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

The water level changing by the rate of -0.0088 feet per hour ( approx )

Explanation:

Given,

The volume of a segment of height h in a hemisphere of radius r is,

`V=\pi h^2(r-(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-(h)/(3))`

`V=9\pi h^2-(\pi h^3)/(3)`

Differentiating with respect to t ( time ),

`(dV)/(dt)=18\pi h(dh)/(dt)-(3\pi h^2)/(3)(dh)/(dt)`

`(dV)/(dt)=\pi h(18-h)(dh)/(dT)`

Here,
`(dV)/(dt)=-2\text{ cubic feet per hour}`

And, h = 6 feet,

Thus,

`-2=\pi 6(18-6)(dh)/(dt)`

`\implies (dh)/(dt)=(-2)/(72\pi)=-0.00884194128288\approx -0.0088`