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Show that 6Hu^2/L^2 <= A.

  1. Jul 11, 2017 #1

    s3a

    User Avatar

    1. The problem statement, all variables and given/known data
    Hello, everyone. :)

    I'm having trouble with problem 2.26(b) from the attached PDF file.

    2. Relevant equations
    d^2 y/dt^2 = 6Hu^2/L^2 - 12Hxu^2/L^3
    d^2 y/dt^2 <= A
    6Hu^2/L^2 - 12Hxu^2/L^3 <= A
    6Hu^2/L^2 <= A

    3. The attempt at a solution
    I don't understand why the fact that x > 0 means that the term -12Hxu^2/L^3 in d^2 y/dt^2 = 6Hu^2/L^2 - 12Hxu^2/L^3 can be ignored.

    To me, it seems that removing it makes the inequality go from some_quantity <= A to some_larger_quantity <= A. How does one justify that?

    Any input would be greatly appreciated!
     

    Attached Files:

  2. jcsd
  3. Jul 11, 2017 #2

    Mark44

    Staff: Mentor

    It doesn't make any sense to me, either.
    If you have x - b <= A, with b being positive, it doesn't necessarily follow that x <= A.
    Simple example: 12 - 3 <= 10, but 12 > 10.
     
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