Wikipedia:Reference desk/Archives/Mathematics/2008 May 22

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May 22

Integrating parametric equations

Recently in Calculus II, we have started doing some integrals of functions defined by parametric equations (as well as in polar coordinates), my question is how does one prove that area under a parametric curve is given by the formula.

$A=\pm \int _{\alpha }^{\beta }g(t)f'(t)dt$

A math-wiki (talk) 06:23, 22 May 2008 (UTC)

f is the x-coordinate and g is the y-coordinate. Substitute $dx=f'(t)\cdot dt$ to get $\int _{\alpha }^{\beta }g(t)\cdot f'(t)\cdot dt=\int _{f(\alpha )}^{f(\beta )}y\cdot dx.$ Bo Jacoby (talk) 08:20, 22 May 2008 (UTC).

I suppose that works, but I really want to see a constructionist type proof (much like the Fundamental theorem of Calculus). 69.54.143.177 (talk) 23:22, 22 May 2008 (UTC)

Solution for Tan(x)=2x. {0<x<Pi/2}

Heyo, I am trying to solve Tan(x)=2x for x, but I have never encountered a function such as this before. The domain is {0<x<Pi/2}. My current efforts in solving it have given me an approximate value of x=1.1655612, but an exact value would be wonderful. Mathematica can't solve it, either, it would seem. I'm only a high school student, so it would be perfectly reasonable to assume that such a solution is beyond my capabilities (for now), so can anyone help me out, please? I assure you that this is not homework, but rather a curiousity I am following. Many thanks Vvitor (talk) 09:02, 22 May 2008 (UTC)

Well, there ic clearly only one solution in the range 0 < x < π/2, but I doubt that there is a closed-form expression for this solution - if there was, Mathematica would probably have found it. So a numerical approximation is the best that you will get. Newton's method works nicely if you take a starting value between 1 and 1.5. Starting further away, say between 0.7 and 0.9, Newton's method sometimes converges to other roots, sometimes cycles, and sometimes does not seem to converge at all. Perhaps someone might be interested in imaging the Newton fractal for this equation ? Gandalf61 (talk) 10:26, 22 May 2008 (UTC)

It's interesting you mention Newton's method. I was attempting find a root for Sin(x) between -Pi/2 and Pi/2, which is clearly zero, using Newton's method. I was investigating, however, the possibly of taking an initial value of x₁ in which the x₂ value would be -x₁. As such, the approximation would continually jump from negative to positive to negative again, etc.. The value of x that would seem to work is such that Tan(x)=2x. I know this sounds silly, and probably because it is, but I was awfully bored at the time :P. Thank you very much for your insight, I'll try and get a fairly accurate numerical approximation :) Vvitor (talk) 10:39, 22 May 2008 (UTC)

Silly, perhaps, but in a good way :). I agree that this problem is intriguing. Note that Mathematica will be very happy to provide a numerical approximation if you ask it nicely - The following command:

N[x /. FindRoot[Tan[x] - 2 x, {x, 1}, AccuracyGoal -> 55, WorkingPrecision -> 65], 50]

gives 1.1655611852072113068339179779585606691345388476931. -- Meni Rosenfeld (talk) 11:20, 22 May 2008 (UTC)

Perhaps it is worth mentioning that if the equation you wanted to solve was $\log x=2x$, there is a non-elementary function called Lambert W function which was invented for just this purpose. You could just as well invent a function that gives the solution to $\tan x=\alpha x$, but it seems no-one has deemed it important enough to give it a name. -- Meni Rosenfeld (talk) 11:33, 22 May 2008 (UTC)

Well, tan is actually composed of (complex) exponentials, so one might be able to use the Lambert W Function for that purpose. Paxinum (talk) 21:22, 22 May 2008 (UTC)

I think I've thought about that once and concluded that it is not actually possible. Too lazy to look at it again. -- Meni Rosenfeld (talk) 00:14, 23 May 2008 (UTC)

Your Mathematica-Fu is very impressive. Thank you very much! Vvitor (talk) 11:36, 22 May 2008 (UTC)

Someone has invented such a function: the tanc function. To follow the naming convention for inverse trig/hyperbolic function, the sought value would be atanc(2). I've only seen the name tanc used on MathWorld, and the name atanc was made up on the spot here, but both really should be standard since equations similar to tan(x)=x are so common in physics. Maybe if I implement both in mpmath and sympy, the world will follow... - Fredrik Johansson 10:56, 23 May 2008 (UTC)

Flux across surface.

