19. [-/2 Points] DETAILS MY NOTES SCALCET9 2.6 .068 . A tank contains \( 9,000 \mathrm{~L} \) of pure water. Brine that contains 35 g of salt per liter of water is pumped into the tank at a rate of \( 25 \mathrm{~L} / \mathrm{min} \). (a) Find the concentration of salt after \( t \) minutes (in \( \mathrm{g} / \mathrm{L} \) ). \[ \mathrm{C}(\mathrm{t})=\square \mathrm{g} / \mathrm{L} \] (b) What happens to the concentration (in \( \mathrm{g} / \mathrm{L} \) ) as \( t \rightarrow \infty \) ? \( \mathrm{g} / \mathrm{L} \)

We start by letting S(t) be the total mass of salt (in grams) in the tank after t minutes. Since the tank initially contains pure water, S(0) = 0. Brine with a salt concentration of 35 g/L is entering at a rate of 25 L/min, so the salt is entering at a rate of   Rate in = 25 L/min × 35 g/L = 875 g/min. Because there is no mention of the mixture leaving the tank, the salt simply accumulates. Thus, the differential equation is   dS/dt = 875. Integrate with respect to time:   S(t) = 875t + C. Using the initial condition S(0) = 0, we find C = 0, so   S(t) = 875t (g). While salt is accumulating, the volume of fluid in the tank is also increasing. Initially, there are 9,000 L of water, and brine is entering at 25 L/min, so the volume at time t is   V(t) = 9,000 + 25t (L). The concentration of salt in the tank after t minutes is then   C(t) = S(t) / V(t) = 875t / (9000 + 25t) (g/L). This is the answer for part (a). We can simplify C(t) by factoring the denominator (notice 25 is a common factor):   9000 + 25t = 25(t + 360), and since 875 = 35 × 25, we can write   C(t) = (35 × 25t) / (25(t + 360)) = (35t) / (t + 360) (g/L). For part (b), we want to find the behavior of C(t) as t → ∞. Consider   lim₍t→∞₎ (35t)/(t + 360). Divide numerator and denominator by t:   lim₍t→∞₎ [35/(1 + 360/t)] = 35/1 = 35 (g/L). Thus, as t becomes very large, the concentration of salt in the tank approaches 35 g/L. Final Answers: (a) C(t) = (35t)/(t + 360) g/L. (b) As t → ∞, the concentration approaches 35 g/L.