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Summersnow8 Group Title

I was wondering if I did this problem right: find an equation for the graph shown in the form y= A sin (Bx)

  • one year ago
  • one year ago

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  1. Summersnow8 Group Title
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    • one year ago
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  2. amistre64 Group Title
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    no

    • one year ago
  3. amistre64 Group Title
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    3 is good, but we have an upside down sine wave to start with, so -3 sin(kx) unless you are adjusting it: 3 sin(k(x-p))

    • one year ago
  4. amistre64 Group Title
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    it looks like a normal period of 2pi is being squished into an interval of 1 1period = 2pi 1/2pi period = 1\[y=-3sin(\frac{1}{2pi}x)\]

    • one year ago
  5. amistre64 Group Title
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    pfft, figures ... my period is incorrect

    • one year ago
  6. Summersnow8 Group Title
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    ok, i will wait

    • one year ago
  7. amistre64 Group Title
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    -3 sin(kx) = 0, when x=1/2 might suit us better sin(0) or sin(pi) = 0 k/2 = 0, when k=0 ... not a good choice k/2 = pi when k=2pi, thats better y = -3 sin(2pi x)

    • one year ago
  8. phi Group Title
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    when I do these problems, sin (k x) match 2pi/T = k where T is the length of 1 period from the graph, T=1, and we find k= 2 pi/1 = 2 pi

    • one year ago
  9. amistre64 Group Title
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    the solution on your paper produces: http://www.wolframalpha.com/input/?i=y%3D3sin%283pi+x%2F2%29 a period of 4/3

    • one year ago
  10. Summersnow8 Group Title
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    i dont understand

    • one year ago
  11. amistre64 Group Title
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    i think i see what you did: the graph you are given is marked of in intervals is: 1,2,3,... you mistakenly attributed this to 1pi, 2pi, 3pi

    • one year ago
  12. Summersnow8 Group Title
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    okay..... so what does that mean

    • one year ago
  13. amistre64 Group Title
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    it means you did something wrong ...

    • one year ago
  14. Summersnow8 Group Title
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    could you walk me through what i did wrong then.... I don't understand how to solve for Bx

    • one year ago
  15. amistre64 Group Title
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    a period is conventionally defined as the "shortest" interval that is repeated ... in this case; the normal period of sin(x) is 2pi |dw:1370963313143:dw|

    • one year ago
  16. Summersnow8 Group Title
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    yeah i understand that 2pi is the normal period of the sine function

    • one year ago
  17. amistre64 Group Title
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    your solution attributed this to: 2pi/k = 3pi instead of 2pi/k = 1

    • one year ago
  18. amistre64 Group Title
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    there was also some other mismathing going on

    • one year ago
  19. Summersnow8 Group Title
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    well... that didn't help

    • one year ago
  20. whpalmer4 Group Title
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    The normal period of a sine function is \(2\pi\) but your graph has a period going by in 1, rather than \(2\pi \approx 6.28\). Think of multiplying the argument to the sine function as speeding up or slowing the wiggling. If you want twice as many wiggles per unit length of the x-axis, you need to multiply the argument by 2, right? Well, here you are trying to get \(2\pi\) wiggles in the space of 1, so you need to multiply the argument by \(2\pi\) to speed up the wiggling.

    • one year ago
  21. whpalmer4 Group Title
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    Some pictures will help, I think. First, \(y = \sin x\)

    • one year ago
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  22. Summersnow8 Group Title
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    is there a certain equation that I could use to find this?

    • one year ago
  23. whpalmer4 Group Title
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    multiply argument by a factor of \(\pi\) closer, but still not enough

    • one year ago
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  24. whpalmer4 Group Title
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    Multiplying argument by \(2\pi\) gets the right frequency: \(y = \sin 2\pi x\)

    • one year ago
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  25. amistre64 Group Title
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    the equation/formula is the same one that weve been addressing: sin(kx), such that 2pi/k = period stated.

    • one year ago
  26. whpalmer4 Group Title
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    But comparing with problem graph, we still don't have the right amplitude, and our graph is inverted. Fix that by multiplying by -3, which scales and inverts:

    • one year ago
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  27. whpalmer4 Group Title
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    The period business is just a simple proportion of the desired period and \(2\pi\)...

