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zaphod

Please help...

  • one year ago
  • one year ago

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  1. zaphod
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    http://screencast.com/t/N9H6p6CGYc

    • one year ago
  2. zaphod
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    @satellite73 @Shadowys @RadEn

    • one year ago
  3. satellite73
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    \(a\) is the amplitude. since this starts and 3 and goes up do 9, then comes back down to 3 and then to -3, the amplitude is 6

    • one year ago
  4. satellite73
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    that is, the range is of length 12 from -3 to 9, so the amplitude is half of that, therefore \(a=6\)

    • one year ago
  5. zaphod
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    IS THERE ANY OTHER METHOD TO SOLVE IT WITH EQUATIONS.

    • one year ago
  6. satellite73
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    from your eyes you see that the period is \(\pi\) the period of \(\sin(bx)\) is \(\frac{2\pi}{b}\) so set \(\frac{2\pi}{b}=\pi\) and solve for \(b\)

    • one year ago
  7. satellite73
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    no there are no equations here, you have to visualize, since you are given a picture

    • one year ago
  8. satellite73
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    well there is an equation to find \(b\) . it is \[\frac{2\pi}{b}=\pi\] but you only know that the period is \(\pi\) from looking at the graph

    • one year ago
  9. zaphod
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    and i know c, now can i substitute it in the main equation and find a?

    • one year ago
  10. satellite73
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    that is the entire point of this exercise, not to use equations, but to visualize the period, and amplitude from the picture

    • one year ago
  11. satellite73
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    we know \(c=3\) because this is the graph of sine lifted up 3 units

    • one year ago
  12. zaphod
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    can u explain why period of sin(bx) = 2pi/b

    • one year ago
  13. satellite73
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    it is always the case that the period of \(\sin(bx)\) is \(\frac{2\pi}{b}\) we can think of it this way. since is periodic with period \(2\pi\) so it does everything on the interval \([0,2\pi)\) now if \(bx=0\) that means \(x=0\) and if \(bx=2\pi\) that means \(x=\frac{2\pi}{b}\) so that gives you the period

    • one year ago
  14. allamiro
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    sub x = pi then x = 2pi when x = pi y =3 when x = 2pi y = 9 then solve the equations

    • one year ago
  15. zaphod
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    @satellite73 period is for one complete wave right, how come its 2 pi, it has to be pi

    • one year ago
  16. allamiro
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    when x = 0 y = 0 then solve the equations for a b and c

    • one year ago
  17. zaphod
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    cn u show the working @allamiro

    • one year ago
  18. allamiro
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    dsregard the 2 pi thing = 9 just x = pi and x = 0

    • one year ago
  19. allamiro
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    what you mean show the work x = pi y = 3 3 = a sin ( b* pi ) + c x= 0 y = 0 0 = a sin ( b * 0 ) + c

    • one year ago
  20. zaphod
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    yes?

    • one year ago
  21. allamiro
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    sorry again y = 3 when x = 0 I didnt focus at the graph

    • one year ago
  22. allamiro
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    so c = 3

    • one year ago
  23. zaphod
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    ok how do u find b now.

    • one year ago
  24. allamiro
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    yes so now lets say 9 = a sin ( bx) + 3 the highest value for sin when sinx b = 1 so bx = pi /2 so from there a =6

    • one year ago
  25. zaphod
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    b?

    • one year ago
  26. allamiro
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    y = 3 when x = pi /2 3 = 6 sin ( b 2 pi ) + 3 6 sin ( 2 b pi) = 0 sin ( 2 b pi ) = sin ( 2b pi ) = sin ( pi ) b = 1/2

    • one year ago
  27. allamiro
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    y = 3 when x = 2pi * correctiion

    • one year ago
  28. allamiro
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    :)

    • one year ago
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