Day: December 7, 2022

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202212071606 Solution to 1976-AL-AMATH-I-8

During an epidemic in a country, the rate of spread of the disease is proportional to the number of people who are healthy; the rate of cure is proportional to the number currently sick. Let P be the total population:

(a) Show that

\displaystyle{\frac{\mathrm{d}W}{\mathrm{d}t}=kP-(k+k_1)W}

where W is the number of sick people; k and k_1 are constants.
(b) Let the total population be growing at a steady rate, so that (independent of the epidemic)

P=a+bt

where a and b are positive. Show that the number of sick people is given by

W=\alpha +\beta t+ce^{\mu t}.

Calculate \alpha, \beta, and \mu in terms of the given constants a, b, k, and k_1.
What further information would you require to determine c?
(c) After a very long time, what is the proportion of people who will be in a state of sickness?

Roughwork.

\begin{aligned} \frac{\mathrm{d}W}{\mathrm{d}t}+(k+k_1)W & = k(a+bt) \\ \dots\,\textrm{ by integrating factor } I=e^{\int (k+k_1)\,\mathrm{d}t} & =e^{(k+k_1)t}\,\dots \\ e^{(k+k_1)t}\bigg(\frac{\mathrm{d}W}{\mathrm{d}t}\bigg) + \Big( e^{(k+k_1)t} (k+k_1)\Big) W & = e^{(k+k_1)t} k(a+bt) \\ e^{(k+k_1)t}W & = \int e^{(k+k_1)t}k(a+bt)\,\mathrm{d}t \\ \end{aligned}

\begin{aligned} \textrm{RHS} & = \int e^{(k+k_1)t}k(a+bt)\,\mathrm{d}t \\ & = ka\int e^{(k+k_1)t}\,\mathrm{d}t + kb\int te^{(k+k_1)t}\,\mathrm{d}t \\ & = \frac{ka}{k+k_1}\Big( e^{(k+k_1)t}\Big) + \frac{kb}{(k+k_1)^2}\int t'e^{t'}\,\mathrm{d}t' \\ & = \frac{ka}{k+k_1}\Big( e^{(k+k_1)t}\Big) + \frac{kb}{(k+k_1)^2}\Big( e^{(k+k_1)t}((k+k_1)t-1)+C\Big) \\ W &= \textrm{RHS}\Big/ e^{(k+k_1)t} \\ & = \bigg(\frac{ka}{k+k_1}-\frac{kb}{(k+k_1)^2}\bigg) + \bigg(\frac{kb}{k+k_1}\bigg) t + \bigg(\frac{kbC}{(k+k_1)^2}\bigg) e^{(-(k+k_1))t} \\ & = \alpha +\beta t+ce^{\mu t} \\ \end{aligned}

\begin{aligned} \frac{W}{P} & = \frac{\alpha +\beta t+ce^{\mu t}}{a+bt} \\ \lim_{t\to \infty}\frac{W}{P} & = \lim_{t\to\infty}\frac{\beta t}{a+bt} \\ & = \frac{\beta}{b} \\ & = \frac{k}{k+k_1} \\ \end{aligned}

This problem is not to be attempted.

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202212071217 Solution to 1971-HL-PHY-I-7

The figure below shows a three-dimensional network in the form of a pyramid, in which A is the apex, and BCDE the square base.

Each of the eight edges of the pyramid is a wire of resistance 1\,\mathrm{\Omega}. A 12\,\mathrm{V} battery with internal resistance of 0.1\,\mathrm{\Omega} is connected across B and D. Calculate

(a) the power input of the network,
(b) the terminal voltage across the battery when current flows through the network, and
(c) the potential at the points B, C, D, and E if the apex A is earthed.

Roughwork.

Draw the circuit.

Label the potential.

Straighten the main.

Calculate the equivalent.

\begin{aligned} R_{\textrm{eq}} & = \big( R\parallel R\parallel R\big) + \big( R\parallel R\parallel R\big) \\ & = \frac{1}{\frac{1}{R}+\frac{1}{R}+\frac{1}{R}}\times 2 \\ & = \frac{2R}{3} \\ & = \frac{2(1)}{3} \\ & = \frac{2}{3}\,\mathrm{\Omega} \\ \end{aligned}

This problem is not to be attempted.