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) 52.5 minutes have elapsed from the time when the hour-glass was full in the upper chamber?
Help me! I don't know What should I do. I tried one's best but my answer always are wrong
So, what did you try? It'd be nice if you showed your work on the problem.
http://www.jiskha.com/display.cgi?id=1476782673
The answer are 9.26 in/hr and 13.33 in/hr
Well, I made a boo-boo. Clearly you did not bother to read my work, or you would have found it.
I was just as blind, because if the glass empties 10 in. in one hour, clearly the .15 in/hr I got for an answer was way too small.
So, it all goes back to this: I swapped y and r in the 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, -9.26 in/hr
Since it is negative, the level is falling at 9.26 in/hr
No worries! Let's break down the problem and solve it step by step.
First, let's define some variables:
- Let h(t) represent the depth of the sand in inches at time t in minutes.
- Let V(t) represent the volume of the sand in the upper chamber at time t.
- Let dV/dt represent the rate of change of the volume of the sand with respect to time, which is equivalent to the rate at which the sand is falling.
Now, let's find an equation relating the depth of the sand to the volume of the sand.
We have an upper chamber in the shape of a cone. The volume of a cone is given by the formula V = (1/3)πr²h, where r is the radius and h is the height.
In this case, the radius is constant at 3 inches, and the height is decreasing over time as the sand falls. So, we can write the volume equation as V(t) = (1/3)π(3)²h(t).
Next, we need to find an equation for h(t). The sand is falling through the aperture at a constant rate. Since it takes exactly one hour (60 minutes) for the upper chamber to empty, the sand is falling at a rate of (10 inches)/(60 minutes) = 1/6 inches per minute. Therefore, we can write the equation h(t) = 10 - (1/6)t.
Now, let's find the equation for V(t) using the given equation for h(t):
V(t) = (1/3)π(3)²[10 - (1/6)t]
= π(9 - (1/18)t²)
Now, let's find dV/dt, the rate of change of volume with respect to time:
dV/dt = d/dt [π(9 - (1/18)t²)]
= -π(1/9)t
a) To find the rate at which the level is falling when the depth of the sand is 6 inches (t = 6 minutes), substitute t = 6 into dV/dt:
dV/dt = -π(1/9)(6)
= -2π/3 inches per minute
So, the level is falling at a rate of -2π/3 inches per minute when the depth of the sand is 6 inches.
b) To find the rate at which the level is falling 52.5 minutes after the hour-glass was full in the upper chamber (t = 52.5 minutes), substitute t = 52.5 into dV/dt:
dV/dt = -π(1/9)(52.5)
= -5π/6 inches per minute
So, the level is falling at a rate of -5π/6 inches per minute, approximately, 2.61799 inches per minute, 52.5 minutes after the hour-glass was full in the upper chamber.
Remember to always double-check your calculations, and make sure you're using consistent units throughout!