What does R^sup 2^ mean when talking about curvature?
822597
in the plane,\nthere are two kinds of curvature,\nthe regular one, which is the norm of the second derivative, if the curve has unit tangent.\nand the signed curvature, which in addition measures if the first derivative and the second derivative preserve orientation.\ns
307198
It is linear on intervals, on (- infinite, 0) and on (0, +infinite) but not on the whole R. \n\nTry plotting it, you get a reversed V, not a straight line.
465096
From memory (so you should double check it):\n\nRa = Rb Sqrt(1-(V/c)^2) \n\n^2 = squared\n\nRa=dilated time\nRb=reference time\nV= velocity\nc=speed of light\n\nSo this is one of the problems with luminal or superluminal travel. If v=c then Ra=0 no matter what Rb is.\n\nAnd if v>c then you get an imaginary number... what could that mean?
521925
You find it in the normal coordinates way; i.e.\n((x1+x2)/2, (y1+y2)/2)\n\nThe curvature of our Earth does not affect it as the coordinates we are using is 2-dimensional.
392390
it means you have to derive it twice ,which means you have to derive its second order derivative ,for example derivig the distance formula gives you the speed,then deriving the speed which is the 1st derivative of distance gives you the acceleration.
684813
I had to refresh myself on the method of Lagrange multipliers. The following theorem is paraphrased from a problem in Spivak's "Calculus on Manifolds", an excellent calculus text for mathematicians.\nLet A be a subset of R^n. If g:A -> R^m is differentiable and has rank m on the kernel of g, and if f:R^n -> R is also differentiable and the extremum of f on the kernel of g occurs at a, then there exist m real multipliers L1, ..., Lm such that df(a) = (L1)d(g1)(a) + ... + (Lm)d(gm)(a). (d is the differential operator)\nThe proof is not that difficult, but a concise one would require the introduction of too much machinery, so we'll leave it as something to look up.\nIf g has only one component function (if g:A -> R), then the above is very simple with only one multiplier. If n=2, the kernel of g is just a curve in R^2 and the graph of f is a surface in R^3. In this case, you can visualize the constraint on the surface as a curve in the surface of f whose projection on R^2 considered as a subset of R^3 is the kernel of g.\nIn your case, we want to minimize the distance function f(x, y, z) = Sqrt[x^2 + y^2 + z^2] over the two surfaces above, or equivalently the kernel of the differentiable function g(x, y, z) = (2y + 4z - 5, 4(x^2 + y^2) - z^2) (g is thus a map from R^3 -> R^2, so m = 2. The kernel of g is trivially the intersection of the two surfaces. You should show that the rank of g is 2 whenever g(x, y, z) = 0 to validate the use of Lagrange multipliers). \nWe simplify this by noting that the points corresponding to the minimum of f over the kernel of g coincides with the points corresponding to the minimum of [f(x, y, z)]^2 = F(x, y, z) over the kernel of g which is easier to deal with. The theorem tells us we will have to deal with two Lagrange multipliers, one for each component of g. We thus have the system of 5 equations:\n2x = (L1)*0 + (L2)*8x\n2y = (L1)*2 + (L2)*8y\n2z = (L1)*4 + (L2)*(-2z)\ng(x, y, z) = (2y + 4z - 5, 4(x^2 + y^2) - z^2) = (0, 0)\nThese are 5 equations in 5 unknowns, so it is easily solved. Note that the actual values of the multipliers are seldom important.\n\nIn case you are not familiar with the differential operator, if f(x, y, z) = x^2 + y^2 + z^2 + xy, then df(x, y, z) is a covector (actually a cotangent bundle in more formal terms) such that df(x, y, z) = 2x dx + 2y dy + 2z dz. It is similar to notation for implicit differentiation but has a much more rigorous foundation and is used in differential geometry. The dx, dy and dz terms are treated as basis vectors, thus the above is a 3-dimensional vector, normally called the gradient of f in multivariable calculus, where it is treated as if it were a vector in R^3: (2x, 2y, 2z). You should now see where the first 3 equations come from.
41743
It's "circumference", and it's the boundry of a circular area, and actually, it's not that hard. \n\nThe circumference is the length of the outside boundry of a circle expressed as 2*pi*r ("pi" = the pi constant, sometimes abreviated as 3.14 and "r" is the radius or half of the distance across the circle. You can also do simply ). It's the length you'd get if you were to wrap a string around a balloon and measure it.
