Water is leaking out the bottom of a hemispherical tank of radius 88 feet at a r

Question

Water is leaking out the bottom of a hemispherical tank of radius 88 feet at a rate of 22 cubic feet per hour. The tank was full at a certain time. How fast is the water level changing when its height h is 66 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 bracketh2[r(h/3)].The water level is changing at a rate of nothing ft divided by hrft/hr when its height h is 66 feet.

(Simplify your answer. Round to the nearest thousandth as needed.)

Explanation / Answer

Radius of hemisphere= 88 feet

Rate of change of volume= dV/dt = 22 ft/hr

Now

V= pi(h)2(r-(h/3))

dV/dt={2pi*h(r-h/3)+pi*h2(-1/3)}dh/dt

Now dV/dt=22 ft/hr

h=66

r=88

Putting in the equation

22=dh/dt{(2pi(66)(88-66/3))-pi(662*(-1/3))}

Upon solving

dh/dt = 9.6*10-4

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