Page 2 of 3 FirstFirst 123 LastLast
Results 16 to 30 of 43

Math Help - Amusement park problem.

  1. #16
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,132
    Thanks
    1012

    Re: Amusement park problem.

    Quote Originally Posted by explodingtoenails View Post
    Mhm, I know that equation.

    If the y-axis is h and the x-axis is t, would the equation be:

    h(t) = 19cos24t - 19

    I did the - 19 because the graph starts in the negatives.
    attached is the graph of your position function ... period is not 15 seconds and the position is always negative. ???

    I recommend you take some time to review the basics of trig graphs ...

    Graphing Trigonometric Functions
    Attached Thumbnails Attached Thumbnails Amusement park problem.-cosine1.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Junior Member
    Joined
    Aug 2011
    Posts
    73

    Re: Amusement park problem.

    Why isn't the period 15 seconds?
    Follow Math Help Forum on Facebook and Google+

  3. #18
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,132
    Thanks
    1012

    Re: Amusement park problem.

    Quote Originally Posted by explodingtoenails View Post
    Why isn't the period 15 seconds?
    the period of motion is 15 seconds ... the graph of your function does not reflect that fact.
    Follow Math Help Forum on Facebook and Google+

  4. #19
    Junior Member
    Joined
    Aug 2011
    Posts
    73

    Re: Amusement park problem.

    That's the problem I had from the beginning...I didn't know how to relate the motion to the height in an equation. I need help.
    Follow Math Help Forum on Facebook and Google+

  5. #20
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,132
    Thanks
    1012

    Re: Amusement park problem.

    graph this equation and compare it to the first graph I posted ...

    h = -19\cos\left(\frac{2\pi t}{15}\right)

    ... then see if you can determine why it models the motion of a seat on the wheel.
    Follow Math Help Forum on Facebook and Google+

  6. #21
    Junior Member
    Joined
    Aug 2011
    Posts
    73

    Re: Amusement park problem.

    I'll give my best go at it...

    Uh 19 is half of 38, so it would go up 19 and down 19. Would it matter if it was -19 or positive 19?

    Also, why is it cosine instead of sine? Would it matter since they are just the same graph with different translations?

    2pi is one revolution, times t number of seconds, and it's divided by 15 because a revolution happens every 15 seconds.
    Last edited by explodingtoenails; August 18th 2011 at 12:38 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #22
    Junior Member
    Joined
    Aug 2011
    Posts
    73

    Re: Amusement park problem.

    19sin((2pi/15)t - pi/2) produces the same graph as the one you gave to me. I think I'm finally getting this!
    Follow Math Help Forum on Facebook and Google+

  8. #23
    Junior Member
    Joined
    Aug 2011
    Posts
    73

    Re: Amusement park problem.

    a) 19sin((2pi/15)t - (pi/2))

    b) 18.6 feet

    c) I used the equation 27 = 19sin((2pi/15)t - (pi/2)) + 3 but apparently you can't arcsine anything greater than 1 so I'm not sure how to do this one.
    Follow Math Help Forum on Facebook and Google+

  9. #24
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,132
    Thanks
    1012

    Re: Amusement park problem.

    Quote Originally Posted by explodingtoenails View Post
    I'll give my best go at it...

    Uh 19 is half of 38, so it would go up 19 and down 19. Would it matter if it was -19 or positive 19?

    in this case, -19, because the graph of motion in a cosine curve reflected over the t-axis

    Also, why is it cosine instead of sine? Would it matter since they are just the same graph with different translations?

    easier to use cosine in this case ... no horizontal shift is necessary

    2pi is one revolution, times t number of seconds, and it's divided by 15 because a revolution happens every 15 seconds.

    that's right
    hope you learned something
    Follow Math Help Forum on Facebook and Google+

  10. #25
    Junior Member
    Joined
    Aug 2011
    Posts
    73

    Re: Amusement park problem.

    Can you help me out with part c, please? I don't know what I'm doing wrong.
    Follow Math Help Forum on Facebook and Google+

  11. #26
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,132
    Thanks
    1012

    Re: Amusement park problem.

    as mentioned in the problem statement, the t-axis represents the middle of the wheel.

    where would the ground be relative to the t-axis?

    where would 27 ft above the ground be?
    Attached Thumbnails Attached Thumbnails Amusement park problem.-ferris2.jpg  
    Follow Math Help Forum on Facebook and Google+

  12. #27
    Junior Member
    Joined
    Aug 2011
    Posts
    73

    Re: Amusement park problem.

    The ground would be 3 feet below the 19 foot radius of the wheel.

    There's no way to specify where 27 feet above the ground would be.
    Follow Math Help Forum on Facebook and Google+

  13. #28
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,132
    Thanks
    1012

    Re: Amusement park problem.

    Quote Originally Posted by explodingtoenails View Post
    The ground would be 3 feet below the 19 foot radius of the wheel.

    There's no way to specify where 27 feet above the ground would be.
    if the bottom of the curve represents 3 ft above the ground, where would 27 ft above the ground be? come on ... THINK!
    Follow Math Help Forum on Facebook and Google+

  14. #29
    Junior Member
    Joined
    Aug 2011
    Posts
    73

    Re: Amusement park problem.

    Uhhh it would be somewhere in the upper part of the curve. The maximum height is 41, so 27 is somewhere above the t-axis.
    Follow Math Help Forum on Facebook and Google+

  15. #30
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,132
    Thanks
    1012

    Re: Amusement park problem.

    Quote Originally Posted by explodingtoenails View Post
    Uhhh it would be somewhere in the upper part of the curve. The maximum height is 38, so 27 is somewhere above the t-axis.
    now ... where exactly?
    Attached Thumbnails Attached Thumbnails Amusement park problem.-ferris3.jpg  
    Follow Math Help Forum on Facebook and Google+

Page 2 of 3 FirstFirst 123 LastLast

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: March 29th 2009, 03:59 PM
  2. Circular park
    Posted in the Geometry Forum
    Replies: 3
    Last Post: August 20th 2008, 12:09 AM
  3. amusement park ride
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 6th 2008, 06:06 AM

Search Tags


/mathhelpforum @mathhelpforum