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To find the formula for the trigonometric function [tex]\( D(t) \)[/tex] that models the distance between Earth and the moon [tex]\( t \)[/tex] days after January 1, 2016, we need to follow these steps:
1. Identify the Mean Distance:
The mean distance is the average of the perigee and apogee.
[tex]\[ \text{Mean Distance} = \frac{\text{Perigee} + \text{Apogee}}{2} = \frac{363000 \, \text{km} + 406000 \, \text{km}}{2} = 384500 \, \text{km} \][/tex]
2. Calculate the Amplitude:
The amplitude is half the difference between the apogee and the perigee.
[tex]\[ \text{Amplitude} = \frac{\text{Apogee} - \text{Perigee}}{2} = \frac{406000 \, \text{km} - 363000 \, \text{km}}{2} = 21500 \, \text{km} \][/tex]
3. Determine the Angular Frequency:
The period of the moon's orbit is 27.3 days. The angular frequency [tex]\(\omega\)[/tex] is given by:
[tex]\[ \omega = \frac{2\pi}{\text{Period}} = \frac{2\pi}{27.3} \approx 0.2301533079552962 \, \text{radians per day} \][/tex]
4. Incorporate the Phase Shift:
Since the moon reaches its apogee on January 2, 2016, and we are measuring [tex]\( t \)[/tex] days after January 1, 2016, the amplitude part of the cosine function will be at its maximum at [tex]\( t = 1 \)[/tex]. This simplifies our equation as no phase shift (or phase shift [tex]\( \phi = 0 \)[/tex]) is needed. Hence, the function will model [tex]\( \cos(\omega t) \)[/tex] directly.
Putting all these components together, we get the distance function [tex]\( D(t) \)[/tex]:
[tex]\[ D(t) = 384500 + 21500 \cos(0.2301533079552962 \, t) \][/tex]
So, the trigonometric function that models the distance between Earth and the moon [tex]\( t \)[/tex] days after January 1, 2016, is:
[tex]\[ D(t) = 384500 + 21500 \cos(0.2301533079552962 \, t) \][/tex]
1. Identify the Mean Distance:
The mean distance is the average of the perigee and apogee.
[tex]\[ \text{Mean Distance} = \frac{\text{Perigee} + \text{Apogee}}{2} = \frac{363000 \, \text{km} + 406000 \, \text{km}}{2} = 384500 \, \text{km} \][/tex]
2. Calculate the Amplitude:
The amplitude is half the difference between the apogee and the perigee.
[tex]\[ \text{Amplitude} = \frac{\text{Apogee} - \text{Perigee}}{2} = \frac{406000 \, \text{km} - 363000 \, \text{km}}{2} = 21500 \, \text{km} \][/tex]
3. Determine the Angular Frequency:
The period of the moon's orbit is 27.3 days. The angular frequency [tex]\(\omega\)[/tex] is given by:
[tex]\[ \omega = \frac{2\pi}{\text{Period}} = \frac{2\pi}{27.3} \approx 0.2301533079552962 \, \text{radians per day} \][/tex]
4. Incorporate the Phase Shift:
Since the moon reaches its apogee on January 2, 2016, and we are measuring [tex]\( t \)[/tex] days after January 1, 2016, the amplitude part of the cosine function will be at its maximum at [tex]\( t = 1 \)[/tex]. This simplifies our equation as no phase shift (or phase shift [tex]\( \phi = 0 \)[/tex]) is needed. Hence, the function will model [tex]\( \cos(\omega t) \)[/tex] directly.
Putting all these components together, we get the distance function [tex]\( D(t) \)[/tex]:
[tex]\[ D(t) = 384500 + 21500 \cos(0.2301533079552962 \, t) \][/tex]
So, the trigonometric function that models the distance between Earth and the moon [tex]\( t \)[/tex] days after January 1, 2016, is:
[tex]\[ D(t) = 384500 + 21500 \cos(0.2301533079552962 \, t) \][/tex]
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