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-   -   Area, find value of k. (https://www.askmehelpdesk.com/showthread.php?t=184928)

  • Feb 17, 2008, 06:49 AM
    galactus
    1 Attachment(s)
    Area, find value of k.
    I am posing this problem for kicks. For those out there who like a nice math challenge problem. Since this site rarely gets them, I thought I would pose one if anyone would like a go.

    "Looking at the graph, find the value of k so that areas of region A and region B are equal"
  • Feb 17, 2008, 07:59 AM
    xphelper
    Quote:

    Originally Posted by galactus
    I am posing this problem for kicks. For those out there who like a nice math challenge problem. Since this site rarely gets them, I thought I would pose one if anyone would like a go.

    "Looking at the graph, find the value of k so that areas of region A and region B are equal"

    0.2
  • Feb 17, 2008, 01:03 PM
    galactus
    No, I am sorry to say that is incorrect. I will leave it up until tomorrow evening to see if anyone else wants a go.
  • Feb 18, 2008, 01:02 PM
    ebaines
    How about: k=0.724611?
  • Feb 18, 2008, 01:57 PM
    galactus
    Yes, indeed, ebaines. I knew you could get it. :)
  • Feb 18, 2008, 02:19 PM
    ebaines
    Quote:

    Originally Posted by galactus
    Yes, indeed, ebaines. I knew you could get it. :)

    Thanks! However, I must admit that I was unable to find a closed-form solution for this problem, but rather used an approximation method to solve a nasty equation:

    1 = b*sin(b) + cos(b)

    where b = arcsin(k)-pi.

    Is there a more elegant way?
  • Feb 18, 2008, 02:59 PM
    galactus
    Yes, that's exactly what I done. I used Newton's method.
  • Feb 23, 2008, 10:17 AM
    galactus
    SOLUTION:


    Let (a,k) be the point of intersection of y=k with y=sin(x).

    Then, k=sin(a) and if the areas are equal:



    Solve for 'a' using Newton's method or some other estimation techniques and get: , so

    The horizontal line which intersects sin(x) creating two equal regions is y=0.724611

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