Question

A conical tank has height 3 m and radius 2 m at the top. Water flows in at a rate of 1.5 m3/min. How fast is the water level rising when it is 1.1 m from the bottom of the tank? (Round your answer to three decimal places.)

Answer 1

Step-by-step explanation:

It is known that general formula to calculate the volume of a cone is as follows.

Volume of the cone =
`(1)/(3) \pi r^(2)h`

and here, r =
`(2)/(3)h`

So, V =
`(1)/(3) \pi * ((4)/(9)) * (h^(3))`

On putting the given values into the above formula we will calculate the value f h as follows.

V =
`(1)/(3) \pi * ((4)/(9)) * (h^(3))`

=
`(1)/(3) * \pi * ((4)/(9)) * (h^(3))`

=
`(4 \pi)/(27) h^(3)`

Now, we will differentiate w.r.t t as follows.

`(dV)/(dt) = (d)/(dt)((4 \pi)/(27)h^(3))`

=
`(4\pi)/(27)(3h^(2)) (dh)/(dt)`

It is given that water here flows at a rate of 1.5
`m^(3)/min` and h = 1.1 m.

`(dV)/(dt) = (4 \pi)/(27)(3h^(2)) (dh)/(dt)`

1.5 =
`(4 \pi)/(27)(3h^(2))(dh)/(dt)`

1.5 =
`(4 \pi)/(27)(3(1.1)^(2))(dh)/(dt)`

`(dh)/(dt)` = 0.888

= 0.9 m/min (approx)

Thus, we can conclude that the water level is rising at a rate of 0.9 m/min.