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Question
Water is dripping out from a conical funnel of semi-verticle angle `pi/4` at the uniform rate of `2 cm^2/sec`in the surface, through a tiny hole at the vertex of the bottom. When the slant height of the water level is 4 cm, find the rate of decrease of the slant height of the water.
Solution
Let r be the radius, h be the height and V be the volume of the funnel at any time t
`V = 1/3 pir^2h` ... (i)
Let I be the slant height of the funnel
Given: Semi-vertical angle = 45° in the triangle ADE:
`sin 45^@ = (DE)/(AE) => 1/sqrt2 = r/l`
`cos 45^@ = (AD)/(AE) => 1/sqrt2 = h/l`
`r = 1/sqrt2` and h = `1/sqrt2` ...(ii)
therefore the equation (i) can be rewritten as :
`V = 1/3 pi xx (I/sqrt2)^2 xx I/sqrt2 = pi/(3xx2xxsqrt2) xx I^3``
`V = pi/(6sqrt2) I^3` ...(iii)
Differentiate w.r.t. t :
`(dV)/(dt) = pi/(6sqrt2) xx 3l^2 xx (dl)/(dt)`
`(dv)/(dt) = pi/(2sqrt2) xx l^2 xx (dl)/(dt)`
`(dl)/(dt) = (2sqrt2)/(pil^2) xx (dV)/(dt)` ...(iv)
Since it is given that rate of change (decrease) of volume of water w.r.t. t is
`(dV)/(dt) = -2cm^3"/"sec`
therefore
`(dl)/(dt) = (2sqrt2)/(lambdal^2) xx (-2) = -(4sqrt2)/(lambdal^2)`
`(dl)/(dt)|_"at l = 4" = - (4sqrt2)/(pixx(4)^2) = - (sqrt2)/(4pi) "cm/sec"`
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