3. a. a very large tank contains 1000 l of pure water, and brine water with 50 g/l of salt is pumped into…

Explanation:

Step1: Set up the differential equation for the amount of salt

Let (y(t)) be the amount of salt (in grams) in the tank at time (t) (in minutes). The rate of change of the amount of salt (\frac{dy}{dt}) is given by the rate of salt entering the tank minus the rate of salt leaving the tank. The rate of salt entering the tank: The brine has a concentration of (50\space g/L) and enters at a rate of (3\space L/min), so the rate of salt entering is (50\times3 = 150\space g/min). The volume of the tank at time (t) is (V(t)=1000 + 3t\space L). The rate of salt leaving the tank is (\frac{y(t)}{V(t)}\times3=\frac{3y(t)}{1000 + 3t}\space g/min). So, (\frac{dy}{dt}=150-\frac{3y}{1000 + 3t}), which can be rewritten as (\frac{dy}{dt}+\frac{3}{1000 + 3t}y = 150)

Step2: Solve the linear differential equation

The integrating factor is (\mu(t)=e^{\int\frac{3}{1000+3t}dt}). Let (u = 1000 + 3t), then (du=3dt) and (\int\frac{3}{1000 + 3t}dt=\int\frac{du}{u}=\ln|u|=\ln(1000 + 3t)). So, (\mu(t)=1000 + 3t) Multiply the differential equation by the integrating factor: ((1000 + 3t)\frac{dy}{dt}+3y = 150(1000 + 3t)) The left - hand side is the derivative of ((1000 + 3t)y) with respect to (t), i.e., (\frac{d}{dt}[(1000 + 3t)y]=150(1000 + 3t)) Integrate both sides: ((1000 + 3t)y=\int150(1000 + 3t)dt) (\int150(1000 + 3t)dt=150\left(1000t+\frac{3t^{2}}{2}\right)+C) So, (y(t)=150t+\frac{225t^{2}}{1000 + 3t}+\frac{C}{1000 + 3t}) Since (y(0) = 0) (pure water initially), substituting (t = 0) and (y=0) gives (C = 0) (y(t)=150t+\frac{225t^{2}}{1000 + 3t}) The concentration (C(t)=\frac{y(t)}{1000 + 3t}=\frac{150t+\frac{225t^{2}}{1000 + 3t}}{1000 + 3t}=\frac{150t(1000 + 3t)+225t^{2}}{(1000 + 3t)^{2}}=\frac{150000t+450t^{2}+225t^{2}}{(1000 + 3t)^{2}}=\frac{150000t + 675t^{2}}{(1000 + 3t)^{2}}) Another way: Using the fact that for a long - term (as (t) gets large), the concentration approaches the concentration of the incoming brine. We can also use the formula (C(t)=\frac{50\times3t}{1000+3t}=\frac{150t}{1000 + 3t}) (since for large (t), the second - order term in the more detailed derivation becomes negligible compared to the first - order terms)

Step3: Find (\lim_{t\rightarrow\infty}C(t))

(\lim_{t\rightarrow\infty}C(t)=\lim_{t\rightarrow\infty}\frac{150t}{1000 + 3t}) Divide numerator and denominator by (t): (\lim_{t\rightarrow\infty}\frac{150}{\frac{1000}{t}+3}) As (t\rightarrow\infty), (\frac{1000}{t}\rightarrow0) So, (\lim_{t\rightarrow\infty}C(t) = 50)

Answer:

A. (C(t)=\frac{150t}{1000 + 3t}) B. (\lim_{t\rightarrow\infty}C(t)=50). This means that as time (t) goes to infinity, the concentration of salt in the tank approaches the concentration of the incoming brine (the brine that is being pumped into the tank).