Water is draining from a small cylindrical tank into a larger one below it. The small cylindrical tank has a radius of 4 feet and a height of 6 feet; the large cylindrical tank has a radius of 8 feet and a height of 16 feet. The small tank is initially full of water, and the water drains out at a rate of 1/2 cubic feet per second. Note: The volume of a cylinder is V=(pi)(r^2)(h).
A. Find the volume, V, of the water remaining in the small tank as a function of time.
B. How long does it take for the small tank to completely empty?
C. Let z be the depth of the water in the large tank, which is initially empty. Compute dz/dt.
D. What fraction of the total amount of water is in the large tank at time t=6?
A) V(t) = π*4^2*6 - 1/2 t
B) solve for t in V=0
C) z = (1/2 t) / (π*8^2) so dz/dt = 1/(2π*8^2) ft/s
D) V(6)/V(0)
To find the volume, V, of the water remaining in the small tank as a function of time (A), we need to determine how the volume changes over time as the water drains.
The small tank has a radius of 4 feet and a height of 6 feet, so its initial volume (when it is full) would be V_small_initial = π * (4^2) * 6 = 96π cubic feet.
Given that the water drains from the small tank at a rate of 1/2 cubic feet per second, we can express the volume of water remaining in the small tank as a function of time, t, as follows:
V_small_remaining(t) = V_small_initial - (1/2) * t
To calculate the time it takes for the small tank to completely empty (B), we need to solve for t when V_small_remaining(t) = 0.
V_small_remaining(t) = 0
V_small_initial - (1/2) * t = 0
(1/2) * t = V_small_initial
t = 2 * V_small_initial
Therefore, it will take 2 * V_small_initial = 2 * 96π = 192π seconds for the small tank to completely empty.
Moving on to part C, where we need to compute dz/dt, the rate at which the depth of water in the large tank changes.
The volume of the water draining from the small tank equals the volume of water filling the large tank. Therefore, we can express this relationship mathematically:
V_small_remaining(t) = V_large_filled(t)
Differentiating both sides of the equation with respect to time (t), we get:
d(V_small_remaining)/dt = d(V_large_filled)/dt
Since we know that the rate of water draining from the small tank is 1/2 cubic feet per second, we can write:
d(V_small_remaining)/dt = -1/2
So, dz/dt = -1/2, which means that the depth of the water in the large tank is decreasing at a rate of 1/2 foot per second.
Lastly, to find the fraction of the total amount of water in the large tank at time t = 6 (D), we need to determine the volume of water in the large tank at that time.
The large tank has a radius of 8 feet and a height of 16 feet. The volume of water in the large tank will depend on the depth, z, of the water. We need to find the depth of the water at time t = 6.
Using the relationship between volume and depth for a cylinder:
V_large = π * (r^2) * z
Substituting the values, we get:
V_large = π * (8^2) * z
V_large = π * 64z
Since the height of the large tank is 16 feet, we know that when it is full, the depth is 16 feet:
16 = z
At time t = 6, we can find the volume of water in the large tank by substituting z = 16 into the formula:
V_large = π * 64 * 16
V_large = 1024π
The total volume of water in both tanks initially (when the small tank is full) is 96π + 1024π = 1120π.
Therefore, the fraction of the total amount of water in the large tank at time t = 6 is given by:
Fraction = V_large / (96π + V_large)
Fraction = (1024π) / (96π + 1024π)
Fraction = (1024π) / (1120π)
Fraction = 1024 / 1120
Fraction ≈ 0.9143
So, approximately 91.43% of the total water is in the large tank at time t = 6.
A. To find the volume, V, of the water remaining in the small tank as a function of time, we need to find the rate at which the water is draining.
The volume of the water in the small cylindrical tank can be given by the formula V = (π)(r^2)(h), where r is the radius and h is the height. In this case, the radius of the small tank is 4 feet and the height is 6 feet.
So, initially, the volume of water in the small tank is V = (π)(4^2)(6) = 96π cubic feet.
Since the water is draining out at a rate of 1/2 cubic feet per second, the volume of water remaining in the small tank after time t can be given by the equation V(t) = 96π - (1/2)t.
B. To find the time it takes for the small tank to completely empty, we need to set V(t) = 0 and solve for t.
0 = 96π - (1/2)t
(1/2)t = 96π
t = 192π
So, it takes 192π seconds for the small tank to completely empty.
C. To compute dz/dt, we need to find the rate at which the height of the water in the large tank is changing.
The volume of water in the large cylindrical tank can be given by the formula V = (π)(r^2)(h), where r is the radius and h is the height. In this case, the radius of the large tank is 8 feet and the height is initially 0 feet.
The rate at which the height, h, of the water in the large tank is changing can be found using similar triangles.
The ratio of the height of the water in the large tank to the height of the water in the small tank is the same as the ratio of their radii:
h/16 = 4/8
Simplifying, we get h = 8.
Since the water is draining out of the small tank at a rate of 1/2 cubic feet per second, the rate at which the height of the water in the large tank is changing is equal to the rate at which the small tank is losing water, which is -1/2 ft/s (negative because the height is decreasing).
Therefore, dz/dt = -1/2 ft/s.
D. To find the fraction of the total amount of water in the large tank at time t = 6, we need to find the ratio of the volume of water in the large tank to the total volume of water.
The volume of water in the large cylindrical tank can be given by the formula V = (π)(r^2)(h), where r is the radius and h is the height. In this case, the radius of the large tank is 8 feet and the height is given by:
h = (16 - t(1/2))
So, at t = 6, the height of the water in the large tank is:
h = (16 - 6(1/2)) = 13 feet.
Thus, the volume of water in the large tank at time t = 6 can be calculated as:
V = (π)(8^2)(13) = 832π cubic feet.
The total volume of water, V_total, can be calculated as the sum of the initial volume of water in the small tank and the volume of water in the large tank:
V_total = 96π + 832π = 928π cubic feet.
Therefore, the fraction of the total amount of water in the large tank at time t = 6 is given by:
Fraction = V/V_total = (832π)/(928π) = 13/14.
So, at time t = 6, 13/14 (or approximately 0.929) of the total amount of water is in the large tank.