Tutorial Week 3
From UBCMATH WIKI
Worksheet Questions
- 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).
- Write a differential equation with initial conditions for the total amount $Q(t)$ of $\mathrm{CO}_2$ (in grams) in the glass at time $t$.
- What would the steady state be if the glass were infinitely tall?
- Solve the initial value problem from part (a).
- 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.)
- Does the height of the foam $F(t)$ reach a steady state? If so, what is the limiting value?
- 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 $4ty_2'(t)$ 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.
- Rename $v'(t)=w(t)$. Simplify and solve the resulting equation for $w(t)$.
- Find $v(t)$ and substitute back to get $y_2(t)$.
Solutions
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- $\dfrac{dQ}{dt}=(0.005 \cdot 200)-0.5Q$ (3 points), $\ Q(0)=0$ (1 point)
- $Q(\infty)=2$g (1 point)
- $Q(t)=2 - 2e^{-t/2}$. Integrating factor (1 point), antiderivatives (2 points), find Q(t) (1 point), solve for initial condition (1 point)
- $\dfrac{dF}{dt} = 40\cdot0.5 Q(t) - 0.1F$ (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)
- $F(\infty) = 400$mm (1 point)
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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)
- $t^4 v''=0$ (2 points)
- $w' =0$ (1 point), $w(t)= C$ (obvious 1 point).
- $v(t)=Ct + D$ (D=0 is ok) (1 point) $ \quad y_2(t) = t^3$ (1 point)