Answer:
a) ![s(t) =400- 385 e^{-\frac{1}{10} t}](https://tex.z-dn.net/?f=s%28t%29%20%3D400-%20385%20e%5E%7B-%5Cfrac%7B1%7D%7B10%7D%20t%7D)
b) ![s(t=25min) =400- 385 e^{-\frac{1}{10}25}=368.397](https://tex.z-dn.net/?f=s%28t%3D25min%29%20%3D400-%20385%20e%5E%7B-%5Cfrac%7B1%7D%7B10%7D25%7D%3D368.397%20)
Step-by-step explanation:
Part a
Assuming that the original concentration of salt is 8 oz/gal and that the rate of in is equal to the rate out = 5 gal/min.
For this case we know that the rate of change can be expressed on this way:
![Rate change = In-Out](https://tex.z-dn.net/?f=Rate%20change%20%3D%20In-Out)
And we can name the rate of change as ![\frac{ds}{dt}=rate change](https://tex.z-dn.net/?f=%5Cfrac%7Bds%7D%7Bdt%7D%3Drate%20change)
And our variable s would represent the amount of salt for any time t.
We know that the brine containing 8oz/gal and the rate in is equal to 5 gal/min and this value is equal to the rate out.
For the concentration out we can assume that is ![\frac{s}{50gal}](https://tex.z-dn.net/?f=%5Cfrac%7Bs%7D%7B50gal%7D)
And now we can find the expression for the amount of salt after time t like this:
![\frac{dS}{dt}= 8 \frac{oz}{gal}(5\frac{gal}{min}) -\frac{s}{50gal} 5 \frac{gal}{min} =40\frac{oz}{min}- \frac{s}{10} \frac{oz}{min}](https://tex.z-dn.net/?f=%5Cfrac%7BdS%7D%7Bdt%7D%3D%208%20%5Cfrac%7Boz%7D%7Bgal%7D%285%5Cfrac%7Bgal%7D%7Bmin%7D%29%20-%5Cfrac%7Bs%7D%7B50gal%7D%205%20%5Cfrac%7Bgal%7D%7Bmin%7D%20%3D40%5Cfrac%7Boz%7D%7Bmin%7D-%20%5Cfrac%7Bs%7D%7B10%7D%20%5Cfrac%7Boz%7D%7Bmin%7D)
And we have this differential equation:
![\frac{dS}{dt} +\frac{1}{10} s = 40](https://tex.z-dn.net/?f=%5Cfrac%7BdS%7D%7Bdt%7D%20%2B%5Cfrac%7B1%7D%7B10%7D%20s%20%3D%2040)
With the initial conditions y(0)=15 oz
As we can see we have a linear differential equation so in order to solve it we need to find first the integrating factor given by:
![\mu = e^{\int \frac{1}{10} dt }= e^{\frac{1}{10} t}](https://tex.z-dn.net/?f=%5Cmu%20%3D%20e%5E%7B%5Cint%20%5Cfrac%7B1%7D%7B10%7D%20dt%20%7D%3D%20e%5E%7B%5Cfrac%7B1%7D%7B10%7D%20t%7D)
And then in order to solve the differential equation we need to multiply with the integrating factor like this:
![e^{\frac{1}{10} t} s = \int 40 e^{\frac{1}{10} t} dt](https://tex.z-dn.net/?f=e%5E%7B%5Cfrac%7B1%7D%7B10%7D%20t%7D%20s%20%3D%20%5Cint%2040%20e%5E%7B%5Cfrac%7B1%7D%7B10%7D%20t%7D%20dt)
![e^{\frac{1}{10} t} s = 400 e^{\frac{1}{10} t} +C](https://tex.z-dn.net/?f=e%5E%7B%5Cfrac%7B1%7D%7B10%7D%20t%7D%20s%20%3D%20400%20e%5E%7B%5Cfrac%7B1%7D%7B10%7D%20t%7D%20%2BC)
Now we can divide both sides by
and we got:
![s(t) =400 + C e^{-\frac{1}{10} t}](https://tex.z-dn.net/?f=s%28t%29%20%3D400%20%2B%20C%20e%5E%7B-%5Cfrac%7B1%7D%7B10%7D%20t%7D)
Now we can apply the initial condition in order to solve for the constant C like this:
![15 = 400+C](https://tex.z-dn.net/?f=15%20%3D%20400%2BC)
![C=-385](https://tex.z-dn.net/?f=C%3D-385)
And then our function would be given by:
![s(t) =400- 385 e^{-\frac{1}{10} t}](https://tex.z-dn.net/?f=s%28t%29%20%3D400-%20385%20e%5E%7B-%5Cfrac%7B1%7D%7B10%7D%20t%7D)
Part b
For this case we just need to replace t =25 and see what we got for the value of the concentration:
![s(t=25min) =400- 385 e^{-\frac{1}{10}25}=368.397](https://tex.z-dn.net/?f=s%28t%3D25min%29%20%3D400-%20385%20e%5E%7B-%5Cfrac%7B1%7D%7B10%7D25%7D%3D368.397%20)