Question

A satellite circles the earth in such a manner that it is y miles from the equator​ (north or​ south, height from the surface not​ considered) t minutes after its​ launch, where y is given by the equation below.

(picture)
At what times t in the interval ​, the first 4​ hr, is the satellite mi north of the​ equator?

Answer 1

9514 1404 393

Answer:

t ∈ {19, 85, 109, 175, 199}

or

t ∈ {40, 64, 130, 154, 220}

Explanation:

A graphing calculator relieves the tedium of solving for t. The attachments show the times the satellite is 4000 miles from the equator. Since y may be either north or south, the satellite may be 4000 north of the equator for y = 4000 or for y = -4000.

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If y is miles north of the equator, then ...

4000 = 6000cos(z) . . . . . where z=(π/45)(t-7)

z = arccos(4000/6000) ≈ 0.84107 radians

There is another solution at 2π -0.84107 radians, about 5.44212 radians

The corresponding time values are ...

t = 7 +(45/π)(z) ≈ 19.047 and 84.953 . . . . minutes

The period of the function is 90 minutes, the times will be the above times and at 90-minute intervals after each.

The satellite is 4000 miles north of the equator at ...

t ≈ 19, 85, 109, 175, 199 . . . minutes after launch

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The solution is similar if y represents miles south of the equator. Then we have ...

-4000 = 6000cos(z)

z = arccos(-4000/6000) ≈ 2.30052 radians (and 3.98266 radians)

The corresponding time values are ...

t = 7 +(45/π)(z) ≈ 39.952 and 64.047 . . . . minutes

Again, considering the period, the satellite is 4000 miles north of the equator at ...

t ≈ 40, 64, 130, 154, 220 . . . minutes after launch

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Additional comment

We might usually think of y as being positive in the northerly direction. However, in this age of political correctness and bias sensitivity, we have to consider that "y is north or south" means exactly that. So, we have also shown the solutions when y is positive south of the equator.

Usually, we prefer that the variables are defined unambiguously.