anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
change of rates eh....
amistre64
  • amistre64
v' = 1 ft^3 per sec; or simply 1 the rate of h is what we wanna determine
amistre64
  • amistre64
volume of a cone = (1/3)(base area)(height)

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amistre64
  • amistre64
the relation of height to raduis is given by the cross section of the cone
amistre64
  • amistre64
h = -2r+8 right?
anonymous
  • anonymous
can you show the solution?
amistre64
  • amistre64
v = (1/3)(pi 2^2)(-2r +8)
amistre64
  • amistre64
im working on it; havent got to the end yet lol
amistre64
  • amistre64
pi 2^2 was meant to be pi r^2
amistre64
  • amistre64
lets re work that with h as the variabe so we can see the rate of change with respect to h and not r
amistre64
  • amistre64
r = (h-8)/-2 or (8-h)/2
amistre64
  • amistre64
v = (1/3)(pi ((8-h)/2)^2)(h) perhaps?
amistre64
  • amistre64
\[v = \frac{h}{3}*\frac{\pi.(8-h)^2}{4}\]
amistre64
  • amistre64
64hpi +h^3pi -16h^2pi --------------------- = v derive now 12
amistre64
  • amistre64
64pi +3h^2pi -32hpi --------------------- = v' derive now 12
amistre64
  • amistre64
pi(64 +3h^2 -32h) = 12 64 +3h^2 -32h = 12/pi right?
anonymous
  • anonymous
ang gulo wah
amistre64
  • amistre64
maybe i mess that up alittle dv/dh = that up there dv/dt = 1 and dv/dh dh/dt = dv/dt
amistre64
  • amistre64
we wanna find dh/dt sooo dh/dt = dh/dv dv/dt ; gotta take the inverse if thats gonna work
amistre64
  • amistre64
dh/dt is the inverse of dh/dv right :)
amistre64
  • amistre64
on paper i get dh/dt = -1/(9pi)

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