how do i solve a spiral curve?
149957
Maybe the question is, given a spiral curve, how do I find the equation that generated it?  If that is the question, you must be given, or able to measure, some points on it.  Fit those points into the form of one of rt11guru's equations by adjusting coefficients as needed.
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Example: Find the length of a helix which winds three times around a cylinder of height 12" and radius 3"\n\nAnswer: divide the cylinder into three sections. The helix goes once around a cylinder of height 4" and radius 3"\n\nThe surface of the cylinder can be "peeled" into a rectangle 4" by 2 * pi * 3" = 18.850". The helix is a diagonal of that rectangle. So its length is sqt(4^2 + 18.850^2) = 19.27"\n\nIt goes around the cylinder three times and its total length is therefore 57.81"
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A spiral is just a monotonically increasing function that defines the radius (r) as a function of the angle (a).  The angle is normally in radians.  The simplest spiral is a linear one:\n\nr = k*a\n\nYou can use any increasing function of r.  \n\nAn exponential spiral: r = e^(k*z) will grow faster for each revolution.\n\nA logrithmic spiral: r = ln(k*a + 1)  grow more slowly with each revolution. \n\nYou can define spirals that are lumpy and maybe not technically spirals such as:\n\nr = a - sin(5a)\n\nWhich makes pentagon shaped loops.
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For a fixed a and n, and allowing b and c to vary, the curve would be a hyperbolar centered at the origin. You will only be taking the rational points on that curve, though.
389156
actually, you slolve each of the three parenthesis and that tells you when the equation equals zero...\n\nX=-2, x=1 and x=5.\n\nNow look at each interval to determine whether the curve is greater than or less than zero in that interval.  For example, \nless than -2: x=-3 is less than -2.  So substitute x=-3 and solve... (-1)(-4)(-8)=-24 which is less than 0.\nbetween -2 and 1: x=0 is between these numbers.  So substitute x=0 and solve...(2)(-1)(-5)=+10 which is greater than 0.\nbetween 1 and 5: x=2 is between these numbers.  So substitute x=2 and solve....(4)(1)(-3)=-12 which is less than 0. \ngreater than 5: use x=6: (8)(5)(1)=+40\n\nSo the number line should show an open circle at -2 and an arrow pointing left, and open circles at 1 and 5 with a line connecting the two circles at 1 and 5.
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f(x) = x^m(1-x)^n\nm zeros at  x = 0 and n zeros at x = 1 \nSo x-axis is tangential to the curve at x = 0 and x = 1\n\nf'(x) = mx^(m - 1)(1 - x)^n + x^m * -n(1 - x)^(n - 1)\n= x^(m - 1) * (1-x)^(n - 1) * (m(1 - x) - nx)\n=  x^(m - 1) * (1-x)^(n - 1) * [m - (m + n)x]\n= 0 for stationary points.\n\nSo (m - 1) zeros at x = 0, (n - 1) zeros at x = 1 and a zero at x = m / (m + n)\n\nNow there has to be a turning point between x = 0 and x = 1 as f(x) is a polynomial and therefore continuous and differentiable for all x\n\nSo examine the possibilities\n... Domain → ... x < 0 ... 0<x<1 ... x>1 \nm .... odd    \nn ..... odd    \nx^m ..................... - .......... + ........... + ... Stationary inflexion at (0, 0)\n(1 - x)^n .............. + .......... + ........... - ... Local maximum at (m/(m+n), f(m/(m+n))\nf(x) ....................... - .......... + ........... - ... Stationary inflexion at (1, 0)\n\n\n... Domain → ... x < 0 ... 0<x<1 ... x>1 \nm .... odd    \nn ..... odd    \nx^m ..................... - .......... + ........... + ... Stationary inflexion at (0, 0)\n(1 - x)^n .............. + .......... + ........... + ... Local maximum at (m/(m+n), f(m/(m+n))\nf(x) ...................... - .......... + ........... + ... Local minimum at (1, 0)\n\n\n... Domain → ... x < 0 ... 0<x<1 ... x>1 \nm .... odd    \nn ..... odd    \nx^m ..................... + .......... + ........... + ... Local minimum at (0, 0)\n(1 - x)^n ............... + .......... + ........... - ... Local maximum at (m/(m+n), f(m/(m+n))\nf(x) ....................... + .......... + ........... - ... Stationary inflexion at (1, 0)\n\n\n... Domain → ... x < 0 ... 0<x<1 ... x>1 \nm .... odd    \nn ..... odd    \nx^m .................... + .......... + ........... + ... Local minimum at (0, 0)\n(1 - x)^n .............. + .......... + ........... + ... Local maximum at (m/(m+n), f(m/(m+n))\nf(x) ...................... + .......... + ........... + ... Local minimum at (1, 0)
939024
Draw yourself a cartesian graph, vertical line is the y axis, horizontal is the x axis.\n\npick points on the x axis like 1, 2, 3, etc.  plug thos values into the equation and solve for y. then pair the numbers, place points in the graph by taking the x value going straight up (or down if y is a negative number) and place the dot accross from the y value.  do this for each set of points then connect the dots with a line or a curve. for instance in your equationstart with x=1\n\ny= 7(1-5)(1-2)^4(1+.5)(1+2.5)^3\ny= 7(-4)(-1)^4(1.5)(3.5)^3\ny=7(-4)(1)(1.5)(42.875)\ny=-1800.75\n\nso in XY format the first point would go at (1, -1800.75)
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find the slopes of both lines and if there answers are opposite recipricals, then they are parrallel\n\nheres adn ex:\nslope = 2 and -1/2\n\nin other workds, if you multiply the two slopes, u should get -1
