Question

at any time i. Also find the limiting amount of salt in the tank - A tank contains 100 gal of water and 50 oz of salt. Water containing a salt concentration of (1 + i sin t) oz/gal flows into the tank at a rate of 2 gal/min, and the mixture in the tank flows out at the same rate. (a) Find the amount of salt in the tank at any time. (b) Plot the solution for a time period long enough so that you see the ultimate behavior of the graph. (c) The long-time behavior of the solution is an oscillation about a certain constant level. What is this level? What is the amplitude of the oscillation?

Answer 1

9 50 25- 1- t→ © The amplitude is the (absolute value of lost the) Coefficient of cost : is amplitude = - 1250 à à 0.24 0.2499. 5002

wer Qlt) be the amount of Solt of time t. Q(O) = 50 0Z, dt 100 Rate Equation & dQ = Rate in - Roke out. => dende = (salt concentraties in) (firw pote )– (tank Soft Concentration) (fitow roote out). ► den = # (1+sint). 2. - see 3 ade t = ž (1+& sint) - ► den + = £ (4+ sost) This is a differential eq? of the forme n +b la y aceptes -- - Integracy factora IF= Skoda poft & Sat60 The general solution is then =gree) → IF = +150 QC) = 1S (4+4582) 2 450 at 3&49 = 4o [5 esto (2* *.*me) at +c]

+501 +-750 → QLV)= -lita sist) dt + 7150 160 e t50 +150 - 25€ (50 cost-sint) + 25 € (t) = 1501 5002 7150 -1250 cost +255 int +25 + - → Q(u) = 5002 etto e At t=0, Q10)=50 i 50 = -- 1250 + 25 + c 5002 * CE 69150 2501 Hence IQ (t) = --1250 cost +255 int - 5002 +25 + e 150 [2501) --- I. from . After long time, the average value of cost and Sint becomes zero, and 150 quen to zero. Hence, the graph appears to temes off at about the rolue of Q=25

9 50 25- 1- t→ © The amplitude is the (absolute value of lost the) Coefficient of cost : is amplitude = - 1250 à à 0.24 0.2499. 5002