1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Random problem from vector cal

  1. Oct 12, 2004 #1
    Can anyone point me in the right direction for this problem? I can't seem to approach it from the right angle:


    Let A, B, C be real numbers such that [tex]A>0[/tex], [tex]B>0[/tex], [tex]AC-B^2>0[/tex].

    a. Prove that a number [tex]Q>0[/tex] exists such that [tex]Ax^2+2Bxy+Cy^2 \geq Q(x^2+y^2)[/tex]

    b. Find the largest possible Q.


    I don't need answers (or rather, I do, but, you know what I mean; my main concern is figuring out how to approach this without my head exploding. Thanks in advance.
     
  2. jcsd
  3. Oct 12, 2004 #2

    NateTG

    User Avatar
    Science Advisor
    Homework Helper

    Are you familiar with the quadratic formula?
     
  4. Oct 12, 2004 #3
    Solving for x? Solving for y? Solving for Q? I don't understand how that solves anything. Not being a smart-alec; I just don't understand.
     
  5. Oct 13, 2004 #4

    NateTG

    User Avatar
    Science Advisor
    Homework Helper

    I'm just trying to give you hints that you might be able to use to find an answer.

    The LHS of the inequality looks a lot like:
    [tex](ax+cy)^2=a^2x^2+2acxy+c^2y^2[/tex]

    Similarly, the quadratic formula:
    [tex]\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
    has
    [tex]b^2-4ac[/tex]
    which looks a lot like
    [tex]B^2-AC[/tex]
     
  6. Oct 13, 2004 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Not necessarily the best way but here's how I would do it:
    You want
    [tex]Ax^2+2Bxy+Cy^2 \geq Q(x^2+y^2)[/tex]
    which is the same as
    [tex](A-Q)x^2+ 2Bxy+ (C-Q)y^2\geq 0[/tex].
    The will be true as long as the "discriminant" is not positive:
    [tex]4B^2- 4(A-Q)(C-Q)\leq 0[/tex]
    or
    [tex]4B^2- 4AC+ 4AQ+4CQ- 4Q^2\leq 0[/tex]
    That, of course, is the same as
    [tex]Q^2+ (A+C)Q+ AC- B^2\geq 0[/tex]

    You can use the quadratic formula to find values of Q for which that is true.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Random problem from vector cal
  1. AP Cal DE problem. (Replies: 12)

Loading...