Question

Question: Solve Problem 21 assuming that pure water is pumped into the tank.

Short answer

The number of grams of salt in the tank at time is $A=\; 30e^{-\frac{1}{50}t}$.

Step-by-step solution

Define mixtures.

The mixing of two fluids sometimes gives rise to a linear first-order differentiale quation.

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

Solve for first order number of grams.

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

$$\begin{gathered} \frac{\text{dA}}{\text{dt}}\text{= Input rate of salt - Input rate of salt} \\ {\text{=F}}_{\text{in}}{\text{C}}_{\text{in}}{\text{-F}}_{\text{out}}{\text{C}}_{\text{out}} \end{gathered}$$… (1)

Substitute the values${\mathrm{F}}_{\mathrm{in}}={\mathrm{F}}_{\mathrm{out}}=4\mathrm{L}/\min ,{\mathrm{C}}_{\mathrm{in}}=0\mathrm{gram}/\mathrm{L},$ and${\text{C}}_{\text{out}}\text{=}\frac{\text{Amount of salt}}{\text{Volume of tank}}\text{=}\frac{\text{A}}{\text{200}}\text{gram/L}$ in the equation (1).

$$\begin{gathered} \frac{\mathrm{dA}}{\mathrm{dt}}=4\mathrm{L}/\min \times 0\mathrm{gram}/\mathrm{L}-4\mathrm{L}/\min \times \frac{\mathrm{A}}{200}\mathrm{gram}/\mathrm{L} \\ \frac{\mathrm{dA}}{\mathrm{dt}}=-\frac{\mathrm{A}}{50} \end{gathered}$$… (2)

As the equation (2) is linear and separable, so integrate the equation and separate the variables.

$\begin{array}{rcl}& & \frac{\text{dA}}{\text{-}\frac{\text{A}}{\text{50}}}\text{= dt}\\ & & \int \frac{\text{- 50}}{\text{A}}\text{dA =}\int \text{dt}\\ & & \text{- 50}\int \frac{\text{1}}{\text{A}}\text{dA =}\int \text{dt}\\ & & {\text{- 50ln(A) = t +c}}_{\text{1}}\\ & & \text{ln(A) = -}\frac{\text{1}}{\text{50}}{\text{t +c}}_{\text{2}}\\ & & {\text{e}}^{\text{ln(A)}}{\text{=e}}^{\left(\text{-}\frac{\text{1}}{\text{50}}{\text{t +c}}_{\text{2}}\right)}\end{array}$

Then, the equation becomes,

$\begin{array}{rcl}& & {\text{A =e}}^{{\text{c}}_{\text{2}}}{\text{e}}^{\text{-}\frac{\text{1}}{50}\text{t}}\\ & & {\text{= ce}}^{\text{-}\frac{\text{1}}{\text{50}}\text{t}}\end{array}$… (3)

Obtain the values of constants.

To find the values of constants, apply the point$\text{(A,t) = (30 grams,0 minutes)}$ in the equation (3), then

$\begin{array}{rcl}& & {\text{30grams = ce}}^{\text{0}}\\ & & \text{c = 30}\end{array}$

Substitute the value of in the equation (3)

${\text{A = 30e}}^{\text{-}\frac{\text{1}}{50}\text{t}}$.