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We model the tank problem the following way of \begin{equation*} \frac{dx}{dt}=R_{in}-R_{out} \end{equation*} Then \begin{equation*} \frac{dx}{dt}=-\frac{2x}{200}=-\frac{x}{100} \end{equation*} We know the original concentration of \(1g/L\), then apply the separable equations method \begin{equation*} \int \frac{100}{x} dx = \int -1 dt \end{equation*} Solving it yields \begin{equation*} x = 200 e^{-\frac{t}{100}} \end{equation*} We need to find the time that will elapse before the concentration of dye in the tank reaches \(1\%\). So \begin{equation*} \frac{x(t)}{x(0)} = 0.01 \implies 0.01 = e^{-\frac{t}{100}} \end{equation*} Solving for \(t\) returns that \(t = 100 \ln 100\) min \(\approx 460.5\) min.

So our kinetic and potential energies are equal. Then \begin{equation*} mgh = \frac{1}{2}mv^2 \implies v = \sqrt{2gh} \end{equation*}

Recall \begin{equation*} \frac{dv}{dt} = A(h) \frac{dh}{dt} \quad\text{and}\quad \frac{dv}{dt} = av \end{equation*} So then because constant \(\alpha\) is contracting, the change is negative \begin{equation*} A(h)\frac{dh}{dt} = -\alpha a \sqrt{2gh} \end{equation*}

Recall \(A(r) = \pi r^2\). Let \(h=3\) and \(\alpha = 0.6\). Then solving for radius of 1m and the circular outlet radius of 0.1m, we get an equation \begin{equation*} A(1)\frac{dh}{dt} = -(0.6) \times A(0.1) \sqrt{2gh}\\ \implies \pi\frac{dh}{dt} = -0.006 \pi \sqrt{2gh}\\ \implies \frac{dh}{dt} = -0.006 \sqrt{2gh}\\ \end{equation*} Solving for \(t\) yields \(\approx 130.41\).

See that \begin{equation*} \frac{Q(5730)}{Q_0} = 0.5\\ \implies \frac{Q_0 e^{-r(5730)}}{Q_0} = 0.5\\ \implies e^{-r(5730)} = 0.5\\ \end{equation*} Solve for \(r\) to get \(r = 0.00012097 yr^{-1}\)

Trivially, it would be \(Q_0 \exp{(-0.00012097t)}\), where \(t\) is measured in years.

Solve \(e^{-rt} = 0.2\) to get \(13,305yr\).

\begin{equation*} y_{1,2} = \frac{K + T \pm \sqrt{K^2 - KT + T^3}}{3} \end{equation*}