POSSIBLE SOLUTION(S)

Situation 1

If the tank is y feet full, then the volume of water is given by
V = Pi 3^2 y = 9 Pi y, so dV/dt = 9 Pi dy/dt. On the other hand, the change in volume with respect to time is the area of the hole times the velocity of the water squirting through the hole, therefore,

dV/dt = -(area)(velocity) = -1/ 144 ft ^2 (0.6) Sqrt(2 32 y) ft/sec or

dy/dt = dV/dt/ (9 Pi) = -1/144 (0.6) 8 Sqrt(y) / (9 Pi)
= -4.8 Sqrt(y)/ (144*9 Pi) .

We solve this initial value problem, with y(0) = 8:

Input := 

sol = DSolve[{y'[t]== -4.8Sqrt[y[t]]/ (144*9 Pi),
			y[0]==8}, y[t],t]

Output =

                        0.00185185 t 2
{{y[t] -> (-2 Sqrt[2] - ------------) }, 
                             Pi
 
                        0.00185185 t 2
  {y[t] -> (2 Sqrt[2] - ------------) }}
                             Pi

Since y decreases, the second solution is the one we want:

Input := 

s[t_] = y[t]/.sol[[2]]

Output =

             0.00185185 t 2
(2 Sqrt[2] - ------------)
                  Pi

To find when the tank is empty, we solve:

Input := 

solt = Solve[s[t]==0,t]

Output =

{{t -> 4798.31}, {t -> 4798.31}}

We examine a plot of the graph of s(t).

Input := 

sPlot = Plot[s[t],{t,0,5000}]

Output =

-Graphics-

Therefore the tank is empty in 1.33 hrs.

Input := 

4798.31/3600

Output =

1.33286

Situation 2

Fix a certain time, and suppose the height of the water in the tank is y feet deep at that time.

Let x be a variable which varies from 0 to y, and consider a small piece of the slit dx high. That is, consider a hole at height x of area dx*1/(120) square feet.

Then the change in volume through that hole is

±0.6 * Sqrt( 232(y-x)) /120 dx,

where y - x is the height of the water above the hole.

To get the total change in volume through the entire slit, we integrate the last expression with respect to x from 0 to y:

Input := 

Integrate [-3/5*Sqrt[2*32*(y-x)]/120,{x,0,y}]

Output =

    3/2
-2 y
-------
  75

That is, dV/dt = -2/75 y ^(3/2). As in Situation 1, we replace dV/dt with 9Pi dy/dt to get the initial value problem:

dy/dt = -2/75 y ^(3/2)/(9 Pi), y(0) = 8

And we solve the initial value problem.

Input := 

sol1 = DSolve[{y'[t]==-2/75 y[t]^(3/2)/(9Pi),y[0]==8},
				y[t],t]

Output =

              -1         t    -2
{{y[t] -> (--------- + ------)  }, 
           2 Sqrt[2]   675 Pi
 
                1         t    -2
  {y[t] -> (--------- + ------)  }}
            2 Sqrt[2]   675 Pi

We plot both solutions so we can determine which is valid:

Input := 

s1[t_] = y[t]/.sol1[[1]]; s2[t_] = y[t]/.sol1[[2]];

Input := 

Plot[s1[t],{t,0,5000}]

Output =

-Graphics-

Input := 

s2Plot = Plot[s2[t],{t,0,5000},
				PlotStyle->Thickness[.01]]

Output =

-Graphics-

Since the height of the liquid, y, must decrease, we choose s2(t) and we notice that s2(t) is asymptotic to y = 0, but never yields a solution to s2(t) = 0.

Input := 

Solve[s2[t]==0,t]

Output =

{}

Input := 

Plot[s2,{t,0,10000}]

Output =

-Graphics-

Indeed, using the Limit command we see the asymptotic behaviour.

Input := 

Limit[s2,t->Infinity]

Output =

0

We now examine the plots from Situation 1 (light line) and Situation 2 (dark line) to compare them.

Input := 

Show[sPlot,s2Plot]

Output =

-Graphics-

It appears that in Situation 2 the liquid drops faster at first but then slows down its flow when we are near the bottom of the tank.

Situation 3

There is no practical answer to this problem.

The average pressure and the total area of the hole both decrease, so the rate of change of volume (and height) must necessarily decrease.

We can demonstrate this mathematically as follows.

The slit in Situation 2 was 1/120 ft wide. We replace this number by an unknown function w(y) which tells the width of the slit at height y.

