Problem 3, Pg 469

The following data were obtained for the growth of a sheep population introduced into a new environment on the island of Tasmania (adapted from J. Davidson, “On the Growth of the Sheep Population in Tasmania,” Trans. R. Soc. S. Australia 62(1938): 342-346).

Problem 3, Pg 469

Problem 3, Pg 469

a. Make an estimate of \(M\) by graphing \(P(t)\)

year = c(1814, 1824, 1834, 1844, 1854, 1864)
pop = c(125, 275, 830, 1200, 1750, 1650)
sheep_pop = data.frame(year, pop)
ggplot(sheep_pop, aes(x = year, y = pop)) +
     geom_point() +  geom_smooth(se = FALSE)
## `geom_smooth()` using method = 'loess'

From the graph, \(M\) appears to be ~1800 sheep

b. Plot \(ln[P/(M-P)]\) against \(t\). If a logistic curve seems reasonable, estimate \(rM\) and \(t*\).

ggplot(sheep_pop, aes(x = year, y = log(pop/(1800 - pop)))) +
    geom_point() +  geom_smooth(method = "lm", se = FALSE)

The graph does approximate a straight line, so we can assume a logistic growth for the population of sheep.

sheep_pop_lm = lm(log(pop/(1800 - pop))~year)
#Slope
(slope = sheep_pop_lm$coefficients[2])
##      year 
## 0.1189136

With our slope, \(rM\) being 0.1189136 and \(M\) approximated at 1800, \(r\) would be 6.606309110^{-5}

For \(t^*\), i.e. the time at when the population reaches M/2, and using the equation: \[t^* = t_0 - \frac{1}{rM}ln \frac{P_0}{M-P_0}\]

\[t^* = 1814 - \frac{1}{0.1189136} ln \frac{125}{1800-125} = 1836\] So, in 1836, the sheep population reached half of its limiting value.

Problem 1, Pg 481

a. Using the estimate that \(d_b = 0.054 v^2\), where 0.054 has dimension \(ft\cdot hr^2/mi^2\), show that the constant \(k\) in Equation (11.29) has the value 19.9 \(ft/sec^2\).

With the understanding that there are 3600 seconds per hour and 5281 ft in a mile. \[0.054v^2 ft\cdot hr^2/mi^2 \times \frac{3600^2 sec^2}{hr^2}\frac{mi^2}{5281^2 ft^2} = 0.0250938v^2 sec^2/ft\] Therefore \(\frac{1}{2k} = 0.0250938 sec^2/ft \rightarrow k = 19.9252408 ft/sec^2\)

b. Using the data in Table 4.26, plot \(d_b\) in ft versus \(v^2/2\) in ft2/sec2 to estimate \(1/k\) directly.

Table 4.26

Table 4.26

The original quesiton stated that the data was in Table 4.4. The only braking Table in Chapter 4 is Table 4.26. Therefore, Table 4.26 was used instead.

#Convert to ft/sec
speed = seq(20, 80, 5)*5281/60^2
b_dist = c(42, 56, 73.5, 91.5, 116, 142.5, 173, 209.5, 248, 292.5, 343, 401, 464)
b_dist_df = data.frame(speed, b_dist)

ggplot(b_dist_df, aes(x = speed^2/2, y = b_dist)) +
     geom_point() +  geom_smooth(method = "lm", se = FALSE)

v2 = speed^2/2
brake_lm = lm(b_dist~v2)
(slope = brake_lm$coefficients[2])
##         v2 
## 0.06451653

Since the slope is defined as \(\frac{1}{k} = 0.0645165sec^2/ft\), \(k = 15.4999046ft/sec^2\)

Problem 21, pg 522

Oxygen flows through one tube into a liter flask filled with air, and the mixture of oxygen and air (considered well stirred) escapes through another tube. Assuming that air contains 21% oxygen, what percentage of oxygen will the flask contain after 5 L have passed through the intake tube?

Let:

\[dP = dV - PdV\] \[\frac{dP}{dV} = 1 - P\] \[\cdot \int_{0.28}^{x}\frac{1}{1-P}dP = \cdot \int_{0}^{5}dV\] \[-ln(|x - 1|) + ln(|0.28-1|) = ln(\frac{0.72}{|x-1|}) = 5\]

Raising each side by \(e\)

\[\frac{0.72}{|x-1|} = e^5 \rightarrow 1 -\frac{0.72}{e^5} = 0.9951487\]

There would be 99.5148678% \(O_2\) after 5 L had passed.

Problem 22, pg 522

If the average person breathes 20 times per minute, exhaling each time 100 \(in^3\) of air containing 4% carbon dioxide. Find the percentage of carbon dioxide in the air of a 10,000-\(ft^3\) closed room 1 hr after a class of 30 students enters. Assume that the air is fresh at the start, that the ventilators emmit 1000 \(ft^3\) of fresh air per minute, and that the fresh air contains 0.04% carbon dioxide.

Convert

Assume

Let:

\[dP/dt = 34.72222(0.04-P) + 1000(0.0004 - P)\] \[dP/dt = (1.388889 - 34.72222P) + (0.4 - 1000P) = 1.788889 - 1034.72222P\] \[\cdot \int_{0.0004}^{x}\frac{1}{1.788889 - 1034.72222P}dP = \cdot \int_{0}^{60} \space dt\] \[\frac{ln(|1.788889 - 1034.72222*0.0004|)}{-1034.72222} - \frac{ln(|1.788889 - 1034.72222*x|)}{-1034.72222} = ln \frac{|1.788889 - 1034.72222*0.0004|}{|1.788889 - 1034.72222*x|} = 60\]

Raising each side by \(e\) and isolating x

\[\frac{|1.788889 - 1034.72222*0.0004|}{e^{60}} = 1.20402\times 10^{-26} = |1.788889 - 1034.72222*x|\]

\[x = 0.001728859\]

There would be 0.1728859% \(CO_2\) after 60 minutes had passed