Respuesta :
Answer:
The amount of salt in the tank at any moment [tex]t[/tex] is
[tex]f(t)=-\frac {4\times 10^6}{(200+t)^2}+200+t[/tex]
The concentration of salt in the tank when it is at the point of overflowing is [tex]0.968[/tex].
The theoretical limiting concentration of an infinite tank is [tex]1[/tex] lb per gallon.
Step-by-step explanation:
Let [tex]f(t)[/tex] be the amount of salt in the tank at any time [tex]t.[/tex]
Then, its time rate of change, [tex]f'(t)[/tex], by (balance law).
Since three gallons of salt water runs in the tank per minute, containing [tex]1[/tex]lb of salt, the salt rate is
[tex]3.1=3[/tex]
The amount of water in the tank at any time [tex]t[/tex] is.
[tex]200+(3-2)t=200+t,[/tex]
Now, the outflow is [tex]2[/tex] gal of the solution in a minute. That is [tex]\frac 2{200+t}[/tex] of the total solution content in the tank, hence [tex]\frac 2{200+t}[/tex] of the salt salt content [tex]f(t)[/tex], that is [tex]\frac{2f(t)}{200+t}[/tex].
Initially, the tank contains [tex]100[/tex] lb of salt,
Therefore we obtain the initial condition [tex]f(0)=100[/tex]
Thus, the model is
[tex]f'(t)=3-\frac{2f(t)}{200+t}, f(0)=100[/tex]
[tex]\Rightarrow f'(t)+\frac{2}{200+t}f(t)=3, f(0)=100[/tex]
[tex]p(t)=\frac{2}{200+t} \;\;\text{and} \;\;q(t)=3[/tex] Linear ODE.
so, an integrating factor is
[tex]e^{\int p dt}=e^{2\int \frac{dt}{200+t}=e^{\ln(200+t)^2}=(200+t)^2[/tex]
and the general solution is
[tex]f(t)(200+t)^2=\int q(200+t)^2 dt+c[/tex]
[tex]\Rightarrow f(t)=\frac 1{(200+t)^2}\int 3(200+t)^2 dt+c[/tex]
[tex]\Rightarrow f(t)=\frac c{(200+t)^2}+200+t[/tex]
Now using the initial condition and find the value of [tex]c[/tex].
[tex]100=f(0)=\frac c{(200+0)^2}+200+0\Rightarrow -100=\frac c{200^2}[/tex]
[tex]\Rightarrow c=-4000000=-4\times 10^6[/tex]
[tex]\Rightarrow f(t)=-\frac {4\times 10^6}{(200+t)^2}+200+t[/tex]
is the amount of salt in the tank at any moment [tex]t.[/tex]
Initially, the tank contains [tex]200[/tex] gal of water and the capacity of the tank is [tex]500[/tex] gal. This means that there is enough place for
[tex]500-200=300[/tex] gal
of water in the tank at the beginning. As concluded previously, we have one new gal in the tank at every minute. hence the tank will be full in [tex]30[/tex]min.
Therefore, we need to calculate [tex]f(300)[/tex] to find the amount of salt any time prior to the moment when the solution begins to overflow.
[tex]f(300)=-\frac{4\times 10^6}{(200+300)^2}+200+300=-16+500=484[/tex]
To find the concentration of salt at that moment, divide the amount of salt with the amount of water in the tank at that moment, which is [tex]500[/tex]L.
[tex]\text{concentration at t}=300=\frac{484}{500}=0.968[/tex]
If the tank had an infinite capacity, then the concentration would be
[tex]\lim\limits_{t \to \infty} \frac{f(t)}{200+t}= \lim\limits_{t \to \infty}\left(\frac{\frac{3\cdot 10^6}{(200+t)^2}+(200+t)}{200+t}\right)[/tex]
[tex]= \lim\limits_{t \to \infty} \left(\frac{4\cdot 10^6}{(200+t)^3}+1\right)[/tex]
[tex]=1[/tex]
Hence, the theoretical limiting concentration of an infinite tank is [tex]1[/tex] lb per gallon.
