(last updated: 2:10:14 PM, October 08, 2020)
Population groww at rate \(r\) with initial population size \(N_0\): \(N(t)=N_0(1+r)^t\).
A city has a population of 3.5 million in 2020 and is growing 1.8% per year. (1) Project the population size in 2035. (2) In what year would it be projected to double to 7 million?
Solution: The model equation is \(N(t)=3.5(1.018)^t\).
We plug in \(t=15\) to get the population size in 2035, \(N(15)=3.5(1.018)^{15}=4.57\) million.
We want to find \(t\) that gives \(3.5(1.018)^t=7\). You can use logarithms if you know how, otherwise you can graph \(y=3.5(1.018)^x\) and \(y=7\) and see qhere they intersect. This gives \(x=\frac{\ln2}{\ln1.018}=38.9\). So it will happen during the calendar year 2020+38=2058.
\(A(t)=A_0\left(\frac12\right)^{t/\lambda}\)
\(A_0=\) initial amount, usually in units of mass such as grams, milligrams
\(\lambda=\) half life, the time it takes for the sample to decay by 50%
\(t=\) time, measured in the same units as half-life \(\lambda\)
Example: Carbon-14 has a half-life of about 5,700 years. (1) If a sample initially has 240 grams, how much is left after 11,400 years? (2) How much is left after 20,000 years? (3) How long does it take to decay to 30 grams? (4) How long does it take to decay to 25 grams?
Solution:
If you can see that 11,400=$2$5,700, then we just cut the sample in half twice: \(240\rightarrow120\rightarrow60\). So there are 60 grams left. Here it is worked out with the equation explicitly: \(A(11400)=240\left(\frac12\right)^{11400/5700}=240\left(\frac12\right)^{2}=\frac{240}{4}=60\)
\(A(20000)=240\left(\frac12\right)^{20000/5700}=240\left(\frac12\right)^{3.508772}=240\cdot 0.08785256=21.1\) so there are 21.1 grams left.
\[D=10\log I + 160\]
Example:
Find the decibel rating for a sound with intensity \(I=10^{-2} W/cm^2\).
\(D=10\log 10^{-2} + 160=10\cdot(-2)+160=-20+160=140\) dB
If a seismograph is distance \(D\) within 200 km of the epicenter of an earthquake, then the Richter magnitude \(R\) can be estimated by \[M=\log A -2.48+2.76\log(D)\] where \(A\) is the seismograph amplitude reading in \(\mu m\) (micrometers).
Example:
Find the Richter magnitude for an earthquake if a seismograph registers an amplitude of \(123\ \mu m\) and is a distance of \(20\ km\) from the epicenter.
Solution: \(M=\log (123) -2.48+2.76\log(20) = 3.2\)
Example:
Find the seismograph amplitude reading for an earthquake with a Richter magnitude of \(M=6.7\) given that the seismograph is a distance of \(30\ km\) from the epicenter.
Solution:
\(6.7=\log (A)-2.48+2.76\log(30)\)
\(6.7=\log (A)+1.6\)
\(5.1=\log (A)\)
So \(A=10^{5.1} \approx 125,893 \ \mu m\)
Note that this is about \(126 \ mm\) or \(0.126\ cm\).
\[ \diamond \S \diamond \]