20. the population of mosquitoes in a given area is primarily influenced by precipitation, humidity, and…

20. the population of mosquitoes in a given area is primarily influenced by precipitation, humidity, and temperature. in tropical regions, these tend to fluctuate sinusoidally in the course of a year. using trap counts and statistical projections, fairly accurate estimates of a mosquito population can be obtained. suppose the population in a certain region was modeled by the function p(t)=50 cos(\\frac{\\pi}{26}t)+950, where p(t) was the mosquito population (in thousands) in week t of the year. use the model to (a) find the period of the function; (b) graph the function over one period; (c) find the maximum and minimum population values; and (d) estimate the number of weeks the population is less than 915,000.

20. the population of mosquitoes in a given area is primarily influenced by precipitation, humidity, and temperature. in tropical regions, these tend to fluctuate sinusoidally in the course of a year. using trap counts and statistical projections, fairly accurate estimates of a mosquito population can be obtained. suppose the population in a certain region was modeled by the function p(t)=50 cos(\\frac{\\pi}{26}t)+950, where p(t) was the mosquito population (in thousands) in week t of the year. use the model to (a) find the period of the function; (b) graph the function over one period; (c) find the maximum and minimum population values; and (d) estimate the number of weeks the population is less than 915,000.

Answer

Explanation:

Step1: Recall period formula for cosine function

For $y = A\cos(Bt)+C$, the period $T=\frac{2\pi}{B}$. Here $B = \frac{\pi}{26}$, so $T=\frac{2\pi}{\frac{\pi}{26}}$.

Step2: Calculate the period

$T=\frac{2\pi}{\frac{\pi}{26}}=2\pi\times\frac{26}{\pi}=52$ weeks.

Step3: Analyze maximum and minimum of cosine - based function

The range of $\cos(\frac{\pi}{26}t)$ is $[- 1,1]$. For $P(t)=50\cos(\frac{\pi}{26}t)+950$, when $\cos(\frac{\pi}{26}t)=1$, $P_{max}=50\times1 + 950=1000$ (in thousands). When $\cos(\frac{\pi}{26}t)=-1$, $P_{min}=50\times(-1)+950 = 900$ (in thousands).

Step4: Solve for when $P(t)<915$

Set $P(t)=50\cos(\frac{\pi}{26}t)+950<915$. Then $50\cos(\frac{\pi}{26}t)<915 - 950=-35$, so $\cos(\frac{\pi}{26}t)<-\frac{35}{50}=-0.7$. Using the inverse - cosine function, $\frac{\pi}{26}t>\cos^{-1}(-0.7)$ and $\frac{\pi}{26}t<2\pi-\cos^{-1}(-0.7)$. $\cos^{-1}(-0.7)\approx2.346$ and $2\pi-\cos^{-1}(-0.7)\approx3.937$. So $t > \frac{2.346\times26}{\pi}\approx19.4$ and $t<\frac{3.937\times26}{\pi}\approx32.6$. The number of weeks in one period when $P(t)<915$ is approximately $32.6 - 19.4 = 13.2\approx13$ weeks.

Answer:

(a) The period is 52 weeks. (b) To graph $P(t)=50\cos(\frac{\pi}{26}t)+950$, start with the basic cosine function $y = \cos(x)$ with amplitude 50, vertical shift 950, and period 52. Mark key points like maximums, minimums, and x - intercepts within one period $[0,52]$. (c) The maximum population is 1000000 and the minimum population is 900000. (d) Approximately 13 weeks.