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Question
A conical water tank with vertex down of 12 metres height has a radius of 5 metres at the top. If water flows into the tank at a rate 10 cubic m/min, how fast is the depth of the water increases when the water is 8 metres deep?
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Solution
From the figure `"r"/"h" = 5/12`
r = `(5"h")/12`
Given rate of change of volume `"dV"/"dt"` = 10
When h = 8 to find `"dh"/"dt"`
Volume of cone V = `1/3 pi"r"^2"h"`
V = `pi/3((5"h")/12)^2"h"`
V = `pi/3((25"h"^3)/144)`
= `(25pi)/432 "h"^3`
DIfferentiating w.r.t. 't'
`"dV"/"dt" = (25pi)/432 (3"h")^2 "dh"/"dt"`
10 = `(25pi)/432 (3(8)^2) "dh"/"dt"` ......[∵ Given h = 8]
∴ `"dh"/"dt" = 1/pi (432 xx 10)/(3 xx 25 xx 64)`
= `4320/(4800pi)`
= `0.9/pi`
= 9/(10pi)`
The depth of the water increasing at the rate of `9/(10pi)` m/min
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