Respuesta :
Answer:
16.52 lb NaCl will be in the tank at the end of 3 hours.
Explanation:
Let s(t) = gal of NaCl in a container at time t (in minutes).
s'(t) = The rate at which the amount of gal of NaCl in the container is changing.
s'(t) = Rate of NaCl going in - Rate of NaCl going out
[tex]s'(t)=0-\dfrac{6}{600}s[/tex]
[tex]\dfrac{ds}{dt}=-\dfrac{1}{100}s[/tex]
By variable separable method we get
[tex]\dfrac{ds}{s}=-\dfrac{1}{100}dt[/tex]
Integrate both sides.
[tex]\int \dfrac{ds}{s}=\int -\dfrac{1}{100}dt[/tex]
[tex]\ln (s)=-\dfrac{t}{100}+ln C[/tex]
[tex]e^{\ln (s)}=e^{-\dfrac{t}{100}+ln C}[/tex]
[tex]s=e^{-\dfrac{t}{100}}\times e^{ln C}[/tex]
[tex]s=Ce^{-\dfrac{t}{100}}[/tex] [tex][\because e^{\ln x}=x][/tex]
100 Ib of NaCl are initially dissolved in 600 gal of a NaCl solution in a container. it means s(0)=100.
[tex]100=Ce^{-\dfrac{(0)}{100}}[/tex]
[tex]100=C[/tex]
Therefore the required function is
[tex]s=100e^{-\dfrac{t}{100}}[/tex] .... (1)
We need to find the amount of NaCl will be in the tank at the end of 3 hours.
3 hours = 180 minutes.
Substitute t=180 in equation (1).
[tex]s=100e^{-\dfrac{(180)}{100}}[/tex]
[tex]s=16.52988[/tex]
[tex]s\approx 16.52[/tex]
Therefore, 16.52 lb NaCl will be in the tank at the end of 3 hours.
