Water is leaking out the bottom of a hemispherical tank of radius 9 feet at a rate of 2 cubic feet per hour. The tank was full at a certain time. How fast is the water level changing when its height h is 6 feet? Note: the volume of a segment of height h in a hemisphere of radius r is pi h squared left bracket r minus left parenthesis h divided by 3 right parenthesis right bracket.
Accepted Solution
A:
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]