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Oksi-84 [34.3K]
4 years ago
14

Water is pouring into a conical tank at the rate of 8 cubic feet per minute. If the height of the tank is 10 feet and the radius

of its circular opening is 4 feet, how fast is the water level rising when the water is 4 feet deep
Physics
1 answer:
iren [92.7K]4 years ago
3 0

Answer:

\frac{25}{8\pi}ft/min

Explanation:

We are given that

\frac{dV}{dt}=8 ft^3/min

Height of tank=h=10 ft

Radius of tank, r=4 feet

We have to find the dh/dt when the water is 4 feet deep.

\frac{r}{h}=\frac{4}{10}=\frac{2}{5}

r=\frac{2}{5}h

Volume of cone , V=\frac{1}{3}\pi r^2 h

Substitute the values

Volume of cone , V=\frac{1}{3}\pi(\frac{2}{5}h)^2h=\frac{4}{75}\pi h^3

Differentiate w.r.t t

\frac{dV}{dt}=\frac{12}{75}\pi\times h^2\times \frac{dh}{dt}

Substitute the values

8=\frac{12}{75}\times\pi (4)^2\times \frac{dh}{dt}

\frac{dh}{dt}=\frac{8\times 75}{12\times 16\pi}=\frac{25}{8\pi}ft/min

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