Tutorial Week 3
From UBCMATH WIKI
Worksheet Questions
- 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.
- Write a differential equation with initial condition for the depth of foam at time t, $\dfrac{dF}{dt}$.
- What would the steady state foam depth be if the glass were infinitely tall?
- Solve the initial value problem.
- 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 $t$?
- 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.
- 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.
- Assume the second solution is of the form $y_2(t)=y_1(t)v(t)$ and calculate $ty_2'(t)$ and $2t^2y_2''(t)$.
- 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 $w(t)$.
- Find $v(t)$ and substitute back to get $y_2(t)$.
Solutions
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- $\dfrac{dF}{dt}=2-0.1F$ (3 points), $\ F(0)=0$ (1 point)
- $F(t)=20mm$ (2 points)
- $F(t)=20 -20e^{-0.1t}$. Integrating factor (1 point), antiderivatives (2 points), find F(t) (1 point), find C (1 point)
- $H(t)=F(t)+B(t)=20(1-e^{-0.1t})+10t$ (1 point)
- $H(t)=20(1-e^{-0.1t})+10t=150$ (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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- $ty_2'(t)=ty_1'v+ty_1v'=-\dfrac{v}{t}+v'$ (2 points), $ \quad 2t^2y_2''=2t^2y_1''v+4t^2y_1'v'+2t^2y_1v'' = \dfrac{4v}{t}-4v'+2tv''$ (2 points)
- $v''-\dfrac{3}{2t}v'=0$ (2 points)
- $w'-\dfrac{3}{2t} w=0$ (separable equation 1 point), $w(t)=Ct^{3/2}$ (take the antiderivative 2 points).
- $v(t)=Ct^{5/2}+D$ (D=0 is ok) (2 points) $ \quad y_2(t) = t^{3/2}$ (1 point)