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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 ?

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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.

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  • $\begingroup$ 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 $ $\endgroup$ – Out Of Bounds Nov 17 '13 at 0:12

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