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:
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?
To find the rate at which the sand level is falling in the hourglass, we can use the concept of related rates and the geometry of the cone.
Let's start by finding the volume of the upper cone chamber. The volume of a cone is given by the formula V = (1/3)πr^2h, where r is the radius and h is the height.
a) When the depth of the sand is 6 inches, we need to find the rate at which the depth is changing, dh/dt.
The volume of the upper cone chamber at this depth can be calculated as follows:
V = (1/3)π(3^2)(6)
V = 18π cubic inches
Now, we need to differentiate both sides of the equation with respect to time t to find the rate of change of volume (dV/dt) and relate it to the rate of change of depth (dh/dt).
dV/dt = (dV/dh) * (dh/dt)
We know that dV/dh = (1/3)πr^2, which is the rate at which the volume changes with respect to the depth.
Substituting the known values:
dV/dt = (1/3)π(3^2) * (dh/dt)
dV/dt = 9π * (dh/dt)
We are given that if the upper chamber is full, it requires exactly one hour (3600 seconds) for it to empty. This means that dV/dt = -V/t = -18π / 3600.
Now, we can solve for dh/dt.
-18π / 3600 = 9π * (dh/dt)
dh/dt = (-18π / 3600) / (9π)
dh/dt = -1 / 200 inches per second
Therefore, when the depth of the sand is 6 inches, the rate at which the level is falling is -1/200 inches per second.
b) To find the rate at which the level is falling when 105/2 minutes have elapsed, we need to convert the time to seconds.
105/2 minutes = (105/2) * 60 = 3150 seconds
Using a similar approach as in part a, we know that dV/dt = -V/t = -18π / 3600.
Substituting the known values:
-18π / 3600 = 9π * (dh/dt)
dh/dt = (-18π / 3600) / (9π)
dh/dt = -1 / 200 inches per second
Therefore, when 105/2 minutes have elapsed, the rate at which the level is falling is -1/200 inches per second.