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The moon's distance from Earth varies in a periodic way that can be modeled by a trigonometric function.
When the moon is at its perigee (closest point to Earth), it's about 
363,000km away. When it's at its apogee (farthest point from Earth), it's about 
406,000km away. The moon's apogees occur 27.3 days apart. The moon will reach its apogee on January 2, 2016.
Find the formula of the trigonometric function that models the distance 
D between Earth and the moon 
t days after January 1 , 2016. Define the function using radians.

D(t)=◻

The moon's distance from Earth varies in a periodic way that can be modeled by a trigonometric function.\newlineWhen the moon is at its perigee (closest point to Earth), it's about 363,000 km 363,000 \mathrm{~km} away. When it's at its apogee (farthest point from Earth), it's about 406,000 km 406,000 \mathrm{~km} away. The moon's apogees occur 2727.33 days apart. The moon will reach its apogee on January 22, 20162016.\newlineFind the formula of the trigonometric function that models the distance D D between Earth and the moon t t days after January 11 , 20162016. Define the function using radians.\newlineD(t)= D(t)=\square

Full solution

Q. The moon's distance from Earth varies in a periodic way that can be modeled by a trigonometric function.\newlineWhen the moon is at its perigee (closest point to Earth), it's about 363,000 km 363,000 \mathrm{~km} away. When it's at its apogee (farthest point from Earth), it's about 406,000 km 406,000 \mathrm{~km} away. The moon's apogees occur 2727.33 days apart. The moon will reach its apogee on January 22, 20162016.\newlineFind the formula of the trigonometric function that models the distance D D between Earth and the moon t t days after January 11 , 20162016. Define the function using radians.\newlineD(t)= D(t)=\square
  1. Determine Amplitude and Period: To find the trigonometric function that models the distance between Earth and the moon, we need to determine the amplitude, period, phase shift, and vertical shift of the function.
  2. Calculate Vertical Shift: The amplitude AA of the function is half the distance between the maximum and minimum values. The maximum distance is at apogee 406,000406,000 km and the minimum distance is at perigee 363,000363,000 km.A=406,000 km363,000 km2A = \frac{406,000 \text{ km} - 363,000 \text{ km}}{2}A=43,000 km2A = \frac{43,000 \text{ km}}{2}A=21,500 kmA = 21,500 \text{ km}
  3. Convert Period to Radians: The vertical shift DD is the average of the maximum and minimum values, which is also the midline of the function.\newlineD=(406,000km+363,000km)/2D = (406,000 \, \text{km} + 363,000 \, \text{km}) / 2\newlineD=769,000km/2D = 769,000 \, \text{km} / 2\newlineD=384,500kmD = 384,500 \, \text{km}
  4. Find Angular Frequency: The period TT of the function is the time it takes for the moon to go from one apogee to the next, which is 27.327.3 days. Since we are defining the function using radians, we need to convert the period into radians. The moon completes a full cycle 2π2\pi radians in one period TT.\newlineT=27.3T = 27.3 days
  5. Determine Phase Shift: To find the angular frequency (ω\omega), we use the formula ω=2πT\omega = \frac{2\pi}{T}.ω=2π27.3\omega = \frac{2\pi}{27.3}
  6. Define Trigonometric Function: The phase shift will be determined by the fact that the moon reaches its apogee on January 22, 20162016, which is 11 day after January 11, 20162016. Since the trigonometric function starts at its maximum at t=0t = 0, we need to shift the function to the right by 11 day.\newlinePhase shift = 11 day
  7. Define Trigonometric Function: The phase shift will be determined by the fact that the moon reaches its apogee on January 22, 20162016, which is 11 day after January 11, 20162016. Since the trigonometric function starts at its maximum at t=0t = 0, we need to shift the function to the right by 11 day.\newlinePhase shift = 11 dayNow we can define the trigonometric function using the cosine function, which starts at its maximum value. The general form of the function is:\newlineD(t)=Acos(ωtφ)+DD(t) = A \cdot \cos(\omega t - \varphi) + D\newlineWhere φ\varphi is the phase shift in radians. Since the phase shift is 11 day, we need to convert this to radians using the angular frequency ω\omega.\newlineφ=ω1\varphi = \omega \cdot 1 day\newlineφ=(2π/27.3)1\varphi = (2\pi / 27.3) \cdot 1
  8. Define Trigonometric Function: The phase shift will be determined by the fact that the moon reaches its apogee on January 22, 20162016, which is 11 day after January 11, 20162016. Since the trigonometric function starts at its maximum at t=0t = 0, we need to shift the function to the right by 11 day.\newlinePhase shift = 11 dayNow we can define the trigonometric function using the cosine function, which starts at its maximum value. The general form of the function is:\newlineD(t)=Acos(ωtφ)+DD(t) = A \cdot \cos(\omega t - \varphi) + D\newlineWhere φ\varphi is the phase shift in radians. Since the phase shift is 11 day, we need to convert this to radians using the angular frequency ω\omega.\newlineφ=ω1\varphi = \omega \cdot 1 day\newlineφ=(2π/27.3)1\varphi = (2\pi / 27.3) \cdot 1Putting it all together, we get the function:\newlinet=0t = 000

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