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Math Help - Find the times when you're 65 feet off the ground.

  1. #1
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    Find the times when you're 65 feet off the ground.

    So, there's a Ferris wheel.
    I have to find out the times when you are 65 feet off the ground within the first four minutes, with t being the time in minutes, with this equation:
    h(t)=50-42sin(\frac{\pi}{2}t)

    I also had to solve it graphically, which was easy, so I know there are two points in time at which you are 65 feet off the ground(something like ~2.23 and ~3.77 minutes). What I'm having trouble with is solving it algebraically.

    This is my thought process:
    50-42sin(\frac{\pi}{2}t) = 65
    -42sin(\frac{\pi}{2}t) = 15
    sin(\frac{\pi}{2}t) = -\frac{15}{42}
    arcsin(sin(\frac{\pi}{2}t)) = arcsin(-\frac{15}{42})
    \frac{\pi}{2}t = arcsin(-\frac{15}{42})
    t = \frac{arcsin(-\frac{15}{42})}{\frac{\pi}{2}}

    I'm pretty sure I'm doing this right.. but when I get to the last step, it gives me something like t = -.2324981381

    I've plugged that in and it is obviously not the right answer. It gives me 65 as it should, but it doesn't correspond to the answers I get when solving graphically. Any help with what I'm doing wrong?
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  2. #2
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    Quote Originally Posted by qleeq View Post

    I'm pretty sure I'm doing this right.. but when I get to the last step, it gives me something like t = -.2324981381
    You might want to check if your calculator is in the right mode? degree/radians. Also could this be an answer just in the worng quadrant?
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  3. #3
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    Quote Originally Posted by pickslides View Post
    You might want to check if your calculator is in the right mode? degree/radians. Also could this be an answer just in the worng quadrant?
    Find the times when you're 65 feet off the ground.-30na.png
    That is the graph of the function. the straight line is y = 65
    the curve is h(t) = 50-42sin((pi/2)t)

    scale is y [0,100,10] and x [0,4,1]
    as you can see, it crosses in both places. my calculator is in radians mode. the graph is only in the first quadrant
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  4. #4
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    Something I figured out..
    This number: t = -.2324981381
    is significant to the answer somehow.. when finding the places where y = 65 on the graph, one of them is positive 2.2324981


    edit: okay WOW, the t=-.232(...) is an answer in the second quadrant for x = -.232(...)
    how do I find the other values of x?
    Last edited by qleeq; October 14th 2010 at 01:57 PM. Reason: wtf
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  5. #5
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    \arcsin\left(-\frac{15}{42}\right) is in quads III and IV ...

    when the point is in quad III ...

    t = \frac{2[\pi - \arcsin\left(-\frac{15}{42}\right)]}{\pi} =\approx 2.23 \, s

    quad IV ...

    t = \frac{2[\arcsin\left(-\frac{15}{42}\right) + 2\pi]}{\pi} \approx 3.77 \, s
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  6. #6
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    Quote Originally Posted by skeeter View Post
    \arcsin\left(-\frac{15}{42}\right) is in quads III and IV ...

    when the point is in quad III ...

    t = \frac{2[\pi - \arcsin\left(-\frac{15}{42}\right)]}{\pi} =\approx 2.23 \, s

    quad IV ...

    t = \frac{2[\arcsin\left(-\frac{15}{42}\right) + 2\pi]}{\pi} \approx 3.77 \, s
    thanks for that, but could you explain to me where you did \pi - \arcsin\left(-\frac{15}{42} and the one with +2pi? I don't quite understand that..
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  7. #7
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    Trigonometric functions with multiple zeros

    I've been posting stuff on modelling temperatures lately, and I understand it now. I have a test coming up on Tuesday, and I need a little extra help with something.

    I'm given a function, and I can set it equally to zero and solve it no problem. Then there is another zero that I need to find, and I'm having a hard time doing it. This has seemed to slip my mind.

    Can anybody explain to me in a few steps how I can go beyond finding the first zero of a sinusoidal graph? Algebraically of course.

    Say I have something like: 50-42sin(\frac{\pi}{6}t) and I need to find values of t where y = 65.

    I can solve this by setting the equation equal to 65, but after this I get stuck. Any help please?
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  8. #8
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    Quote Originally Posted by qleeq View Post
    I've been posting stuff on modelling temperatures lately, and I understand it now. I have a test coming up on Tuesday, and I need a little extra help with something.

    I'm given a function, and I can set it equally to zero and solve it no problem. Then there is another zero that I need to find, and I'm having a hard time doing it. This has seemed to slip my mind.

    Can anybody explain to me in a few steps how I can go beyond finding the first zero of a sinusoidal graph? Algebraically of course.

    Say I have something like: 50-42sin(\frac{\pi}{6}t) and I need to find values of t where y = 65.

    I can solve this by setting the equation equal to 65, but after this I get stuck. Any help please?
    I suggest that you draw the graphs of \displaystyle y = 50 - 42 \sin \left(\frac{\pi}{6}t\right) and y = 65 on the one set of axes to see what's happening. After doing this, it should be more apparent what to do.

    The values of t you want (suggested by a unit circle) satisfy \displaystyle \frac{\pi}{6} t = -\sin^{-1} \left(\frac{5}{13}\right) + 2n \pi and \displaystyle \frac{\pi}{6} t = \pi + \sin^{-1} \left(\frac{5}{13}\right)+ 2 n \pi.
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  9. #9
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    Quote Originally Posted by mr fantastic View Post
    I suggest that you draw the graphs of \displaystyle y = 50 - 42 \sin \left(\frac{\pi}{6}t\right) and y = 65 on the one set of axes to see what's happening. After doing this, it should be more apparent what to do.
    Yes, I've done that, and I know that for the first four values of x there are two zeros. Someone showed me in another thread but I didn't understand what he was doing.
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  10. #10
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    Quote Originally Posted by qleeq View Post
    Yes, I've done that, and I know that for the first four values of x there are two zeros. Someone showed me in another thread but I didn't understand what he was doing.
    Did you draw the unit circle like I said to do in my previous reply? I'm not sure what else can be said until you have put more thought into what I posted.
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