Given
- $\mbox{A circular stain grows}$
- $\mbox{Such that the rate of increase of the radius}$
- $\mbox{Is inversely proportional to the square of the radius.} $
- $\mbox{Another stain area grows quicker. Its equation is:} $
- $\dfrac{ds}{dt} = \dfrac{2 e^{2t} }{\sqrt{s}} $
Questions
- $\mbox{What is equation for the change in area of the stain over time?} $
- $\mbox{Show} \; \dfrac{da}{dt} \varpropto \dfrac{1}{\sqrt{a}} $
- $\mbox{For second stain, for s = 81 at t=0}$
- $\mbox{Find the time when s = 100}$
Rationale
- $\dfrac{dr}{dt} = k \dfrac{1}{r^2} $
- $\dfrac{da}{dt} = \dfrac{da}{dr} * \dfrac{dr}{dt} $
- $ a = \pi r^2 $
- $\dfrac{da}{dr} = 2 \pi r $
- $\dfrac{da}{dt} = ( 2 \pi r ) * (k \dfrac{1}{r^2} )$
- $\dfrac{da}{dt} = \dfrac{2 \pi k}{r} $
- $\mbox{And da/dt in terms of area is}$
- $a = \pi r^2 $
- $r^2 = \dfrac{a}{\pi} $
- $r = \sqrt{ \dfrac{a}{\pi}} $
- $\mbox{Substituting for r in previous solution}$
- $\dfrac{da}{dt} = \dfrac{2 \pi k}{ \sqrt{ \dfrac{a}{\pi}}} $
- $\dfrac{da}{dt} = \dfrac{2 \pi^{(\dfrac{3}{2})} k} {\sqrt{a}}$
- $\dfrac{da}{dt} \varpropto \dfrac{1}{\sqrt{a}} $
- $\dfrac{ds}{dt} = \dfrac{2 e^{2t} }{\sqrt{s}} $
- $s^{\dfrac{1}{2}} ds = 2 e^{2t} dt $
- $\int{s^{\dfrac{1}{2}} ds} = \int{2 e^{2t} dt} $
- $\dfrac{s^{\dfrac{3}{2}}}{\dfrac{3}{2}} = e^{2t} + C$
- $\dfrac{2}{3} s^{\dfrac{3}{2}} = e^{2t} + C $
- $s^{\dfrac{3}{2}} = \dfrac{3}{2} e^{2t} + K $
- $\mbox{Solve for K using s=81 at t=0} $
- $81^{\dfrac{3}{2}} = \dfrac{3}{2} e^{2 * 0} + K $
- $9^3 = \dfrac{3}{2} e^0 + K $
- $729 = \dfrac{3}{2} + K $
- $K = 727.5 $
- $\mbox{Solve for t when s is 100} $
- $s^{\dfrac{3}{2}} = \dfrac{3}{2} e^{2t} + 727.5 $
- $100^{\dfrac{3}{2}} = \dfrac{3}{2} e^{2t} + 727.5 $
- $10^3 = \dfrac{3}{2} e^{2t} + 727.5 $
- $1000 = \dfrac{3}{2} e^{2t} + 727.5 $
- $1000 - 727.5 = \dfrac{3}{2} e^{2t} = 272.5 $
- $\dfrac{2}{3} 272.5 = e^{2t} $
- $181.6666667 = e^{2t} $
- $\ln{181.6666667} = 2t $
- $t = \dfrac{\ln{181.6666667}}{2} = 2.601086753 \approx 2.6$
Source
Given
- $\mbox{A radioactive isotope has a half-life of} \; 18 \; \mbox{days} $
- $\mbox{You wish to have 25 grams after 33 days.} $
Question
- $\mbox{How much radioactive isotope should you start with?}$
Rationale
- $\dfrac{dm}{dt} = - k * m $
- $ \dfrac{dm}{m} = - k * dt $
- $\int \dfrac{dm}{m} = - k \int dt $
- $\ln{(m)} = - k t + C$
- $ m = e^{-kt+C} = C e^{-kt} = m_0 e^{-kt}$
- $\mbox{Substituting the half-life conditions} $
- $\dfrac{1}{2} m_0 = m_0 e^{-18k}$
- $\dfrac{1}{2} = e^{-18k}$
- $2 = e^{18k} $
- $\ln(2) = 18k$
- $k = \dfrac{\ln(2)}{18}$
- $\mbox{Updating our decay equation for k} $
- $ m = m_0 e^{-\dfrac{\ln(2)}{18}t}$
- $\mbox{Determine answer to question} $
- $ 25 = m_0 e^{-\dfrac{\ln(2)}{18}33}$
- $ m_0 = 25 e^{\dfrac{\ln(2)}{18}33} = 89.08987181 \approx 89.1$
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
