The upper chamber of an hour-glass is a cone of radius 3 inches and height 10 inches and, if full, it requires exactly one hour to empty. Assuming that the sand falls through the aperture at a constant rate, how fast is the level falling when:

Question

a) the depth of the sand is 6 inches?
b) 105/2 minutes have elapsed from the time when the hour-glass was full in the upper chamber?

Answer 1

v = π/3 * 3^2 * 10 = 30π
so, dv/dt = -30π in^3/hr = -π/2 in^3/min

When the sand has a depth of y inches, the surface has a radius of r = (3/10) y

So, at that point,

v = π/3 y^2 r = π/10 y^3

dv/dt = 3π/10 y^2 dy/dt
-π/2 = 3π/2 * 6^2 dy/dt
dy/dt = -1/108 in/min

For the 2nd part, find r, using
v(t) = 30π - π/2 t
then use it as in the first part.

Answer 2

Why you have v(t) = 30π - π/2 t???

Answer 3

I have that formula because, as I showed at the very first, the volume starts at 30π ion^3, and drops at a rate of π/2 in^3/min.

-----------------------------------
For those who followed a link to get here,
-----------------------------------

I swapped y and r in the volume formula. So, doing it right,

v = π/3 r^2 y = 3π/100 y^3

dv/dt = 9π/100 y^2 dy/dt
-π/2 = 9π/100 * 6^2 dy/dt
dy/dt = -25/162 in/min
Or, -250/27 = -9.26 in/hr
Since it is negative, the level is falling at 9.26 in/hr