{dt}$$ and the second: $$\frac{d}{dt}h=\frac{dh}{dt}.$$ Now apply the product rule.
• can you explain to me how to get the derivative of this function because i'm confused. I mean do i start with product rule for $\frac{\pi}{3} \; and \; r^2 \; or \; i \; start \; with \;r^2 \; and \; h$ – Out Of Bounds Nov 17 '13 at 0:12