query_04

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course MTH 279

query 042.5.

1. A 3% saline solution flows at a constant rate into a 1000-gallon tank initially full of a 5% saline solution.

The solutions remain well-mixed and the flow of mixed solution out of the tank remains equal to the flow into the tank.

What constant rate of flow is necessary to dilute the solution in the tank to 3.5% in 8 hours?

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Your solution:

We know that the Concentration in, C_i = 3%. The volume of the tank, v(t) = 1000 gallons.

The amount of salt initially is at 5%, Q(0) = 5%. The rate in is equal to the rate out, r_i = r_o.

We want to find what the flow rate is necessary to dilute the solution to 3.5% in 8 hours, rather what is the rate equal to when Q(8) = 3.5%.

We are looking for dQ/dt = (C_i)(r_i) - (Q(t)/V(t))(r_o). Plugging in our known facts, and assuming r_i = r_o = 1, we can simplify the equation to

dQ/dt = (3%)(1) - (Q(t)/1000)(1). Simplifying further we get dQ/dt = 3% - Q(t)/1000.

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Good so far.

For comparison:

Let q be the amount of salt in the tank, r the common inflow and outflow rate.

Then the rate at which salt flows into the tank is .03 r, the concentration is q / 1000 and the rate at which salt flows out is r * (q / 1000), giving us a net rate of

rate of change of amount of salt in tank = .03 r - (r / 1000) * q

The rate of change of the amount of salt in the tank is dq/dt, giving us the equation

dq/dt = .03 r - (r/1000) * q.

The only difference between this and your equation is the factor r on the right-hand side, which is easily dealt with.

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Our IVP is now: dQ/dt = 3% = dQ/dt + Q(t)/1000, with initial condition Q(0) = 5%.

dQ/dt = 3% - 3.5%/1000, so our rate, dQ/dt = 0.29965

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Now you have to solve the equation, which is a first-order linear nonhomogeneous equation.

The equation is rearranged to

dq/dt + (r/1000) * q = .03 r.

Can you complete the solution?

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): Something seems wrong...I thought I was doing okay up until I found the IVP..

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Self-critique rating:

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Question: 2. Solve the preceding question if the tank contains 500 gallons of 5% solution,

and the goal is to achieve 1000 gallons of 3.5% solution at the end of 8 hours.

Assume that no solution is removed from the tank until it is full, and that once the tank is full,

the resulting overflow is well-mixed.

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Your solution:

Now our volume v = 500 and the in concentration c_i = 5%. We are looking for the rate, dQ/dt that takes for

Q(8) = 3.5%.

Now knowing our initial conditions, we plug in: dQ/dt = (5%)(1) - (Q(t)/500)(1) = 5 - Q(t)/500.

Solving dQ/dt = 5% - 3.5%/500, so our rate dQ/dt = 3.493.

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Once the tank becomes full, which it does in time 500 / r (it requires 500 additional gallons, with solution flowing in at rate r), the equation will again be as before:

q ' + (r / 1000) * q = .03 r.

However the initial conditions will be different, since 500 gallons of one solution has been added to the original.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): Wouldn't it be easier if the concentration was in weight (pounds) instead of percentages?

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Units are a little more straightforward if all quantities are measured in the same units, in this case gallons.

Of course gallons can convert easily enough to pounds, if you have the right information, so it could be done either way.

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Question: 3. Under the conditions of the preceding question, at what rate must 3% solution be pumped into the tank,

and at what rate must the mixed solution be pumped from the tank, in order to achieve 1000 gallons of 3.5% solution at

the end of 8 hours, with no overflow?

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Your solution:

c_i = 3%

V = 1000

Q(8) = 3.5%

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): I'm not sure where to even start....

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Question: 4. Under the conditions of the first problem in this section, suppose that the overflow from the first tank

flows into a second tank, where it is mixed with 3% saline solution. At what constant rate must the 3% solution flow

into that tank to achieve a 4% solution at the end of 8 hours?

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Your solution:

The percentages are confusing me...How do we figure out what is flowing in?!

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Figure it in gallons.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Question: 5. In the situation of Problem #1, suppose that solution from the first tank is pumped at a constant

rate into the second, with overflow being removed, and that the process continues indefinitely. Will the concentration

in the second tank approach a limiting value as time goes on? If so what is the limitng value? Justify your answer.

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Yes there will be a limiting value. Solving the IVP dQ/dt = 3% = dQ/dt + Q(t)/1000, with initial condition Q(0) = 5%, the limiting value will be whatever the limit as t->infinity of Q(t) is.

So first we will solve for Q(t). p(t) = 1/1000, e(t) = e^t/1000.

The (integral of)(e^t/1000 * Q(t))'dt = (integral of)e^t/1000 dt. This simplifys to e^t/1000(Q(t)) = 1000e^t/1000 + C.

Using our initial condition Q(0) = 5, we can solve for C: 1000 + Ce^-t/1000 = 1000 + Ce^0 = 5 -> C = -955.

Now to calculate the limit, we do the limit(t->infinity)(1000-955e^-t/1000) = 1000.

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Now suppose that the flow from the first tank changes hour by hour, alternately remaining at a set constant rate for

one hour, and dropping to half this rate for the next hour before returning to the original rate to begin the two-hour

cycle all over again. Will the concentration in the second tank approach a limiting value as time goes on?

If so what is the limiting value? Justify your answer.

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?!?

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Answer the same questions, assuming that the rate of flow into (and out of) the tank is 10 gallons / hour * ( 3 - cos(t) ),

where t is clock time in hours.

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Your solution:

NO IDEA HOW TO SOLVE THIS QUESTION!?!

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): Confused how we incorporate the flow rate change...

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Question: 6. When heated to a temperature of 190 Fahrenheit a tub of soup, placed in a room at constant temperature

80 Fahrenheit, is observed to cool at an initial rate of 0.5 Fahrenheit / minute.

If at the instant the tub is taken from the oven the room temperature begins to fall at a constant rate of 0.25

Fahrenheit / minute, what temperature function T(t) governs its temperature?

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Your solution:

The temperature of the object at time t = O(t).

The temperature of the surroundings at time t = S(t).

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The room is at constant temperature, which simplifies this.

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Therefore from our problem we know that O(t) = 190 and S(t) = 80. The rate of cooling is 0.5 Fahrenheit/minute. We know that

O'(t) = k(S(t)-O(t)) where k is a constant. Since S < O, we know that k > 0. and the temperature of the object is decreasing,

so O'(t) < 0.

Here we need to find what happens what the rate of cooling is 0.25 Fahrenheit/minute.

Using our initial problem, we can find k. 0.5 = k(80-190) simplifys to 0.5 = k(-110) shows us that k = 0.0045.

Using our new conditions, we find that 0.25 = 0.0045(190-n), our new room temperature is 134 degrees.

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This will work.

So our equation is

dT / dt = .0045 * (T - S),

where S is the constant temperature of the room.

Can you solve this equation for T(t)?

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): I'm not sure that I am solving for what you are asking for?

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Self-critique rating:

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Question: "

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The mixing problems increase in complexity. You have a good start on the first problem.

Check my notes and see what progress you can make.

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