Differences of cartesian and polar coordinates.?
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Please use the search function. This question was asked 5 days ago:\n\nA proper polar coordinate system (one that assigns each point in the plane a unique coordinate) will always leave out a ray of points. For example, the standard polar system that identifies the open set (0, oo) X (0, 2Pi) with R^2 - {0} by the function f(r, t) = (rcos(t), rsin(t)) ( = (x, y) ) does not correspond to any points on the positive x-axis and necessarily leaves out the origin (there is no way to include the origin uniquely). This is usually glossed over in a basic calculus or physics course as one actually uses an atlas of polar coordinate charts (this atlas never includes the origin, but at least covers the plane minus the origin); you use whichever chart is convenient for the problem at hand. Some definitions will disregard the need for a homeomorphism and include the positive x-axis by considering a domain that is not open, (0, oo)X[0, 2Pi)
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1. For two given points P(p1, p2, p3) and Q(q1, q2, q3), the vector v between them is v = (x, y, z) = (q1 - p1, q2 - p2, q3 - p3).\n\nThe magnitude or length of v, denoted by ||v|| is\n||v|| = sqrt[ x^2 + y^2 + z^2 ]\n\nJust do that for every line that you have.\n\n2. This one's a little harder. Let's define the first line as the vector u, and your second line that you're using as the vector v.\n\nYou first need to write each one as a unit vector, which means dividing all of the values through by a value so that the magnitude will be 1. For each of these vectors, this means dividing by the length of the vector.\n\nAs an example, if I wanted to turn w = (1, 1, 1) into a unit vector, I would notice that the magnitude is ||w|| = sqrt[3] and divide all the values through by that to get (1 / sqrt[3], 1 / sqrt[3], 1 / sqrt[3]).\n\nSo, let us say that the new unit vectors are u / ||u|| = (a, b, c) and v / ||v|| = (x, y, z)\n\nWhat these values give you now are called the direction cosines of each vector. If A is the angle between u and the x axis, then a = cos A, and similarly for b with the y-axis and c with the z-axis.\n\nSo, to find the angles between each vector and the x, y, and z axes (also called the direction angles), you take the inverse cosine (cos -1) of the components (values) of the vector.\n\nLet us say that after taking the inverse cosine of u / ||u|| = (a, b, c) = (cos A, cos B, cos C), you get (A, B, C), and the inverse cosine of v / ||v|| gives you (X, Y, Z).\n\nThe angles between the two vectors in the x, y, and z directions are (X - A, Y - B, Z - C).\n\nSo the main steps are finding the magnitude of each vector, normalizing each vector (making them into unit vectors), taking the inverse cosine of each vector, and subtracting the two vectors.\n\nHope that helps.
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y = r sin theta, x = r cos theta (by drawing a triangle or otherwise.)\nSubstitute them in:\n2r cos theta + 3r sin theta = 6.\nOr r(2 cos theta + 3 sin theta) = 6.
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the 1st one is simple\n(x+y) dy/dx + y =0\nor xdy/dx + ydy/dx + y =0\nor (xdy/dx + y)  + ydy/dx = 0\nxdy/dx + y = d/dx(xy)\n\nand d(y^2/2)/dx \n\nso d/dx(xy+y^2/2) = 0\n\nor xy+y^2/2 = c\nor 2xy+y^2 = 2c or C\n\nproved\nfor the second one\n\ndy/dx -(1/x)y = xy^2\n\nlet y/x = t\ny = xt\ndy/dx -y/x = t+ x dt/dx -t = x dt/dx\nso\nx dt/dx = - x (xt)^2\nor dt/dx = x^2t^2\ndt/t^2 = x^2 dx\nyou should be able to proceed
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I've been studying this stuff recently, and I'm pretty sure this is how it goes:\n\nK is now (-2,1) and D is now (-3,-2).\n\nFor future reference on determining rotations, simply turn the graph counter-clockwise one time for 90 degrees, upside down for 180 degrees, and for 270 degrees, just turn the graph clockwise once to see what the new cooridnates are.