    • one year ago
  28. zepdrix Group Title
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    An equation? Yes it's of this form: \(\large Period \qquad=\qquad \dfrac{2\pi}{B}\) We recognize that the graph shows a period of 1. \(\large 1\qquad=\qquad \dfrac{2\pi}{B}\) And from here you can solve for your missing B term. This is pretty much the same thing Ami said, but I wanted to include the B variable so you can match it up with your equation.

    • one year ago
  29. Summersnow8 Group Title
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    how is the period 1?

    • one year ago
  30. whpalmer4 Group Title
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    Look at the graph. the function crosses through x=0, y=0, and the next time it goes through y=0 at the same point on the graph is at x=1, right?

    • one year ago
  31. whpalmer4 Group Title
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    "at the same point on the graph" means "the corresponding spot in the next cycle of the periodic function"

    • one year ago
  32. zepdrix Group Title
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    One full period of Sine cosists of 2 "loops". :) See how it loops down once then back up, and crosses x=1.

    • one year ago
  33. Summersnow8 Group Title
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    no i don't understand.

    • one year ago
  34. whpalmer4 Group Title
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    Look at the graph in your problem. Start at x = -1. From there to x = 0 is one complete cycle of the sine function they want you to make, right? After that, from x = 0 to x = 1 is another identical cycle.

    • one year ago
  35. Summersnow8 Group Title
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    okay so there are 4 cycles

    • one year ago
  36. Summersnow8 Group Title
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    what now? I mean... you have told me the answer, but if I am given another problem like this, there is no way I would be able to solve it because all I know is what you told me..... its the first number where the second cycle starts

    • one year ago
  37. zepdrix Group Title
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    Don't worry about how many cycles there are. The question is, how much distance along the x axis does ONE cycle cover?

    • one year ago
  38. Summersnow8 Group Title
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    this isn't making any sense.....

    • one year ago
  39. zepdrix Group Title
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    |dw:1370965653907:dw|Here is a different sine curve. What is the length of one cycle?

    • one year ago
  40. Summersnow8 Group Title
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    4?

    • one year ago
  41. zepdrix Group Title
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    Yes good :) Now look back at your graph, what is the length of one cycle? That is your period.

    • one year ago
  42. whpalmer4 Group Title
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    Well, to get one cycle of \sin x, your x value has to go from 0 to \(2\pi\), right? If you need that to happen in more or less space along the x axis, you just have to multiply or divide to make that happen. Say the next problem asked you to make a sine function with period \(\pi/2\). That means that your new multiplier and the old multiplier (which is 1 by default, as in \(y = \sin 1x\)) have to satisfy a proportion: \[\frac{1}{B} = \frac{T}{2\pi}\] where \(T\) is the new period, and \(B\) is the multiplier. If our period we want is \(2\pi\) then our value of \(B\) must be 1. That's just our starting point with \(y=\sin 1x\). If we want two cycles in the space normally occupied by 1, that means our period will be \(\pi\) instead of \(2\pi\) and \[\frac{1}{B} = \frac{T}{2\pi} = \frac{\pi}{2\pi} = \frac{1}{2}\] so \(B=2\). And if you graph \(y = \sin x\) and \(y = \sin 2x\) together you'll see that it does the right thing.

    • one year ago
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  43. Summersnow8 Group Title
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    so [4=2\Pi/B\]

    • one year ago
  44. zepdrix Group Title
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    In the example I gave yes, that would be the correct way to find our missing \(\large B\) term summer. But as I said, that was just an example. That doesn't match the graph you were given.

    • one year ago
  45. Summersnow8 Group Title
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    @whpalmer4 you have confused me. I don't know how to these..... sorry for wasting your time

    • one year ago
  46. zepdrix Group Title
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    What is the length of one cycle on the graph you were given? Look at the x-axis.

    • one year ago
  47. Summersnow8 Group Title
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    1, because you told me

    • one year ago
  48. zepdrix Group Title
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    Hmm because I told you? :\ But here's the thing.. you recognized that there are 4 cycles on your own. So can't you measure the length of one of those cycles?

    • one year ago
  49. phi Group Title
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    See http://www.khanacademy.org/math/trigonometry/basic-trigonometry/trig_graphs_tutorial/v/we-amplitude-and-period it might help

    • one year ago
  50. Summersnow8 Group Title
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    @zepdrix what if the tick marks were in the middle of the x axis, no number actually given though. then i would be screwed

    • one year ago
  51. zepdrix Group Title
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    If no numbers were given, you would be unable to determine the period.

