Wikipedia:Reference desk/Archives/Mathematics/2007 August 15

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August 15

7th grade math

Can any 1 find out what kids mite learn in 7th grade in KPS thx i want to get redy for wen schoool starts and i dont no wat im gona learn so i cant study it thx agin --Sivad4991 04:47, 15 August 2007 (UTC)

The easiest way would be to contact the teacher. I'm not sure what KPS is, but given that you seem to have personal internet access it's very likely your school has a website with contact details for your teacher, if not your teacher's website which should have numerous details on their course work. If your teacher doesn't have a website, you will probably have to call them or email them and ask- they will probably be thrilled that anyone cares. You could also recommend that they use a blog site that offers a free service (like wordpress) as a make-shift webhost, so that other students can keep track of their upcoming assignments. It's impossible to know without asking your teacher though, which is what it all comes down to. --Lucid 04:54, 15 August 2007 (UTC)

Your teacher may be available this and next week before school starts. Hope that you will have a lucky school year. :) --Mayfare 13:43, 22 August 2007 (UTC)

The Height of the Curve of Water

If a rope was attached to a stake on the shore of Lake Michigan, in Sheboygan, Wisconsin, and stretched across and attached to a stake on the beach in Michigan, a distance of around 50 miles (assuming the rope is perfectly straight and level) how far under water would the rope be in the middle?--ChesterMarcol 17:56, 15 August 2007 (UTC)

Forgive me if I point out that this sounds uncannily like a homework problem. But in any case, take a look at the articles on trigonometry and the Pythagorean theorem. Baccyak4H (Yak!) 18:04, 15 August 2007 (UTC)

You might also find circular segment and Earth#Shape helpful. --Sean 18:45, 15 August 2007 (UTC)

By the way, depending on your definition of "level", it may not be possible for such a rope to be both "perfectly straight and level": if it's straight, its midpoint will be lower (i.e. closer to the center of the Earth) than its ends. Presumably what is meant here by "level" is that the ends are at the same elevation, which one would further presume to be approximately the lake's surface level. But this is definitely nitpicking. —Ilmari Karonen (talk) 20:09, 15 August 2007 (UTC)

For a rope, see Catenary. Alternatively you might wish to answer the question with a laser beam which would be (for practical purposes) straight (but not level). Your bonus to ponder with this question is "Is the lake flat" or are the beach levels different on the two sides. Unless flow affects it, I suspect both sides are at the same altitude but that isn't the case with the English Channel, a fact needed to be taken into account when drilling tyhe tunnel. -- SGBailey 21:04, 15 August 2007 (UTC)

Sorry this isnt a homework question. I work at a golf course on the shore of Lake Michigan and this is the kind of stuff that you start to think about after standing on a fairway holding a hose for hours at a time. Yesterday we had at least 10 people, some of them engineering and math majors, trying to figuere this out. I think the best we could figure out was around 300'. Im really bad at math so I dont think I could do any better.--ChesterMarcol 01:28, 16 August 2007 (UTC)

The answer h should be the smaller root of the quadratic equation

h² − 2Rh + ¹/₄c² = 0,

where R and c are as in the diagram. If c is much less than R, this can be approximated by

h = c² / 8R.

Using R = 6373 km and c = 80.47 km, I get h = 0.127 km, or 417'.  --Lambiam 03:02, 16 August 2007 (UTC)

Greens Theorem

I'm reading the article about Green's Theorem, but it does not make any sense.

In physics and mathematics, Green's theorem gives the relationship between a line integral around a simple closed curve C and a double integral over the plane region D bounded by C. It is a special two-dimensional case of the more general Stokes' theorem, and is named after British scientist George Green.

