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?
The large cylinder has a cross-section 4 times that of the smaller one, so its water level rises 1/4 as fast as it falls in the smaller one.
A. To find the volume, V, of the water remaining in the small tank as a function of time, we need to determine how the volume changes over time as the water drains out.
The initial volume of the small tank is given by the formula for the volume of a cylinder: V_small = (pi)(r_small^2)(h_small).
Since the water is draining out at a constant rate of 1/2 cubic feet per second, the volume of water in the small tank at time t is given by:
V_remaining = V_small - (1/2)(t)
B. To find how long it takes for the small tank to completely empty, we set V_remaining = 0 and solve for t:
0 = V_small - (1/2)(t)
(1/2)(t) = V_small
t = 2V_small
So, it takes t = 2V_small seconds for the small tank to completely empty.
C. To find dz/dt, we need to relate the change in the depth of the water in the large tank to the rate at which the water is draining from the small tank.
The volume of water draining from the small tank is given by (1/2)(t). This volume is equal to the volume of water entering the large tank, which is also equal to the change in the volume of water in the large tank.
The volume of the large tank is given by the formula for the volume of a cylinder: V_large = (pi)(r_large^2)(h_large).
Since the radius of the small tank is half the radius of the large tank and the height of the small tank is half the height of the large tank, we have:
(r_small/r_large) = 1/2
(h_small/h_large) = 1/2
Using these ratios, we can express the volume of the small tank in terms of the volume of the large tank:
V_small = (1/4)(pi)(r_large^2)(h_large)
Now, we can equate the volume of water draining from the small tank to the change in the volume of water in the large tank and differentiate with respect to time:
(1/2)(t) = dV_large/dt
Simplifying, we find:
dz/dt = (1/2)(t) / (pi)(r_large^2)
Note that dz/dt represents the rate at which the depth of the water in the large tank is changing.
D. To find the fraction of the total amount of water in the large tank at time t=6, we need to determine the volume of the water in the large tank at that time and compare it to the total volume of the large tank.
The volume of the water in the large tank at time t is given by:
V_water = (1/2)(t) = (1/2)(6) = 3 cubic feet
The total volume of the large tank is given by:
V_large = (pi)(r_large^2)(h_large) = (pi)(8^2)(16)
The fraction of the total amount of water in the large tank at time t=6 is:
Fraction = V_water / V_large = 3 / [(pi)(8^2)(16)]