Question

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.

Answer 1

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,

$V=\pi h^2(r-\frac{h}{3})$

Where, r is the radius of the hemispherical tank,

h is the water level, ( in feet )

Here, r = 9 feet,

$\implies V=\pi h^2(9-\frac{h}{3})$

$V=9\pi h^2-\frac{\pi h^3}{3}$

Differentiating with respect to t ( time ),

$\frac{dV}{dt}=18\pi h\frac{dh}{dt}-\frac{3\pi h^2}{3}\frac{dh}{dt}$

$\frac{dV}{dt}=\pi h(18-h)\frac{dh}{dT}$

Here, $\frac{dV}{dt}=-2\text{ cubic feet per hour}$

And, h = 6 feet,

Thus,

$-2=\pi 6(18-6)\frac{dh}{dt}$

$\implies \frac{dh}{dt}=\frac{-2}{72\pi}=-0.00884194128288\approx -0.0088$