A tank initially contains 120 L of pure water. A mixture containing a concentration of γ g/L of salt enters the tank at a rate of 2 L/min, and the well-stirred mixture leaves the tank at the same rate. Find an expression in terms of γ for the amount of salt in the tank at any time t. Also ﬁnd the limiting amount of salt in the tank as t→∞

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diskusje5 · Accepted Answer

Step 1Let&amp;nbsp;Q(t)&amp;nbsp;salt content of tank at time t, bethe initial salt content is&amp;nbsp;Q(0)=0The differential equation that describes this issue is&amp;nbsp;dQ&amp;nbsp;dt&amp;nbsp;=rate∈−rateoutrate∈=2γrateout=Q(t)120⋅2Consequently, the problem of beginning value isdQ&amp;nbsp;dt&amp;nbsp;=2γ−Q60Q(0)=0Step 2Division with differential equation&amp;nbsp;Q−120γ&amp;nbsp;to obtain the standard separable equation.This is acceptable for everyone.&amp;nbsp;Q≠120&amp;nbsp;dy&amp;nbsp;&amp;nbsp;dx&amp;nbsp;Q−120γ=−160Integrate both parties in relation to tKeeping in mind the chain rule, you should be aware that the integration of the left side may be expressed as simply &quot;integration with respect to Q.&quot;∫1Q−120γdQ=∫(−160)&amp;nbsp;dt&amp;nbsp;ln⁡midQ−120γmid+C1=−t60+C2midQ−120γmid=e−t60eC2−C1Since&amp;nbsp;C1&amp;nbsp;and&amp;nbsp;C2&amp;nbsp;are constants&amp;nbsp;eC2−C1&amp;nbsp;is a positive constant. Call it CThe right side of the equation is positive for all t, as you can see.Notice that&amp;nbsp;Q&amp;lt;120γ&amp;nbsp;(because&amp;nbsp;120γ&amp;nbsp;is max. amount of salt in the tank)120γ−Q=Ce−t60Q=120γ−Ce−t60Step 3Calculate constant C using the starting conditions.Q(0)=0120γ−Ce−060=0C=120γConsequently, the salt content&amp;nbsp;Q(t)&amp;nbsp;each moment t is in the tankQ(t)=120γ−120γe−t60Additionally, the tank&#039;s maximum salt content islimt→∞Q(t)=limt→∞(120γ−120γe−t60)=120γ

Vasquez · Accepted Answer

At time t=0, the tank contains 120 L of pure water, which means there is no salt in the tank initially. Therefore, S(0) = 0 g.Now, let&#039;s consider the rate at which salt enters the tank. The mixture entering the tank has a concentration of γ g/L of salt, and the rate of flow is 2 L/min. This means that the amount of salt entering the tank per minute is *2γ g/min*.Since the mixture entering the tank is well-stirred and leaves the tank at the same rate, the amount of salt leaving the tank per minute is also 2γg/min. Therefore, the rate of change of the amount of salt in the tank is given by:dSdt=(rate&amp;nbsp;of&amp;nbsp;salt&amp;nbsp;entering&amp;nbsp;the&amp;nbsp;tank)−(rate&amp;nbsp;of&amp;nbsp;salt&amp;nbsp;leaving&amp;nbsp;the&amp;nbsp;tank)dSdt=2γ−2γ=0We obtained 0 as the derivative of S(t) with respect to t. This implies that the amount of salt in the tank is constant over time. Therefore, the expression for the amount of salt in the tank at any time t is:S(t)=S(0)+∫0tdSdtdtS(t)=0+∫0t0dtS(t)=0Hence, the amount of salt in the tank at any time t is 0 g.Next, let&#039;s find the limiting amount of salt in the tank as t approaches infinity (t→∞). In this case, we can evaluate the integral from 0 to infinity:S(∞)=S(0)+∫0∞dSdtdtS(∞)=0+∫0∞0dtS(∞)=0Therefore, the limiting amount of salt in the tank as t approaches infinity is 0 g.

RizerMix · Accepted Answer

Answer:- The expression for the amount of salt in the tank at any time t is S(t)=120γ+2γt.- The limiting amount of salt in the tank as t→∞ is infinity.Explanation:Let t be the time in minutes.Let V(t) be the volume of water in the tank at time t in liters.Let S(t) be the amount of salt in the tank at time t in grams.Given that the tank initially contains 120 L of pure water, we can express V(t) as follows:V(t)=120+(2Lmin)·tThe concentration of salt in the mixture entering the tank is γ g/L. Since the mixture enters and leaves the tank at the same rate, the concentration of salt in the tank remains constant over time.Therefore, the amount of salt in the tank at any time t is given by:S(t)=γ·V(t)Substituting the expression for V(t), we have:S(t)=γ·(120+(2Lmin)·t)Simplifying this expression, we get:S(t)=120γ+2γtThus, the amount of salt in the tank at any time t is given by the expression 120γ+2γt.Now, let&#039;s determine the limiting amount of salt in the tank as t→∞. In the long run, the tank will continuously receive the mixture with concentration γ, while an equal volume of mixture leaves the tank. Therefore, the concentration of salt in the tank will stabilize at γ g/L.As t approaches infinity, the amount of salt in the tank can be calculated as:limt→∞S(t)=limt→∞(120γ+2γt)Since the term 2γt becomes infinitely large as t approaches infinity, the limiting amount of salt in the tank as t→∞ is infinity.

karton · Accepted Answer

To solve the given problem, we&#039;ll first find an expression for the amount of salt in the tank at any time t, and then determine the limiting amount of salt as t approaches infinity.Let&#039;s denote the amount of salt in the tank at time t as Q(t). Initially, the tank contains 120 L of pure water, so there is no salt present: Q(0)=0.The rate at which the mixture enters the tank is 2 L/min, and the concentration of salt in the entering mixture is γ g/L. This means that the amount of salt entering the tank per minute is 2γ g/min.The mixture is well-stirred, so the concentration of salt remains uniform throughout the tank. Since the mixture is leaving the tank at the same rate of 2 L/min, the amount of salt leaving the tank per minute is also 2γ g/min.Therefore, the rate of change of the amount of salt in the tank can be expressed as:dQdt={ratein}−{rate&amp;nbsp;out}=2γ−2γ=0This implies that the amount of salt in the tank remains constant over time, as there is no net change in the salt content.Hence, the expression for the amount of salt in the tank at any time t is:Q(t)=Q(0)=0The limiting amount of salt in the tank as t→∞ can be found by taking the limit of Q(t) as t approaches infinity:limt→∞Q(t)=limt→∞0=0Therefore, the limiting amount of salt in the tank as t→∞ is 0 g.

A tank initially contains 120 L of pure water. A mixture containing a concentrat

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