Then as in Situation 2 we have:

The change in volume though that hole is ±0.6 * Sqrt( 232(y-x))w(x) dx,
where y-x is the height of the water above the hole. To get the total change in volume through the entire slit, we integrate the last expression with respect to x from 0 to y, but since w(x) is unkown, we suppose that F(x) is an anti-derivative of the integrand, so

Integrate [-3/5Sqrt[232(y-x)]/120,{x,0,y}] = F(y) - F(0),

That is, dV/dt = 9 Pi dy/dt = F(y) - F(0). We want dy/dt to be constant,
so we set dy/dt = (F(y) - F(0)) / (9 Pi) =k. Now if we differentiate with
respect to t, we have (F'(y) dy/dt ) / (9 Pi) = 0, but since dy/dt = k, we have F'(y) = 0. Since y varies from 0 to 8, and F(x) is an antiderivative for
±0.6 Sqrt( 232(y-x))w(x), we conclude that the latter equals zero.

POSSIBLE SOLUTION(S)

Situation 1

If the tank is y feet full, then the volume of water is given by V = Pi 3^2 y = 9 Pi y, so dV/dt = 9 Pi dy/dt. On the other hand, the change in volume with respect to time is the area of the hole times the velocity of the water squirting through the hole, therefore,

dV/dt = -(area)(velocity) = -1/ 144 ft ^2 (0.6) Sqrt(2 32 y) ft/sec or dy/dt = dV/dt/ (9 Pi) = -1/144 (0.6) 8 Sqrt(y) / (9 Pi) = -4.8 Sqrt(y)/ (144*9 Pi) .

We solve this initial value problem, with y(0) = 8:

Since y decreases, the second solution is the one we want:

To find when the tank is empty, we solve:

We examine a plot of the graph of s(t).

Therefore the tank is empty in 1.33 hrs.

Situation 2

Fix a certain time, and suppose the height of the water in the tank is y feet deep at that time.

Let x be a variable which varies from 0 to y, and consider a small piece of the slit dx high. That is, consider a hole at height x of area dx*1/(120) square feet.

Then the change in volume through that hole is ±0.6 * Sqrt( 2*32*(y-x)) /120 dx, where y - x is the height of the water above the hole.

To get the total change in volume through the entire slit, we integrate the last expression with respect to x from 0 to y:

That is, dV/dt = -2/75 y ^(3/2). As in Situation 1, we replace dV/dt with 9Pi dy/dt to get the initial value problem: dy/dt = -2/75 y ^(3/2)/(9 Pi), y(0) = 8

And we solve the initial value problem.

We plot both solutions so we can determine which is valid:

Since the height of the liquid, y, must decrease, we choose s2(t) and we notice that s2(t) is asymptotic to y = 0, but never yields a solution to s2(t) = 0.

Indeed, using the Limit command we see the asymptotic behaviour.

We now examine the plots from Situation 1 (light line) and Situation 2 (dark line) to compare them.

It appears that in Situation 2 the liquid drops faster at first but then slows down its flow when we are near the bottom of the tank.

Situation 3

There is no practical answer to this problem.

The average pressure and the total area of the hole both decrease, so the rate of change of volume (and height) must necessarily decrease.

We can demonstrate this mathematically as follows.

The slit in Situation 2 was 1/120 ft wide. We replace this number by an unknown function w(y) which tells the width of the slit at height y.

Then as in Situation 2 we have:

Therefore we must have w(x) = 0, which is the only function that allows a constant rate of drainage, that is, none.

If the tank is y feet full, then the volume of water is given by
V = Pi 3^2 y = 9 Pi y, so dV/dt = 9 Pi dy/dt. On the other hand, the change in volume with respect to time is the area of the hole times the velocity of the water squirting through the hole, therefore,

dV/dt = -(area)(velocity) = -1/ 144 ft ^2 (0.6) Sqrt(2 32 y) ft/sec or

dy/dt = dV/dt/ (9 Pi) = -1/144 (0.6) 8 Sqrt(y) / (9 Pi)
= -4.8 Sqrt(y)/ (144*9 Pi) .

Then the change in volume through that hole is

±0.6 * Sqrt( 232(y-x)) /120 dx,

where y - x is the height of the water above the hole.

That is, dV/dt = -2/75 y ^(3/2). As in Situation 1, we replace dV/dt with 9Pi dy/dt to get the initial value problem:

dy/dt = -2/75 y ^(3/2)/(9 Pi), y(0) = 8

Therefore we must have w(x) = 0, which is the only function that allows
a constant rate of drainage, that is, none.