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