What is the equation to graph a spiral?
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Logarithmic Spiral -- From MathWorld\nThe logarithmic spiral is a spiral whose polar equation is given by r==ae^(btheta), where r is the distance from the origin, theta is the angle from the x-axis, and a and b are arbitrary constants. The logarithmic spiral is also known as the growth spiral, equiangular spiral, and spira mirabilis. It can b\nhttp://mathworld.wolfram.com/LogarithmicSpiral.html - 29k - 2002-06-04\n\nCornu Spiral -- From MathWorld\nA plot in the complex plane of the points B(t)==S(t)+iC(t), where S(t) and C(t) are the Fresnel integrals (von Seggern 1993, p. 210; Gray 1997, p. 65). The Cornu spiral is also known as the clothoid or Euler's spiral. It was probably first studied by Johann Bernoulli around 1696 (Bernoulli 1967, pp. 1084-\nhttp://mathworld.wolfram.com/CornuSpiral.html - 27k - 2006-01-17\n\nArchimedes' Spiral -- From MathWorld\nArchimedes' spiral is an Archimedean spiral with polar equation r==atheta. This spiral was studied by Conon, and later by Archimedes in On Spirals about 225 BC. Archimedes was able to work out the lengths of various tangents to the spiral.The curvature of Archimedes' spiral is kappa(theta)==(2+theta^2)/(a\nhttp://mathworld.wolfram.com/ArchimedesSpiral.html - 24k - 2003-12-10\n\nArchimedean Spiral -- From MathWorld\nAn Archimedean spiral is a spiral with polar equation r==atheta^(1/n), where r is the radial distance, theta is the polar angle, and n is a constant which determines how tightly the spiral is "wrapped."Values of n corresponding to particular special named spirals are summarized in the following table, tog\nhttp://mathworld.wolfram.com/ArchimedeanSpiral.html - 23k - 2003-12-10\n\nFermat's Spiral -- From MathWorld\nFermat's spiral, also known as the parabolic spiral, is an Archimedean spiral with m==2 having polar equation r^2==a^2theta. This curve was discussed by Fermat in 1636 (MacTutor Archive). For any given positive value of theta, there are two corresponding values of r of opposite signs. The left plot above\nhttp://mathworld.wolfram.com/FermatsSpiral.html - 22k - 2003-10-30\n\nNatural Equation -- From MathWorld\nA natural equation is an equation which specifies a curve independent of any choice of coordinates or parameterization. The study of natural equations began with the following problem: given two functions of one parameter, find the space curve for which the functions are the curvature and torsion.Euler ga\nhttp://mathworld.wolfram.com/NaturalEquation.html - 23k - 2006-01-14\n\nHyperbolic Spiral -- From MathWorld\nAn Archimedean spiral with polar equation r==a/theta. The hyperbolic spiral, also called the inverse spiral (Whittaker 1944, p. 83), originated with Pierre Varignon in 1704 and was studied by Johann Bernoulli between 1710 and 1713, as well as by Cotes in 1722 (MacTutor Archive).It is also a special case o\nhttp://mathworld.wolfram.com/HyperbolicSpiral.html - 21k - 2002-08-27\n\nLogarithmic Spiral Evolute -- From MathWorld\nlogspev.gifFor a logarithmic spiral given parametrically asevolute is given byAs first shown by Johann Bernoulli, the evolute of a logarithmic spiral is therefore another logarithmic spiral, having b^'=b and a^'=ab,In some cases, the evolute is identical to the original, as can be demonstrated by making t\nhttp://mathworld.wolfram.com/LogarithmicSpiralEvolute.html - 28k - 2003-04-10\n\nNielsen's Spiral -- From MathWorld\nNielsen's spiral, also called the sici spiral (von Seggern 1993) is the spiral with parametric equationswhere ci(t) is the cosine integral and si(t) is the sine integral.The arc length and curvature are given bywhere the arc length is measured from t==1.This gives the Cesàro equation as kappa==(e^(s/a))/\nhttp://mathworld.wolfram.com/NielsensSpiral.html - 19k - 2004-05-01\n\nSuperellipse -- From MathWorld\nA superellipse is a curve with Cartesian equation |x/a|^r+|y/b|^r==1, first discussed in 1818 by Lamé. A superellipse may be described parametrically bySu
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Graph the equations. They should look like a big "U". The axis of symmetry is an imaginary line going right through the vertex and splits the graph in half. The point at where the axis crosses the x axis is what the axis of symmetry is. \n\nPlugging in is easy. make a chart for x. Now choose like 4 different easy values to put into the equation as x. 0,1,2,3. Use your calculator to find the answers, and make a separate side of the chart as y. the answers you get are your y's. Then you graph it. \n\nI don't have a calculator right now, so I'm leaving that to you.
