Posted by **Tim** on Saturday, December 17, 2011 at 5:05pm.

A cup has a radius of 2" at the bottom and 6" on the top. It is 10" high. 4 Minutes ago, water started pouring at 10 cubic " per minute. How fast was the water level rising 4 minutes ago? How fast is the water level rising now? What will the rate be when the glass is full?

- Calculus -
**Damon**, Saturday, December 17, 2011 at 5:18pm
r = 2 + 4 (h/10) = 2 + .4 h

surface area = pi r^2

surface area*dh = dV

(volume of slice at surface)

pi r^2 dh = DV

pi (4 + 1.6 h + .16 h^2) dh = dV

thus

pi (4 + 1.6 h + .16 h^2) dh/dt = dV/dt

but we know that dV/dt = 10 in^3/min

so

dh/dt = (10/pi)(1/(4 + 1.6 h + .16 h^2)

when h = 0

dh/dt = 10/(4pi) etc

- Calculus -
**Tim**, Saturday, December 17, 2011 at 5:34pm
Thanks. I think that I understand the 1st question now. How about the other questions?

- Calculus -
**Damon**, Saturday, December 17, 2011 at 6:05pm
You only need solve for h as a function of time

pi(4 + 1.6 h + .16 h^2)dh/dt = dV/dt =10

(4 + 1.6 h + .16 h^2)dh = (10/pi) dt

4 h + .8h^2 + .0533 h^3 = (10/pi) t

1.26 h + .251 h^2 + .0167 h^3 = t

You have to solve this for h and then r when t = 4 min (I would do it by making a few guesses and interpolating)

at the top is easy because then you know that h = 10 and r = 6

- Calculus -
**Tim**, Saturday, December 17, 2011 at 6:13pm
OK, I understand now. Thanks for all of your help.

- Calculus -
**Tim**, Sunday, December 18, 2011 at 4:06pm
In the 2nd question (rate of water rising now), how do you calculate r?

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