Question

Determine the amount of salt in the tank at time tin Example 5if the concentration of salt in the inflow is variable and given by ${\mathbf{c}}_{\mathbf{in}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{+}\mathbf{sin}\mathbf{(}\mathbf{t}\mathbf{/}\mathbf{4}\mathbf{)}\mathbf{lb}\mathbf{/}\mathbf{gal}$. Without actually graphing,conjecture what the solution curve of the IVP should look like.Then use a graphing utility to plot the graph of the solution on the interval $\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{300}\mathbf{]}$. Repeat for the $\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{600}\mathbf{]}$intervaland compare your graph with that in Figure 3.1.6(a).

Short answer

The number of pounds of salt in the tank at time is $\mathrm{A}=600+3\left[\frac{50}{313}\sin \left(\frac{\mathrm{t}}{4}\right)-\frac{1250}{313}\cos \left(\frac{\mathrm{t}}{4}\right)\right]-\frac{168,400}{313}{\mathrm{e}}^{-\frac{\mathrm{t}}{100}}$. The amount of salt in the tank depends on the sine function and so, it will increase and decrease with time locally like a sine wave.

Step-by-step solution

Define mixtures.

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

$\frac{dA}{dt}=\left(\text{input rate of salt}\right)-\text{(output rate of salt})\mathbf{=}{\mathit{R}}_{\mathbf{\text{in}}}\mathbf{-}{\mathit{R}}_{\mathbf{\text{out}}}$

Solve for first order number of pounds.

Let the concentration of the solution entering in the tank be${\mathrm{c}}_{\mathrm{in}}\left(\mathrm{t}\right)=2+\sin (\mathrm{t}/4)\mathrm{lb}/\mathrm{gal}$at the rate of$\text{3gal/min}$. Then the rate at which the salt enters the tank is,

${\mathrm{R}}_{\mathrm{in}}=\left[2+\sin \left(\frac{\mathrm{t}}{4}\right)\right]\left(\frac{\mathrm{lb}}{\mathrm{gal}}\right)\times 3\left(\frac{\mathrm{gal}}{\min }\right)=\left[6+3\sin \left(\frac{\mathrm{t}}{4}\right)\right]\left(\frac{\mathrm{lb}}{\min }\right)$

And, the rate at which the salt leaves the tank is,

${\mathrm{R}}_{\mathrm{out}}=\frac{\mathrm{A}}{100}\left(\frac{\mathrm{lb}}{\min }\right)$

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

$$\begin{gathered} \frac{\text{dA}}{\text{dt}}\text{= Input rate of salt - Input rate of salt} \\ {\text{=F}}_{\text{in}}{\text{C}}_{\text{in}}{\text{-F}}_{\text{out}}{\text{C}}_{\text{out}} \end{gathered}$$

$\frac{\mathrm{dA}}{\mathrm{dt}}=6+3\sin \left(\frac{\mathrm{t}}{4}\right)-\frac{\mathrm{A}}{100}$… (1)

Multiply the equation by the IF.

Let the integrating factor be,

${\mathrm{e}}^{\int \frac{1}{1000}\mathrm{dt}}={\mathrm{e}}^{\frac{\mathrm{t}}{1010}}$

