数学代写|MATH1013 Mathematical Modeling

A coal miner is trapped in a cave-in after an explosion. He lies in a small chamber (volume $50 \mathrm{~m}^3$ ) into which air containing $100 \mathrm{~g} / \mathrm{m}^3$ of carbon monoxide (CO) is seeping at a rate of $0.001 \mathrm{~m}^3$ per minute, and from which the well-mixed $\mathrm{CO}$ and air is leaking at the same rate.
a. State an IVP for the quantity $Q(t) \mathrm{g}$ of $\mathrm{CO}$ in the cave for $t \geq 0$, assuming the chamber was free of $\mathrm{CO}$ at the time of the cave-in.
b. Solve the IVP in a.
c. Given that $\mathrm{CO}$ at a concentration of $1200 \mathrm{ppm}$ ( or $\approx 1.4 \mathrm{~g} / \mathrm{m}^3$ ) is lethal to human beings, how long do his rescuers have to reach him?

In Module 1, Section 1, Lecture B iii a, the video shows the development of Toricelli’s Law. One of the assumptions made is that the stream of water exiting a asmall hole of cross-sectional area $a$ in the bottom of a tank retains that cross-section. If you’ve ever watched the flow of water from a tap, you’ll know that’s not the case – the column of water actually narrows as it exits the tap. This motivates the introduction of a dimensionless contraction coefficient $\alpha$, so that the volume of water leaving the tank per unit time through the hole is assumed to be $\alpha a \sqrt{2 g h}$, rather than just $a \sqrt{2 g h}$, where $h$ is the depth of water in the tank.
a. Suppose a cylindrical tank with constant cross-sectional area $A$ has a small hole of area $a$ at the bottom. If water is pumped into the tank at a constant rate $r \mathrm{~m}^3$ per minute, justify this DE for the depth $h(t)$ of water in the tank:
$$
\frac{d h}{d t}=\frac{r-\alpha a \sqrt{2 g h}}{A} .
$$
b. Find the equilibrium solution $h_e$ of this $\mathrm{DE}$, and give a physical interpretation.
c. If the tank has height $H$, state a condition for the equilibrium solution to be physically meaningful. (Ask yourself what happens if $r$ is too large.)
d. At $t=0$, the pump is turned off (i.e., $r=0$ ) just as $h=3 \mathrm{~m}$. If $A=\pi \mathrm{m}^2$, $a=0.01 \pi \mathrm{m}^2, \alpha=0.6$ for water, and $g=9.8 \mathrm{~ms}^{-2}$, how long will it take the tank to empty?

At $t=0$, a well-circulated pond with $10^6$ litres of water contains a bacterial pollutant at a concentration of $0.01 \mathrm{~kg}$ per litre. Bacteria-free water drains into the pond at a rate of 2400 litres per day, and pure water evaporates from the pond surface (leaving the bacteria behind) at a rate of 1200 litres per day.
a. Find the volume $V(t)$ of fluid in the pond for $t \geq 0$.
b. Find the quantity $Q_0$ of bacterial pollutant in the pond at time $t=0$, and explain why it does not change.
c. Show that the concentration $C(t)=\frac{Q_0}{V(t)}$ of bacterial pollutant decreases very slowly by determining how long it takes to drop to $0.001 \mathrm{~kg}$ per litre.
d. You realize that you forgot to allow for the fact that the bacteria in the pond also reproduce at a rate $r Q(t)$ kg per day. Find the quantity $Q(t)$ of bacteria for $t \geq 0$, using $Q(0)=Q_0$. Then show that if $r \geq 0.0012$ per day, the concentration $C(t)=\frac{Q(t)}{V(t)}$ never decreases (i.e., $\frac{d C}{d t} \geq 0$ for all $t \geq 0$ ).

The intricately rough surface of a grain of pollen, its size, and other features, result in motion subject to viscous damping.
Thus, assuming displacement positive upward, a grain of pollen of mass $m$ falls from rest according to the IVP
$$
m \frac{d v}{d t}=-m g-\gamma v, \quad v(0)=0
$$
with known solution
$$
v(t)=-\frac{m g}{\gamma}\left(1-e^{-\frac{1}{m} t}\right)
$$
If we observe that a grain of pollen of known mass falls to earth from a height of $1 \mathrm{~m}$ after $25 \mathrm{~s}$, can we find the drag coefficient $\gamma$ from this data? Let’s see.
a. Find the displacement $s(t)$ by solving the IVP
$$
\frac{d s}{d t}=-\frac{m g}{\gamma}\left(1-e^{-\frac{\chi}{m} t}\right), s(0)=1
$$
b. Show that, in terms of the ratio $x=\frac{\gamma}{m}$, setting $s(t)=0$ gives the equation
$$
0=-\frac{g}{x^2}\left(e^{-x t}+x t-\frac{x^2}{g}-1\right)
$$
In particular, with $t=25 \mathrm{~s}$ and $g=9.8 \mathrm{~ms}^{-2}$, we see that $s(25)=0$ is equivalent to
$$
\frac{1}{9.8} x^2-25 x+1=e^{-25 x}
$$
c. Obviously one solution of this equation is $x=0$ (not too useful). Use Maple plots to narrow down the possible solution to around $x=245$. Since this clearly occurs at the greater positive root of the quadratic, find its roots, verifying that $x=244.96 \mathrm{~s}^{-1}$ to two decimals.
d. Now comes the fun part. We will use data for the pollen of Amaranthus palmeri, assuming a spherical grain with diameter $30 \mu \mathrm{m}$ and density $1435 \mathrm{~kg} \mathrm{~m}^{-3}$. Find the mass $m$, and use the solution from part c. to find the drag coefficient $\gamma$.
e. The researchers found that for a grain of pollen with the properties of part d., the terminal velocity had magnitude of about $4 \mathrm{~cm} \mathrm{~s}^{-1}$. Do your results from c. and d. confirm that value?