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Question
Water is dripping out at a steady rate of 1 cu cm/sec through a tiny hole at the vertex of the conical vessel, whose axis is vertical. When the slant height of water in the vessel is 4 cm, find the rate of decrease of slant height, where the vertical angle of the conical vessel is `pi/6`
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Solution
Given that `"dv"/"dt"` = 1 cm3/s
Where v is the volume of water in the conical vessel.
From the Fig. 6.2, l = 4 cm
h = `"l" cos pi/6`
= `sqrt(3)/2 "l"` and r = `"l" sin pi/6 = "l"/2`.
Therefore, v = `1/3 pi"r"^2"h"`
= `pi/3 "l"^2/4 sqrt(3)/2 "l"`
= `(sqrt(3)pi)/24 "l"^3`
`"dv"/"dt" = (sqrt(3)pi)/8 "l"^2 "dl"/"dt"`
Therefore, l = `(sqrt(3)pi)/8 16 * "dl"/"dt"`
⇒ `"dl"/"dt" = 1/(2sqrt(3)pi)` cm/s.
Therefore, the rate of decrease of slant height = `1/(2sqrt(3)pi)` cm/s
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