Is the flux of the vector field F=(xz)i+(x)j+(y)k across the surface of the hemisphere of radius 5 oriented in the direction of the positive y-axis equal to Zero? I'm getting up to a string of double-integrals between 0 and pi, but each term has a sin function that results in everything ending up zero, but it seems an unlikely answer. —Preceding unsigned comment added by Damian Eldridge (talk • contribs) 11:39, 22 May 2008 (UTC)

Yes, it is zero. No messy double integrals are necessary, by the way. Algebraist 12:09, 22 May 2008 (UTC)

Let $S$ be your surface, which is symmetric under negation of $x$ or $y$. Let $G(x,y,z)=(xz)i+0j+(y)k$ and $H(x,y,z)=0i+(x)j+0k$. Flipping $G$ across the $x$-$y$ plane, we obtain $(x(-z))i+0j+(y)(-k)=-G$; so $G$ and $-G$ have the same flux, which must be zero. Flipping $H$ across the $y$-$z$ plane, we obtain $0(-i)+(-x)j+0k=-H$, so similarly the flux of $H$ is 0. But the flux of $F$ is the sum of the fluxes of $G$ and $H$, so must be zero. Eric. 144.32.89.104 (talk) 12:23, 22 May 2008 (UTC)

I choose the lazy approach. The divergence theorem is very useful here. Note that $\nabla \cdot {\overrightarrow {\mathbf {F} }}$ is anti-symmetric about the plane z=0. --Prestidigitator (talk) 20:49, 22 May 2008 (UTC)

Yeah, that was my approach. Algebraist 21:19, 22 May 2008 (UTC)

Ah -- I had misread "surface of the hemisphere" as if it were "half of the surface of a sphere", which would be an unclosed surface. Eric. 217.42.199.10 (talk) 10:23, 23 May 2008 (UTC)

Actually, that was my reading too. Fortunately, the flux through the flat side of the hemisphere is obviously zero, so you get the same answer. Algebraist 11:03, 23 May 2008 (UTC)

Four-dimensional pyramid

I know that the four-dimensional equivalent of the tetrahedron is the pentachoron, which would resemble a tetrahedron dwindling to a point as it passed through our three-dimensional space, but what's the four-dimensional equivalent of a square-based Egyptian-type pyramid? Would its cross section(if it fell "face-first") be a cube, a triangular prism, or something else? 69.111.191.122 (talk) 15:00, 22 May 2008 (UTC)

I am not 100% sure about this, but I suspect there is some ambiguity about what the analog actually would be. The tetrahedron's point group T_d has triply degenerate symmetries, meaning that the three dimensions are in some sense equivalent or at least exchangeable. The pyramid (C_4v ) does not, only two of the dimensions (the ones defining the base) are exchangeable. So you could either make the next dimension a second "height" or a third "base" dimension. The resulting two figures would not be the same. If you made specific the type of analog you were considering, it would help. Baccyak4H (Yak!) 15:25, 22 May 2008 (UTC)

As Baccyak says, there are multiple things that could be called a 4-D pyramid. I would go with a cube going to a point (a square pyramid is a square going to a point, add one dimension to a square, you get a cube). I think that might be what Baccyak means by adding a third base dimension. I'm not quite sure what adding a 2nd height dimension would mean... --Tango (talk) 15:30, 22 May 2008 (UTC)

A square pyramid going to a point, maybe. Black Carrot (talk) 15:45, 22 May 2008 (UTC)

Re adding a second height dimension: in general, the join of two shapes S (in d dimensions) and T (in e dimensions) is formed by taking disjoint d-dimensional and e-dimensional affine subspaces of (d + e + 1)-dimensional space, placing S and T within them, and taking the convex hull of S and T as placed together in the larger dimensional space. For the geometry of the resulting shape it matters where S and T are placed but for the combinatorics of what faces the join has it doesn't matter. A three-dimensional pyramid over some base is the join of that base (2-dimensional) with a single point (0-dimensional). The 4-simplex and pyramid-over-cube you're talking about are joins of a regular tetrahedron or cube with a single point. But the 4-simplex is also the join of a triangle and a line segment, and you could consider the shape formed by the join of a square with a line segment as a generalization of a pyramid. This shape has as its facets two square pyramids (the joins of the square with the endpoints of the line segment) and four tetrahedra (the joins of the sides of the square with the whole line segment). Its Schlegel diagram can be formed as a square pyramid with a single vertex inside it connected by edges to all five pyramid corners. And, as Black Carrot suggests, it is also the join of a square pyramid and a single point. —David Eppstein (talk) 15:47, 22 May 2008 (UTC)

My 2c..

Starting with a shape S, there are many ways to increase its dimension. One way is to add a point in the next dimension. This gives a pyramid, whose base is the original shape.

• If S is a line segment, you get a triangle.
• If S is a triangle, you get a tetrahedron.
• If S is a square, you get a square pyramid.
• If S is a circle, you get a cone.

Let's call this operation, Adding a point, A. Another way is to move the shape in the direction of the new dimension, and see what it carves out. This gives you a prism, whose base is the original shape.

• If S is a line segment, you get a square.
• If S is a triangle, you get a triangular prism.
• If S is a square, you get a cube.
• If S is a circle, you get a cylinder.

