Question

Oscillating growth rates Some species have growth rates that oscillate with an (approximately) constant period P. Consider the growth rate function N'(t) = r+A sin 2πt/P, where A and r are constants with units of individuals/yr, and t is measured in years. A species becomes extinct if its population ever reaches 0 after t=0.

a. Suppose P=10, A=20, and r=0. If the initial population is N(0)=10, does the population ever become extinct? Explain.

Accepted Answer

Identify the given growth rate function: $$N'(t) = r + A \sin \left( \frac{2\pi t}{P} ight)$$, with $$P=10$$, $$A=20$$, and $$r=0. $$Write the differential equation explicitly with the given values: $$N'(t) = 0 + 20 \sin \left( \frac{2\pi t}{10} ight) = 20 \sin \left( \frac{\pi t}{5} ight). To $$find the population function $$N(t)$$, integrate the growth rate $$N'(t)$$ with respect to $$t$$:

$$N(t) = N(0) + \$$int_0$$^t N'(s) \, ds = 10 + \$$int_0$$^t 20 \sin \left( \frac{\pi s}{5} ight) ds. $$Compute the integral:

$$\int 20 \sin \left( \frac{\pi s}{5} ight) ds = - \frac{100}{\pi} \cos \left( \frac{\pi s}{5} ight) + C.

$$Apply the limits from 0 to $$t to $$get:

$$N(t) = 10 - \frac{100}{\pi} \left[ \cos \left( \frac{\pi t}{5} ight) - \cos(0) ight]. $$Analyze the expression for $$N(t) to $$determine if it ever reaches zero or below for $$t \u003e 0. $$Since $$\cos(0) = 1$$, rewrite $$N(t) as$$:

$$N(t) = 10 - \frac{100}{\pi} \left( \cos \left( \frac{\pi t}{5} ight) - 1 ight) = 10 + \frac{100}{\pi} \left( 1 - \cos \left( \frac{\pi t}{5} ight) ight).

$$Because $$1 - \cos(	heta) \geq 0$$ for all $$	heta$$, $$N(t) is $$always greater than or equal to 10, so the population never reaches zero and thus does not become extinct.

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Key Concepts

Differential Equations and Population Growth

Oscillatory Functions and Periodicity

Extinction Criteria and Population Thresholds