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Homework Help: Some math Qs

  1. Feb 11, 2004 #1
    Id appreciate some help with these questions:

    (A) [tex]\alpha,\beta[/tex] are roots of the equation [tex]x^2 - 2lx + m = 0[/tex]. Show that [tex]\alpha^2 + \beta^2 = 2(2l^2 - m)[/tex]. Express [tex]\alpha^4 + \beta^4[/tex]in terms of [tex]l and m[/tex].

    Ive shown that [tex]\alpha^2 + \beta^2 = 2(2l^2 - m)[/tex], but im having trouble figuring out what [tex]\alpha^4 + \beta^4[/tex] is - i was thinking [tex](\alpha^2 + \beta^2)^2[/tex], but this didnt seem right. Any ideas?

    (B)The vertices of a rectangle lie on a circle of radius = a.
    Show that the area of the rectangle is not greater than [tex]2a^2[/tex].

    Im lost as to how i should answer this one. Any help is appreciated.
    Last edited: Feb 11, 2004
  2. jcsd
  3. Feb 11, 2004 #2
    as for a) Your answer is wrong. Unfortuanatly I'm not sure as to the correct answer. It might just be because its late and I'm just not working fully so I'll try again in the morning.

    and for b) If all 4 vertices lie on the circle then no one side of the rectangle can be greater than 2a therfore the area cannot be greater than 2a^2
  4. Feb 12, 2004 #3
    Let a = alpha and b = beta (I can't get the hang of this LaTeX code).

    You show a^2 + b^2 to be equal to 2(2l^2 - m) by finding a and b from the quadratic formula.

    You find a to be l + (l^2 - m)^0.5, and b to be l - (l^2 - m)^0.5

    So all you have to do to find a^4 + b^4 is raise a and b each to the power 4, then add. Remember:

    (x + y)^4 = x^4 + 4(x^3)(y) + 6(x^2)(y^2) + 4(x)(y^3) + y^4

    Regarding the rectangle, another way of doing it is to say that the largest area occurs when the rectangle is a square, thus the area of the square is 2(a^2) - the largest possible area.
  5. Feb 12, 2004 #4
    Ok, does it matter that (a^2 + b^2) is 2(2l^2 - m) as opposed to ^2 + b^2 = 2(2l^2 - m) . For part two im supposed to use calculus to solve, not just logic.
    Thanks again
  6. Feb 12, 2004 #5


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    You don't need the quadratic formula for part 1. You already know what [itex]\alpha + \beta[/itex] and [itex]\alpha * \beta[/itex] are just by looking at the formula!

    You thought to start with [itex](\alpha^2 + \beta^2)^2[/itex] in your search for the value of [itex]\alpha^4 + \beta^4[/itex]; well, expand it and see what the difference is, and see if you know what the difference is.

    (incidentally, you can do a similar thing to find [itex]\alpha^2 + \beta^2[/itex])
  7. Feb 12, 2004 #6


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    I don't see any difference between "(a^2 + b^2) is 2(2l^2 - m)" and
    "a^2 + b^2 = 2(2l^2 - m)". Is it just the extra parentheses you are asking about? Of course, that doesn't matter.

    Since we know that a and b (much simpler than using "tex" just to get alpha and beta!) are roots of x2+ 2lx+ m, it follows that (x-a)(x-b)= x2- (a+b)x+ ab= x2+ 2lx+ m so that a+b= -2l and ab= m. Squaring each side of the first equation, we have a2+ b2= a2+ 2ab+ b2= 4l2. From the second equation, 2ab= 2m.
    Subtracting: a2+ b2= 4l2- 2m= 2(2l2-m).

    Now do the same thing to find a4+b4:
    (a+ b)4= a4+4a3b+ 6a2b2+ 4ab3+b4. Since we want to keep a4+b4, we need to "get rid of" 4a3b+ 6a2b2+ 4ab3.
    Okay, we can clearly factor 2ab out of that to get 2ab(2a2+ 3ab+ 2b2)= 2ab(2(a2+b2)+ 3ab). Since we know how to express both a2+b2 and ab in terms of l and m, the rest is simple.

    For the second problem, the earlier posts were suggesting the easy way. The first response, however, was in error. Saying that neither side can be greater than 2a only shows that the area cannot be greater than (2a)2= 4a2, not 2a2. The second post was correct but assumes that you know that the largest rectangle must be a square.

    Of course, you hadn't told us that you could use calculus. Even so there are two ways to do this depending on exactly what calculus you can use.

    a) The "elementary" way. We can (by symmetry) argue that the largest rectangle is symmetric about the x-axis. Let x be the x-coordinate of the vertex in the first quadrant. The y-coordinate is, then, &radic(a2- x2). The lengths of the 2 sides are 2x and 22&radic(a2- x2) and the area is given by A= 4x&radic(a2- x2). That will be a maximum when the derivative of A with respect to x is 0. (Rest of the work, I leave to you.)

    b) Using Lagrange multipliers. Again, we can use symmetry to argue that, taking (x,y) as a vertex in the first quadrant, the lengths of the sides are 2x and 2y and so the area is 4xy. x and y must satisfy the equation of the circle: x2+ y2= a2. At the maximum area, we have 4y= λ(2x) and 4x= λ(2y) for some number λ. From that it follows that x= y so the rectangle really is a square. Thence, x2+ y2= 2x2= a2 while the area of that largest rectangle (square) is 4x2= 2a2.
    Last edited by a moderator: Feb 13, 2004
  8. Feb 12, 2004 #7
    That was a copy/paste mistake - i was referring to infites advice:
    What i had actually found in the first part was (a^2 + b^2) not (a + b)^2, which i think are very different. All i really needed to know was what the expansion of (a + b)^4 was, which ive gotten, thanks.

    I know! But im a "tex" newbie and it just looked cool and more interesting.

    Im not familiar with either of the calculus methods you gave, what we normally use is areas under curves and that kind of thing, so im still a little lost, but thanks for the help.
  9. Feb 12, 2004 #8
    so u have done the area c

    an u find the area of the rectangle if thus formed

    ans maximize it
  10. Feb 12, 2004 #9
    So can somebody explain what the following means:

    What does "&radic" refer to?
    Where does the "22&radic(a2 - x2)" come from?
  11. Feb 13, 2004 #10


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    I think that was my carelessness. &radic if it has a ";" after it will show up as a square root symbol: √ but I keep forgetting the final ; !

    "The lengths of the 2 sides are 2x and 22&radic(a2- x2) and the area is given by A= 4x&radic(a2- x2)"

    Should have been
    "The lengths of the 2 sides are 2x and 2 √(a2- x2) and the area is given by A= 4x√(a2-x2)."

    That is so because (x, √(a2- x2)) is a point on the circle in the first quadrant. The horizontal edge of the rectangle extends an equal distance to the left of the y-axis:
    length 2x. The vertical edge of the rectangle extends an equal distance below the x-axis: width 2√(a2-x2);. Of course, area= length times width giving
    A= 4x√(a2-x2).
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