Mathematics, 20.09.2020 14:01 keiryporthillo

A tank with capactity 500 gal originally contains 200 gal water with 100 lbs of salt mixed into it. Water containing 1 lb of salt per gallon is entering at a rate of 3 gal/min, and is allowed to leave at 2 gal/min. Find the amount of salt in the tank at any time prior to the instant the tank overflows. Find the concentration (in pounds per gallon) of salt in the tank when it is at the point of overflowing. Compare this concentration with the theoretical limiting concentration if the tank had infinite volume.

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Answer from: kids2006

The amount of salt in the tank at any moment is

$f(t)=-\frac {4\times 10^6}{(200+t)^2}+200+t$

The concentration of salt in the tank when it is at the point of overflowing is 0.968 .

The theoretical limiting concentration of an infinite tank is lb per gallon.

Step-by-step explanation:

Let f(t) be the amount of salt in the tank at any time

Then, its time rate of change, f'(t) , by (balance law).

Since three gallons of salt water runs in the tank per minute, containing lb of salt, the salt rate is

3.1=3

The amount of water in the tank at any time is.

200+(3-2)t=200+t,

Now, the outflow is gal of the solution in a minute. That is $\frac 2{200+t}$ of the total solution content in the tank, hence $\frac 2{200+t}$ of the salt salt content f(t) , that is $\frac{2f(t)}{200+t}$ .

Initially, the tank contains 100 lb of salt,

Therefore we obtain the initial condition f(0)=100

Thus, the model is

$f'(t)=3-\frac{2f(t)}{200+t}, f(0)=100$

$\Rightarrow f'(t)+\frac{2}{200+t}f(t)=3, f(0)=100$

$p(t)=\frac{2}{200+t} \;\;\text{and} \;\;q(t)=3$ Linear ODE.

so, an integrating factor is

$e^{\int p dt}=e^{2\int \frac{dt}{200+t}=e^{\ln(200+t)^2}=(200+t)^2$

and the general solution is

$f(t)(200+t)^2=\int q(200+t)^2 dt+c$

$\Rightarrow f(t)=\frac 1{(200+t)^2}\int 3(200+t)^2 dt+c$

$\Rightarrow f(t)=\frac c{(200+t)^2}+200+t$

Now using the initial condition and find the value of .

$100=f(0)=\frac c{(200+0)^2}+200+0\Rightarrow -100=\frac c{200^2}$

$\Rightarrow c=-4000000=-4\times 10^6$

$\Rightarrow f(t)=-\frac {4\times 10^6}{(200+t)^2}+200+t$

is the amount of salt in the tank at any moment

Initially, the tank contains 200 gal of water and the capacity of the tank is 500 gal. This means that there is enough place for

500-200=300 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 min.

Therefore, we need to calculate f(300) to find the amount of salt any time prior to the moment when the solution begins to overflow.

$f(300)=-\frac{4\times 10^6}{(200+300)^2}+200+300=-16+500=484$

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 500 L.

$\text{concentration at t}=300=\frac{484}{500}=0.968$

If the tank had an infinite capacity, then the concentration would be

$\lim\limits_{t \to \infty} \frac{f(t)}{200+t}= \lim\limits_{t \to \infty}\left(\frac{\frac{3\cdot 10^6}{(200+t)^2}+(200+t)}{200+t}\right)$

$= \lim\limits_{t \to \infty} \left(\frac{4\cdot 10^6}{(200+t)^3}+1\right)$

Hence, the theoretical limiting concentration of an infinite tank is lb per gallon.

Answer from: Quest

segment bisector

step-by-step explanation:

line segment bisector. definition: a line, ray or segment which cuts another line segment into two equal parts. try this drag one of the orange dots at a or b and note the the line ab always divides the line pq into two equal parts.

mark brainliest!

Answer from: Quest

(-14+3/2b) - (1+8/2b) = -5 -5/2b(-10+b) + (7b-5) = 8b-15(7/2b-3) - (8+6b) = -5/2b-11(5+2b) + (2b+3/2) = 4b+ 13/2

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A tank with capactity 500 gal originally contains 200 gal water with 100 lbs of salt mixed into it....