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A 50 gallon tank initially contains 10 gallons of fresh water. at t = 0 a brine solution containing 1 pound of salt per gallon is poured into the tank at the rate of 4 gal/min., while the well-stirred mixture leaves the tank at the rate of 0.5 gal/min. find the amount of salt in the tank at the moment of overflow..

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LammettHash
The rate at which the amount of salt in the tank, [tex]A(t)[/tex], changes over time is given by the ODE

[tex]A'(t)=\dfrac{4\text{ gal}}{1\text{ min}}\cdot\dfrac{1\text{ lb}}{1\text{ gal}}-\dfrac{0.5\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{10+(4-0.5)t\text{ gal}}[/tex]

and we're told that at the start there is no salt in the tank (fresh water), so [tex]A(0)=0[/tex]. The amount of solution in the tank is given by [tex]10+(4-0.5)t[/tex], so the tank will overflow once this expression is the same as the total tank volume, i.e.

[tex]10+(4-0.5)t=50\implies t=\dfrac{80}7\approx11.43[/tex]

We find an integrating factor of the form

[tex]\exp\left(\displaystyle\int\frac{\mathrm dt}{20+7t}\right)=\exp\left(\frac17\ln|20+7t|\right)=(20+7t)^{1/7}[/tex]

then distribute this to all terms of the ODE, giving


[tex]A'(t)+\dfrac1{20+7t}A(t)=4[/tex]
[tex]\implies(20+7t)^{1/7}A'(t)+(20+7t)^{-6/7}A(t)=4(20+7t)^{1/7}[/tex]
[tex]\left((20+7t)^{1/7}A(t)\right)'=4(20+7t)^{1/7}[/tex]
[tex](20+7t)^{1/7}A(t)=\dfrac12(20+7t)^{8/7}+C[/tex]
[tex]A(t)=\dfrac{20+7t}2+\dfrac C{(20+7t)^{1/7}}[/tex]

Since [tex]A(0)=0[/tex], we get

[tex]0=\dfrac{20}2+\dfrac C{20^{1/7}}\implies C\approx-15.34[/tex]

so that the amount of salt in the tank is given by

[tex]A(t)=\dfrac{20+7t}2-\dfrac{15.34}{(20+7t)^{1/7}}[/tex]

Then once the tank starts to overflow at [tex]t\approx11.43[/tex], the amount of salt in the tank is

[tex]A(11.43)\approx42.06\text{ lb}[/tex]
MrRoyal

The flow of salt in the tank is an illustration of rates.

The amount of salt in the tank at the moment of overflow is 3 gallons.

The given parameters are:

[tex]\mathbf{V = 50}[/tex] --- the size of the tank

[tex]\mathbf{V_1 = 10}[/tex] --- the initial content of the tank

[tex]\mathbf{Rate = 4gal\ min^{-1}}[/tex]

So, the volume function is:

[tex]\mathbf{V(t) = V_1 +Rate \times t}[/tex]

[tex]\mathbf{V(t) = 10 +4\times t}[/tex]

[tex]\mathbf{V(t) = 10 +4t}[/tex]

Substitute 50 for V

[tex]\mathbf{10 +4t = 50}[/tex]

Subtract 10 from both sides

[tex]\mathbf{4t = 40}[/tex]

Divide both sides by 4

[tex]\mathbf{t = 10}[/tex] ---- this means that, the tank will overflow after 10 seconds

The differential equation that models the amount of salt in the tank is:

[tex]\mathbf{\frac{dx}{dt} = 0.5 - \frac{4}{10 + 4t}x}[/tex]

This gives

[tex]\mathbf{\frac{dx}{dt} = \frac{5 + 2t}{10 + 4t} - \frac{4x}{10 + 4t}}[/tex]

Integrate, and solve for x

[tex]\mathbf{x = \frac{1}{10 + 4t}(5t + t^2 + c)}[/tex]

At x(0) = 0, we have:

[tex]\mathbf{0 = \frac{1}{10 + 4 \times 0}(5(0) + (0)^2 + c)}[/tex]

[tex]\mathbf{0 = \frac{1}{10}(0 + 0 + c)}[/tex]

[tex]\mathbf{0 = \frac{1}{10}(c)}[/tex]

Multiply through by 10

[tex]\mathbf{c = 0}[/tex]

So, we have:

[tex]\mathbf{x = \frac{1}{10 + 4t}(5t + t^2 + c)}[/tex]

[tex]\mathbf{x = \frac{1}{10 + 4t}(5t + t^2 + 0)}[/tex]

[tex]\mathbf{x = \frac{1}{10 + 4t}(5t + t^2)}[/tex]

Recall that: the time of overflow is [tex]\mathbf{t = 10}[/tex]

So, we have:

[tex]\mathbf{x(10) = \frac{1}{10 + 4\times 10}(5\times 10 + 10^2)}[/tex]

[tex]\mathbf{x(10) = \frac{1}{50}(150)}[/tex]

[tex]\mathbf{x(10) = 3}[/tex]

Hence, the amount of salt in the tank at the moment of overflow is 3 gallons.

Read more about rates at:

https://brainly.com/question/4505307

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