\( \definecolor{flatred}{RGB}{192, 57, 43} \definecolor{flatblue}{RGB}{41, 128, 185} \definecolor{flatgreen}{RGB}{39, 174, 96} \definecolor{flatpurple}{RGB}{142, 68, 173} \definecolor{flatlightblue}{RGB}{52, 152, 219} \definecolor{flatlightred}{RGB}{231, 76, 60} \definecolor{flatlightgreen}{RGB}{46, 204, 113} \definecolor{flatteal}{RGB}{22, 160, 133} \definecolor{flatpink}{RGB}{243, 104, 224}\)

March 7, 2021

Inverse Circular Functions (Wrap-Up)

In class we saw a problem about how the populations of rabbits and coyotes fluctuated over a period of \(24\) years.

Qualitatively, you should notice that both populations are roughly sinusoidal, but they're "out of phase" with each other, which is pretty typical for a predator-prey model:

  • At first the rabbits are at a maximum.
  • Then, as the coyotes eat the rabbits, the rabbits decrease in population and the coyotes thrive.
  • At some point, there are no longer enough rabbits to feed the coyotes, and the coyotes start to die off.
  • As the coyote population declines, the rabbit population once again begins to surge.
  • And so the cycle repeats.

To get a better handle on what's going on, it helps to come up with equations that approximately model the populations. We can come up with an equation for the rabbits: \[R=5000\cos(\tfrac{2\pi}{12}t)+10000\] And then another for the coyotes: \[C=300\sin(\tfrac{2\pi}{12}t)+2000\] We can get these models by taking into account the highs and lows of the populations, where along their cycle they start, and the period of \(12\) years.

To find the times when the rabbit population equals \(12000\), we can set \(R=12000\) and solve: \[ \begin{align*} 5000\cos(\tfrac{\pi}{6}t)+10000 &= 12000\\ 5000\cos(\tfrac{\pi}{6}t) &= 2000\\ \cos(\tfrac{\pi}{6}t) &= \dfrac25\\ \end{align*} \] Using the inverse cosine function and the unit circle gives us two cases: \[\tfrac\pi6 t \approx 1.159+2\pi n\text{, }-1.159+2\pi n\] And finally multiplying through by \(\tfrac6\pi\): \[t \approx 2.214+12n\text{, }-2.215+12n\] By plugging in \(n=0,1,2,\ldots\), we find the solutions that fall within the bounds of the study: \[t \approx 2.214, 9.786, 14.214, 21.786\]

Similarly, to find times when the coyote population equals \(1800\), we can set \(C=1800\) and solve: \[ \begin{align*} 300\sin(\tfrac{\pi}{6}t)+2000 &= 1800\\ 300\sin(\tfrac{\pi}{6}t) &= -200\\ \sin(\tfrac{\pi}{6}t) &= -\tfrac23\\ \end{align*} \] Using the inverse sine function and the unit circle gives us two cases: \[\tfrac\pi6t \approx -0.729+2\pi n\text{, }-2.411+2\pi n\] And finally multiplying through by \(\tfrac6\pi\): \[t \approx -1.393+12n\text{, }-4.606+12n\] Plugging in integers for \(n\) gives us the times the population hits \(1800\) during the study: \[t \approx 7.394, 10.606, 19.394, 22.606\] Unfortunately, \(10.606-7.394=3.212\), so the scientists do have reason to be concerned: the population of coyotes is indeed dropping below \(1800\) for more than \(2\) years at a time.

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