Sect2.3a - [SECTION 41-3! Section 2.3, Problem 2 Problem. A...

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Unformatted text preview: [SECTION 41-3! Section 2.3, Problem 2 Problem. A tank initially contains 120 liters of pure water. A mixture containing a concentration of 7% of Salt enters the tank at a rate of 2%, and the well—stirred mixture leaves the tank at the same rate. Find an expression in terms of 7 for the amount of salt in the tank at any time t. Also find the limiting amount ofsaltinthetankastaoo. I ‘ V Solution. One wants a model for the amount M (t) of salt in the tank at a time t. t should be measured in minutes, and M (t) should be measured in grams, as those are the given units. I ‘ Entering the tank are 'y[g] of salt, and leaving the tank are 1% -M[g] of salt. So an equation describing the change % of the amount of salt is: dM 2 7;: = +27“ m1” (1” V To solve this equation, which one notes resembles Example 1 in the book, it may be advantageous to rewrite it in the form 4% + %M = 27. Then one can (as in Section 2.1) use an integrating factor, Mt) = eéi. Multiplying all terms in the equation by u(t) turns the equation into eét% + gldefilfitM = 276%? By the product rule, the left side of the equation ‘is now the derivative, with respect to t, of eélfitM. The equation can be rewritten as $(efilfitM '= 276th. One can then integrate both sides of the equation with respect to t, turning the equation into e%tM (t) = 120‘yt'361_oIt + C, for some constant (3'. Then one divides all terms, including the constant C, by edit and so M (t) = 120 + Ce‘Elfit. C' is determined by using the known initial condition, in this case, that the tank starts with pure water. The amount of salt at time t '= 0 then is zero, or M(0) = 120 + 061160 = 120 + c = 0. So, 0' = ,—120. The amount of salt in the tank is M(t) = 120 —120e‘$t. ’ ' The limiting amount of salt in the tank, how much remains as t —-> 00, is the limit of .M (t) = 120 — 120e'6i0t as t grows infinitely large. As t grows, 120 will be unchanged, but 1206‘6'16t will become infinitessimally small — it will go to Zero. As t ——+ 00, then, the amount of salt in the tank M —> 120[g]. 2.7.39.1} ®= arm/um. a; game (le Volume of +m = [00 341. 0(07= O («en'sh wade» mug) 1‘8 powwd m at flab, Z 01 ' M 0K 3 /mm, and 4kg“wa .6: allow 40 [em/t d0 v = . .l\ / ,t (a 54.0 (2%) - kg; at)“ g; 59 ' dk - i " go ‘ K Vommp 084.4%) " onk 9.0635 125 Now flush wa~1m .03 pawd at at Deck :3 Zjd/mu'») tom/1 ‘1“? mm Manna of "Li/‘0 Same $0016} 3‘ 10 35(7‘ m.) "" fififiz 3.3;) cm. . Use, I.C’. +0 64nd C3: Olo)=‘i.o@3§ = C3 QM = (9.04735) 3%“. /312~Lm {O W [W anmwi cgsdt 4/» W 44111: L‘s: Gum (fioossye‘mA *- "’W <0 7' Sam 5;, deposded ck bank w+ page aimed at annual mule .— (03 Find ‘Hw kme T resouth for oust Sum +0 double . 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Sect2.3a - [SECTION 41-3! Section 2.3, Problem 2 Problem. A...

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