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September 21st, 2015, 09:42 AM   #1
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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.
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September 21st, 2015, 11:16 AM   #2
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$\dfrac{r}{h}=\dfrac{6}{9} \implies h=\dfrac{3r}{2}$

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

$\dfrac{dV}{dt}=\dfrac{3\pi}{2}r^2 \cdot \dfrac{dr}{dt}$

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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