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Thread: Application Problem

  1. February 28th 2009, 01:37 PM #1
    chrozer
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    Application Problem

    The force F (in pounds) on a person's back when he or she bends over at angle (in degrees) is modeled by

    F = \frac {0.6W\sin {(\Theta + 90)}}{\sin{12}}

    where W is the person's weight (in pounds).

    a) Simplify the model.
    Would it just be F = \frac {0.6W\cos {\Theta}}{\sin{12}} since \sin{(\Theta +90)} = \cos{\Theta} ?
    b) Use a graphing utility to graph the model, where W = 185 and 0 < \theta < 90.
    This is done.

    c) At what angle is the force a maximum? At what angle is the force a minimum?
    The angle is at a maximum at 0 and 360 degrees and minumum at 180 degrees. Does that seem right? Because at 180 degrees F is negative.

    I have also attached a picture of the original problem.
    Attached Thumbnails Attached Thumbnails Application Problem-5-ps-8.bmp  
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  2. February 28th 2009, 01:58 PM #2
    skeeter
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    look at values of the force for values of \theta between 0 and 90 degrees only.

    max F at 0

    min F at 90
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  3. February 28th 2009, 02:06 PM #3
    chrozer
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    Quote Originally Posted by skeeter View Post
    look at values of the force for values of \theta between 0 and 90 degrees only.

    max F at 0

    min F at 90
    Oh yeah. Thanks. Btw, part (a) is right, right?
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  4. February 28th 2009, 02:27 PM #4
    skeeter
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    Quote Originally Posted by chrozer View Post
    Oh yeah. Thanks. Btw, part (a) is right, right?
    graph both the original model and the simplified version ... do the graphs match up?
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  5. February 28th 2009, 02:32 PM #5
    chrozer
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    Quote Originally Posted by skeeter View Post
    graph both the original model and the simplified version ... do the graphs match up?
    They do. That's the simplest it can be simplified right?
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