Let's call this operation, Moving the shape, M. Now we can think of what you get if you apply A and M successively, starting with a line segment L. (Or, you could start with a single point if you prefer)

• AL is a triangle,
• ML is a square

Then, in three dimensions...

• AAL is a tetrahedron
• MAL is a triangular prism
• AML is a square pyramid
• MML is a cube

Your question is about the next step...

• AAAL is a pentachoron
• AMAL is a pyramid whose base is a triangular prism. Its "faces" has two tetrahedra, three square pyramids, and the base, a triangular prism.
• AAML is a pyramid whose base is a square pyramid. Its "faces" are four tetrahedra, and two square pyramids (including the base). In fact, once the shape is made, there's no way to identify which of the two square pyramids was the "original" base.
• AMML is a pyramid with a cube for its base. Its "faces" are 6 square pyramids and the cubic base.

Any of the last three could perhaps count as the "equivalent" of the square pyramid, but I'd vote for the last. Then, there are the prisms..

• MAAL is a prism with a tetrahedral base. Its faces are 4 triangular prisms, and the 2 tetrahedra.
• MMAL is a prism with a triangular prism for its base. Its faces include has 4 triangular prisms, in two opposite pairs, and three cubes. Either of the pairs of triangular prisms could count as the base.
• MAML is a prism with pyramids at the top and bottom. The sides consist of 1 cube, and 4 triangular prisms.
• MMML is the good ole' tesseract.

Hope that helps! mike40033 (talk) 02:10, 23 May 2008 (UTC)

Thanks, that's very informative! Are there any images, nets, or animations of those shapes? (Most of the 4D geometry pages I've seen just have the perfectly symmetrical ones and their stellations.) —Preceding unsigned comment added by 69.111.191.122 (talk) 00:44, 24 May 2008 (UTC)

That was a great answer - worth three cents, at least! --Tango (talk) 20:15, 23 May 2008 (UTC)

A simpler way to explain why there is an ambiguity is that we don't know the form of the extra dimension. In the case of a regular tetrahedron, it's reasonable to assume that the 4th dimensional analog is also a regular figure - so you get the tetrahedron that shrinks to a dot as the 4 dimensional shape translated through 3-space.

But a square-based pyramid is an irregular shape - there are any number of irregular 4 dimensional spaces which have pyramidal hyper-faces. It's kinda like if you were a 2D being - you could ask for 3D analogs of things like squares, circles and equilateral triangles - and get answers like "cube", "sphere" and "tetrahedron". But demanding the 3D analog of an irregular 2D figure would be an unanswerable question.

70.116.10.189 (talk) 15:43, 24 May 2008 (UTC)

Figuring "Final Average"

My professor said to figure my final average for the year, without the final exam, to:

1. Take my first quarter grade and double it
2. Take my second quarter grade and double it
3. Take my semester exam
4. Take my third quarter grade and double it
5. Take my fourth quarter grade and double it

Then I am to add these numbers together and divide by 10 to get my final average.

For example Q1=93%, Q2=90%, SE=93%, Q3=100%, and Q4=94%; my final average would be 85%.

What's the logic behind the doubling and dividing. Also how do I go from 90's (as in my example) to 85? Thanks...I'm really lost. §hep • ¡Talk to me! 22:34, 22 May 2008 (UTC)

My best guess would be that the semester exam is out of 200 points, not 100: but has equal weighting as the quarters. Thus to get equal weighting he gave them all a value of 200 points by doubling each quarter.

The key is that you’re dividing by 10 at the end, and so the original sum of (1) to (5) must be 1000. Assuming each quarter is out of 100, that’s how I deduced that the semester exam is probably out of 200.

(If everything were out of 100, then the maximum possible would be 900/10=90) GromXXVII (talk) 22:40, 22 May 2008 (UTC)

Sorry, that's percentage not points. I should have been more specific. §hep • ¡Talk to me! 22:46, 22 May 2008 (UTC)

Ok, the semester exam is worth half as much as the quarter grades, so you count each quarter grade twice. That means you've effectively got 9 marks to count, plus the final exam, makes the 10 you divide by. The reason your average is less than the individual marks is because it's assuming you get 0% on the final exam (which apparently has the same weight as the semester exam). If you want just your average for the bits you've taken so far, not including the final exam at all, just divide by 9 instead of 10. --Tango (talk) 22:56, 22 May 2008 (UTC)

Ahhh. Gotcha. Thanks! §hep • ¡Talk to me! 23:11, 22 May 2008 (UTC)

This type of average is called a weighted mean. Confusing Manifestation(Say hi!) 23:30, 22 May 2008 (UTC)

So we can deduce that the final exam has a weight of 10% in the final average. If you scoe x% in the final exame then your final average will be 85 + (x/10) %. So you can be certain of a final average of at least 85%, but even if you ace the final exam you can't get a final average better than 95%. Gandalf61 (talk) 08:49, 23 May 2008 (UTC)