Answer:
0.36ft/min
Step-by-step explanation:
We are given that
[tex]\frac{dv}{dt}=9ft^3/min[/tex]
Diameter of tank,d=10ft
Radius,[tex]r=\frac{d}{2}=\frac{10}{2}=5ft[/tex]
Height of tank,h=10 ft
We have to find the rate at which radius of the water in the tank increasing when r=2 ft
[tex]\frac{h}{r}=\frac{10}{5}=2[/tex]
[tex]h=2r[/tex]
Volume of conical tank=[tex]V=\frac{1}{3}\pi r^2 h[/tex]
Substitute the values
[tex]V=\frac{1}{3}\pi r^2(2r)=\frac{2}{3}\pi r^3[/tex]
Differentiate w.r.t t
[tex]\frac{dV}{dt}=2\pi r^2\frac{dr}{dt}[/tex]
Substitute the values
[tex]9=2\pi(2)^2\frac{dr}{dt}[/tex]
[tex]\frac{dr}{dt}=\frac{9}{2\pi(2)^2}=\frac{9}{8\pi}=0.36ft/min[/tex]
