# Water flows through a conical tank at a constant rate of 3m^3 per sec. The radius of the cone is 5m and it's height is 6 m. At why rate is the water level rising when the radius is 4m?

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### 1 Answer

let the volume of a cone be, `V`

` ` height of the cone be,` `` `` h `

radius of the cone be,`r`

Then the volume of the cone can be expressed as,

`V = (1/3)Pi(r^2)h --- (1)`

now we have to find a relationship between the radius and the height of the cone

we can draw a right angle with base `r` and height `h`

let the angle between `h` and the hypotenuse be `alpha` (half apex angle)

then,

`tan alpha = r/h`

`r = htan alpha`

we can substitute this in the equation (1)

`V = (1/3)pi((h tan alpha)^2)h`

`V = (1/3)pi((tan alpha)^2)h^3`

here it is given that , r = 5 and h = 6

then `tan alpha = 5/6`

hence,

`V = (1/3)pi((5/6)^2)h^3`

`V = ((25pi)/108)h^3-----(2)`

` `in this question the rate of volume increase is given (`(dV)/(dt)` )

therefore lets differenciate the equation(2) with respect to t

`(dV)/(dt) = ((25pi)/108)3h^2((dh)/(dt))`

` ` it is given that` ``(dV)/(dt) = 3 m^3/s `

`hence`

`3 = ((25pi)/108)3h^2((dh)/(dt))`

`(dh)/(dt) = 108/(25pih^2) `

when r = 4m ,

`h = r/(tan alpha) = 4/(5/6) = 24/5`

then,

`(dh)/(dt) = 108/(25pi((24/5)^2)) = 108/(25pi((576/25))) = 0.0597 m/s`

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**hence when the radius is 4m the rate of water level is increase is 0.0597 m/s**

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