Water Irrigation Distribution Box

Given

• $ \mbox{Water is poured into a container at a constant rate } V = \dfrac{10\, {cm}^3}{second} $
• $\mbox{After t seconds, water is distributed from the container at a rate of } \dfrac{V_{current}}{4} \dfrac{{cm}^3}{second} $
• $V_0 = 100 \;\mbox{at} \; t=0 $

Questions

• What is the equation for the volume in the container at any point in time after t=0? 
• What is the steady state of the container

Rationale

• $\mbox{The rate of  change in volume is } = {Increase} - {Decrease} $
• $\mbox{The increase in volume as a function of time } = \dfrac{10\, {cm}^3}{second} $
• $\mbox{The decrease in volume as a function of time }  = \dfrac{V}{4} \dfrac{{cm}^3}{second} $
• $\dfrac{dv}{dt}  = 10 - \dfrac{V}{4}  $
• $\mbox{Derive separable differential equation} $
• $\mbox{Multiply both sides by minus 4} $
•  $-4\dfrac{dv}{dt} =  -40 + V   $
•  $-4\dfrac{dv}{dt} =  V - 40   $
• $\mbox{Dividing both sides by -4 and (V-40)}  $
• $ \dfrac{1}{V - 40} \dfrac{dv}{dt} = - \dfrac{1}{4}   $
• $\mbox{Separate the dv and dt portions by multiplying by dt}   $
• $\dfrac{1}{V - 40} dv = - \dfrac{1}{4} dt   $
• $\mbox{Integrate both sides of the separated differential equation}    $
• $\int \dfrac{1}{V - 40} dv = - \dfrac{1}{4} \int dt   $
• $\ln {|V-40|}  = ( - \dfrac{1}{4} * t )  +  C   $ 
• $\mbox{Substituting initial conditions of V = 100 when t = 0}  $
• $\ln{(100 - 40)} = \ln{60}  = C    $
• $\mbox{Revisit our integration and substitute C}    $
• $\ln {|V-40|}  = ( - \dfrac{1}{4} * t )  +  \ln{(60)}   $ 
• $\mbox{subtract the ln(60) from both sides}     $
• $\ln {|V-40|} - \ln{(60)} = ( - \dfrac{1}{4} * t )  $
• $\ln\dfrac{(V-40)}{(60)} =  - \dfrac{1}{4} * t    $
• $\mbox{Exponentiate both sides of this result}    $
• $\dfrac{(V - 40)}{60} = e^{ - \dfrac{1}{4} t}    $
• $V  - 40 = 60 e^{ - \dfrac{1}{4} t}     $
• $V =  40 + 60 e^{ - \dfrac{1}{4} t}     $
• $\mbox{Test this final answer for t = 0}    $
• $V =  40 + 60 e^{ - \dfrac{1}{4} t}     $
• $V =  40 + 60 e^{ - \dfrac{1}{4} * 0}     $
• $V =  40 + 60 e^0 = 40 + (60 * 1) = 40  + 60 = 100    $ 
•  $\mbox{Test is successful!}    $
• $\mbox{Determine steady state condition}     $
• $V =  40 + 60 e^{ - \dfrac{1}{4} t}     $
• $  \mbox{As time goes on } \; 60 e^{ - \dfrac{1}{4}} t \; \mbox{gets smaller}$
• $\mbox{Therefore at steady state }\; V= 40 + \approx 0 = 40 $

Source

• YouTube video from www.m4ths.com