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a=c*sin(30)=.5c since it is apposite the 30 degree angle\nb=sqrt(c^2-a^2)= .866c\nor b=c*cos(30)=.866c\n\na=b=c*sin(45)=.707c=
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See the picture:\n\nhttp://mudandmuck.com/str2/sin&cos.jpg\n\nThe two triangles in the picture are equivalent.  (Same angles, and one side of each we know to be the same (hypotenuse=1), so all the sides have the same absolute length, only their signs are different.)
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in a graph, the x coordinate defines distance right or left.\n\nthe y coordinate defines distance up or down.
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you've got your four quadrants\n\n1st: x-y axis plus plus >>>>>> plus\n2nd; negative plus>>>>negative\n3rd; negative negative>>>plus\n4th; plus negative>>>negative\n\nfor trig;  all,sine,tan,cos\n\ni hope that this helps
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Rectangular form means expressing  the x and y coordinates of a point on a grid.  This is already in rectangular form.  Polar form would give the length (magnitude) and angle of a line from the origin to the same point on the grid.   The magnitude or length of the line will be the square root of  the x component squared + the y component squared.  \n= square root{2cos(theta) + 1)^2 + {2sin(theta)-3)^2}\nThe angle of the line will be arcTan (y/x)=arcTan[ (2cos(theta)+1)/2sin(theta) -3)]
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a) Assume the following:\n\nAngles a= 10 deg, b= 80 Deg, c=90 deg \nHypotenuse length = 2.5 m\nheight = 2.5 * cos 80 Deg= .434 m\nlength = 2.5 * sin 80 Deg= 2.462 m\nacceleration = "g" * Sin 10 Degree
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difference of squares rule:\n\nx^2 - y^2 \n\n(x-y)(x+y)
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Polar\nits use to converse from dimensional to polar.\nexp\n(3,4) --> (5;53)\nits mean 5 for radius and 53degree
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Best to plot this on a piece of graph paper; "+i" is  to the right' , "+j" is upwards, "-i" is to the left, "-j" is downwards, all with respect to the origin.\n\n1. Draw the vectors on the graph paper. "u" is in the second quadrant (i is neg and j is pos). "v" is in the fourth quadrant, with (i is pos, j neg) . There are two answers to this question, since you could go from u to v or from u to v in the opposite direction. I always take the angle from zero to "u" and add it to the angle from 0 to "v".  For "u", the tangent of the angle to the vertical axis is 1/4, then angle = 14%. So the angle from zero degrees to the vert axis = 90, and "u"s angle from zero degrees is 90 + 14 = 104 degrees. For "v", the tangent of the angle to the horizontal axis is 4/3 = 1.333.. and the angle = 53 degrees. Therefore the angle from "u" to "v" is 104 + 53 = 157 degrees. (the other answer is 360 degrees - 157 degrees = 203 degrees if you rotate in the other direction).\n\n2. Again use graph paper. 240 degrees is 30 degrees less that 270 degrees (the vertical axis, down) draw a line 10 graph units long. It will be in the third quadrant where i and j will both be negative. At this point you can use a polar to rectangular converter, or do it by hand. By hand: the i side of the triangle formed with the down axis can be found by saying sin(angle) = i/10, and cos(angle) = j/10. The angle is 30 degrees (from vert downwards). so sin(30) = 0.5, and, cos(30) = 0.866; thus i= 0.5 and j = 0.866 (ignoring pos/neg). Since the vector is in the third quadrant, both i and j are negative.\nAnswer = -0.5i -0.866j\n\nHope this helps!
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sin=opposite/hypotenuse\ncos=adjacent/hypotenuse\ntan=opposite/adjacent
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I would change cos 2 theta to 2 cos squared theta -1.\n\nYou then have a quadratic in cos theta.
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(1) Subtitute the damn values for what you have.\n\n(2) Sin theta=opp/hyps,\n    Tan   "  =opp/adj,\n    Cos   "  =adj/hyps.\n\nNow usually a blind man follows his stick now you might know where you are at the moment - get it???