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 tank's area or volume is not 4 times as big though.....

A. Well, since the small cylindrical tank is initially full, the volume of water remaining in the small tank as a function of time, t, can be represented by the equation:

V = (pi)(4^2)(6) - (1/2)t

B. To find out how long it takes for the small tank to completely empty, we can set the volume, V, equal to zero and solve for t:

(pi)(4^2)(6) - (1/2)t = 0

Solving this equation would give you the time it takes for the small tank to completely empty.

C. Now, let's move on to computing dz/dt, which represents the rate of change of the depth of water in the large tank over time. As water is draining from the small tank into the large tank, the rate of change of depth in the large tank is proportional to the rate at which the small tank is being emptied.

Therefore, dz/dt is equal to the rate at which the small tank is being emptied, which is 1/2 cubic feet per second.

D. Finally, to determine what fraction of the total amount of water is in the large tank at t=6, we need to calculate the volume of water in the large tank at that specific time.

To do this, we can use the equation for the volume of a cylinder:

V = (pi)(8^2)(16)

And then, we can divide this volume by the total amount of water in both tanks (which is the sum of the initial volume in the small tank and the volume of the large tank):

Total volume = (pi)(4^2)(6) + (pi)(8^2)(16)

By dividing the volume of water in the large tank at t=6 by the total volume, we can find the fraction of the total amount of water in the large tank.

To solve these problems, we will break them down into step-by-step calculations.

A. To find the volume, V, of the water remaining in the small tank as a function of time, we need to consider the rate at which the water is draining. The rate of drainage is given as 1/2 cubic feet per second. Let's denote time as t.

The initial volume of the small tank is V_small = (pi)(r_small^2)(h_small), where r_small = 4 ft and h_small = 6 ft.

As time passes, the volume of water remaining in the small tank decreases. The volume of water remaining 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 need to find the time when V_remaining = 0.

Setting V_remaining = 0 and solving for t, we have:
V_small - (1/2)t = 0
(1/2)t = V_small
t = (2V_small) / 1

Substituting V_small = (pi)(r_small^2)(h_small), we get:
t = (2(pi)(r_small^2)(h_small)) / 1

C. To compute dz/dt, which represents the rate at which the depth of water in the large tank is changing, we can use the fact that the rate of change in volume is equal to the rate of change in height.

The volume of the large tank, V_large, is given by:
V_large = (pi)(r_large^2)(h_large)

To find the rate at which the depth of water changes in the large tank, dz/dt, we can differentiate both sides of the equation with respect to time:
dz/dt = d/dt(h_large)

Since the radius of the large tank is constant, it does not affect the rate at which the depth of water changes. Therefore, dz/dt is equal to the rate at which the height, h_large, is changing.

C. To find dz/dt, we need to find the relation between the heights of the two tanks at any given time t. Since the small tank is draining into the large tank, the height of the water in the large tank can be expressed as a function of time using the volume equation.

The volume of water in the large tank, V_large, can be written as:
V_large = (pi)(r_large^2)(h_large)

Since the water is draining from the small tank at a rate of 1/2 cubic feet per second, the rate of change of volume of water in the large tank is equal to this rate of drainage.

So, dV_large/dt = 1/2

Differentiating the volume equation with respect to time, we get:
dV_large/dt = (pi)(r_large^2)(dh_large/dt)

Setting this equal to 1/2, we have:
(pi)(r_large^2)(dh_large/dt) = 1/2

Simplifying, we get:
dh_large/dt = (1 / (2(pi)(r_large^2)))

D. To find the fraction of the total amount of water in the large tank at time t=6, we need to calculate the volume of water in the large tank at that time and divide it by the total volume of the large tank.

The volume of water in the large tank at time t is given by:
V_large(t) = (pi)(r_large^2)(h_large(t))

To find the fraction, we divide the volume of water in the large tank at time t=6 by the total volume of the large tank:
fraction = V_large(6) / V_large_total

Substituting the values, we have:
fraction = (pi)(r_large^2)(h_large(6)) / ((pi)(r_large^2)(h_large_total))

Simplifying, we get:
fraction = h_large(6) / h_large_total

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 water drains from the tank.

The initial volume of the water in the small tank can be calculated using the formula for the volume of a cylinder:

V_initial = (pi)(r_small^2)(h_small)

where r_small is the radius of the small tank and h_small is the height of the small tank.

Given the values r_small = 4 feet and h_small = 6 feet, we can calculate V_initial.

V_initial = (pi)(4^2)(6)

To find the volume of the water remaining in the small tank at any time t, we need to subtract the amount of water that has drained out of the tank. Since the water drains out at a rate of 1/2 cubic feet per second, the volume of water drained at time t is:

V_drained = (1/2)(t)

where t is the time in seconds.

Therefore, the volume of water remaining in the small tank at any time t can be given by the equation:

V(t) = V_initial - V_drained

Substituting the values, we have:

V(t) = (pi)(4^2)(6) - (1/2)(t)

This is the volume of water remaining in the small tank as a function of time.

Next, to determine how long it takes for the small tank to completely empty, we need to find the time t when the volume of water remaining in the small tank is zero.

Setting V(t) = 0, we can solve for t:

(pi)(4^2)(6) - (1/2)(t) = 0

Simplifying the equation, we have:

(pi)(4^2)(6) = (1/2)(t)

Solving for t, we get:

t = 48(pi)

Therefore, it takes 48(pi) seconds for the small tank to completely empty.

Moving on to part C, we need to find the rate at which the depth of the water in the large tank, z, changes over time. We can determine this by finding dz/dt.

The volume of water that has entered the large tank can be calculated as the difference between the initial volume of the large tank and the volume of water remaining in the small tank at any time t:

V_entered = V_large - V(t)

Substituting the given values for the large tank's dimensions, we can calculate V_large.

V_large = (pi)(8^2)(16)

To determine dz/dt, we need to consider the change in volume with respect to time. The rate of change of volume in the large tank can be expressed as:

dV_large/dt = dV_entered/dt

Since the water drains from the small tank and enters the large tank, the rate of change of volume in the large tank can also be expressed as the rate of change of height multiplied by the area of the base of the large tank:

dV_large/dt = d(z(pi)(8^2))/dt

Applying the chain rule, we have:

dV_large/dt = (dz/dt)(pi)(8^2)

Simplifying, we get:

dz/dt = (dV_large/dt) / [(pi)(8^2)]

Finally, in part D, we need to determine the fraction of the total amount of water in the large tank at time t=6.

At time t=6, the volume of water remaining in the small tank, V(6), can be computed by substituting t=6 into the equation for V(t):

V(6) = (pi)(4^2)(6) - (1/2)(6)

Similarly, the volume of water in the large tank, V_large(6), can be computed using the formula for the volume of a cylinder:

V_large(6) = (pi)(8^2)(6)

Therefore, the fraction of the total amount of water in the large tank at time t=6 is:

V_large(6) / (V(6) + V_large(6))

the small tank has a volume of 96π ft^3

its depth decreases at a rate of (1/2)/(16π) = 1/(32π) ft/s

So,

A: v(t) = 96π - 1/2 t
B: duh
C: the large tank's area is 4 times as big, so its depth increases 1/4 as fast, or 1/8 ft/s.
D: 6/(192π)