Solving Sinusoidal Equations involving Cosine Inverse

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Homework Help Overview

The discussion revolves around modeling the motion of a spaceship in orbit using sinusoidal equations, specifically focusing on cosine and sine functions to express the spaceship's distance from the equator over time. The problem involves determining the equation for distance based on time elapsed since launch and predicting distances at specific time intervals.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the formulation of the equation expressing the spaceship's distance from the equator in terms of time. There is an exploration of the correct variable to use (t vs. x) and the implications of plugging in specific time values into the equation.

Discussion Status

Some participants have provided guidance on how to correctly substitute values into the equation. There is an ongoing examination of the mathematical steps involved in evaluating the distance at given time points, with some participants confirming the correctness of calculations while others seek clarification on the process.

Contextual Notes

There is a focus on ensuring the correct interpretation of the sinusoidal equations and the assumptions regarding the spaceship's motion and its periodic nature. Participants are also addressing potential misunderstandings in the setup of the equations.

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


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

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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