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December 4[edit]

Rotate a graph 45 degrees[edit]

Ok, so I came across this link which says that in order to rotate any graph 45 degrees to the right, you simply replace y with $(x+y)/sqrt(2)$ and x with $(x-y)/sqrt(2)$ . I'd like to know,

How do you solve for y in an absolute value equation and a quadratic equation when x and y have been replaced, and how do you rotate any graph any number of degrees?

Thanks, Xcfrommars 03:30, 4 December 2006 (UTC)[reply]

The formulas you give are a special case of a rotation matrix. The general form is

{\begin{bmatrix}x\\y\end{bmatrix}}\rightarrow {\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \end{bmatrix}}{\begin{bmatrix}x\\y\end{bmatrix}}={\begin{bmatrix}x\cos \theta -y\sin \theta \\x\sin \theta +y\cos \theta \end{bmatrix}}

for a rotation of θ to the counterclockwise, so to rotate 45° counterclockwise, it's

{\begin{bmatrix}x\\y\end{bmatrix}}\rightarrow {\begin{bmatrix}{\frac {1}{\sqrt {2}}}&-{\frac {1}{\sqrt {2}}}\\{\frac {1}{\sqrt {2}}}&{\frac {1}{\sqrt {2}}}\end{bmatrix}}{\begin{bmatrix}x\\y\end{bmatrix}}={\begin{bmatrix}{\frac {(x-y)}{\sqrt {2}}}\\{\frac {(x+y)}{\sqrt {2}}}\end{bmatrix}},

which is equivalent to your formulas. —Keenan Pepper 06:28, 4 December 2006 (UTC)[reply]

(after edit conflict:) Substituting x := f(x,y), y := g(x,y) in an equation means that you are replacing the "old" (x, y) by "new" (x, y). To avoid confusion between old and new x and y, you might imagine labelling the axes of the new graph by x' and y'. Then the substitution you are performing is x:= f(x',y'), y := g(x',y'). So f and g are the transformations that turn the new coordinates back into the old ones. If you want to have a (clockwise) rotation by α of the graph, you need to apply the formula for the opposite rotation, that is, a rotation by −α, to the pair (x', y'). The formulas for a general rotation in the two-dimensional case can be found in our article Rotation (mathematics). For the special case that α = −45°, use the fact that cos(45°) = sin(45°) = ¹/₂√2 = 1/√2.

If your old graph was the graph of y = ax², the new graph is that of (x+y)/√2 = a((x-y)/√2)². Because this is a quadratic equation in the variable y, you can solve this for y by using any of the well-known formulas for finding the roots of quadratic equations. In general you may find two solutions for y for some values of x, and no solutions for other values of x. So the rotated graph is in general not the graph of a normal function, which never has more than one y for any given x.

If you started with y = a|x|, with a > 0, this can be expressed as: (y = ax OR y = –ax) AND y ≥ 0. With the substitution, you get then: ((x+y)/√2 = a(x-y)/√2 OR (x+y)/√2 = –a(x-y)/√2) AND (x+y)/√2 ≥ 0. Simplifying this a bit gives us: (x+y = a(x-y) OR x+y = –a(x-y)) AND x+y ≥ 0. You can solve for y for each of the two possibilities, and discard those solutions not meeting the final requirement. Depending on the value of a, you may find as before two or no solutions, and if a = 1 you may in fact get infinitely many solutions for y at x = 0. --Lambiam ^Talk 06:36, 4 December 2006 (UTC)[reply]

Solid Geometry>>>Spheres>>>Inscribed Right Triangles...[edit]

A]Given that all spheres have an axis,

~and B] That all spheres can be generated by rotating a two-dimensional circle of the same diameter using the axis as its' diameter,

And C] That an infinite number of Right Triangles can be generated by using a common hypotenuse[A sq + B sq = C sq] where C is, in fact, the diameter of the given circle and the axis of the given sphere simultaneously,

>>>Therefore each and every point on the sphere, excluding the points where the axis intersects the spheres' surface area, is the Right Angle of a Right Triangle opposite to the spheres' axis...

~What is the formula for the relationship and the applications of applying it to the surface area, or specific locations, on the sphere?

