Question

Water runs into a conical tank at the rate of 9 ft 3/min. the tank is standing, inverted, and has a height of 10 feet and a base diameter of 10 feet. At what rate is the radius of the water in the tank increasing when the radius is two feet

Answer 1

0.36ft/min

Explanation:

We are given that

`(dv)/(dt)=9ft^3/min`

Diameter of tank,d=10ft

Radius,
`r=(d)/(2)=(10)/(2)=5ft`

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

`(h)/(r)=(10)/(5)=2`

`h=2r`

Volume of conical tank=
`V=(1)/(3)\pi r^2 h`

Substitute the values

`V=(1)/(3)\pi r^2(2r)=(2)/(3)\pi r^3`

Differentiate w.r.t t

`(dV)/(dt)=2\pi r^2(dr)/(dt)`

Substitute the values

`9=2\pi(2)^2(dr)/(dt)`

`(dr)/(dt)=(9)/(2\pi(2)^2)=(9)/(8\pi)=0.36ft/min`