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Pluto's maximum distance from the sun (aphelion) is 7.47.47, point, 4 billion kilometers. Its minimum distance from the sun (perihelion) is 4.44.44, point, 4 billion kilometers. Pluto last reached its perihelion in the year 198919891989, and will next reach its perihelion in 223722372237. Find the formula of the trigonometric function that models Pluto's distance DDD from the sun (in billion \text{km}kmstart text, k, m, end text) ttt years after 200020002000. Define the function using radians.

Respuesta :

temdan2001

Answer:

A.) D = 5.9 + 1.5sin (0.02534t – 51.96) (billions km)

B.) D = 5.87 billions km

Step-by-step explanation:

Assume a sine/cosine wave.Β 

the period (min to max) is 2237–1989 = 248 years

so the first trial is D = M sin ((2Ο€t/248) + k)

where k is the phase and M is the amplitude.

D = M sin (2Ο€(t–1989)/248)

D = M sin ((2Ο€t/248) – (2Ο€1989/248))

at t = 1989, we have sin(0) = 0

at t = 2237, we have sin(56.67534–50.39216) = sin 2Ο€ = 0

but we want a min at those points, equivalent to sin 270ΒΊ or 3Ο€/2, or –π/2, so we have to add in additional phase shift

D = M sin ((2Ο€t/248) – (2Ο€1989/248) – (Ο€/2))

D = M sin (0.0253354t – 50.39216 – 1.570796)

D = M sin (0.0253354t – 51.96296)

at t = 1989,

D = M sin (50.39216 – 51.96296) = M sin 1.5708 = –1 ok

min to max distance is 7.4 – 4.4 = 3

so the equaton isΒ 

D = 5.9 + 1.5sin (0.02534t – 51.96) (billions km)

in 2000

D = 5.9 + 1.5 sin (0.02534Γ—2000 – 51.96) (billions km)

D = 5.9 + 1.5 sin (-1.28)

D = 5.9 – 1.5 Γ— 0.02234

d = 5.87 billions km

Answer:

D(t)=-1.5cos(2[tex]pi[/tex]/248(t+11))+5.9

b) 4.89

Step-by-step explanation:

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