I'm missing a few pieces to this puzzle, Thank You for your time, Your friend Jon Hutchings AKA the sasquatchbythesea

I'm not sure what the question is. The claims you make all seem correct, but what relationship are you looking for? —Keenan Pepper 06:31, 4 December 2006 (UTC)[reply]

Hmmm... if the sphere's axis runs from a "north pole" to a "south pole", then we can identify each point on the sphere with a latitude and longitude. At that point, one could ask what kind of right triangle corresponds to a point with a certain latitude and longitude. The longitude wouldn't change the shape of the triangle, but the latitude would... if the equator is 0° latitude, then I think the smaller angle in our right triangle anchored at x° latitude will be (90-x)/2 degrees. That gives us degenerate triangles at the poles, which is good. Is that what you were asking, Joh Hutchings? -GTBacchus^(talk) 08:56, 4 December 2006 (UTC)[reply]

Just to confirm, you're right: the angle between the axis and a line from the south pole to a point is half of its colatitude, for the usual reasons of being half of the subtended arc. --Tardis 16:40, 4 December 2006 (UTC)[reply]

Group decomposition / re-composition[edit]

The Jordan-Holder Theorem states that any group's composition factors (the quotient groups of successive terms in its composition series) are "constant", in that they will be the same for any composition series of that group, up to isomorphism and permutation. Also, those compostion factros will be finte, simple groups.

My question is this: given some set of finite simple groups, how does one go about finding the group(s) with that set as the set of composition factors? (What's the technical term for this process? "group recomposition"?) Also, when two distinct groups have the composition factors, is there any sort of relationship between the groups? (Other than being of the same order) Tompw 13:57, 4 December 2006 (UTC)[reply]

See group extension. --KSmrq^T 06:43, 5 December 2006 (UTC)[reply]

Volume Times Time[edit]

Is there any measurement that is "volume times time"?The Anonymous One 23:31, 4 December 2006 (UTC)[reply]

Hmm, don't know, really. Perhaps that would be the size of a region in four-dimensional spacetime, but I'm not sure that is ever called for. If it was, chances are that one would multiply the time dimension by the speed of light so that you have four space dimensions, rather than three space dimensions and one time dimension. Do you have a suggestion? —Bromskloss 00:19, 5 December 2006 (UTC)[reply]

Well, you can always play with dimensional analysis as a starting point. Just try to to use a known derived unit with it and get another derived unit. For example:

(m^{3}\cdot s)\cdot (N\cdot m^{-2})=N\cdot m\cdot s=J\cdot s

Which means that this unit you propose, times pressure, gives the same units for angular momentum... ☢ Ҡi∊ff⌇↯ 01:35, 5 December 2006 (UTC)[reply]

Somewhat related: classical mechanics phase space volume is measured as (length)^2NM(mass)^NM(time)^-NM, where N is the dimensionality of your space (3 for us) and M is the number of particles involved. --Tardis 15:30, 5 December 2006 (UTC)[reply]

In spacetime (also known as 4-space because it's made up of 4 dimensions - x, y, z and t), the "length" of a path through space and time is called the space-time interval (or just an "interval"). A "volume" made of any 3 dimensions out of the 4 (eg: a x*y*z or x*z*t) is called a manifold. A "volume" made of space * time is called a hypervolume. --h2g2bob 19:06, 5 December 2006 (UTC)[reply]

As some of those are lacking in detail, here are some identities for integration in 4-space.

Hypervolume: $d\Omega =cdt\,dx\,dy\,dz$
Interval: $ds={\sqrt {(cdt)^{2}-dx^{2}-dy^{2}-dz^{2}}}$

Or, in Einstein notation:

$d\Omega =dx^{0}dx^{1}dx^{2}dx^{3}$
$ds^{2}=dx^{i}dx_{i}$

Those are from some notes I have (no internet version, sorry), but wikibooks:General relativity seems to have the subject in full, agonising horror. --h2g2bob 19:47, 5 December 2006 (UTC)[reply]

On a more mundane level, I would think that volume times time is a natural unit of rented storage space, at least from the point of view of a large renter with many customers. Melchoir 22:28, 5 December 2006 (UTC)[reply]

Haha, very clever. :-) —Bromskloss 23:02, 5 December 2006 (UTC)[reply]

Note that Planck's constant has units of Joule-seconds. --HappyCamper 04:24, 6 December 2006 (UTC)[reply]