Answer :
Sure, let's go through the steps to find the trigonometric function and Pluto's distance from the sun in 2022.
### Step 1: Determine the period of Pluto's orbit
To find the period of the trigonometric function, we calculate the time between two successive perihelions. Given:
- Pluto's perihelion last occurred in 1989.
- Pluto's next perihelion will occur in 2237.
The period [tex]\( P \)[/tex] is:
[tex]\[ P = 2237 - 1989 = 248 \text{ years} \][/tex]
### Step 2: Calculate average distance and amplitude
Next, we find the average distance and the amplitude of the distance variation.
- Maximum distance (aphelion): 7.4 billion km
- Minimum distance (perihelion): 4.4 billion km
The average distance [tex]\( D_{\text{avg}} \)[/tex] is:
[tex]\[ D_{\text{avg}} = \frac{\text{Max distance} + \text{Min distance}}{2} = \frac{7.4 + 4.4}{2} = 5.9 \text{ billion km} \][/tex]
The amplitude [tex]\( A \)[/tex] is:
[tex]\[ A = \frac{\text{Max distance} - \text{Min distance}}{2} = \frac{7.4 - 4.4}{2} = 1.5 \text{ billion km} \][/tex]
### Step 3: Formulate the trigonometric function
The formula for Pluto's distance from the sun [tex]\( D(t) \)[/tex] can be modeled using the cosine function because the cosine function appropriately represents periodic variations. Given that [tex]\( t \)[/tex] is the time in years after 2000, we use the cosine function with period [tex]\( 248 \)[/tex] years:
[tex]\[ D(t) = \text{Average distance} + \text{Amplitude} \cdot \cos\left(\frac{2\pi t}{\text{Period}}\right) \][/tex]
Substitute the values:
[tex]\[ D(t) = 5.9 + 1.5 \cos\left(\frac{2\pi t}{248}\right) \][/tex]
### Step 4: Calculate Pluto's distance from the sun in 2022
To find the distance in 2022, we need [tex]\( t \)[/tex], which is the number of years from 2000 to 2022:
[tex]\[ t = 2022 - 2000 = 22 \][/tex]
Substitute [tex]\( t = 22 \)[/tex] into the trigonometric function:
[tex]\[ D(22) = 5.9 + 1.5 \cos\left(\frac{2\pi \cdot 22}{248}\right) \][/tex]
First, calculate the angle in radians:
[tex]\[ \frac{2\pi \cdot 22}{248} \approx 0.5574 \text{ radians} \][/tex]
Evaluate the cosine of this angle:
[tex]\[ \cos(0.5574) \approx 0.85 \][/tex]
Now, substitute this value back into the distance function:
[tex]\[ D(22) = 5.9 + 1.5 \cdot 0.85 = 5.9 + 1.275 = 7.175 \][/tex]
Rounding to two decimal places:
[tex]\[ D(22) = 7.17 \text{ billion km} \][/tex]
### Final Answers
The trigonometric function that models Pluto's distance [tex]\( D \)[/tex] from the sun [tex]\( t \)[/tex] years after 2000 is:
[tex]\[ D(t) = 5.9 + 1.5 \cos\left(\frac{2\pi t}{248}\right) \][/tex]
In the year 2022, Pluto will be approximately:
[tex]\[ 7.17 \text{ billion km} \][/tex]
from the sun.
### Step 1: Determine the period of Pluto's orbit
To find the period of the trigonometric function, we calculate the time between two successive perihelions. Given:
- Pluto's perihelion last occurred in 1989.
- Pluto's next perihelion will occur in 2237.
The period [tex]\( P \)[/tex] is:
[tex]\[ P = 2237 - 1989 = 248 \text{ years} \][/tex]
### Step 2: Calculate average distance and amplitude
Next, we find the average distance and the amplitude of the distance variation.
- Maximum distance (aphelion): 7.4 billion km
- Minimum distance (perihelion): 4.4 billion km
The average distance [tex]\( D_{\text{avg}} \)[/tex] is:
[tex]\[ D_{\text{avg}} = \frac{\text{Max distance} + \text{Min distance}}{2} = \frac{7.4 + 4.4}{2} = 5.9 \text{ billion km} \][/tex]
The amplitude [tex]\( A \)[/tex] is:
[tex]\[ A = \frac{\text{Max distance} - \text{Min distance}}{2} = \frac{7.4 - 4.4}{2} = 1.5 \text{ billion km} \][/tex]
### Step 3: Formulate the trigonometric function
The formula for Pluto's distance from the sun [tex]\( D(t) \)[/tex] can be modeled using the cosine function because the cosine function appropriately represents periodic variations. Given that [tex]\( t \)[/tex] is the time in years after 2000, we use the cosine function with period [tex]\( 248 \)[/tex] years:
[tex]\[ D(t) = \text{Average distance} + \text{Amplitude} \cdot \cos\left(\frac{2\pi t}{\text{Period}}\right) \][/tex]
Substitute the values:
[tex]\[ D(t) = 5.9 + 1.5 \cos\left(\frac{2\pi t}{248}\right) \][/tex]
### Step 4: Calculate Pluto's distance from the sun in 2022
To find the distance in 2022, we need [tex]\( t \)[/tex], which is the number of years from 2000 to 2022:
[tex]\[ t = 2022 - 2000 = 22 \][/tex]
Substitute [tex]\( t = 22 \)[/tex] into the trigonometric function:
[tex]\[ D(22) = 5.9 + 1.5 \cos\left(\frac{2\pi \cdot 22}{248}\right) \][/tex]
First, calculate the angle in radians:
[tex]\[ \frac{2\pi \cdot 22}{248} \approx 0.5574 \text{ radians} \][/tex]
Evaluate the cosine of this angle:
[tex]\[ \cos(0.5574) \approx 0.85 \][/tex]
Now, substitute this value back into the distance function:
[tex]\[ D(22) = 5.9 + 1.5 \cdot 0.85 = 5.9 + 1.275 = 7.175 \][/tex]
Rounding to two decimal places:
[tex]\[ D(22) = 7.17 \text{ billion km} \][/tex]
### Final Answers
The trigonometric function that models Pluto's distance [tex]\( D \)[/tex] from the sun [tex]\( t \)[/tex] years after 2000 is:
[tex]\[ D(t) = 5.9 + 1.5 \cos\left(\frac{2\pi t}{248}\right) \][/tex]
In the year 2022, Pluto will be approximately:
[tex]\[ 7.17 \text{ billion km} \][/tex]
from the sun.
