Question

A water tank in the shape of an inverted cone has height 8 m and radius 2 m. Water is being added to the tank at a rate of 7 m3/day. When the depth of the water in the tank is 7 m, what is the instantaneous rate of change of the radius of the top surface of the water? Note: the volume of a cone of radius r and height h is 1 3πr2h.

Answer 1

`(4)/(7\pi) \ m/day`

Step-by-step explanation:

h : r = 8 : 2

h = 4r

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

`=(1)/(3) \pi r^2(4r)`

`=(4)/(3) \pi r^3`

`(dV)/(dr) =(4)/(3) (3)\pi r^2`

`=4\pi r^2`

`(dr)/(dV) =(1)/(4\pi r^2)` ... [1]

`(dV)/(dt) =7` ... [2]

[1] & [2]

`(dr)/(dt) =(dr)/(dV) *(dV)/(dt)`

`=(1)/(4\pi r^2) *7`

`=(7)/(4\pi r^2)`

when h = 7m → r =
`(7)/(4)` m

`(dr)/(dt) ((7)/(4) )=(7)/(4\pi ((7)/(4) )^2)`

`=(4)/(7\pi) \ m/day`