Difference between revisions of "Tutorial Week 3"

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===Worksheet Questions===
 
===Worksheet Questions===
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# As a beer is poured, foam is added at the top of the beer at a constant rate of 2 mm per second and the bubbles burst at a continuous rate of 10 percent per second.
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# Beer is being poured from a tap into a glass at a rate of $200$ml/min. The beer leaving the tap contains $0.005$g/ml of $\mathrm{CO}_2$. It is known that $\mathrm{CO}_2$ escapes beer at a rate proportional to its concentration, and that at a concentration of $0.1$g/ml, the concentration decreases at a rate of $0.05$g/(ml$\cdot$min).
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##Write a differential equation with initial condition for the depth of foam at time t, $\dfrac{dF}{dt}$.  
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## Write a differential equation with initial conditions for the total amount $Q(t)$ of $\mathrm{CO}_2$ (in grams) in the glass at time $t$.
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##What would the steady state foam depth be if the glass were infinitely tall?
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## What would the steady state be if the glass were infinitely tall?
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##Solve the initial value problem.
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## Solve the initial value problem from part (a).
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##Assume the beer is poured into a glass 15cm tall so that the depth of beer increases at a rate of 10 mm per second. How high is the beer and foam at time $t$?
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## The $\mathrm{CO}_2$ leaving the beer forms a foam. Each gram of $\mathrm{CO}_2$ that escapes produces $40$mm of foam. Additionally, the bubbles burst at a rate of $10\%$ per minute. Write and solve an equation for the height of the foam $F(t)$ at time $t$. (Hint: use your solution from (a) to determine the amount of $\mathrm{CO}_2$ escaping.)
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##Write down an equation which when solved would give the time at which the bartender must stop pouring to ensure that no beer nor foam spill out of the glass. Give a rough approximation of the value of t along with an estimate of how close you approximation is.
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## Does the height of the foam $F(t)$ reach a steady state? If so, what is the limiting value?
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#Reduction of Order is a technique for finding a second solution to a homogeneous second order ODE when a first solution has already been found. Consider the equation $2t^2y''+ty'-3y = 0$ where $y_1(t) = \dfrac{1}{t}$ is our first solution.
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#Reduction of Order is a technique for finding a second solution to a homogeneous second order ODE when a first solution has already been found. Consider the equation $t^2y''+ 4ty' + 6y = 0$ where $y_1(t) = t^2$ is our first solution.
−
##Assume the second solution is of the form $y_2(t)=y_1(t)v(t)$ and calculate $ty_2'(t)$ and $2t^2y_2''(t)$.
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##Assume the second solution is of the form $y_2(t)=y_1(t)v(t)$ and calculate $4ty_2'(t)$ and $t^2y_2''(t)$.
 
##Plug these in to the ODE and simplify to find a second order equation in terms of $v''(t)$ and $v'(t)$ only.
 
##Plug these in to the ODE and simplify to find a second order equation in terms of $v''(t)$ and $v'(t)$ only.
−
##Rename $v'(t)=w(t)$. Solve the resulting equation for $w(t)$.
+
##Rename $v'(t)=w(t)$. Simplify and solve the resulting equation for $w(t)$.
 
##Find $v(t)$ and substitute back to get $y_2(t)$.
 
##Find $v(t)$ and substitute back to get $y_2(t)$.
  
 
===Solutions===
 
===Solutions===
 
#
 
#
−
##$\dfrac{dF}{dt}=2-0.1F$  '''(3 points)''', $\ F(0)=0$ '''(1 point)'''
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##$\dfrac{dQ}{dt}=(0.005 \cdot 200)-0.5Q$  '''(3 points)''', $\ Q(0)=0$ '''(1 point)'''
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##$F(t)=20mm$ '''(2 points)'''
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##$Q(\infty)=2$g '''(1 point)'''
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##$F(t)=20 -20e^{-0.1t}$. Integrating factor '''(1 point)''', antiderivatives '''(2 points)''', find F(t) '''(1 point)''', find C '''(1 point)'''
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##$Q(t)=2 - 2e^{-t/2}$. Integrating factor '''(1 point)''', antiderivatives '''(2 points)''', find Q(t) '''(1 point)''', solve for initial condition '''(1 point)'''
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##$H(t)=F(t)+B(t)=20(1-e^{-0.1t})+10t$ '''(1 point)'''
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##$\dfrac{dF}{dt} = 40\cdot0.5 Q(t) - 0.1F$ '''(2 points)''', $F(t) = 100 \left(e^{-\frac{t}{2}}-5 e^{-\frac{t}{10}}+4\right)$ Integrating factor '''(1 point)''', antiderivatives '''(2 points)''', find F(t) '''(1 point)''', solve for initial condition '''(1 point)'''
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## $H(t)=20(1-e^{-0.1t})+10t=150$ '''(1 point)''', arguing that $t\in(13,15)$ (steady state time and 0 foam time) '''(1 point)''', conclusion that $t$ must be closer to 13 '''(1 point)'''
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##$F(\infty) = 400$mm '''(1 point)'''
 
