Question

In Example 5the size of the tank containing the salt mixture was not given. Suppose, as in the discussion following Example 5, that the rate at which brine is pumped into the tank is$\mathbf{3}\mathbf{}\mathbf{gal}\mathbf{/}\mathbf{min}$but that the well-stirred solution is pumped out at a rate of$\mathbf{2}\mathbf{}\mathbf{gal}\mathbf{/}\mathbf{min}$. It stands to reason that since brine is accumulating in the tank at the rate of$\mathbf{1}\mathbf{}\mathbf{gal}\mathbf{/}\mathbf{min}$, any finite tank must eventually overflow. Now suppose that the tank has an open top and has a total capacity of 400gallons. (a) When will the tank overflow? (b) What will be the number of pounds of salt in the tank at the instant it overflows? (c) Assume that although the tank is overflowing, brine solution continues to be pumped in at a rate of$\mathbf{3}\mathbf{}\mathbf{gal}\mathbf{/}\mathbf{min}$and the well-stirred solution continues to be pumped out at a rate of$\mathbf{2}\mathbf{}\mathbf{gal}\mathbf{/}\mathbf{min}$. Devise a method for determining the number of pounds of salt in the tank at$\mathbf{t}\mathbf{=}\mathbf{150}$minutes. (d) Determine the number of pounds of salt in the tank as$\mathbf{t}\mathbf{\to }\mathbf{\infty }$. Does your answer agree with your intuition? (e) Use a graphing utility to plot the graph ofon the interval$\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{500}\mathbf{)}$.

Short answer

(a) The tank begins to overflow after 100 minutes.

(b) The number of pounds of salt in the tank at the instant it overflows is $A=490.625\mathrm{lb}$.

(c) The number of pounds of salt in the tank at time is $A\left(150\right)\approx 587.37\mathrm{lb}$.

(d) The number of pounds of salt in the tank at time is $800\mathrm{lb}$.

(e) The graphing utility is,

Step-by-step solution

Define mixtures

The mixing of two fluids sometimes gives rise to a linear first-order differential equation

$\left.\frac{dA}{dt}=\left(\text{input rate of salt}\right)-\text{(output rate of salt}\right)=R_{\text{in}}-R_{\text{out}}$

Diagram of the situation

Let the diagram of the situation be,

Determine the time when the tank begins to overflow

(a)

Let$V\left(t\right)$ represent the volume of the brine solution in the tank at a given time t.

Rate of liquid entering tank:${\mathrm{v}}_{\mathrm{in}}=3\mathrm{gal}/\min$

Rate of liquid leaving tank:${\mathrm{v}}_{\mathrm{out}}=2\mathrm{gal}/\min$

$$\begin{gathered} V\left(t\right)=V_0+\left(v_{in}-v_{out}\right)t \\ V\left(t\right)=300+(3-2)t \\ V\left(t\right)=300+t \end{gathered}$$

The tank begins to overflow when $V\left(t\right)=400\mathrm{gal}$:

$\begin{array}{rcl}V\left(t\right)& =& 300+t\\ 400& =& 300+t\\ t& =& 100\min \end{array}$

The tank begins to overflow after 100 minutes.

Determine be the number of pounds of salt in the tank at the instant it overflows.

(b)

Let linear equation with the amount of salt in pounds as A be,

$\frac{dA}{dt}=R_{in}-R_{out}$ … (1)

Substitute the values${\mathrm{R}}_{\mathrm{in}}=(3\mathrm{gal}/\min )(2\mathrm{lb}/\mathrm{gal}),{\mathrm{R}}_{\mathrm{out}}=(4\mathrm{gallon}/\min )(\mathrm{A}/\mathrm{Vlb}/\mathrm{gal})$in the equation (1).

$$\begin{gathered} \frac{\mathrm{dA}}{\mathrm{dt}}=\left(3\frac{\mathrm{gal}}{\min }\right)\left(2\frac{\mathrm{lb}}{\mathrm{gal}}\right)-\left(2\frac{\mathrm{gal}}{\min }\right)\left(\frac{\mathrm{A}}{\mathrm{V}}\frac{\mathrm{lb}}{\mathrm{gal}}\right) \\ \frac{\mathrm{dA}}{\mathrm{dt}}=\left(6\frac{\mathrm{lb}}{\min }\right)-\left(\frac{2\mathrm{A}}{\mathrm{V}}\frac{\mathrm{lb}}{\min }\right) \end{gathered}$$

Remember:$V\left(t\right)=300+t$

$\begin{array}{rcl}\frac{\mathrm{dA}}{\mathrm{dt}}& =& 6-\frac{2\mathrm{A}}{300+\mathrm{t}}\\ \frac{\mathrm{dA}}{\mathrm{dt}}+\frac{2}{300+\mathrm{t}}\mathrm{A}& =& 6\end{array}$

Solve using Integrating factor:

