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Math help!

Discussion in 'Off-Topic' started by The Future now, Oct 5, 2006.

  1. The Future now

    The Future now Banned

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    Math help!

    This is, I think, pretty simple. I'm just missing something.

    I'm having a bit of a problem of proving limits of 2 or more variables.

    Ok so . . . (For example)

    lim (x,y)--> (0,0) (2x^2 Y^2) / (X^2 + Y^2)

    So the degree of the numerator terms is 4, and the denominator is 2 so guess that the limit exists and is the origin.

    Let E>0

    ABS [(2x^2 Y^2) / (X^2 + Y^2)-0] < E Whenever 0<Sqroot(X^2 + Y^2) < $

    Because everything is squared, can remove the ABS

    (2x^2 Y^2) / (X^2 + Y^2) < E

    Now here's where I start to get a bit confuzzled. My notes have E being replaced by:

    2(2x^2 Y^2)(2x^2 Y^2) / (2x^2 Y^2) so that the denominator can cancel out. (The inequality remains true because Y^2 is always [0, infinity+)

    I'm just not too sure what to do from here . . .



    One other thing... concerning gradient vectors. I'm in pretty good shape except for this "maximum/ minimum" b.s.

    How do you solve this sort of problem?:

    Find the maximum rate of change of f at the given point and the direction in which it occurs.

    f(x,y) = Y^2 / X (2,4)

    First step is finding the partial derivatives, but what then? :sad2:

    If you help me I'll send naked pictures of my gf!!!!!! Or 1337 haxxed itamz
     
  2. Dondrei

    Dondrei IncGamers Member

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    Man, I haven't done that stuff in ages. When you find the partial derivatives, you put them in a vector (grad? I forget) and dot product it with something else... yep, I forget.

    As for epsilon-delta proofs, no thank you... what does the $ represent? Delta?
     
  3. ragnar_ii

    ragnar_ii IncGamers Member

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    Arg, this reminds me that I have to retake multivariable. I got a C- and needed a minimum of a C for it to count towards my minor.
     
  4. Stompwampa

    Stompwampa IncGamers Member

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    I've never gone beyond Algebra II/Triginomotry. Thank, God.
     
  5. Gibbzilla

    Gibbzilla IncGamers Member

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    When'd they start putting letters in it?
     
  6. The Future now

    The Future now Banned

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    Yeah, it kind of looks like the greek letter . . . best I could find without character map.

    Here's an example gradient problem that I actually know how to solve.

    Find the directional derivative of the function at the given point in the direction of vector V

    F(x,y) = 1 + 2x(y^ (1/2)) at point (3,4) V = [4, -3]

    Fx = 2(y^1/2) @ (3,4) = 4
    Fy = Y ^ (-1/2) X = X/ Y^1/2 @ (3,4) = 3/2



    /v/ = Sqroot ( (3)^2 + 4^2) = 5

    Divide the vector by that to get (4/5 , -3/5) Dot product (4 , 3/2)

    16/5 - 9/10 = 23/10

    The maximum stuff has to something to do with Cos(theta) being largest when theta = 1,

    Dunno though.


    edit: Ragnar so far Multi is turning out to be a bit easier than 116 (Calc 2 for those not in Uconn . . . )
     
  7. Dondrei

    Dondrei IncGamers Member

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    Oh, I remember a bit more... you could of course sub in variables and treat it like a maximisation problem, but there's a quick way...

    Oh, okay - think of it this way. The grad vector (Fx, Fy) is normal to the surface at the point - just try to envisage your surface as some big lumpy carpet, and the normal sticking up at 90 degrees to the carpet at any point you choose on one of the bumps (well, anywhere actually but the bumps are the interesting bits). You can see that the normal leans over in the direction of fastest descent (or if it's negative, turn the whole thing upside down and it points in the direction of fastest ascent).

    So all you need to do is project the grad vector onto the (x,y) plane and that'll give you the direction of fastest descent/ascent. Since your coordinates are (x,y,z), I think that just means lopping off the z-coordinate.
     

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