water is running out a conical funnel at the rate of 1cu in per sec. if the radius of the base of the funnel is 4an and the altitude is 8in. find the rate at which the water level is dropping when it is 2 in from the top.

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water is running out a conical funnel at the rate of 1cu in per sec. if the radius of the base of the funnel is 4an and the altitude is 8in. find the rate at which the water level is dropping when it is 2 in from the top.

Mathematics
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change of rates eh....
v' = 1 ft^3 per sec; or simply 1 the rate of h is what we wanna determine
volume of a cone = (1/3)(base area)(height)

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the relation of height to raduis is given by the cross section of the cone
h = -2r+8 right?
can you show the solution?
v = (1/3)(pi 2^2)(-2r +8)
im working on it; havent got to the end yet lol
pi 2^2 was meant to be pi r^2
lets re work that with h as the variabe so we can see the rate of change with respect to h and not r
r = (h-8)/-2 or (8-h)/2
v = (1/3)(pi ((8-h)/2)^2)(h) perhaps?
\[v = \frac{h}{3}*\frac{\pi.(8-h)^2}{4}\]
64hpi +h^3pi -16h^2pi --------------------- = v derive now 12
64pi +3h^2pi -32hpi --------------------- = v' derive now 12
pi(64 +3h^2 -32h) = 12 64 +3h^2 -32h = 12/pi right?
ang gulo wah
maybe i mess that up alittle dv/dh = that up there dv/dt = 1 and dv/dh dh/dt = dv/dt
we wanna find dh/dt sooo dh/dt = dh/dv dv/dt ; gotta take the inverse if thats gonna work
dh/dt is the inverse of dh/dv right :)
on paper i get dh/dt = -1/(9pi)

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