Solving Sinusoidal Equations involving Cosine Inverse

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The discussion revolves around modeling the altitude of a spaceship in orbit around Earth using sinusoidal equations. The spaceship reaches its maximum distance of 4000 kilometers north of the equator ten minutes after launch and then 4000 kilometers south half a cycle later, completing an orbit every 90 minutes. The equation expressing the spaceship's distance from the equator is confirmed as y = 4000cos(π/45(t - 10)). Participants calculate the distance for specific times, confirming that at t = 25 minutes, the spaceship is 2000 kilometers north of the equator. The conversation emphasizes proper variable usage and plugging values into the equation for accurate predictions.
Court_2013
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Suppose that a spaceship is fired into orbit from Cae Canerveral. Ten minutes after it leaves Cape, it reaches its farthest distance north of the equator, 4000 kilometers. Half a cycle later it reaches its farthest distance south of the equator (on the other side of the Earth, of course!), also 4000 kilometers. The spaceship completes an orbit once every 90 minutes.
Let y be the number of kilometers the spaceship is north of the equator (you may consider distances south of the equator to be negative). Let t be the number of minutes that have elapsed since lift off.

b. write an equation expressing y in terms of t.
c. use your equation to predict the distance of the spaceship from the equator when
(i.) t= 25, (ii.) t= 41, (iii.) t= 163

Relevant equations
y=4000cos∏/45(x-10)
y=-4000sin∏/45(x-32.5)

25=4000cos∏/45(x-10) divide by (x-10) then what?
 
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Court_2013 said:
Suppose that a spaceship is fired into orbit from Cae Canerveral. Ten minutes after it leaves Cape, it reaches its farthest distance north of the equator, 4000 kilometers. Half a cycle later it reaches its farthest distance south of the equator (on the other side of the Earth, of course!), also 4000 kilometers. The spaceship completes an orbit once every 90 minutes.
Let y be the number of kilometers the spaceship is north of the equator (you may consider distances south of the equator to be negative). Let t be the number of minutes that have elapsed since lift off.

b. write an equation expressing y in terms of t.
c. use your equation to predict the distance of the spaceship from the equator when
(i.) t= 25, (ii.) t= 41, (iii.) t= 163

Relevant equations
y=4000cos∏/45(x-10)
I assume you mean
y = 4000\cos \left( \frac{\pi}{45}(t - 10)\right),
and that it should be t, not x.

Court_2013 said:
y=-4000sin∏/45(x-32.5)

25=4000cos∏/45(x-10) divide by (x-10) then what?
Do you mean
25=\frac{4000\cos \left( \frac{\pi}{45}(x - 10)\right)}{x - 10}?
If so, that's wrong. All you need to do is to plug in for t. For t = 25,
y = 4000\cos \left( \frac{\pi}{45}(25 - 10)\right) = ?
 
eumyang said:
I assume you mean
y = 4000\cos \left( \frac{\pi}{45}(t - 10)\right),
and that it should be t, not x.


Do you mean
25=\frac{4000\cos \left( \frac{\pi}{45}(x - 10)\right)}{x - 10}?
If so, that's wrong. All you need to do is to plug in for t. For t = 25,
y = 4000\cos \left( \frac{\pi}{45}(25 - 10)\right) = ?

y=4000cos(∏/45(25-10))
y=4000cos(.0698131701(15))
y=4000cos(1.04719751)
y=4000(.5)
y=2000 is this math correct ?
 
Looks right to me.
 
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