Math  /  Algebra

QuestionA population of animals oscillates sinusoidally between a low of 600 on January 1 and a high of 800 on July 1. Graph the population against time and use your graph to find a formula for the population PP as a function of time tt, in months since the start of the year. Assume that the period of PP is one year. P(t)=P(t)= \square

Studdy Solution

STEP 1

What is this asking? We need to find an equation that describes how an animal population changes throughout the year, knowing the highest and lowest populations and when they occur. Watch out! Remember that time tt is in months, and the period is one year, which is 12 months!
Also, make sure your formula gives the correct population at t=0t=0 (January 1) and t=6t=6 (July 1).

STEP 2

1. Find the midline
2. Find the amplitude
3. Determine the cosine function
4. Write the final equation

STEP 3

The midline is the average of the highest and lowest population values.
It's like the middle of a swing, where the population swings up and down!

STEP 4

Midline=Highest Population+Lowest Population2 \text{Midline} = \frac{\text{Highest Population} + \text{Lowest Population}}{2}

STEP 5

Plugging in our **highest population** of 800 and **lowest population** of 600, we get: Midline=800+6002=14002=700 \text{Midline} = \frac{800 + 600}{2} = \frac{1400}{2} = 700

STEP 6

The amplitude is how far the population swings up or down from the midline.
It's like measuring how high the swing goes!

STEP 7

Amplitude=Highest PopulationLowest Population2 \text{Amplitude} = \frac{\text{Highest Population} - \text{Lowest Population}}{2}

STEP 8

Using our **highest population** of 800 and **lowest population** of 600: Amplitude=8006002=2002=100 \text{Amplitude} = \frac{800 - 600}{2} = \frac{200}{2} = 100

STEP 9

We're using cosine because the population starts at its lowest point.
Think of a cosine wave starting at the bottom of its curve.

STEP 10

Since the period is **one year**, which is **12 months**, our function will look like this: P(t)=Acos(Bt)+C P(t) = A \cdot \cos(Bt) + C where AA is the amplitude, BB relates to the period, and CC is the midline.

STEP 11

We already know A=100A = 100 and C=700C = 700.
Now we need to find BB.
Since the period is 12, we have: 2πB=12 \frac{2\pi}{B} = 12

STEP 12

To solve for BB, we can multiply both sides by BB and divide both sides by 12: B=2π12=π6 B = \frac{2\pi}{12} = \frac{\pi}{6}

STEP 13

Because the population *starts* at its *lowest* value, we need to flip our cosine function.
We do this by multiplying the amplitude by -1.
So, our amplitude becomes 100-100.

STEP 14

Putting it all together, we get our final equation: P(t)=100cos(π6t)+700 P(t) = -100 \cdot \cos\left(\frac{\pi}{6} t\right) + 700

STEP 15

P(t)=100cos(π6t)+700 P(t) = -100 \cdot \cos\left(\frac{\pi}{6} t\right) + 700

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