Let C be a positively oriented, piecewise smooth, simple closed curve in the plane and let D be the region bounded by C. If L and M have continuous partial derivatives on an open region containing D, then

$\int _{C}(L\,dx+M\,dy)=\iint _{D}\left({\frac {\partial M}{\partial x}}-{\frac {\partial L}{\partial y}}\right)\,dA$

Bang! I got lost when I hit the text area

If L and M have continuous partial derivatives on an open region containing D

Now what is this strange L and M . The text did not mention anything about them previously. Do I just pick any random undefined variable L and M ? Does L stands for Length and M stands for Mass ?

202.168.50.40 23:49, 15 August 2007 (UTC)

L and M are arbitrary (real-valued) functions defined on an open region containing D, and having continuous partial derivatives everywhere in this region. Algebraist 00:14, 16 August 2007 (UTC)

By arbitrary, do you mean I can just simply choose L= Sin(x) and M = Cos(x) or even L = 0 and M = 0 to make the calculations easier for myself. 202.168.50.40 00:43, 16 August 2007 (UTC)

Well yes you can, but the point is that if you're trying to apply Green's Theorem to an integral you presumably already have your L and M determined by the context. And as long as L and M obey the restrictions imposed, Green's Theorem will work. Confusing Manifestation 02:08, 16 August 2007 (UTC)

SO if I try to apply Green's Theorem to integral f(x,y) = x^2 along any closed curve of my choosing then L and M would be. L=x^2 and M=0 and I just substitute them into the equation. 202.168.50.40 03:10, 16 August 2007 (UTC)

Sure! Note that in this case ${\frac {\partial M}{\partial x}}$ is 0 since M = 0, and ${\frac {\partial L}{\partial y}}$ is 0 since L does not depend on y, and so your integral will be 0. This is not a mistake, and in fact it tells us a wonderful fact: let p be a path from (x₁,y₁) to (x₂,y₂), and let f(x,y) be a function that only depends on x. Then the integral $\int _{p}fdx\,\!$ is the same no matter what path p you choose. See if you can see why. (In fact, we can say more: this integral depends only on x₁ and x₂, the x-coordinates of the points! This requires a little more argument, though.) Tesseran 04:18, 16 August 2007 (UTC)

You should be careful with what you mean by "integrate f along a closed curve, though. The way you've divided it up, you're integrating $\int _{C}x^{2}dx\,\!$, which is very different from $\int _{C}x^{2}dy\,\!$ (note the dx and dy). Hopefully someone else can give a good explanation for what these integrals mean and why they're different (I've always had trouble explaining this). Tesseran 04:26, 16 August 2007 (UTC)

Green's theorem is like many mathematical facts: it tells you that a certain equivalence is available, but like "a+b = b+a" it does not tell you when or how to use it. In most cases we use it when the boundary integral is simpler than the region integral; but not always, and even then we have choices.

Suppose we wish to find the area of a diamond shape,

$D=\{(x,y)\colon |x|+|y|\leq 1\}.\,\!$

The boundary of the region is a simple polygon, and the area integral is merely

$\iint _{D}dA.\,\!$

The theorem is applicable, but it is up to us to choose functions L and M such that

${\frac {\partial M}{\partial x}}-{\frac {\partial L}{\partial Y}}=1.\,\!$

We should get the same result no matter what the choice. Three possibilities are:

${\begin{aligned}L&=0&M&=x\\L&=-y&M&=0\\L&=-{\tfrac {1}{2}}y&M&={\tfrac {1}{2}}x\end{aligned}}$

In the first quadrant, between the positive x and y axes, the boundary has the equation x+y = 1, which we may rewrite as x = 1−y for convenience, to parameterize this piece of the boundary by y. Then Green's theorem using our third possibility rewrites the integral as

${\begin{aligned}\iint _{D}dA&=\iint _{D}\left({\frac {\partial ({\tfrac {1}{2}}x)}{\partial x}}-{\frac {\partial (-{\tfrac {1}{2}}y)}{\partial y}}\right)\,dA\\&=\int _{0}^{1}-{\tfrac {1}{2}}y\,(-dy)+{\tfrac {1}{2}}(1-y)dy\\&=\int _{0}^{1}{\tfrac {1}{2}}\,dy,\end{aligned}}$

where we have used dx = −dy. As expected, we get an area of ¹⁄₂ for the quadrant. By symmetry, we know that the total area must be four times that. It may be illuminating to integrate along each side, to try the different choices for L and M, and to explicitly consider the first quadrant triangle with its axis boundary sides. Other experiments, such as using a circular boundary and computing the center of mass and moment of inertia, will also suggest themselves to the motivated reader.

