Math / Trigonometry

Question1. Interpret a Sinusoidal Model
The average observed population of woodland caribou in Wood Buffalo Provincial Park is modeled by the function: $P(t)=800 \cos \left(\frac{\pi}{3} t\right)+3000$ where $t$ is in years from $2012(t=0$ for 2012). a) Find the period of the population cycle: b) Interpret this function in the problem context in a few well-stated sentences. A sketch would help. c) Without a calculator, determine the average population in the year 2014. d) Use a calculator to determine the years to the nearest tenth (from 2012) in which the population fell below 2400. Clearly show your process. Confirm your answer by sketching the function on DESMOS and finding the points of intersection with the line $y=2400$ .

Studdy Solution

STEP 1

What is this asking? We're looking at how the number of caribou changes over time, figuring out how long a cycle takes, what the equation means in real life, and when the population dips below a certain number. Watch out! Remember that $t=0$ represents the year 2012, not the year 0.
Also, make sure your calculator is in radian mode when working with trigonometric functions!

STEP 2

1. Find the period.
2. Interpret the function.
3. Calculate the population in 2014.
4. Determine when the population falls below 2400.

STEP 3

To find the period of our caribou population cycle, we look at the coefficient of $t$ inside the cosine function.
That's $\frac{\pi}{3}$ .
The general form of a cosine function is $A \cos(Bt + C) + D$ , where the period is given by $\frac{2\pi}{|B|}$ .

STEP 4

In our case, $B = \frac{\pi}{3}$ .
So, the period is $\frac{2\pi}{|\frac{\pi}{3}|} = \frac{2\pi}{\frac{\pi}{3}} = 2\pi \cdot \frac{3}{\pi}$ .
The $\pi$ values divide to one, leaving us with a period of $2 \cdot 3 = \textbf{6}$ years.
This means the caribou population cycle repeats every 6 years!

STEP 5

Our function $P(t) = 800 \cos(\frac{\pi}{3}t) + 3000$ tells us how the caribou population changes over time.
The 3000 represents the average population, the 800 represents how much the population swings above and below this average, and the $\frac{\pi}{3}$ determines how quickly the population cycles through these changes.

STEP 6

Imagine a wave going up and down around a middle point.
The middle point is 3000, the highest point is $3000 + 800 = 3800$ , and the lowest point is $3000 - 800 = 2200$ .
The wave completes a full cycle every 6 years, as we found in the previous step.

STEP 7

We want to find the population in 2014.
Since $t=0$ represents 2012, the year 2014 corresponds to $t = 2014 - 2012 = \textbf{2}$ .

STEP 8

Now we substitute $t=2$ into our function: $P(2) = 800 \cos(\frac{\pi}{3} \cdot 2) + 3000 = 800 \cos(\frac{2\pi}{3}) + 3000$ .

STEP 9

We know that $\cos(\frac{2\pi}{3}) = -\frac{1}{2}$ .
So, $P(2) = 800 \cdot (-\frac{1}{2}) + 3000 = -400 + 3000 = \textbf{2600}$ caribou.

STEP 10

We want to find the values of $t$ for which $P(t) < 2400$ .
That means $800 \cos(\frac{\pi}{3}t) + 3000 < 2400$ .

STEP 11

Subtract 3000 from both sides: $800 \cos(\frac{\pi}{3}t) < -600$ .

STEP 12

Divide by 800: $\cos(\frac{\pi}{3}t) < -\frac{600}{800} = -\frac{3}{4}$ .

STEP 13

Using a calculator, we find the reference angle by taking the inverse cosine of $\frac{3}{4}$ , which is approximately 0.723 radians.
Since cosine is negative in the second and third quadrants, we have $\frac{\pi}{3}t \approx \pi - 0.723$ and $\frac{\pi}{3}t \approx \pi + 0.723$ .
This gives us $\frac{\pi}{3}t \approx 2.419$ and $\frac{\pi}{3}t \approx 3.865$ .

STEP 14

Now, we multiply both sides by $\frac{3}{\pi}$ to isolate $t$ : $t \approx 2.419 \cdot \frac{3}{\pi} \approx 2.3$ and $t \approx 3.865 \cdot \frac{3}{\pi} \approx 3.7$ .
Since the period is 6, we can add or subtract multiples of 6 to these values.
So, between $t=0$ and $t=6$ , the population is below 2400 when $2.3 < t < 3.7$ .

STEP 15

The caribou population cycles every 6 years.
The function shows the population fluctuating around an average of 3000.
In 2014, the population was 2600.
The population falls below 2400 between approximately 2.3 and 3.7 years after 2012.

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Studdy Solution

STEP 1

STEP 2

1. Find the period.
2. Interpret the function.
3. Calculate the population in 2014.
4. Determine when the population falls below 2400.

STEP 3

To find the period of our caribou population cycle, we look at the coefficient of $t$ inside the cosine function.
That's $\frac{\pi}{3}$ .
The general form of a cosine function is $A \cos(Bt + C) + D$ , where the period is given by $\frac{2\pi}{|B|}$ .

