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Question
Water is dripping out from a conical funnel of semi-vertical angle `pi/4` at the uniform rate of 2cm2/sec in the surface area, through a tiny hole at the vertex of the bottom. When the slant height of cone is 4 cm, find the rate of decrease of the slant height of water.
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Solution
If s represents the surface area, then `"ds"/"dt" = (2"cm"^2)/sec`
s = π r.l
= `pi"l" . sin pi/4 "l"`
= `pi/sqrt(2) "l"^2`
Therefore, `"ds"/"dt" = (2pi)/sqrt(2) "l" . "dl"/"dt"`
= `sqrt(2)pi"l" * "dl"/"dt"`
When l = 4cm
`"dl"/"dt" = 1/(sqrt(2)pi4)*2`
= `1/(2sqrt(2)pi)`
= `sqrt(2)/(4pi) "cm"/"s"`
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