641426
In general, there are two important types of curvature: extrinsic curvature and intrinsic curvature. The extrinsic curvature of curves in two- and three-space was the first type of curvature to be studied historically, culminating in the Frenet formulas, which describe a space curve entirely in terms of its "curvature," torsion, and the initial starting point and direction.\n\nAfter the curvature of two- and three-dimensional curves was studied, attention turned to the curvature of surfaces in three-space. The main curvatures that emerged from this scrutiny are the mean curvature, Gaussian curvature, and the Weingarten map. Mean curvature was the most important for applications at the time and was the most studied, but Gauss  was the first to recognize the importance of the Gaussian curvature.\n\nBecause Gaussian curvature is "intrinsic," it is detectable to two-dimensional "inhabitants" of the surface, whereas mean curvature and the Weingarten map are not detectable to someone who can't study the three-dimensional space surrounding the surface on which he resides. The importance of Gaussian curvature to an inhabitant is that it controls the surface area of spheres around the inhabitant.\n\nRiemann  and many others generalized the concept of curvature to sectional curvature, scalar curvature, the Riemann tensor, Ricci curvature, and a host of other intrinsic and extrinsic curvatures. General curvatures no longer need to be numbers, and can take the form of a map, group, groupoid, tensor field, etc.\n\nThe simplest form of curvature and that usually first encountered in calculus is an extrinsic curvature. In two dimensions, let a plane curve be given by Cartesian parametric equations.\nThe curvature of a two-dimensional curve is related to the radius of curvature of the curve's osculating circle.\n\nThe curvature at a point on a surface takes on a variety of values as the plane through the normal varies. As  varies, it achieves a minimum and a maximum (which are in perpendicular directions) known as the principal curvatures.The curvature  is sometimes called the first curvature and the torsion  the second curvature. In addition, a third curvature (sometimes called total curvature).
9490
It's the symbol of exponents. For example 6^2 is 6x6, 5^7 is 5x5x5x5x5x5x5
569932
the radius of curvature of a plane mirror is infinity( iam dead sure)\n\n\nThe term radius of curvature has specific meaning and sign convention in optical design. A spherical lens or mirror surface has a center of curvature located in (x,y,z) either along or decentered from the system local optical axis. The vertex of the lens surface is located on the local optical axis at a distance equal to the radius of curvature from the center of curvature. The sign convention for the optical radius of curvature is as follows:\n\nIf the vertex lies to the left of the center of curvature, the radius of curvature is positive. \nIf the vertex lies to the right of the center of curvature, the radius of curvature is negative. \nThus when viewing a biconvex lens from the side, the left surface radius of curvature is positive, and the right surface has a negative radius of curvature.\n\nNote however that in some areas of optics other than design, other sign conventions are sometimes used.
602701
it means "raised to the power of".\nthat is, 2^3 means "2 raised to the power of 3" or "2 cubed" \n(2^3 = 2*2*2 = 8)\n\nwhile writing, the index is a superscript of the base. In the above example, 3 will be written as the superscript of 2.\nBut when such a representation is not possible, say while typing, a circumflex (^) is used as explained above.
1137727
I like the joke!  To play along, I define:\nr = metric r  and \nd = metric d\n\nV = (pi/3)*(d^2)*(3r - d)\n\nThis will work for the tank being any fraction of a sphere, not just 1/2
184873
Yes.\ngravity is inversely proportional to radius squared, and radius is inversely proportional to the curvature of the same surface, so your statement is true.
768242
it means that as a (linear) transformation from the plane to the plane it reverses the orientation.\nthe way we calculate the area, is really a signed area, in the sense that it takes into consideration the orientation of the 2 vectors.\nit is a bit subtle and that is why most teachers do not really explain it.\n'
293949
Here is one condition:\n\nR1^2  +  R2^2  =  AB^2 (Pythagorean theorem),\n\nwhere R1 and R2  are the two intersecting circles radii and AB is the distance between their centers (AB < R1 + R2)
40559
Just what are you asking? \n\nIf this were the Baking section, then you owe someone two pies. \n\nIf you are talking Math (this section), then I presume you are talking about PI (the value used in Circumference = 2 (PI) * radius, Area = PI * r^2, Volume = (1/3) PI * r^3). Following these derivations, I have no clue where PI^5 lies in Geometry.\n\nOr, finally, if you are thinking 3*PI - 5*PI, you get -2 * PI, or -2 * 3.14159..., = -6.28318...
889821
Actually, solutions to Einstein's field equation indicate that the Universe can be either (1) flat, (2) positively curved (like a sphere)or (3) negatively curved (like a saddle). All three curvatures of spacetime equally satisfies Einstein's equation. Also, I believe the solution indicated that the Universe is finite, but "unbounded".\n\nCurrent data from the WMAP satellite tells us that the curvature of spacetime of our Universe is FLAT.