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First, look at where the curve intersects the line:\n\ncurve y=1/x\nline y=x/4\n\n1/x=x/4\nx=+/- 2\n\nNote that the curve is symmetrical about the origin for x&y positive and x&y negative, so the line intersects twice.\n\nThat's actually somewhat irrelevant, but it helped me visualize the line and the curve.\n\nTake the first derivative of the curve\ndy/dx=-1/x^2\nThis is the instantaneous slope of the curve at any point, which is the tangent by formal definition.\n\nA line perpendicular to the line y=.25x will have a slope of -4.\nSo set the derivative equal to -4 and solve for x\n-4=-1/x^2\nx=+/- 1/2\nsolve for y\nand find the points\n(1/2,2)\n(-1/2,-2)\nThis is where the two lines that are both tangent to the curve and perpendicular to the line intersect the line.\n\none line is y=-4x+4\nthe other is y=-4x -4\n\nj
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you have to \n\n1. Determine the lengthy upto which the helical reinforcement is to be provided Let=L\n2. Determine the pitch =X\n3. Determine Y=L/X\n4.LENGTH OF HELICAL REINFORCEMENT=Y*3.1415*(SQRT  ( X SQ + D SQ ) )  WHERE D IS DIA OF HELICAL REINF.IN MM
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Find the coordinates of the points where the straight line meets 'y=1' meets the circle 'x^2+y^2-14x+2y+45=0'.\n\n'x^2+1^2-14x+2(1)+45=0\nx^2 -14x +48 = 0\nthe solutions are:\nx=(14 +- sqrt( 14^2 -4(48) ) /2\nx=14+- sqrt(4) )/2\nx= (14 +2)/2 or x= (14-2)/2\nx=8 or x = 6\n.
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cool! I'll think about it\n\n---\n\nhere I am.\nUse the two perpenducular segments as a coordinate system with O on their intersection.\n\nSay: the 4-3 segment defines the x axis, and the 2-6 segment defines the y axis.\nYou know four points belonging to the circle: (-4;0),(0;-2),(3;0) and (0,6)\nThe center of the circle has the mean value between -4 and 3 as its x and the mean value between -2 and 6 as its y (property of chords: the center is on the chord's axis)\n\nso the center's coordinates are C(-1/2;2)\n\nNow you can easily calculate the distance between C and anyone of the four points you know to get the radius.\n\nThen finish finding the diameter\n\n----
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uh, find the derivative mathematically and then graph it.\n\nfor example to graph derivative of x^2.\n\nfind the derivative equation f(x)=2x\nand then graph y=2x.\n\nooops, didnt realize it was a picture.\n\nfind the zeroes (a high or low point i.e. tangent is a horizontal line)\nyou should be able to track poitive and negative slopes.  not too hard, you'll figure it out.
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I struggled on my spiral at first too but it got TONS better so don't worry. Just...\n\n(1) Stretch every morning and night.\n\n(2) Hold on to the back of a chair or a windowsill and bring your leg up into spiral postion stretch it as high as you can and HOLD it there for about 30 seconds then let it down, rest for a second and do it again. Repeat about 5-6 times, try and work up to 10 times. \n\n(3) When you do it just keep thinking "Stretch Stretch Stretch" and push yourself to get your leg higher and higher and eventualy your muscle will get strong enough to lift your leg into postion everytime.
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If you have the equation of the curve then integrate it on the lower and upper limits of the variable. In some cases you can find devide the identical parts and find out area of a single identical part and multiply by the number of parts made (sine wave).\n\nIf shape is a simple square circle etc you can find area by simple formula.
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The potential energy at the top is mgR = 2 x10 x R = 20 R.\n\nKinetic energy at the bottom of the curve = 0.5 m U ^2 = 0.5 x 2 x U^2. = U^2.\n\nAs there is no friction  K.E = P.E\n\nU^2 = 20 R.\n\nOn the horizontal path, \n\nInitial speed is U.\n\nFinal speed is V = 0.\n\nDistance traveled = 8 m.\n\nAcceleration = Fricttional Force / mass = -8/ 2 = - 4 m/s^2. ( Negative sign since the force acts opposite to the motion.\n\nUsing V^2 - U^2 = 2as, and also noting that V=0,\n\nU^2 = 2 x 4 x 8 = 64.\n\nU = 8m/s.\n\nThe time of travel is found using V= U+ at.\n\nt = -8/ (-4) = 2 second.\n\nUsing U^2 = 20 R.\n\nR = 64/ 20 = 3.2 m.
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x=r(t-sint) and y= r(1-cost) are the parametric equations of the cycloid where r is the radius of the rolling circle and t is the angle through which the radius of the circle has turned from its starting point. The angle t must be measured in radians.\n\nFor simplicity let r =1 so that we have the unit circle.\nwith t=60 = pi/3 we get\nx = pi/3-sin(pi/3) = pi/3 -(sqrt 3)/2\ny = 1-cos (pi/3) = 1- 1/2 = 1/2\n\ndy/dx= sint/(1-cost)\n=[(sqrt 3)/2]/(1/2) = 3^0.5 (at the point where t= 60= pi/3)\n\nFrom this , you can find the slope at any point define by the parameter t.\n\nb) The tangent is horizontal when t= pi and is perpendicular at the cusp which occurs at 2pi and multiples thereof. The maximum value  is 2r (or 2 for the unit circle). The length of the base of the arch is 2pi *r or just 2pi for the unit circle.