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
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?
5 answers
Thanks. I think that I understand the 1st question now. How about the other questions?
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
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
OK, I understand now. Thanks for all of your help.
In the 2nd question (rate of water rising now), how do you calculate r?