    • one year ago
  52. Summersnow8 Group Title
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    alright :/

    • one year ago
  53. zepdrix Group Title
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    The numbers along the axis are a requirement :) Or the problem isn't solvable. Is the fact that they're nice easy numbers 1,2,3 the part that's confusing you? :3

    • one year ago
  54. whpalmer4 Group Title
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    @zepdrix probably that the don't involve pi, and the problems worked in class did is the confusing part, whether realized or not

    • one year ago
  55. zepdrix Group Title
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    ah, yes maybe :)

    • one year ago
  56. whpalmer4 Group Title
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    I know that was for me, years ago!

    • one year ago
  57. Summersnow8 Group Title
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    |dw:1370966450574:dw| so what is the period of that?

    • one year ago
  58. whpalmer4 Group Title
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    don't you mean pi/2, pi, 3pi/2, 2pi? You've got numbers of decreasing size as you go to the right there...

    • one year ago
  59. zepdrix Group Title
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    Ah yes your labels are a little silly :) hehe

    • one year ago
  60. Summersnow8 Group Title
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    okay sure...... ugh!

    • one year ago
  61. zepdrix Group Title
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    |dw:1370966654243:dw|How bout this one instead :D

    • one year ago
  62. Summersnow8 Group Title
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    but i have trouble with the fraction ones....

    • one year ago
  63. zepdrix Group Title
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    |dw:1370966692448:dw|This length is the period. See how the pattern starts to repeat itself after this length?

    • one year ago
  64. zepdrix Group Title
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    Oh the fractions are the troublesome ones? ok my bad.

    • one year ago
  65. whpalmer4 Group Title
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    @zepdrix be a mensch and do one with fractions for the pretty girl, eh? :-) I gotta run, good luck, all!

    • one year ago
  66. zepdrix Group Title
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    |dw:1370966770302:dw|

    • one year ago
  67. zepdrix Group Title
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    here are some tricky fractions for ya :U

    • one year ago
  68. zepdrix Group Title
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    So what is the period of this sine function? How far do we travel before we've completed one loop up and one loop down?

    • one year ago
  69. Summersnow8 Group Title
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    2pi, pi/3?

    • one year ago
  70. zepdrix Group Title
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    yah pi/3 sounds right. We started at x=0 ended at x=pi/3 after we did 2 loops.

    • one year ago
  71. zepdrix Group Title
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    So in this little example, if we wanted to find the \(\large B\) coefficient we would go back to our formula, now with our period value in hand. \[\large Period \qquad=\qquad \dfrac{2\pi}{B}\] \[\large \frac{\pi}{3} \qquad=\qquad \dfrac{2\pi}{B}\]

    • one year ago
  72. Summersnow8 Group Title
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    6pi?

    • one year ago
  73. Summersnow8 Group Title
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    1/6 i mean

    • one year ago
  74. zepdrix Group Title
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    Solving for B? Hmmm let's see. Multiplying both sides by B,\[\large B\frac{\pi}{3} \qquad=\qquad 2\pi\]Divide each side by pi/3,\[\large B \qquad=\qquad 2\cancel{\pi}\frac{3}{\cancel{\pi}}\]

    • one year ago
  75. zepdrix Group Title
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    I think it's going to be \(\large B=6\) Do those steps make sense?

    • one year ago
  76. Summersnow8 Group Title
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    |dw:1370967157840:dw|

    • one year ago
  77. zepdrix Group Title
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    \[\large \frac{\pi}{3}=\frac{2\pi}{B}\]Dividing both sides by 2pi gives us,\[\large \frac{\pi}{3}\cdot\frac{1}{2\pi}=\frac{1}{B}\] See how the B should still be in the denominator?

    • one year ago
  78. Summersnow8 Group Title
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    |dw:1370967349882:dw|

    • one year ago
  79. zepdrix Group Title
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    ya looks good :)

    • one year ago
  80. Summersnow8 Group Title
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    okay, thank you!

    • one year ago
  81. zepdrix Group Title
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    So in our example, we would be able to write a sine function like this: \(\large y=A\sin6x\)

    • one year ago
  82. Summersnow8 Group Title
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    yeah that makes sense. and if the sine function went up to 10 and down to -10, then it would be y= 10 sin 6x

    • one year ago
  83. zepdrix Group Title
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    Yes, good.

    • one year ago
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