I apologize, I really hate to bounce but the problem seems pretty basic and I would really like to get it figured out so I can get some sleep. I'm still trying to figure it out but can't get anywhere. Any help would be greatly appreciated! Thanks!
A cone-shaped coffee filter of radius 6cm and depth 10cm contains water, which drips out through a hole at the bottom at a constant rate of 1.5cm^3 per second.
a) If the filter starts out full, how long does it take to empty?
v = (1/3)(pi)(r^2)h - 1.5t
0 = (1/3)(pi)(6^2)(10) - 1.5t
0 = 376.99 - 1.5t
376.99 = 1.5t
t = 251.33 sec
b) Find the volume of water in the filter when the depth of the water is h cm.
v = (1/3)(pi)(r^2)h - 1.5t
v + 1.5t = (1/3)(pi)(r^2)h
h = (v + 1.5t) / ((1/3)(pi)r^2)
c) How fast is the water level falling when the depth is 8cm?
8 = (v + 1.5t) / ((1/3)(pi)r^2)
How do I solve this with so many unknown variables (volume, time and radius)? My suspicion is that I did part (b) incorrectly. Can anyone please show me where I went wrong? Thanks!
I do believe that (1/3)(6^2)(10) = 120, not 1200. Which means that t=251.33s
I've never learned this 'similar triangles' thing, but I see that you're just finding the ratio between r and h of the cone so we can express r in terms of h. Sounds good, this is what I was missing.
The height of the water should be 8cm, but this was also very helpful. I got -0.0164 = dh/dt for yours (using h = 9cm), and I got dh/dt = -0.021 (using h = 8cm).
Thanks for all your help!