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3 years ago

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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 (per

ihelion) 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.

2 answers:

Nonamiya [84]3 years ago

8 0

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.

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

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

where k is the phase and M is the amplitude.

D = M sin (2π(t–1989)/248)

D = M sin ((2πt/248) – (2π1989/248))

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

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

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

D = M sin ((2πt/248) – (2π1989/248) – (π/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 is

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

in 2000

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

D = 5.9 + 1.5 sin (-1.28)

D = 5.9 – 1.5 × 0.02234

d = 5.87 billions km

vichka [17]3 years ago

7 0

Answer:

D(t)=-1.5cos(2 $pi$ /248(t+11))+5.9

b) 4.89

Step-by-step explanation:

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