A conical container with a diameter of 30 cm with a height of 45 cm being filled with water at a rate if 16 cm^3.Find the rate of change at which the water is rising in the cylinder when volume reaches 8(Pi)cm^3.
1 Answer | Add Yours
The container is an upside down cone.
The height, `H = 45 cm`
The area of the top of cone, `A = pi(15)^2 = 225pi`
If the water height at time t is h and diameter is 2r``.
The water volume filled at time `v = 1/3pir^2h`
But according to geometrical analysis we know,
`r/h = 15/45`
`r/h = 1/3`
`h = 3r`
So we get, `v = 1/3pir^2(3r) = pir^3`
`v = pir^3`
We know `(dv)/(dt) = 16` (The filling rate)
`16 = (dv)/(dt) = 3pir^2(dr)/(dt)`
`(dr)/(dt) = 16/(3pir^2)`
The radius on top of water when `v = 8pi` is,
`8pi = pir^3`
`r = 2` cm.
Therefore when `v = 8pi` , r is equal to 2 cm.
Therefore `(dr)/(dt) = 16/(3pi2^2)`
`(dr)/(dt) = 0.424` cm per second
But we know, `h =3r`
`(dh)/(dt) = 3(dr)/(dt)`
`(dh)/(dt) = 3(0.424)` cm per second
Therefore the rate of height increase or the rate at water level changes is 1.272 cm per second.
We’ve answered 318,944 questions. We can answer yours, too.Ask a question