I need to compute for L,T,E,M,C,R and stations of PC and PT for the circular curves . Using the chord& arc def
967408
Don't you have a book with a diagram of all those points?
358834
KNOW YOUR RELATED RATES FOR CALC!! thats one thing that killed me and know your 2-D rotation around a line intergrals.
537751
e-chords.com or olga.net
549477
http://www.purplemath.com/modules/percntof.htm\n\nhope this helps you.
855200
that sent me round in circles.\nhere have a link c if u find what u need.\n:)
843247
You have enough information.\n\nIf I understand you correctly, you have the (N,E) coordinates for points A and F.  So given what you know so far, you should be able to compute coordinates for points B, C, and E.  All you need for a complete picture is coordinates for D.  Lets call them (N(D), E(D)).\n\nNow you know the distance DE. Using the distance formula,\n\n(DE)^2 = [N(E) - N(D)]^2 + [E(E) - E(D)]^2 \nthis is one equation with two unknowns.\n\nAlso, you know the azimuth for CD.  Call it θ.\nThen tan(θ) =   [E(D) - E(C)] / [N(D) - N(C)]\nthis is a second equation with the same two unknowns.\n\nTwo equations, two unknowns.  Solve simultaneously to get\nN(D) and E(D).\n\nOnce you have the coordinates you can compute missing azimuth and missing distance.\nIf you would like to email me the numbers I'll take a look.
882948
Just make the following substitution:\nr=sqrt(x^2+y^2)\ntheta=arctan(y/x)\nand then simplify.  Bring everything over to one side and you get your answer.\n\nIf tan(theta)=y/x, then sec(theta)=sqrt(x^2+y^2)/x.\n\nThe problem is a little ugly, but it is doable and then at the end everything ends up canceling nicely.
966696
Arc=r.theta=8*pi/8=pi
726975
Use sin(arcsinx) = x...\n\ncos(arcsinB) = sqrt(1-sin^2(arcsinB))\n= sqrt(1-B^2)\n\nSimilarly, sin(arccosA) = sqrt(1-A^2).\n\nUse the cos(x+y) equality to compute the last one...
144396
Matlab, Gauss, S+ and probably a lot more.\n\nIn a pinch, you can also use MS-Excel.\nDownload this pdf file from Cliff Ragsdale´s book site:\nhttp://swlearning.com/quant/ragsdale/chapter12.exe\nIt actually works better than you´d think.\n\nOr go to http://swlearning.com/quant/ragsdale/fourth_edition/ragsdale.html\npick previous edition and student resource.
913559
0.5 (x) arctan(x) - 0.25 log (x^2+1)
470658
The above answer is basically correct.  Plug in a few more values, such as (21,1543.5), (22,1694), and so on for a few more values of the graph.
1155862
to many to post.....they are all on this site \n\nhttp://www.geom.uiuc.edu/docs/reference/CRC-formulas/
900462
It really depends on how the equation of the path/curve is originally specified.\n\nIf it's specified in terms of x and y, an easy parameterization is:\n\nx = t\ny = f(x) = f(t)\n\nBut there are many other ways to do it.
70364
Well, it would depend what graphics toolkit you're using, what operating system, etc. It would probably involve:\n\n1. creating a window\n2. creating a canvas or drawing area\n3. creating all the elements of a clock on the canvas.\n\nThe "hard" part is 3, where you probably have to use trigonometry to turn your circular clock coordinates into the usual cartesian coordinates.
663777
SOrry, I can't help you. If I had the formulas I could do it, but there is no way I remember that off the top of my head! Good luck
93089
This seems like a homework problem, so I'll give a few hints.\n\nThe arclength is radius times the angle subtended.\n\nTo find the angle subtended, consider the isosceles triangle created inside the circle where the chord is the hypotenuse and the radius of the circle is the two other legs. \n\nSince the isosceles triangle created is not a right triangle you need to use a specific formula to find the angle "C" subtended by the chord.