Chapter 7.3, Problem 48E

a.

To determine

To find: How much salt in the tank after $t$ minutes.

Answer to Problem 48E

After $t$ minutes, the amount of salt in the tank is $y=\frac{130}{3}\left(1-e^{\frac{-3t}{200}}\right)$

Explanation of Solution

Concept used :

A tank contains $1000$ L of pure water. Brine that contains $0.05$ Kg of salt per liter of water enters the tank at a rate of $5\text{L}/\text{min}\text{.}$ . Brine that contains $0.04$ Kg of salt per liter of water enters the tank at a rate of $10\text{L}/\text{min}\text{.}$ The solution kept thoroughly mixed and drains form the tank at a rate of $15\text{L}/\text{min}\text{.}$ .

Formula used :

  $\frac{dy}{dt}=\left(\text{rate in}\right)-\left(\text{rate out}\right)$

Calculation:

Let, $y\left(t\right)$ is the amount of salt after $t$ minutes.

  $\frac{dy}{dt}=\left(\text{rate in}\right)-\left(\text{rate out}\right)$

(Rate in) is the rate in which salt enters the tank and (rate out) is the rate at which salt leaves the tank.

  $\begin{array}{c}\text{rate in }=0.05\frac{Kg}{L}\times 5\frac{L}{\min }+0.04\frac{Kg}{L}\times 10\frac{L}{\min }\\ =0.25\frac{Kg}{\min }+0.4\frac{Kg}{\min }\\ =0.65\frac{Kg}{\min }\end{array}$

  $\begin{array}{c}\text{rate out }=\frac{y}{1000}\frac{Kg}{L}\times 15\frac{L}{\min }\\ =\frac{15y}{1000}\frac{Kg}{\min }\end{array}$

So, $\frac{dy}{dt}=0.65-\frac{15y}{1000}$

  $\begin{array}{l}\frac{dy}{dt}=0.65-\frac{15y}{1000}\\ \Rightarrow \frac{dy}{dt}=0.65-\frac{3y}{200}\\ \Rightarrow \frac{dy}{dt}=\frac{130-3y}{200}\end{array}$

  $\begin{array}{l}\Rightarrow \frac{dy}{dt}=\frac{130-3y}{200}\\ \Rightarrow \frac{dy}{130-3y}=\frac{dt}{200}\end{array}$

Integrating both sides, we get

  $\begin{array}{l}{\displaystyle \int \frac{dy}{130-3y}}={\displaystyle \int \frac{dt}{200}}\\ \Rightarrow -\frac{1}{3}\ln \left|130-3y\right|=\frac{t}{200}+C\\ \Rightarrow \ln \left|130-3y\right|=\frac{-3t}{200}-3C\\ \Rightarrow \ln \left|130-3y\right|=\frac{-3t}{200}+D\left[D=-3C\right]\end{array}$

Now, $y\left(0\right)=0$

So, $D=\ln \left|130\right|$

So, we can write,

  $\begin{array}{l}\ln \left|130-3y\right|=\frac{-3t}{200}+\ln \left|130\right|\\ \Rightarrow \left|130-3y\right|=130e^{\frac{-3t}{200}}\end{array}$

Since, $y\left(t\right)$ is continuous, $y\left(0\right)=0$ and right side is never $0$ , we deduce that $\left(130-3y\right)$ is always positive. Thus, $\left|130-3y\right|=130-3y$

  $\begin{array}{l}130-3y=130e^{\frac{-3t}{200}}\\ \Rightarrow 3y=130-130e^{\frac{-3t}{200}}\\ \Rightarrow y=\frac{130}{3}\left(1-e^{\frac{-3t}{200}}\right)\end{array}$

Therefore, after $t$ minutes, the amount of salt in the tank is $y=\frac{130}{3}\left(1-e^{\frac{-3t}{200}}\right)$

b.

To determine

To find: How much salt in the tank after one hout.

Answer to Problem 48E

After one hour the amount of salt in the tank is $25.715Kg$ .

Explanation of Solution

Given information:

After $t$ minutes, the amount of salt in the tank is $y=\frac{130}{3}\left(1-e^{\frac{-3t}{200}}\right)$

Concept used:

We use the given equation and put $t=60$

Calculation:

  $\begin{array}{c}y=\frac{130}{3}\left(1-e^{\frac{-3\times 60}{200}}\right)\\ \approx 25.715\end{array}$

Therefore, after one hour the amount of salt in the tank is $25.715Kg$ .