$\begin{array}{rcl}\mu & =& e^{\int \frac{2}{300+t}dt}\\ \mu & =& e^{2ln|300+t|}\\ \mu & =& e^{ln{(300+t)}^2}\\ \mu & =& (300+t{)}^2\\ \frac{d}{dt}\left[A\times {(300+t)}^2\right]& =& 6(300+t{)}^2\\ A(300+t{)}^2& =& 6\int {(300+t)}^{2}dt\\ A(300+t{)}^2& =& 6\int \left({300}^2+600t+t^2\right)dt\end{array}$

$A(300+t{)}^2=6\left({300}^{2}t+300t^2+\frac{1}{3}{t}^3+C\right)$… (3)

To find the values of constants, apply the point $A\left(0\right)=50$in the equation (3), then

$\begin{array}{rcl}50(300+0{)}^2& =& 6\left({300}^2\times 0+300\times 0^2+\frac{1}{3}\times 0^3+C\right)\\ 50\times {300}^2& =& 6C\\ 50\times 300\times 300& =& 6C\\ C& =& {50}^2\times 300\end{array}$

Substitute the value of in the equation (3).

$A(300+t{)}^2=6\left({300}^{2}t+300t^2+\frac{1}{3}{t}^3+{50}^2\times 300\right)$… (4)

Substitute the value of$\text{t = 100 minutes}$ in the equation (4).

$\begin{array}{rcl}50(300+0{)}^2& =& 6\left({300}^2\times 0+300\times 0^2+\frac{1}{3}\times 0^3+C\right)\\ 50\times {300}^2& =& 6C\\ 50\times 300\times 300& =& 6C\\ C& =& {50}^2\times 300\end{array}$

Determine be the number of pounds of salt in the tank at the given time.

(c)

Let the volume of liquid remains a constant $V\left(t\right)=400$. If the liquid is being pumped in at a rate of then the liquid will be pumped out at a rate of $3\mathrm{gal}/\min$.

$V\left(t\right)=400+(3-3)t$

$$\begin{gathered} \frac{\mathrm{dA}}{\mathrm{dt}}={\mathrm{R}}_{\mathrm{in}}-{\mathrm{R}}_{\mathrm{out}}\backslash \mathrm{hfill} \\ \frac{\mathrm{dA}}{\mathrm{dt}}=\left(3\frac{\mathrm{gal}}{\min }\right)\left(2\frac{\mathrm{lb}}{\mathrm{gal}}\right)-\left(3\frac{\mathrm{gal}}{\min }\right)\left(\frac{\mathrm{A}}{\mathrm{V}}\frac{\mathrm{lb}}{\mathrm{gal}}\right) \\ \frac{\mathrm{dA}}{\mathrm{dt}}=6-\frac{3\mathrm{A}}{400} \\ \frac{\mathrm{dA}}{\mathrm{dt}}=\frac{2,400-3\mathrm{A}}{400} \end{gathered}$$

Let use the substitution method to solve the DE.

$\int \frac{1}{2,400-3A}dA=\int \frac{1}{400}dt$

Let,

$\begin{array}{rcl}U& =& 2,400-3A\\ du& =& -3dA\\ dA& =& -\frac{du}{3}\\ & & \end{array}$

$\begin{array}{rcl}-\frac{1}{3}{\int }_1\frac{1}{u}du& =& {\int }_t\frac{1}{400}dt\\ -\frac{1}{3}ln\left|u\right|& =& \frac{C}{400}+C\\ -\frac{1}{3}ln|2,400-3A|& =& \frac{t}{400}+C\\ ln|2,400-3A|& =& -\frac{3t}{400}+C\\ ln|2,400-3A|& =& -0.0075t+C\end{array}$

$\begin{array}{rcl}2,400-3A& =& e^{-0.0075t+C}\\ 2,400-3A& =& e^{-0.0075t}{e}^C\\ 2,400-3A& =& C_1{e}^{-0.0075t}\\ A& =& 800-C_1{e}^{-0.0075t}\end{array}$

Apply the condition$A\left(100\right)=490.625$ to find the value of constant.

$\begin{array}{rcl}490.625& =& 800-C_1{e}^{-0.0075\left(100\right)}\\ -309.375& =& -C_1{e}^{-0.75}\\ C_1& =& 309.375e^{0.75}\\ C_1& \approx & 654.947\end{array}$

Then, the equation becomes,

$\begin{array}{rcl}A\left(t\right)& =& 800-654.947e^{-0.0075t}\\ A\left(150\right)& =& 800-654.947e^{-0.0075\left(150\right)}\\ A\left(150\right)& =& 800-654.947e^{-1.125}\\ A\left(150\right)& \approx & 587.37lb\end{array}$

(d)

Let,

$$\begin{gathered} =\underset{t\to \infty }{\lim }\left[800-654.947e^{-0.0075t}\right] \\ =\underset{t\to \infty }{\lim }\left[800-\frac{654.947}{e^{0.0075t}}\right] \\ =800\mathrm{lb} \end{gathered}$$

Determine the graphing utility

Let the graph be,