$\begin{array}{rcl}\frac{\mathrm{dA}}{\mathrm{dy}}{\mathrm{e}}^{\frac{\mathrm{t}}{100}}+\frac{\mathrm{A}}{100}{\mathrm{e}}^{\frac{\mathrm{t}}{100}}& =& \left[6+3\sin \left(\frac{\mathrm{t}}{4}\right)\right]{\mathrm{e}}^{\frac{\mathrm{t}}{100\mathrm{I}}}\\ \frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{Ae}}^{\frac{\mathrm{t}}{100}}\right)& =& 6{\mathrm{e}}^{\frac{1}{100}}+3\sin \left(\frac{\mathrm{t}}{4}\right){\mathrm{e}}^{\frac{\mathrm{t}}{100}}\\ \int \frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{Ae}}^{\frac{\mathrm{t}}{100}}\right)& =& \int 6{\mathrm{e}}^{\frac{\mathrm{t}}{100}}+\int 3\sin \left(\frac{\mathrm{t}}{4}\right){\mathrm{e}}^{\frac{\mathrm{t}}{100}}\\ {\mathrm{Ae}}^{\frac{\mathrm{t}}{100}}& =& \frac{6{\mathrm{e}}^{\frac{\mathrm{t}}{100}}}{\frac{1}{100}}+3\int \sin \left(\frac{\mathrm{t}}{4}\right){\mathrm{e}}^{\frac{\mathrm{t}}{100}}\\ & =& 600{\mathrm{e}}^{\frac{\mathrm{t}}{100}}+3\left[100{\mathrm{e}}^{\frac{\mathrm{t}}{100}}\sin \left(\frac{\mathrm{t}}{4}\right)\right.\left.-\int 25{\mathrm{e}}^{\frac{\mathrm{t}}{100}}\cos \left(\frac{\mathrm{t}}{4}\right)\mathrm{dt}\right]\\ & =& 600{\mathrm{e}}^{\frac{\mathrm{t}}{100}}+3\left[100{\mathrm{e}}^{\frac{\mathrm{t}}{100}}\sin \left(\frac{\mathrm{t}}{4}\right)\right.\left.-2500\left(\frac{25}{626}{\mathrm{e}}^{\frac{\mathrm{t}}{\mathrm{I}00}}\sin \left(\frac{\mathrm{t}}{4}\right)+\frac{1}{626}{\mathrm{e}}^{\frac{\mathrm{t}}{100}}\cos \left(\frac{\mathrm{t}}{4}\right)\right)\right]\\ {\mathrm{Ae}}^{\frac{\mathrm{t}}{100}}& =& 600{\mathrm{e}}^{\frac{\mathrm{t}}{100}}+3\left[\frac{50}{313}{\mathrm{e}}^{\frac{\mathrm{t}}{100}}\sin \left(\frac{\mathrm{t}}{4}\right)-\frac{1250}{313}{\mathrm{e}}^{\frac{\mathrm{t}}{\mathrm{I}00}}\cos \left(\frac{\mathrm{t}}{4}\right)\right]+\mathrm{C}\end{array}$

$\mathrm{A}=600+3\left[\frac{50}{313}\sin \left(\frac{\mathrm{t}}{4}\right)-\frac{1250}{313}\cos \left(\frac{\mathrm{t}}{4}\right)\right]+{\mathrm{Ce}}^{-\frac{\mathrm{t}}{100}}$… (3)

Obtain the values of constants.

To find the values of constants, apply the point$\mathrm{A}\left(0\right)=30$in the equation (3), then

$\begin{array}{rcl}50& =& 600+3\left[\frac{50}{313}\sin \left(\frac{0}{4}\right)-\frac{1250}{313}\cos \left(\frac{0}{4}\right)\right]+{\mathrm{Ce}}^{-\frac{0}{100}}\\ -550& =& 3\left[0-\frac{1250}{313}\right]+\mathrm{C}\\ \mathrm{C}& =& -550-\frac{3750}{313}\\ & =& -\frac{168,400}{313}\end{array}$

Substitute the value of $c$in the equation (3).

$\mathrm{A}=600+3\left[\frac{50}{313}\sin \left(\frac{\mathrm{t}}{4}\right)-\frac{1250}{313}\cos \left(\frac{\mathrm{t}}{4}\right)\right]-\frac{168,400}{313}{\mathrm{e}}^{-\frac{\mathrm{t}}{100}}$

Obtain the graph of the obtained solution.

Let the graph of the solution on the interval $(0,300)$be,

Let the graph of the solution on the interval $(0,600)$be,