A cone-shaped coffee filter is draining at a rate of 20 in^3 /min. The filter has a diameter of 8 inches. the radius of the filter is 2/3 the height. how fast is the coffee level falling when the coffee is 3 inches deep?
v = 1/3 pi r^2 h
r = 2h/3, so
v = pi/3 (4h^2/9) h
= 4pi/27 h^3
dv/dt = 4pi/9 h^2 dh/dt
-20 = 4pi/9 * 9 dh/dt
-5/pi = dh/dt
idk what that is hehe
To find the rate at which the coffee level is falling, we need to apply related rates of change.
Let's denote:
V = volume of coffee in the cone-shaped filter (in^3)
r = radius of the coffee level in the cone (in)
h = height of the coffee level in the cone (in)
x = rate at which the coffee level is dropping (in/min)
We are given that the coffee filter is draining at a rate of 20 in^3/min. This means that dV/dt = -20.
Since the filter has a diameter of 8 inches, the radius is half of the diameter. Therefore, r = 8/2 = 4 inches.
We are also given that the radius of the filter is 2/3 the height. So, r = (2/3)h.
First, let's find an equation relating V, r, and h. The volume of a cone can be calculated using the formula V = (1/3)πr^2h.
Substitute the values of r and h: V = (1/3)π(2/3h)^2h = (4/27)πh^3.
Differentiate both sides of this equation with respect to time (t) to get dV/dt in terms of h and dh/dt:
dV/dt = (4/27)π(3h^2)(dh/dt) = (4/9)πh^2(dh/dt)
Now we substitute the known values into the equation:
-20 = (4/9)πh^2(dh/dt)
To solve for dh/dt, we need to find h when the coffee is 3 inches deep:
r = (2/3)h
4 = (2/3)h
h = (3/2) * 4
h = 6 inches
Now we substitute this value of h into the equation to solve for dh/dt:
-20 = (4/9)π(6^2)(dh/dt)
-20 = (4/9)π(36)(dh/dt)
-20 = (4/9)π(36)(dh/dt)
-20 = (4/9) * 36 * π * (dh/dt)
-20 = (16/9) * 36 * π * (dh/dt)
-20 = 20.12 * π * (dh/dt)
dh/dt = -20 / (20.12 * π)
dh/dt ≈ -0.0314 in/min
Therefore, the coffee level is falling at a rate of approximately 0.0314 inches per minute when it is 3 inches deep.