1 / 19

Sinusoidal Models 6.7

Sinusoidal Models 6.7. We can apply Sinusoidal models in many situations. Ex: 1. The water depth in a harbor is 21 m at high tide and 11 m at low tide. One cycle is completed approximately every 12 h. a) Find an equation for the water depth as a function of the time, t hours, after low tide.

nerina
Download Presentation

Sinusoidal Models 6.7

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Sinusoidal Models 6.7

  2. We can apply Sinusoidal models in many situations • Ex:1. • The water depth in a harbor is 21 m at high tide and 11 m at low tide. One cycle is completed approximately every 12 h. a)Find an equation for the water depth as a function of the time, t hours, after low tide. b)State the times, in a 48 hour period, at which the water depth was: • i) a maximum ii) a minimum iii) at its average value

  3. How do we tackle this problem? • Start with the easy stuff • Amplitude and Equation of axis • Next move to the tougher stuff period and potential phase shift if necessary.

  4. What will be the amplitude? Amp = (max-min)/2 =(21-11)/2 =(10)/2 =5m

  5. What is Equation of Axis? E.O.A = (max+min)/2 = (21+11)/2 =(32)/2 =16m What does this represent in the context of the problem?

  6. What is the K- value Since one cycle is completed every 12 hours we can assume that the period is 12h. What will the k value be? K=(360)/Period =360/12 =30

  7. How to set up the equation. • Sketch a rough draft • Now that we know the basic stuff we have some decisions to make. • Do we use Cos or Sine? • Will there be any phase shift?

  8. Cos or Sin? • Assess what the question says again. • Look at the graph Any Phase shift?

  9. Now answer the other questions. • Maximums occur every 12 hours starting at hour 6. So 6,18,30,42 • Minimums occur every 12 hours starting at 0. So 0,12,24,36,48 • Normal water level occurs at quarter points. So 3,9,15,21,27,33,39,45

  10. Try one more! • The number of hours of daylight measured in one year in Lewisville can be modeled by a sinusoidal function. During 2006, (not a leap year), the longest day occurred on June 21 with 15.7 h of daylight. The shortest day of the year occurred on December 21 with 8.3 h of daylight. Write a sinusoidal equation to model the hours of daylight in Lewisville.

  11. How do we tackle this problem? • Start with the easy stuff • Amplitude and Equation of axis • Next move to the tougher stuff period and potential phase shift if necessary.

  12. What will be the amplitude? Amp = (max-min)/2 =(15.7-8.3)/2 =(7.4)/2 =3.7hrs

  13. What is Equation of Axis? E.O.A = (max+min)/2 = (15.7+8.3)/2 =(24)/2 =12 hrs What does this represent in the context of the problem?

  14. What is the K- value Since the number of days between June 21 and Dec 21 is 183 days, we have to assume that our cycle is 366 days because in any sinusoidal graph, the distance between max and min represents half of the cycle. What will the k value be? K=360/366 =180/183

  15. How to set up the equation. • Sketch a rough draft • Now that we know the basic stuff we have some decisions to make. • Do we use Cos or Sine? • Will there be any phase shift?

  16. Cos or Sin? • Assess what the question says again. • Look at the graph The choice is up to you! Let’s try cos.

  17. Let’s try cos. • For a cos function, we would expect the graph to start out at a maximum. But our graph does not reach maximum till June 21. • In other words the graph has been shifted 172 days to the right.

  18. Let’s try sin • With sin the max should occur at first quarter or 366/4 =91.5. • But as we see the max occurs on day 172. The difference is 172-91.5=80.5

  19. Conclusion • Yes, this stuff makes you think! • Do the easy stuff first, then move on to the harder stuff!

More Related