Answer:
The water level changing by the rate of -0.0088 feet per hour ( approx )
Step-by-step explanation:
Given,
The volume of a segment of height h in a hemisphere of radius r is,
[tex]V=\pi h^2(r-\frac{h}{3})[/tex]
Where, r is the radius of the hemispherical tank,
h is the water level, ( in feet )
Here, r = 9 feet,
[tex]\implies V=\pi h^2(9-\frac{h}{3})[/tex]
[tex]V=9\pi h^2-\frac{\pi h^3}{3}[/tex]
Differentiating with respect to t ( time ),
[tex]\frac{dV}{dt}=18\pi h\frac{dh}{dt}-\frac{3\pi h^2}{3}\frac{dh}{dt}[/tex]
[tex]\frac{dV}{dt}=\pi h(18-h)\frac{dh}{dT}[/tex]
Here, [tex]\frac{dV}{dt}=-2\text{ cubic feet per hour}[/tex]
And, h = 6 feet,
Thus,
[tex]-2=\pi 6(18-6)\frac{dh}{dt}[/tex]
[tex]\implies \frac{dh}{dt}=\frac{-2}{72\pi}=-0.00884194128288\approx -0.0088[/tex]
