Tutorial Week 2

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 $T_{n = m}(t)$.

Solution

1. 1. 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.
   2. Linearly independent (same as above) 1 pt
   3. Linearly independent (same as above) 1 pt
   4. 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.
2. 1. 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)
   2. 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).
   3. 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).