Question

A water tank has the shape of an inverted right circular cone with base radius 3 meters and height 6 meters. Water is being pumped into the tank at the rate of 12 meters3/sec. Find the rate, in meters/sec, at which the water level is rising when the water is 2 meters deep. Give 2 decimal places for your answer. Type your answer in the space below. If your answer is a number less than 1, place a leading "0" before the decimal point (ex: 0.35).

Answer 1

3.82 meters/sec

Explanation:

The guy above is correct he just forgot to put the third decimal place

Answer 2

the rate of change is 3.8m/s

Explanation:

The volume of the right circular cone is

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

The rate of change of volume is

`(dV)/(dt) =(d)/(dt)( (\pi*r^2h)/(3))=(\pi)/(3) (d)/(dt)(r^2h)=(\pi)/(3)(2rh(dr)/(dt)+r^2(dh)/(dt))`

In order to proceed further we have to define r in terms of h so tht we can compute the derivative above.

The ratio between h and r is

`(h)/(r)=(6)/(3) =2`

Therefore
`r=(h)/(2)`.

We plug that into the derivative above and get:

`2rh(dr)/(dt)+r^2(dh)/(dt)=(h^2)/(2)(dh)/(dt)+(h^2)/(4)(dh)/(dt)=(3h^2)/(4)(dh)/(dt).`

Thus

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

Now for the numerical part.

The rate of change of volume
`(dV)/(dt)` is
`12(m^3)/(s)`, so when the water is 2 meters deep
`h=2`, therefore:

`12=(\pi*3*2^2)/(4*3) (dh)/(dt) \\\\\therefore \boxed{ (dh)/(dt)=3.8m/s.}`