Question

A conical water tank with vertex down has a radius of 11 feet at the top and is 20 feet high. If water flows into the tank at a rate of 10 ft3/minft3/min, how fast is the depth of the water increasing when the water is 12 feet deep?

Answer 1

Answer:

0.073 ft/min

Step-by-step explanation:

The radius of the tank = 11 feet and the height of the tank = 20 feet. The depth of water = h = 12 feet, let r be the radius of the water, hence:

$\frac{r}{h}=\frac{11}{20}\\ \\r=\frac{11}{20}h\\ \\The \ volume\ of\ the \ tank\ is:\\\\V = \frac{1}{3} \pi r^2 h=\frac{1}{3}\pi (\frac{11}{20}h)^2.h=\frac{121}{1200}\pi h^3 \\\\Given\ that\ rate=\frac{d}{dt}(V)= 10ft^3/min\\\\\frac{dV}{dt}= \frac{d}{dt}( \frac{121}{1200}\pi h^3)\\\\\frac{dV}{dt}=\frac{121}{1200}\pi (3h^2\frac{dh}{dt})\\\\\frac{dV}{dt}=\frac{121}{400}\pi h^2(\frac{dh}{dt})\\\\Given\ water\ is\ 12ft\ deep, i.e.\ h=12.Substituting:$

$10=\frac{121}{400}\pi (12^2)\frac{dh}{dt} \\\\\frac{dh}{dt} =0.073\ ft/min$