#
 
#
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##$ty_2'(t)=ty_1'v+ty_1v'=-\dfrac{v}{t}+v'$ '''(2 points)''',  $ \quad 2t^2y_2''=2t^2y_1''v+4t^2y_1'v'+2t^2y_1v'' = \dfrac{4v}{t}-4v'+2tv''$ '''(2 points)'''
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## $4 t y_2' = 4t(2 t v + t^2 v') = 8 t^2 v + 4 t^2 v'$ '''(2 points)''' and $t^2 y_2'' = t^2(2v + 4tv' + t^2 v'')$ '''(2 points)'''
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## $v''-\dfrac{3}{2t}v'=0$ '''(2 points)'''
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## $t^4 v''=0$ '''(2 points)'''
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## $w'-\dfrac{3}{2t} w=0$ (separable equation '''1 point'''),  $w(t)=Ct^{3/2}$ (take the antiderivative '''2 points''').
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## $w' =0$ '''(1 point)''',  $w(t)= C$ (obvious '''1 point''').
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## $v(t)=Ct^{5/2}+D$ (D=0 is ok) '''(2 points)'''  $ \quad y_2(t) = t^{3/2}$ '''(1 point)'''
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## $v(t)=Ct + D$ (D=0 is ok) '''(1 point)'''  $ \quad y_2(t) = t^3$ '''(1 point)'''

Revision as of 23:39, 24 January 2017

Worksheet Questions

  1. Beer is being poured from a tap into a glass at a rate of $200$ml/min. The beer leaving the tap contains $0.005$g/ml of $\mathrm{CO}_2$. It is known that $\mathrm{CO}_2$ escapes beer at a rate proportional to its concentration, and that at a concentration of $0.1$g/ml, the concentration decreases at a rate of $0.05$g/(ml$\cdot$min).
    1. Write a differential equation with initial conditions for the total amount $Q(t)$ of $\mathrm{CO}_2$ (in grams) in the glass at time $t$.
    2. What would the steady state be if the glass were infinitely tall?
    3. Solve the initial value problem from part (a).
    4. The $\mathrm{CO}_2$ leaving the beer forms a foam. Each gram of $\mathrm{CO}_2$ that escapes produces $40$mm of foam. Additionally, the bubbles burst at a rate of $10\%$ per minute. Write and solve an equation for the height of the foam $F(t)$ at time $t$. (Hint: use your solution from (a) to determine the amount of $\mathrm{CO}_2$ escaping.)
    5. Does the height of the foam $F(t)$ reach a steady state? If so, what is the limiting value?
  2. Reduction of Order is a technique for finding a second solution to a homogeneous second order ODE when a first solution has already been found. Consider the equation $t^2y''+ 4ty' + 6y = 0$ where $y_1(t) = t^2$ is our first solution.
    1. Assume the second solution is of the form $y_2(t)=y_1(t)v(t)$ and calculate $4ty_2'(t)$ and $t^2y_2''(t)$.
    2. Plug these in to the ODE and simplify to find a second order equation in terms of $v''(t)$ and $v'(t)$ only.
    3. Rename $v'(t)=w(t)$. Simplify and solve the resulting equation for $w(t)$.
    4. Find $v(t)$ and substitute back to get $y_2(t)$.

Solutions

    1. $\dfrac{dQ}{dt}=(0.005 \cdot 200)-0.5Q$ (3 points), $\ Q(0)=0$ (1 point)
    2. $Q(\infty)=2$g (1 point)
    3. $Q(t)=2 - 2e^{-t/2}$. Integrating factor (1 point), antiderivatives (2 points), find Q(t) (1 point), solve for initial condition (1 point)
    4. $\dfrac{dF}{dt} = 40\cdot0.5 Q(t) - 0.1F$ (2 points), $F(t) = 100 \left(e^{-\frac{t}{2}}-5 e^{-\frac{t}{10}}+4\right)$ Integrating factor (1 point), antiderivatives (2 points), find F(t) (1 point), solve for initial condition (1 point)
    5. $F(\infty) = 400$mm (1 point)
    1. $4 t y_2' = 4t(2 t v + t^2 v') = 8 t^2 v + 4 t^2 v'$ (2 points) and $t^2 y_2'' = t^2(2v + 4tv' + t^2 v'')$ (2 points)
    2. $t^4 v''=0$ (2 points)
    3. $w' =0$ (1 point), $w(t)= C$ (obvious 1 point).
    4. $v(t)=Ct + D$ (D=0 is ok) (1 point) $ \quad y_2(t) = t^3$ (1 point)
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