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Summersnow8

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

  • 10 months ago
  • 10 months ago

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  1. Summersnow8
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    • 10 months ago
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  2. amistre64
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    no

    • 10 months ago
  3. amistre64
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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))

    • 10 months ago
  4. amistre64
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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\[y=-3sin(\frac{1}{2pi}x)\]

    • 10 months ago
  5. amistre64
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    pfft, figures ... my period is incorrect

    • 10 months ago
  6. Summersnow8
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    ok, i will wait

    • 10 months ago
  7. amistre64
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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)

    • 10 months ago
  8. phi
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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

    • 10 months ago
  9. amistre64
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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

    • 10 months ago
  10. Summersnow8
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    i dont understand

    • 10 months ago
  11. amistre64
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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

    • 10 months ago
  12. Summersnow8
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    okay..... so what does that mean

    • 10 months ago
  13. amistre64
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    it means you did something wrong ...

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

    • 10 months ago
  15. amistre64
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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|

    • 10 months ago
  16. Summersnow8
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    yeah i understand that 2pi is the normal period of the sine function

    • 10 months ago
  17. amistre64
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    your solution attributed this to: 2pi/k = 3pi instead of 2pi/k = 1

    • 10 months ago
  18. amistre64
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    there was also some other mismathing going on

    • 10 months ago
  19. Summersnow8
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    well... that didn't help

    • 10 months ago
  20. whpalmer4
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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.

    • 10 months ago
  21. whpalmer4
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    Some pictures will help, I think. First, \(y = \sin x\)

    • 10 months ago
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  22. Summersnow8
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    is there a certain equation that I could use to find this?

    • 10 months ago
  23. whpalmer4
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    multiply argument by a factor of \(\pi\) closer, but still not enough

    • 10 months ago
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  24. whpalmer4
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    Multiplying argument by \(2\pi\) gets the right frequency: \(y = \sin 2\pi x\)

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

    • 10 months ago
  26. whpalmer4
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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:

    • 10 months ago
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  27. whpalmer4
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    The period business is just a simple proportion of the desired period and \(2\pi\)...

    • 10 months ago
  28. zepdrix
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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.

    • 10 months ago
  29. Summersnow8
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    how is the period 1?

    • 10 months ago
  30. whpalmer4
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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?

    • 10 months ago
  31. whpalmer4
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    "at the same point on the graph" means "the corresponding spot in the next cycle of the periodic function"

    • 10 months ago
  32. zepdrix
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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.

    • 10 months ago
  33. Summersnow8
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    no i don't understand.

    • 10 months ago
  34. whpalmer4
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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.

    • 10 months ago
  35. Summersnow8
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    okay so there are 4 cycles

    • 10 months ago
  36. Summersnow8
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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

    • 10 months ago
  37. zepdrix
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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?

    • 10 months ago
  38. Summersnow8
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    this isn't making any sense.....

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

    • 10 months ago
  40. Summersnow8
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    4?

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

    • 10 months ago
  42. whpalmer4
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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.

    • 10 months ago
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  43. Summersnow8
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    so [4=2\Pi/B\]

    • 10 months ago
  44. zepdrix
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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.

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

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

    • 10 months ago
  47. Summersnow8
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    1, because you told me

    • 10 months ago
  48. zepdrix
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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?

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

    • 10 months ago
  50. Summersnow8
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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

    • 10 months ago
  51. zepdrix
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    If no numbers were given, you would be unable to determine the period.

    • 10 months ago
  52. Summersnow8
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    alright :/

    • 10 months ago
  53. zepdrix
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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

    • 10 months ago
  54. whpalmer4
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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

    • 10 months ago
  55. zepdrix
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    ah, yes maybe :)

    • 10 months ago
  56. whpalmer4
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    I know that was for me, years ago!

    • 10 months ago
  57. Summersnow8
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    |dw:1370966450574:dw| so what is the period of that?

    • 10 months ago
  58. whpalmer4
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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...

    • 10 months ago
  59. zepdrix
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    Ah yes your labels are a little silly :) hehe

    • 10 months ago
  60. Summersnow8
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    okay sure...... ugh!

    • 10 months ago
  61. zepdrix
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    |dw:1370966654243:dw|How bout this one instead :D

    • 10 months ago
  62. Summersnow8
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    but i have trouble with the fraction ones....

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

    • 10 months ago
  64. zepdrix
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    Oh the fractions are the troublesome ones? ok my bad.

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

    • 10 months ago
  66. zepdrix
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    |dw:1370966770302:dw|

    • 10 months ago
  67. zepdrix
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    here are some tricky fractions for ya :U

    • 10 months ago
  68. zepdrix
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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?

    • 10 months ago
  69. Summersnow8
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    2pi, pi/3?

    • 10 months ago
  70. zepdrix
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    yah pi/3 sounds right. We started at x=0 ended at x=pi/3 after we did 2 loops.

    • 10 months ago
  71. zepdrix
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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. \[\large Period \qquad=\qquad \dfrac{2\pi}{B}\] \[\large \frac{\pi}{3} \qquad=\qquad \dfrac{2\pi}{B}\]

    • 10 months ago
  72. Summersnow8
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    6pi?

    • 10 months ago
  73. Summersnow8
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    1/6 i mean

    • 10 months ago
  74. zepdrix
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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,\[\large B \qquad=\qquad 2\cancel{\pi}\frac{3}{\cancel{\pi}}\]

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

    • 10 months ago
  76. Summersnow8
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    |dw:1370967157840:dw|

    • 10 months ago
  77. zepdrix
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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?

    • 10 months ago
  78. Summersnow8
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    |dw:1370967349882:dw|

    • 10 months ago
  79. zepdrix
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    ya looks good :)

    • 10 months ago
  80. Summersnow8
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    okay, thank you!

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

    • 10 months ago
  82. Summersnow8
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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

    • 10 months ago
  83. zepdrix
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    Yes, good.

    • 10 months ago
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