Question

Water is flowing into a spherical tank with 6 foot radius at the constant rate of 30pi cu ft per hour. When the water is h feet deep, the volume of water in the tank is given by
V=(pih^2)/3*(18-h).
What is the rate at which the depth of the water in the tank is increasing at the moment when the water is 2 feet deep?
Answer: 1.5 ft per hr
Please explain how to come to this answer.

Answer 1

\[given~\frac{ dV }{dt} =30\pi ft^3/hr \]
\[V=\frac{ \pi h^2 }{3 }*\left( 18-h \right) =\frac{ \pi }{3 }\left( 18 h^2-h^3 \right)\]
\[\frac{ dV }{ dt }= \frac{ \pi }{3 }\left( 36h-3h^2 \right)\frac{ dh }{dt }=\pi \left( 12h-h^2 \right)\frac{ dh }{ dt }\]
h=2 ft
\[substitute~the~values~of~h~and~\frac{ dV }{dt }\]

Answer 2

Thanks! :)

Answer 3

yw