Given:
The radius of the tank is: R = 7.5 feet.
Volume rate is: dV/dt = 60 cubic feet per hour.
h = 2.8 feet.
To find:
V, R, h, dV/dt, dR/dt, dh/dt
Explanation:
The given expression for volume is:
[tex]V=\pi h^2(R-\frac{1}{3}h)[/tex]
Substituting the values in the above equation, we get:
[tex]V=3.14\times2.8^2\times(7.5-\frac{1}{3}\times2.8)=161.74\text{ cubic feet}[/tex]
The volume is 161.74 cubic feet.
The radius of the tank is given as: R = 7.5 feet
The height of the water in the tank is given as: h = 2.8 feet
The volume rate of water is given as: dV/dt = 60 cubic feet per hour
As the radius of the tank is constant, its derivative will be zero. Thus, dR/dt = 0.
dh/dt can be calculated by differentiating the given volume equation with respect to time as:
[tex]\begin{gathered} \frac{dV}{dt}=\pi^2\times\frac{dh^2}{dt}\times(R-\frac{1}{3}\frac{dh}{dt}) \\ \\ \frac{dV}{dt}=\pi^2\times\frac{2dh}{dt}\times(R-\frac{1}{3}\frac{dh}{dt}) \\ \\ \frac{dV}{dt}=2\pi^2R\times\frac{dh}{dt}-\frac{2\pi^2}{3}\times(\frac{dh}{dt})^2 \end{gathered}[/tex]
Substituting the values in the above equation, we get:
[tex]\begin{gathered} 60=2\pi^2\times7.5\times\frac{dh}{dt}-\frac{2\pi^2}{3}\times(\frac{dh}{dt})^2 \\ \\ 60=148.044\frac{dh}{dt}-6.580(\frac{dh}{dt})^2 \\ \\ 6.580(\frac{dh}{dt})^2-148.044\frac{dh}{dt}+60=0 \end{gathered}[/tex]
Solving the above quadratic equation, we get:
dh/dt = 0.412 ft/hr or dh/dt = 22.086 ft/hr.
Final answer:
V = 161.74 cubic feet
R = 7.5 feet
h = 2.5 feet
dV/dt = 60 cubic feet per hour
dR/dt = 0
dh/dt = 22.086 ft/hr or 0.412 ft/hr