I'm strongly tempted to try to give some intuition for the theorem; perhaps later. --KSmrq^T 22:16, 16 August 2007 (UTC)

Your explanation is excellent, KSmrq, but you make an invalid assumption when you apply symmetry to this problem. Certainly the area in the first quadrant is one-quarter the total, but you must then integrate over $\partial (D\cap Q)$, not $\partial D\cap Q$ (where of course $Q:=\{(x,y)\colon x\geq 0,y\geq 0\}$). It worked for your choice of L and M, but no such luck for, say, $L={xy\color {red}+1},M={\frac {x^{2}}{2}}+x$ (the red is a correction). --Tardis 17:00, 17 August 2007 (UTC)

I don't follow your complaint. Using your L and M, the integral along that same edge of the polygon is

$\int _{0}^{1}\left((1-y)y\,(-dy)+{\big (}{\tfrac {1}{2}}(1-y)^{2}+(1-y){\big )}\,dy\right).$

Once again the result is ¹⁄₂. If we introduce internal boundaries, such as the horizontal and vertical sides of a quadrant, they will aways cancel in the final area because we traverse each twice, in opposite directions. Perhaps I should point out that I chose y as the parameter for the edge so that traversal would put the area on the left; it is important to pay attention to orientation! --KSmrq^T 00:26, 19 August 2007 (UTC)

Oops; my example was bad (because L is 0 on the x-axis and M is 0 on the y-axis). (When I checked it I very cleverly did $\int {\frac {3}{2}}y^{2}dy={\frac {y^{3}}{3}}$, so I thought it was a counterexample.) Try $L=xy+1$ (which allows the same M, of course) instead; in general, the ability to add to L and M arbitrary constants (or even functions only of x or y, respectively) forces us to use only closed contours of integration. --Tardis 20:34, 20 August 2007 (UTC)

Ah, I believe I understand the point you want to bring out. We can do it more simply using L = 1, M = x. The region of integration has the same symmetry as before, but with this choice, L does not. Still, we need only integrate along the boundaries, only now we cannot use the same result four times.

These area integrals lend themselves to a geometric interpretation, which may help clarify the theorem as well. Take the simple choice L = 0, M = x, and consider any region whose boundary is a simple polygon. The integral for a single edge is the signed area of the trapezoid between that edge and its perpendicular projection onto the y axis. For, let the endpoints of the edge be (x₀,y₀) and (x₁,y₁), with the region on the left; and parameterize the edge as

$x=x_{0}+{\frac {y-y_{0}}{y_{1}-y_{0}}}(x_{1}-x_{0}).\,\!$

Thus the edge integral is merely

$\int _{y_{0}}^{y_{1}}\left(x_{0}+{\frac {y-y_{0}}{y_{1}-y_{0}}}(x_{1}-x_{0})\right)\,dy={\frac {x_{0}+x_{1}}{2}}(y_{1}-y_{0}),$

consistent with the promised trapezoid. If we instead let L be a constant, c, then we project onto the line x−c = 0 rather than the y axis (x = 0). This adds a positive contribution for some edges but a negative contribution for others, and these effects cancel, yielding the same total area.

A moment's thought convinces us that symmetry of the region is useless without symmetry of the function as well; and now we see that a Green's theorem transformation need not preserve the symmetry. Thanks, Tardis, for highlighting this detail. --KSmrq^T 09:14, 21 August 2007 (UTC)