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Trig/constructing sine function from graph

  1. Dec 21, 2009 #1
    1. The problem statement, all variables and given/known data

    Construct or "work backwards" from a sine graph to a sine formula. The formula should be in:

    y = A sin(Bx+C) +D

    Where A is amplitude, D is Vertical Shift (on y access) and Bx+C are arguments of the function

    The graph shows that the Phase Shift is 1 and Period is 6 (meaning the basic cycle begins at 1 and ends at 6)


    2. Relevant equations

    0 <= x <= 2pi (pi as in 3.14..) ("<=" is less than or equal too)

    3. The attempt at a solution

    1 <= x <= 6

    0 <= x-1 <= 5 /** subtract 1 from all terms **/

    0 <= pi(x)-1 <= 5pi /** multiply all terms by pi **/

    0 <= (pi(x)-1)/2 <= 5pi/2 /** divide all terms by 2 **/

    STUCK

    The solution should be in 0 <= Yx+-Z <= 2pi (Where Y and Z can be any integer, the only thing that matters is that the 0 is at the beginning and that 2pi is the end)
    ________________________________

    This is not the only solution I tried, just the last one, so unless I have missed something rather fundamental I can't seem to solve this one. Please note that I solved 14 other problems that were basically of the same thing using the same methods that I have used to try to solve this problem and they all worked out and check out in the answers section (I am not a student, just picking up where I left off in high school).

    If I have done something wrong (in the formatting of this posting) or you need something that I neglected to mention just let me know as this is my first time posting anything on this forum.

    Also just to clarify all i am looking for is a solution the equalities statement not the whole y = equation as I understand that entirely.

    thanks immensely!
     
    Last edited: Dec 21, 2009
  2. jcsd
  3. Dec 21, 2009 #2

    HallsofIvy

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    Then the period is 6-1= 5, not 6. The "standard" graph goes from 0 to [itex]2\pi[/itex] and has period [itex]2\pi - 0= 2\pi[/itex]. Basically, whatever your "Bx+ C" is you must have B(1)+ C= B+ C= 0 and [itex]B(6)+ C= 6B+ C= 2\pi[/itex]. Subtracting the first equation from the second, [itex]6B+ C- (B+ C)= 5B= 2\pi[/itex] so [itex]B= 2\pi/5[/itex]. Then B+ C= 0 becomes [itex]2\pi/5+ C= 0[/itex] so [itex]C= -2\pi/5[/itex]. Your equation should look like [itex]A sin((2\pi/5)x- 2\pi/5)+ D[/itex] or, equivalently, [itex]A sin(2\pi/5(x- 1))+ D[/itex]


    You are going the wrong way here!

    Why do you say they should be integers?

     
  4. Dec 21, 2009 #3

    HallsofIvy

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    To find the "Bx+ C" for any sine or cosine with a period from [itex]x_0[/itex] to [itex]x_1[/itex], Solve [itex]Bx_0+ C= 0[/itex] and [itex]Bx_1+ C= 2\pi[/itex].
     
  5. Dec 21, 2009 #4
    THANKS!

    I looked over the other problems and noticed that they were all with phase shift 0, I don't know how I missed that....

    Anyway I appreciate your replies

    @HallsOfIvy

    I tried it your way and it makes sense but when I solve the equation using period of 5 (6-1) I end with an equation of 0 <= ((pi)x-1)/2 <= 2pi which is markedly different from your procedure and end result. My equation checks out in the answers section, so I guess my question is why did you end up with something different for the same problem?
     
    Last edited: Dec 21, 2009
  6. Dec 22, 2009 #5

    HallsofIvy

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    Since you don't say what you got or what the answer in the book was, I can't answer that.
     
  7. Dec 22, 2009 #6
    @HallsOfIvy

    I did say what my end result was with "0 <= ((pi)x-1)/2 <= 2pi" but if you want the sine equation it was y = 2 sin(((pi)x-1)/2)

    EDIT sorry for all the parentheses, I really need to learn LaTeX
     
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