#$&* course Mth 279 2/19/14 query 042.5.
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: time to fill the take is t_1 = 500 / r saline in the tank will be Q(t_1) = Q(500 / r) = .05 * 500 + .035 * r * t_1 = .05 * 500 + .03 * r * 500 / r = .05 * 500 + .03 * 500 = 40 concentration is .04 .05 * 500 + .03 * 500
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):Not sure if I did it right ------------------------------------------------ Self-critique rating:3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: r_1 the rate at which solution flows in and r_2 the rate of outflow 500 gal * (r_1 - r_2) after 8 hours tank reaches 1000 gal capacity don’t know where to go from here
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: 4. Under the conditions of the first problem in this section, suppose that the overflow from the first tank flows into a large 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 3.5% solution at the end of 8 hours? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: dQ/dt = 0.03r(t) - Q(t)r(t)/1000 = r(t) (0.03 - Q(t)/1000) = (r(t)/1000) * (30 - Q(t)) = (-r(t)/1000) * (Q(t) - 30) dQ/(Q(t) - 30) = -r(t)/1000 dt integrating both sides yeilds ln(Q(t) - 30) = -r(t)*t/1000 + c Q(t) - 30 = C * e^(-r(t)*t/1000) Q(t) = 30 + Ce^(-r(t)*t/1000) Q(0) = 30 + Ce^0 = 0.05(1000) = 50 30 + C(1) = 50 C = 20 Changing this last part a bit: Q(t) = 30 + 20e^(-r(t) * t/1000) Q(8*60) = 30 + 20e^(-r(t) * 480/1000) = 0.035(1000) = 40 20e^(-r(t) * 480/1000) = 10 e^(-r(t) * 480/1000) = 10/20 e^(-r(t) * 480/1000) = 0.5 ln(0.5) = -r(t) * 480/1000 r(t) = - ln(0.5) / 0.48 r(t) = 1.444 gal/min approx.
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating: ********************************************* 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. **** As the time advances, the concentration would increase until they both reach the concentration of the amount being pumped. If not then one tank would reach it before the other. #$&* 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. **** The first tank would reach the concentration of what is being pumped into it first, the second would take twice as long #$&* Answer the same questions, assuming that the rate of flow into (and out of) the tank is 10 gallons / hour * ( 3 - cos(t) ), where t is clock time in hours. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): not sure how to set up ------------------------------------------------ Self-critique rating: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: dT / dt = (.05 / 110) / min * T_diff T '(t) = k(T_room - T(t)) T '(t) = k(80 - T(t)) integrate: T(t) = 80 + Ce^(-kt) T(o) = 190 190 = 80 + C C = 110 T(t) = 80 + 110e^(-kt)
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):Think I was headed in the right direction ------------------------------------------------ Self-critique rating:2 ********************************************* Question: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!