Question

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

Answer 1

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

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

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