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    Lightning55's Avatar
    Lightning55 Posts: 97, Reputation: 7
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    #1

    Oct 29, 2009, 11:46 AM
    Semicircle Area Inscribed in a Parabola
    Okay, I know I should have some work for my question, but I really have nothing.
    We're supposed to find the area of a semicircle inscribed between the parabola 12-3x^2 and the x-axis. My teacher said we're supposed to use triangles to represent x and y values and the hypotenuse as our radius. I'm not really getting how to even begin solving it.
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #2

    Oct 29, 2009, 12:03 PM

    Well, I'm not sure what you have to find either. If you know how to sketch the graph of y = -3x^2 + 12, you'll have a general idea of how the semicircle is. If the semicircle has such a radius such that its centre is on the x axis, and it's circumference touches the curve, then you have a semicircle with centre (0, 0) with radius 2.

    That seems too easy though...

    y = -3x^2 + 12

    So you get your maximum at (0, 12) and the x intercepts at -2 and 2.
    Lightning55's Avatar
    Lightning55 Posts: 97, Reputation: 7
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    #3

    Oct 29, 2009, 12:29 PM

    No, it isn't a radius of 2. A radius of 2 actually crosses the parabola.
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #4

    Oct 29, 2009, 01:19 PM
    If we solve for x using the quadratic formula, we end up with a discriminant of

    Setting this to 0 and solving for r gives

    Just slightly less than 2.

    I have attached the graph of the radius 2 case and this case to show the difference.
    Attached Images
      
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #5

    Oct 29, 2009, 11:09 PM

    Oh, I get it now! Thanks galactus!

    I had a feeling that was too easy the way I did ;)
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