Posted by Tim on Sunday, December 18, 2011 at 4:08pm.
Still not done?
You have 40 in^3 of water in the cup at four minutes after start.
How deep is it?
r = 2 + (6-2)h/10 = 2 + .4 h
v = integral pi r^2 dh from 0 to h
= pi int (4 + 1.6 h + .16 h^2) dh
= pi (4h + .8 h^2 + .0533 h^3)
= 40 at t = 4
40/pi = 12.9 = 4 h + .8 h^2 + .0533 h^3)
find h where v = 3.23
3.23 = h + .2 h^2 + .0133 h^3
guess h = 1 then v =1.21
h = 2.0 , v = 2.90
h = 2.5 , v = 3.96
h = 2.4 , v = 3.74
h = 2.3 , v = 3.51
h = 2.2 , v = 3.31 very close
h = 2.15 , v = 3.21 close enough to 3.23
so do the problem with h = 2.15 deep
I should use some other letter than v because it is v/4pi = 40/4pi = 3.23
actually I think closer to 3.18
3.18 = h + .2 h^2 + .0133 h^3
guess h = 1 then v =1.21
h = 2.0 , v = 2.90
h = 2.5 , v = 3.96
h = 2.4 , v = 3.74
h = 2.3 , v = 3.51
h = 2.2 , v = 3.31
h = 2.15 , v = 3.21
h = 2.12 , v = 3.14
h = 2.13 , v = 3.156
h = 2.14 , v = 3.186 very close to 3.18
so do the problem with h = 2.14 deep
I am pretty sure I showed you how to do the rest before
dv/dt = pi r^2 dh/dt = 40
so
dh/dt = 40/(pi r^2)
but r = 2 + .4 h and at 4 minutes we said h = 2.14
so
r = 2 + .4 (2.14) = 2.86
r^2 = 8.16
dh/dt = 40/(8.16pi) = 1.56 in/min
Thanks again for responding. When we earlier did h=0, the formula was dh/dt=5/2pi=.8 in/min. Intuitively, it makes sense that the rate would reduce as the cup is filling up since the radius is increasing. So, should the answer above have 10 in the numerator instead of 40. Then, the answer would be dh/dt = 10/(8.16pi) = .39 in/min.
What do you think?
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