Respuesta :
Answer:
(a) Amount of salt as a function of time
[tex]A(t)=100-80e^{-0.02t}[/tex]
(b) The time at which the amount of salt in the tank reaches 50 lb is 23.5 minutes.
(c) The amount of salt when t approaches to +inf is 100 lb.
Step-by-step explanation:
The rate of change of the amount of salt can be written as
[tex]\frac{dA}{dt} =rate\,in\,-\,rate\,out\\\\\frac{dA}{dt}=C_i*q_i-C_o*q_o=C_i*q_i-\frac{A(t)}{V}*q_o\\\\\frac{dA}{dt}=0.5*4-\frac{A(t)}{200}*4\\\\\frac{dA}{dt} =\frac{100-A(t)}{50}=-(\frac{A(t)-100}{50})[/tex]
Then we can rearrange and integrate
[tex]\frac{dA}{dt}= -(\frac{A(t)-100}{50})\\\\\int \frac{dA}{A-100}=-\frac{1}{50} \int dt\\\\ln(A-100)=-\frac{t}{50}+C_0\\\\ A-100=Ce^{-0.02*t}\\\\\\A(0)=20 \rightarrow 20-100=Ce^0=C\\C=-80\\\\A=100-80e^{-0.02t}[/tex]
Then we have the model of A(t) like
[tex]A(t)=100-80e^{-0.02t}[/tex]
(b) The time at which the amount of salt reaches 50 lb is
[tex]A(t)=100-80e^{-0.02t}=50\\\\e^{-0.02t}=(50-100)/(-80)=0.625\\\\-0.02*t=ln(0.625)=-0.47\\\\t=(-0.47)/(-0.02)=23.5[/tex]
(c) When t approaches to +infinit, the term e^(-0.02t) approaches to zero, so the amount of salt in the solution approaches to 100 lb.
