Question

A cone has a radius of 5 ft and a height of 15 ft. It is empty and is being filled with water at a constant rate of 15 15 ft 3 ft3 / sec sec . Find the rate of change of the radius of the surface of the water when the radius of the surface of the water is 2 2 ft. (You must also include the units)

Answer 1

`(dr)/(dt)=0.11062ft/sec`

Explanation:

From the question we are told that:

Radius
`r=5ft`

Height
`H=15ft`

Rate
`R=15ft/3sec =5ft/s`

Surface Radius
`R_(surf)=2.2ft`

Generally the equation for Volume is mathematically given by

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

Since radius to height ratio gives

`(r)/(h)=(5)/(15)`

`(r)/(h)=(1)/(3)`

`h=3r`

Therefore

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

`V=\pi r^3`

Generally the equation for Change of Volume is mathematically given by

`(dv)/(dt)=\pi (d)/(dt)(r^3)`

`(dv)/(dt)=\pi 3*r^2 (dr)/(dt)`

`(dv)/(dt)=\pi 3*(2.2)^2 (dr)/(dt)`

`(dr)/(dt)=(5)/(45.62)`

`(dr)/(dt)=0.11062ft/sec`