September 21st, 2015, 09:42 AM  #1 
Newbie Joined: Sep 2015 From: mexico Posts: 10 Thanks: 0  water leaks out of cone
Water leaks from the bottom of a cone at a rate of 1ft^3/min. (radius is 6 and height is 9) A) at what rate is the level of the water changing when the water level is 6ft deep? This is 1/4pi ft/min B) at what rate is the radius of the water changing when the water is 6 ft deep? This is 1/12pi so this is where I'm stuck (left a & b maybe could be helpful idk). Assume that at time 0 the tank is full. At what rate is the radius of the water changing at t=6? Last edited by skipjack; September 21st, 2015 at 06:27 PM. 
September 21st, 2015, 11:16 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,883 Thanks: 1503 
$\dfrac{r}{h}=\dfrac{6}{9} \implies h=\dfrac{3r}{2}$ $V=\dfrac{\pi}{3}r^2 \cdot \dfrac{3r}{2}=\dfrac{\pi}{2}r^3$ $\dfrac{dV}{dt}=\dfrac{3\pi}{2}r^2 \cdot \dfrac{dr}{dt}$ substitute 1 for $\dfrac{dV}{dt}$ and separate variables ... $\dfrac{2}{3\pi} \, dt = r^2 \, dr$ integrate ... $\dfrac{2t}{3\pi} + C = \dfrac{r^3}{3}$ when $t=0$, $r=6$ ... $C=72$ solve for $r$ in terms of $t$ ... $r=\sqrt[3]{\dfrac{2t}{\pi}+216}$ from this point, you should be able to determine $\dfrac{dr}{dt}$ when $t=6$ 

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