STEP 4

In our case, $B = \frac{\pi}{3}$ .
So, the period is $\frac{2\pi}{|\frac{\pi}{3}|} = \frac{2\pi}{\frac{\pi}{3}} = 2\pi \cdot \frac{3}{\pi}$ .
The $\pi$ values divide to one, leaving us with a period of $2 \cdot 3 = \textbf{6}$ years.
This means the caribou population cycle repeats every 6 years!

STEP 5

STEP 6

Imagine a wave going up and down around a middle point.
The middle point is 3000, the highest point is $3000 + 800 = 3800$ , and the lowest point is $3000 - 800 = 2200$ .
The wave completes a full cycle every 6 years, as we found in the previous step.

STEP 7

We want to find the population in 2014.
Since $t=0$ represents 2012, the year 2014 corresponds to $t = 2014 - 2012 = \textbf{2}$ .

STEP 8

Now we substitute $t=2$ into our function: $P(2) = 800 \cos(\frac{\pi}{3} \cdot 2) + 3000 = 800 \cos(\frac{2\pi}{3}) + 3000$ .

STEP 9

We know that $\cos(\frac{2\pi}{3}) = -\frac{1}{2}$ .
So, $P(2) = 800 \cdot (-\frac{1}{2}) + 3000 = -400 + 3000 = \textbf{2600}$ caribou.

STEP 10

We want to find the values of $t$ for which $P(t) < 2400$ .
That means $800 \cos(\frac{\pi}{3}t) + 3000 < 2400$ .

STEP 11

Subtract 3000 from both sides: $800 \cos(\frac{\pi}{3}t) < -600$ .

STEP 12

Divide by 800: $\cos(\frac{\pi}{3}t) < -\frac{600}{800} = -\frac{3}{4}$ .

STEP 13

STEP 14

STEP 15

The caribou population cycles every 6 years.
The function shows the population fluctuating around an average of 3000.
In 2014, the population was 2600.
The population falls below 2400 between approximately 2.3 and 3.7 years after 2012.

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Studdy Solution

STEP 1

STEP 2

1. Find the period.2. Interpret the function.3. Calculate the population in 2014.4. Determine when the population falls below 2400.

STEP 3

STEP 4

STEP 5

STEP 6

STEP 7

We want to find the population in **2014**.Since t=0t=0t=0 represents **2012**, the year **2014** corresponds to t=2014−2012=2t = 2014 - 2012 = \textbf{2}t=2014−2012=2.

STEP 8

Now we substitute t=2t=2t=2 into our function: P(2)=800cos⁡(π3⋅2)+3000=800cos⁡(2π3)+3000P(2) = 800 \cos(\frac{\pi}{3} \cdot 2) + 3000 = 800 \cos(\frac{2\pi}{3}) + 3000P(2)=800cos(3π​⋅2)+3000=800cos(32π​)+3000.

STEP 9

We know that cos⁡(2π3)=−12\cos(\frac{2\pi}{3}) = -\frac{1}{2}cos(32π​)=−21​.So, P(2)=800⋅(−12)+3000=−400+3000=2600P(2) = 800 \cdot (-\frac{1}{2}) + 3000 = -400 + 3000 = \textbf{2600}P(2)=800⋅(−21​)+3000=−400+3000=2600 caribou.

STEP 10

We want to find the values of ttt for which P(t)<2400P(t) < 2400P(t)<2400.That means 800cos⁡(π3t)+3000<2400800 \cos(\frac{\pi}{3}t) + 3000 < 2400800cos(3π​t)+3000<2400.

STEP 11

**Subtract 3000** from both sides: 800cos⁡(π3t)<−600800 \cos(\frac{\pi}{3}t) < -600800cos(3π​t)<−600.

STEP 12

**Divide by 800**: cos⁡(π3t)<−600800=−34\cos(\frac{\pi}{3}t) < -\frac{600}{800} = -\frac{3}{4}cos(3π​t)<−800600​=−43​.

STEP 13

STEP 14

STEP 15

The caribou population cycles every **6** years.The function shows the population fluctuating around an average of **3000**.In **2014**, the population was **2600**.The population falls below **2400** between approximately **2.3** and **3.7** years after **2012**.

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1. Find the period.
2. Interpret the function.
3. Calculate the population in 2014.
4. Determine when the population falls below 2400.

We want to find the population in 2014.
Since $t=0$ represents 2012, the year 2014 corresponds to $t = 2014 - 2012 = \textbf{2}$ .

Now we substitute $t=2$ into our function: $P(2) = 800 \cos(\frac{\pi}{3} \cdot 2) + 3000 = 800 \cos(\frac{2\pi}{3}) + 3000$ .

We know that $\cos(\frac{2\pi}{3}) = -\frac{1}{2}$ .
So, $P(2) = 800 \cdot (-\frac{1}{2}) + 3000 = -400 + 3000 = \textbf{2600}$ caribou.

We want to find the values of $t$ for which $P(t) < 2400$ .
That means $800 \cos(\frac{\pi}{3}t) + 3000 < 2400$ .

Subtract 3000 from both sides: $800 \cos(\frac{\pi}{3}t) < -600$ .

Divide by 800: $\cos(\frac{\pi}{3}t) < -\frac{600}{800} = -\frac{3}{4}$ .

The caribou population cycles every 6 years.
The function shows the population fluctuating around an average of 3000.
In 2014, the population was 2600.
The population falls below 2400 between approximately 2.3 and 3.7 years after 2012.