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Trigonometric Functions- Grade 11

  1. Jan 2, 2004 #1
    Ok I have two questions both i have done but not 100% sure I have done them correct. I am good with math but not so well at Trigonemetric stuff. Ok here it is.

    1) Convert the following angles to radian measure, leave in simpilest rational form. 350 degrees.
    Ok what i have done is this:
    35pie/18 radians. Is that correct and simpilest form?

    2) A ferris wheel with a radius of 20m rotates once every 40 seconds. At the bottom of the ride, the passenger is 1m above the ground. a) Determine a function that represents height, h, above ground and at time t, if h=41 at t=0.

    Ok so i went and drew a graph of what i think it would looks like and from it got this information. Verticle translation of 21, phase shift of 10 to the right, amplitude of 20, period is 40 which equals pie/20. So from this I got the equation to be:

    Would this be correct for that question?

    The follow up question part b) is Determine the passenger's height above the ground after 15 seconds. So i just took 15 and put it into my equation as t and got an answer of 35.142m. Did I do this correct and get the correct answer? Thank you very much it is much appreciated.
  2. jcsd
  3. Jan 2, 2004 #2
    1) Looks good to me.

    2) If you calculate h for t=0 you won't get 41 so your formula isn't correct (phase is wrong). That would also make your followup answer incorrect. I would use the cosine because the formula would be a bit simpler.

  4. Jan 2, 2004 #3
    Oh well the problem is i dont know what i have done wrong so that is where i get stuck, i wasnt sure if it looked right and this is one thing i am not very good at so any help would be appricated.
  5. Jan 3, 2004 #4
    You shouldn't give up so easily. Unless you haven't figured it out after a couple of weeks like a certain fundamental problem I have been working on. Also, I'm not sure why you say you are not very good at this. You have everything in this problem exactly right except for one thing, the phase. Like I said before.

    If you are going into the physical sciences or engineering you will see this kind of problem appear on a regular basis so it is a good idea to learn to do it well now.

    Once you have your function it is always a good idea to check it for a few values of t. Because this one is periodic, the beginning and the middle of the period are good choices. Because the period is 40 seconds, you should check the function for t=0 and t=20. The problem states that h=41 at t=0. Thus h=1 at t=20. Your function gives h=1 at t=0 and h=41 at t=20. This is exactly the opposite of what you want. This means the phase if off by [tex]\pi[/tex] radians. You have the sign of the phase wrong.

    When you looked at the graph of [tex]sin(\theta)[/tex] you noticed that it is at maximum when [tex]\theta=\pi/2[/tex]. You probably reasoned that you needed to compensate with a phase of [tex]-\pi/2[/tex] where the phase you chose should have been [tex]\pi/2[/tex]. So, if your equation of motion is:


    you correctly chose:

    [tex]A=20, \omega=\frac{2\pi}{40}, c=21[/tex]

    but chose:


    instead of


    So your resulting equation should be:


    Lecture over.

  6. Jan 3, 2004 #5
    Oh ok ya i had tried and checked if it would worked befour and got an answer of 1 but thought i might had put a calculator error. Ok so i just had the sign wrong. Thanks for the help as that makes more sence. So than if the time was 15 seconds using the formula you would get approximently 6.858 correct?
    Last edited: Jan 3, 2004
  7. Jan 4, 2004 #6

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