Problem 75 - Algebra and Trigonometry with Analytic Geometry

Question

Many animal populations, such as that of rabbits, fluctuate over ten-year cycles. Suppose that the number of rabbits at time \(t\) (in years) is given by $$ N(t)=1000 \cos \frac{\pi}{5} t+4000 . $$ (a) Sketch the graph of \(N\) for \(0 \leq t \leq 10\). (b) For what values of \(t\) in part (a) does the rabbit population exceed 4500 ?

Step-by-Step Solution

Answer

The rabbit population exceeds 4500 for \( 0 < t < 3.33 \) and \( 6.67 < t < 10 \).

Step 1: Identify the Function

The function given for the rabbit population over time is \( N(t) = 1000 \cos \frac{\pi}{5} t + 4000 \). This is a cosine function with a vertical shift of 4000 and an amplitude of 1000.

Step 2: Determine Key Features of the Graph

The function \( N(t) \) oscillates between 3000 and 5000 since its amplitude is 1000. The period of the function is found using the formula \( \frac{2\pi}{b} \), where \( b = \frac{\pi}{5} \). Thus, the period is 10 years.

Step 3: Sketch the Graph

For \( 0 \leq t \leq 10 \), sketch the cosine wave starting at \( t=0 \) at 5000, decreasing to 4000 at \( t=2.5 \), 3000 at \( t=5 \), back to 4000 at \( t=7.5 \), and ending at 5000 at \( t=10 \). Extend this smoothly to complete the cycle.

Step 4: Solve for Population Exceeding 4500

We set the inequality \( N(t) > 4500 \) which simplifies to \( 1000 \cos \frac{\pi}{5} t + 4000 > 4500 \). This gives \( \cos \frac{\pi}{5} t > 0.5 \).

Step 5: Calculate the Time Interval from Inequality

The cosine of \( \pi/3 \) is 0.5, hence solve \( \frac{\pi}{5} t < \frac{\pi}{3} \) and \( \frac{\pi}{5} t > -\frac{\pi}{3} \). Converting, we find \( 3.33 > t > -3.33 \) or in the positive interval from 0 to 10, \( t \in (0, 3.33) \) and \( t \in (6.67, 10) \).

Key Concepts

Understanding the Cosine Function

The cosine function is a fundamental trigonometric function that describes oscillatory behavior. In the context of our rabbit population example, it helps model the fluctuations observed over time. The basic form of a cosine function is \( y = a \cos(bx + c) + d \). Here:

• \(a\) represents the amplitude.
• \(b\) affects the period of the wave.
• \(c\) shifts the wave horizontally.
• \(d\) shifts the wave vertically.

In the equation \(N(t) = 1000 \cos \frac{\pi}{5} t + 4000\), the function oscillates around the line \(y = 4000\), due to the vertical shift. The role of the cosine function here is to capture the cyclic nature of rabbit population growth and decline.

Exploring Amplitude

Amplitude is the measure of how much a wave fluctuates above and below its central axis. For the cosine function, it defines half the distance between the maximum and minimum values of the function. In our example of rabbit populations, the amplitude is 1000. This means the population will vary by 1000 rabbits above and below the equilibrium value of 4000. Thus, the highest population point reaches 5000 rabbits, while the lowest point dips to 3000 rabbits. Amplitude is crucial for determining how significant these fluctuations are in practical terms.

Identifying the Period

The period refers to the length of one complete cycle of a wave. It is the time taken for the function to return to its starting value and is calculated using the formula \( \frac{2\pi}{|b|} \) for the cosine function \( y = \cos(bx) \).In our function, \(b = \frac{\pi}{5}\), so the period is \( \frac{2\pi}{\pi/5} = 10 \) years. Therefore, the rabbit population completes one full cycle of rising and falling every 10 years. This period reflects the natural fluctuation cycle promised by the cosine model.

Solving Inequality Solutions

To find when the rabbit population exceeds 4500, we solve the inequality \(N(t) > 4500\). This simplifies to \(1000 \cos \frac{\pi}{5} t + 4000 > 4500\), leading to \(\cos \frac{\pi}{5} t > 0.5\).The key is knowing that \(\cos \theta = 0.5\) at \(\theta = \pm \frac{\pi}{3}\). Solving \(\frac{\pi}{5} t < \frac{\pi}{3}\) and \(\frac{\pi}{5} t > -\frac{\pi}{3}\), we find critical intervals where \(t\) satisfies this inequality: \(t \in (0, 3.33)\) and \(t \in (6.67, 10)\).Therefore, the rabbit population surpasses 4500 rabbits during these intervals. This approach helps determine crucial time frames in population increases for biological or environmental management.