Some math Qs

  • #1
mcintyre_ie
66
0
Hey
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 I am 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:

Answers and Replies

  • #2
pedestrian
8
0
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 therefore the area cannot be greater than 2a^2
 
  • #3
[ infinite ]
6
0
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.
 
  • #4
mcintyre_ie
66
0
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 I am supposed to use calculus to solve, not just logic.
Thanks again
 
  • #5
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,967
19
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])
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
43,021
971
Originally posted by mcintyre_ie
Ok, does it matter that (a^2 + b^2) is 2(2l^2 - m) as opposed to a^2 + b^2 = 2(2l^2 - m) . For part two I am supposed to use calculus to solve, not just logic.
Thanks again

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:
  • #7
mcintyre_ie
66
0
Hallsofivy:
quote:
--------------------------------------------------------------------------------
Originally posted by mcintyre_ie
Ok, does it matter that (a^2 + b^2) is 2(2l^2 - m) as opposed to a^2 + b^2 = 2(2l^2 - m) . For part two I am supposed to use calculus to solve, not just logic.
Thanks again
--------------------------------------------------------------------------------



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.

That was a copy/paste mistake - i was referring to infites advice:
You show a^2 + b^2 to be equal to 2(2l^2 - m) by finding a and b from the quadratic formula.
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 I've gotten, thanks.

Since we know that a and b (much simpler than using "tex" just to get alpha and beta!)
I know! But I am 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 I am still a little lost, but thanks for the help.
 
  • #8
himanshu121
654
1
so u have done the area c

an u find the area of the rectangle if thus formed

ans maximize it
 
  • #9
mcintyre_ie
66
0
So can somebody explain what the following means:

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.)

What does "&radic" refer to?
The lengths of the 2 sides are 2x and 22&radic(a2- x2) and the area is given by A= 4x&radic(a2- x2)
Where does the "22&radic(a2 - x2)" come from?
 
  • #10
HallsofIvy
Science Advisor
Homework Helper
43,021
971
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).
 

Suggested for: Some math Qs

  • Last Post
2
Replies
49
Views
869
Replies
1
Views
231
Replies
25
Views
442
Replies
11
Views
484
Replies
15
Views
414
Replies
6
Views
498
Replies
2
Views
574
Top