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http://www.graphcalc.com/
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lol.\nfind formula for area of spiral.  subtract smaller from bigger.  sry no more help.
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Equation of a line:  y = mx + b\n\nm = 3 (slope)\ny = -4 (y-intercept)\n\ntherefore:  y = 3x - 4 \n\nCrosses x-axis at y = -4 (0,-4)\nCrosses y-axis at x = 4/3 (4/3,0)
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There are two equations there, not one.\n\nFor the first equation, plug the numbers on your X-axis into "x", calculate "y", then draw the points on your graph.  I'll get you started:\n\nx=0, y = -3\nx=1, y = -1\nx=2, y = 1\netc.\n\nThe second equation simply means that x = 1.5 and y = 3.  This is just a single dot on a graph.
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first you find the slope using formulae: \nm = (y2-y1)/(x2-x1)\nm = (-1-7)/(5+3)\nm = -8/8\nm = -1\n\nthen you substitute this into the formulae:\ny = mx + b\ny = -x + b (use one of the coordinates to find b)\n7 = -(-3) + b\n7 = 3 + b\n7 - 3 = b\nb = 4 (substitute back into the equation)\n\ny = -x + 4 (there goes your answer!)
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To find the slope, you always have to set the equation as y=mx+b.  That equation is called the slope-intercept form.  "m" is the slope and b is the y-intercept.\nNow, to find the slope, try to get y alone.\n1. Subtract 2x from both sides.  You'll get: -8y=-2x+7\n2. Divide both sides by -8.  You'll get: y=(1/4)x+(-7/8)\nSo, the equation is y=(1/4)x-(7/8) and the slope is (1/4).
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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.
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hi bob!\nDo you mean ellipse? If so then its equation is: (x/a)^2+(y/b)^2=1, a=4, b=3;\nHammer nails in ellipse focuses with coordinates [-f 0] & [+f 0], now attach the rope ends to these nails, let the rope length be L, keep the rope taught with pencil and draw.\nLet y=0, then x=a (or x=-a which does not matter here), then L=(a+f)+(a-f)=2*a\nLet x=0, then y=b, then f^2+b^2=(L/2)^2, f=sqrt(a*2-b^2)
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My dear, just go to www.quickmath.com and on the left hand side click on 'Equations' under 'Graphs' and thewn type in your equation u will get the graph. I have been using it for years....It's a great site!! I hope this helps :)
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Basically, its a straight line going through (1,3), with gradient, m=-3. we have y=mx+c. putting the point (1,3) into here, \n\n3=(-3)(1)+c, so c=6, which is the y-intercept. so the line equation is,\n\ny=-3x + 6, which is a straight line with y-intercept at y=6. the line will be sloping downwards since the gradient (m) is negative. if you plot a few values of x and y which satisfy the equation you can see the line. to check axis intercepts, put y=0 for the x-intercept and x=0 for the y-intercept.
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just pick two points from the graph\nif you wanna check plug them in the equation\n\n\n\nand when the line crosses the Y axis Y does NOT =0\nX=0 when line crosses Y axis\n\nchoose any set of #s that will not give you the square root of any negative. because that would result in an imaginary #, not a real
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if your equation is y=x^2, then the graph that you draw is going to be a parabola. You can see this by having a table first. The table should consist of 2 values...x and y. So you say when \nx=-3, y=9,\nx=-2, y=4,\nx=-1, y=1\nx=0,  y=0\nx=1,  y=1\nx=2,  y=4\nx=3,  y=9\nso now if you have a square line paper, just plot it and you will see that it will give you a parabola. Another thing is that when you draw it, first just plot the points and then when you connect the point connect them using curves and not a line,...that's why u get a u shaped graph and not just a combo of lines.
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(0,0)\n(3,0)\ndraw the graphs and then find them out!\nif u can't, then mail me!
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It is the same as the graph of -cos(2 theta).\n\nIt starts at (0, -1) , has an amplitude of 1, and has a period of pi radians which means it passes through (pi/2, 1)\n\nOr in degrees, it has a period of 180 and passes through (90, 1)
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y=mx+b\n\n"m" is the slope, and "b" is the y axis. Best wishes in your studies!