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.

Answer 1

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"`