Question

A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep.

Answer 1

Answer:

$\frac{dh}{dt}$≅$0.286\frac{ft^{3} }{min}$

Step-by-step explanation:

$V=\frac{\pi }{3}r^{2}h$; rate of change $\frac{dV}{dt}=10\frac{ft^{3} }{min}$, we must find the rate of change of the depth $\frac{dh}{dt} =?;h=8ft$

5h=12r; $V=\frac{\pi }{3}\({\((5h}/12} )} ^{2}h=\frac{\pi }{3}(\frac{25h^{2} }{144})h; V=\frac{25\pi h^{3}}{432}$; deriving $\frac{dV}{dt} = \frac{25\pi }{432}(3h^{2})\frac{dh}{dt}$ → $10=\frac{25\pi h^{2}}{144} \frac{dh}{dt}$ → h=8 then $\frac{dh}{dt}=\frac{1440}{25\pi 64}=\frac{9}{10\pi}$≅ 0.286$\frac{ft^{3} }{min}$