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[Bouchart]

I haven't done this sort of thing in a while but try differentiating

B2= (27 + X)² + H² with respect to H:

2B(dB/dH) = (2)(27 + X)(dX/dH) + 2H(dH/dH)

and I think dB/dH = 0 (since the length of the board, B, is fixed, it doesn't change with respect to anything) and dH/dH = 1.

It's been 8 years since I've done this, though.

[Christes]

The board length does change with respect to H, since H controls where the board hits the wall.  (Remember that the board length here isn't fixed - we are trying to minimize it)

The first thing I would note is that x and H are not independent of each other.  You can write one in terms of the other.  (Think of similar triangles)  This will kill a variable and hopefully make the problem simpler.

[Bouchart]

Yes that is correct about dB/dH.

And H = ((x+27)/x

[Christes]

indeed

[RedWarrior0]

My intuition says it's where the board is at a 45 degree angle. I'm not in calculus yet, but I semilogicked my way through it and it feels right. Now to wait for Vector to come and tell me I'm wrong or something.

[KaminaSquirtle]

Hmmm, that's a tricky one.

Okay.  Every vertex in G is going to be connected to at least n/2 vertices in either G or G bar.  Either G or G bar is going to have at least half of the possible edges in its set.

We know the chromatic number for a fully connected graph (X(G) = n), X(G bar) = 0).  We don't know the number for a half-connected graph (X(G) ~= X(G bar) = ).  If my intuition is right, X(G) for a half-connected G is going to be >= 2*square root(n) by itself...which will make your life a lot easier if it's true.

What's the lowest possible chromatic number for a graph that has only half of its possible edges, rounded up?

Firstly, one small detail, χ(G bar) where G is a complete graph is still one, since you need a color to color all the disconnected vertices.

Secondly, I'm going to guess that your intuition is right.  Actually, one of the big problems here will be establishing that χ(G) + χ(G bar) for a half connected graph is minimal for a Graph with n vertices.  If we can pull it off, another problem will immediately fall (χ(G)+χ(G bar) <= n+1), since in proving the above I would guess we would also prove that a complete graph would have maximal χ(G)+χ(G bar).

As for the last question, the only minimums on chromatic number that I know are the maximal complete subgraph, and the number of vertices in the graph over the size of the largest group of independent vertices.

Wait a second, though.  I'm not sure we can say χ(G)=χ(G bar).  In fact I'm quite sure we can't.  Dang...

I'll keep brainstorming...

Wait again, though.  If we can show that χ(G) >= square root(n), then we immediately know that χ(G bar) >= square root (n), since if we can show it in general for a set number of edges, and G bar will have the same number of edges, χ(G bar) will be be part of the general group.

Hmmmmm

[Sowelu]

Okay!  I think I've got some traction here.

I'm getting the terminology wrong, but let's say that G₃ is a graph with a chromatic number of 3.

In G_C, each color x from 1..C is represented by n_x nodes.  What's important here is the size of the LARGEST color group.  The largest group can never be smaller than n/C.  So, in G_C, there is always a group of the same color of at least size n/C.

Now, no node can be connected to a node of the same color of it.   There is a subset in G_C of at least size n/C in which NO node in the subset is connected to any other node in the subset.  Which means that in Soviet G_C G_C bar, there is a subset of at least size n/C which is fully connected.  Therefore, if there are n nodes and X(G) = C, X(G bar) >= n/C.

How can we get from here to saying that X(G) + X(G bar) >= 2 * sqrt(n)?  Well...  Uh...  Tripping over myself here...

Spoiler (click to show/hide)

At minimum, X(G bar) = n / X(G).  Assign C = X(G).
C + n/C >= 2 * sqrt(n)...
C^2 + n + n^2/C^2 >= 4 * n...
C^2/n + 1 + n/C^2 >= 4...
C^2/n + n/C^2 >= 3...  Is it enough to assert that n / X(G)^2 >= 3?  If it is, we can stop here.
(n + C^2) / (n * C^2) >= 3...
This isn't going well.

[Darvi]

Spoiler: stuff (click to show/hide)

Okay, so my Calculus class is doing a section on Optimization. I have it down fairly well, but this type of problem is giving me trouble. I need to find the shortest possible straight plank (the orange line) that can touch both the building and the ground to the left of the 8 foot high wall that is 27 feet away.



Here is what I know.

B²= (27 + X)² + H²
B= ((27 + X)² + H²)^.5
H= (B² - (27 + X)²)^.5
X= [(B² - H²)^.5] - 27

Well , according to Thales, we also have x/(x+27)=8/H <=>H*x=216+8x <=> (H-8)*x=216

The angle between the x and B lines would be arcsin(H/B) so x=8*tan(arcsin(H/B))  and since tan(arcsin(α))=,
we can say that x=

Yeah, got nothing more to say for now :/

[Argembarger]

Hey guys,

apparantly the work required to pump water one meter over the top of a 3m-radius hollow sphere filled with water (62.5kg/m^3) is actually 28,274.334 Joules and not 68,918.6889 Joules.

You know, in case you were wondering.

bluh.

Spoiler (click to show/hide)

totally set up the integral as integrate 562.5*pi*sin^2(pi*y/6)*(7-y) from y = 0 to 6. It was TERRIBLE. Later I went back and realized it was simply integrate 62.5*pi*(4-y)(9-y^2) from y = -3 to 3. MUCH MUCH better. Like hopy shit.

Anyway just sharing my maths derp.

I'mma get back to this assignment now.

[Darvi]

Shouldn't water be 1000 kg/m^3?

[Argembarger]

Not when they specifically tell you that it isn't.

It's, uh

really special water I guess??

EDIT: ohai wait no the units were imperial not metric.

I still got the right values, but derped up my units.

At least now the problem makes more sense.

[Jim Groovester]

I think 62.5 is the weight of water in pounds per cubic foot.

You have a units issue.

[Argembarger]

Yeah, after fixing the units I still got the right answer. It's just in foot-pounds and not joules.

Mostly because I accidentally omitted gravity when I thought the units were in metric. Hah. Good thing THAT didn't come back to bite me, or anything.

Fun fact: If it WAS a 3-meter radius sphere with 1000 kg/m^3 water, the answer comes out roughly to 4,433,415.553 J

[Heron TSG]

Spoiler: stuff (click to show/hide)

Okay, so my Calculus class is doing a section on Optimization. I have it down fairly well, but this type of problem is giving me trouble. I need to find the shortest possible straight plank (the orange line) that can touch both the building and the ground to the left of the 8 foot high wall that is 27 feet away.



Here is what I know.

B²= (27 + X)² + H²
B= ((27 + X)² + H²)^.5
H= (B² - (27 + X)²)^.5
X= [(B² - H²)^.5] - 27

Well , according to Thales, we also have x/(x+27)=8/H <=>H*x=216+8x <=> (H-8)*x=216

The angle between the x and B lines would be arcsin(H/B) so x=8*tan(arcsin(H/B))  and since tan(arcsin(α))=,
we can say that x=

Yeah, got nothing more to say for now :/

Arcsin, arctan... what are those? What kind of variation of trig functions is that? I don't really understand what is meant by these terms.

[Vector]

Inverse trigonometric functions.