Answer:
[tex]\frac{dh}{dt}[/tex]≅[tex]0.286\frac{ft^{3} }{min}[/tex]
Step-by-step explanation:
[tex]V=\frac{\pi }{3}r^{2}h[/tex]; rate of change [tex]\frac{dV}{dt}=10\frac{ft^{3} }{min}[/tex], we must find the rate of change of the depth [tex]\frac{dh}{dt} =?;h=8ft[/tex]
5h=12r; [tex]V=\frac{\pi }{3}\({\((5h}/12} )} ^{2}h=\frac{\pi }{3}(\frac{25h^{2} }{144})h; V=\frac{25\pi h^{3}}{432}[/tex]; deriving [tex]\frac{dV}{dt} = \frac{25\pi }{432}(3h^{2})\frac{dh}{dt}[/tex] → [tex]10=\frac{25\pi h^{2}}{144} \frac{dh}{dt}[/tex] → h=8 then [tex]\frac{dh}{dt}=\frac{1440}{25\pi 64}=\frac{9}{10\pi}[/tex]≅ 0.286[tex]\frac{ft^{3} }{min}[/tex]
