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: Find the amount of salt in the tank at time (t)

The rate at which salt is entering the tank is (50\times3 = 150) g/min. The amount of salt (S(t)) in the tank at time (t) (using the formula for the amount of a substance in a mixing problem (S(t)=S_{in}(t)), since there is no outflow in this case as the tank is "very large" and we assume no - fluid outflow for the concentration formula derivation) is (S(t)=150t). The volume of the fluid in the tank at time (t) is (V(t)=1000 + 3t).

Step2: Find the concentration formula (C(t))

The concentration (C(t)=\frac{S(t)}{V(t)}). Substitute (S(t)=150t) and (V(t)=1000 + 3t) into the formula: (C(t)=\frac{150t}{1000 + 3t}).

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

We use the fact that for a rational function (\lim_{t\rightarrow\infty}\frac{at + b}{ct + d}=\frac{a}{c}) (when (c\neq0)). For (C(t)=\frac{150t}{1000+3t}), we can rewrite it as (C(t)=\frac{150}{3+\frac{1000}{t}}) (divide numerator and denominator by (t)). Then (\lim_{t\rightarrow\infty}C(t)=\lim_{t\rightarrow\infty}\frac{150}{3+\frac{1000}{t}}). As (t\rightarrow\infty), (\frac{1000}{t}\rightarrow0). So (\lim_{t\rightarrow\infty}C(t)=\frac{150}{3}=50).

Answer:

A. (C(t)=\frac{150t}{1000 + 3t}) B. (\lim_{t\rightarrow\infty}C(t) = 50). This means that as (t) (time) gets very large, the concentration of salt in the tank approaches the concentration of the brine water that is being pumped into the tank. In other words, after a long - period of time, the concentration of salt in the tank will be approximately equal to the concentration of the incoming brine.