Question

salt water of concentration 0.5 pound of salt per gallon flows into a large tank that initially contains 40 gallons of pure water. (a) if the flow rate of salt water into the tank is 5 gal/min, find the volume v(t) of water and the amount a(t) of salt in the tank after t minutes. v(t) = gal a(t) = lb (b) find a formula for the salt concentration c(t) (in lb/gal) after t minutes. c(t) = lb/gal (c) discuss the variation of c(t) as t → ∞. as t → ∞, c(t) → 0.5 (lb/gal). as t → ∞, c(t) → ∞ (lb/gal). as t → ∞, c(t) → 0 (lb/gal). as t → ∞, c(t) → 2.5 (lb/gal).

Explanation:

Step1: Find volume formula

The tank initially has 40 gallons of water and salt - water flows in at a rate of 5 gal/min. So the volume of water $V(t)$ is the initial volume plus the volume that has flowed in. Using the formula $V(t)=V_0 + rt$, where $V_0 = 40$ gallons and $r = 5$ gal/min, we get $V(t)=40 + 5t$.

Step2: Find amount of salt formula

The salt - water has a concentration of 0.5 lb/gal and flows in at a rate of 5 gal/min. The rate of salt entering the tank is $0.5\times5=2.5$ lb/min. Since there is no salt initially in the tank ($A(0) = 0$), the amount of salt $A(t)$ as a function of time $t$ is given by the rate of salt entering times $t$, so $A(t)=2.5t$.

Step3: Find concentration formula

The concentration $c(t)$ is the amount of salt divided by the volume of water in the tank. So $c(t)=\frac{A(t)}{V(t)}=\frac{2.5t}{40 + 5t}$.

Step4: Analyze limit as $t\to\infty$

We find $\lim_{t\to\infty}c(t)=\lim_{t\to\infty}\frac{2.5t}{40 + 5t}$. Divide both the numerator and denominator by $t$: $\lim_{t\to\infty}\frac{2.5}{\frac{40}{t}+5}$. As $t\to\infty$, $\frac{40}{t}\to0$. So $\lim_{t\to\infty}c(t)=\frac{2.5}{5}=0.5$.

Answer:

$V(t)=40 + 5t$ gal
$A(t)=2.5t$ lb
$c(t)=\frac{2.5t}{40 + 5t}$ lb/gal
As $t\to\infty$, $c(t)\to0.5$ (lb/gal)