# Thread: Rate of Change: Volume of Water in a Cone

1. ## Rate of Change: Volume of Water in a Cone

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!

2. ## Re: Rate of Change: Volume of Water in a Cone

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)
Your answer to part b) is pretty simple: v = (1/3)(pi)(r^2)h. However, we do have information that ties r and h together. The ratio is

r/h = 6/10 = 3/5

Or in other words,

r = 3h/5

So we can rewrite v = (1/3)(pi)(r^2)h as

v = (1/3)(pi)((3h/5)^2)h = [3pi(h)^2]/25

When we are asked, “How fast is the water level falling?”, it means, “What is dh/dt?”

v = [3pi(h)^2]/25

dv/dt = (6pi(h)/25)(dh/dt)

(dh/dt) = (dv/dt)/ (6pi(h)/25)

Hope that helps.

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