Thursday

February 11, 2016
Posted by **Tim** on Sunday, December 18, 2011 at 4:08pm.

- Calculus -
**Damon**, Sunday, December 18, 2011 at 5:34pmStill 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

- Calculus -
**Damon**, Sunday, December 18, 2011 at 5:45pmI 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

- Calculus -
**Damon**, Sunday, December 18, 2011 at 5:51pmI 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

- Calculus -
**Tim**, Sunday, December 18, 2011 at 10:18pmThanks 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?