Difference between revisions of "Tutorial Week 2"

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## (Optional - mostly to think about later on your own) Notice that the solution in the case of $n = m$ ($T_{n = m}(t)$) does not have the same form as the solution in the case of $n \ne m$ ($T_{n \ne m}(t)$). Show that as $m$ approaches $n$, the function $T_{n \ne m}(t)$ approaches $T_{n = m}(t)$.
 
## (Optional - mostly to think about later on your own) Notice that the solution in the case of $n = m$ ($T_{n = m}(t)$) does not have the same form as the solution in the case of $n \ne m$ ($T_{n \ne m}(t)$). Show that as $m$ approaches $n$, the function $T_{n \ne m}(t)$ approaches $T_{n = m}(t)$.
  
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===Solution===
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[[Tutorial Week 2 Solutions]]
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## Linearly independent (Show Wronskian calculation to prove independence, Reference: http://mathworld.wolfram.com/LinearlyDependentFunctions.html). Points: '''1 pt''' for setting up the determinant, '''1 pt''' for calculating the value, '''1 pt''' for correct conclusion.
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## Linearly independent (same as above) '''1 pt'''
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## Linearly independent (same as above) '''1 pt'''
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## Linearly dependent. Works with $c_1 = 1$, $c_2 = 1$ and $c_3 = -2$. Points: '''1 pt''' for figuring out the c values that work, '''1 pt''' for correct conclusion.
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## We have $\frac{dU}{dt} = -mU$ and $\frac{dT}{dt} = -nT+mU$ ('''3 pt''' - subtract one for each error, e.g. - instead of +, missing T dependence, writing $U(t)$ instead of $dU/dt$, etc.) with initial conditions $U(0)=50$ grams '''(1 pt)''' and $T(0)=0$ '''(1 pt)'''
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## Solution: $T_{n \ne m}(t) = \frac{50m}{n-m}e^{-mt} - \frac{50m}{n-m}e^{-nt}$ ('''1 pt''' for getting U(t), '''1 pt''' for using the correct integrating factor to solve the T(t)) equation, '''1 pt''' for getting T(t)) and $T_{n = m}(t) = 50nte^{-nt}$ ('''1 pt''' for getting the general solution, '''1 pt''' for using the IC to determine the value of the arbitrary constant).
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## We have:$\lim_{m \to n} \frac{50m}{n-m}(e^{-mt} - e^{-nt}) = -50n \lim_{m \to n} \frac{e^{-mt}-e^{-nt}}{m-n} = -50n \frac{d}{dn} e^{-nt} = 50nte^{-nt}$. ('''1 pt''' for recognizing the limit has the form of a derivative, '''1 pt''' for expressing the argument correctly).
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Latest revision as of 22:59, 30 December 2020

Workshop 2

  1. Recall that two functions $f$ and $g$ are linearly independent if in order to have $c_1f(x) + c_2g(x) = 0$ for all $x$, we must have $c_1 = c_2 = 0$. (Likewise, three functions $f,g$ and $h$ are linearly independent if $c_1f(x) + c_2g(x) + c_3h(x) = 0$ implies $c_1 = c_2 = c_3 = 0$.) Are the following sets of functions linearly independent or linearly dependent for all $x$? If they are independent, use the Wronskian to show that. If they are not independent, give nonzero values of the constants $c_i$ such that the weighted sum of the functions is zero.
    1. $f(x) = x$, $g(x) = x+1$
    2. $f(x) = x$, $g(x) = x-1$
    3. $f(x) = x+1$, $g(x) = x-1$
    4. $f(x) = x+1$, $g(x) = x-1$, $h(x)=x$
  2. An atom of a radioactive substance typically decays into an atom of some other radioactive substance with its own decay constant. Suppose we have $50$ grams of pure Uranium-238 with a decay constant $m$. Each atom of Uranium-238 decays to a single atom of Thorium-234 which has a decay constant $n$. We have no Thorium-234 initially.
    1. Write down the differential equations and initial conditions for the radioactive decay of Uranium-238 ($U(t)$) and production and decay of Thorium-234 ($T(t)$).
    2. Solve the differential equations and initial conditions to find the amount of Thorium-234 after time $t$ in terms of $m$ and $n$. You should find that there are two cases that must be treated separately, $m \ne n$ and $m = n$.
    3. (Optional - mostly to think about later on your own) Notice that the solution in the case of $n = m$ ($T_{n = m}(t)$) does not have the same form as the solution in the case of $n \ne m$ ($T_{n \ne m}(t)$). Show that as $m$ approaches $n$, the function $T_{n \ne m}(t)$ approaches $T_{n = m}(t)$.

Tutorial Week 2 Solutions

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