Convert the following coordinates...  polar r =15, theta=2.75radians...(x , y)= ?
601534
To convert from polar to cartesian co-ordinates, yousimply use:\n\nx = r cos (theta)\n\nand:\n\ny = r sin (theta)\n\nYou have radians - so make sure your calculator is set to radians.  This gives:\n\nx = 15 × cos (2.75) = -13.864...\ny = 15 × sin (2.75) = 5.7249...
755943
If you are assuming "uniformly accelerated rotation" then the following apply;  (please excuse the crude variables)\n\nw = dθ / dt\na = dw / dt = d^2 θ / dt^2\n\nwhere "w" stands for omega <-- angular velocity\nand "a" stands for alpha <--  angular acceleration.\n\nIf you have had any calculus you can take the first derivitive of your θ expression to get the angular velocity, and then the second derivitive to get the angular acceleration.\n\nsince we have a a simple polynomial the following says\n\nd/dx [x^n] = nx^(n-1)  , however in our case we will be deriving t\n\nTo yeild       w = 8t + 6t^2   , for angular velocity\nand              a = 8 + 12t      , for angular acceleration\n\n\n(a) if you plug in 0 for t then you get θ = 2radians\n(b) its angular velocity is zero when t = 0\n(c) w = 8(4) + 6(4^2) = 128 radians/second\n(d) a = 8 + 12(2) = 32radians/second^2\n(e)  I don't believe its angular acceleratin is constant.  Because if you do some testing with plugging in times for acceleratin and velocity they don't quite add up...   \n\nBut that could be because I am totally wrong.  Anyway that is my stab at it.  Hope it helps
694637
tan(theta)=x/x^3=1/x^2. So theta=arctan(1/x^2)
890202
in trigonomertic form\n\nz = r cos t + ir sin t = 1 + i\n\nso r cos t = 1\n r sin t = 1\n\nsquare and add r^2 = 2\nso r = sqrt(2)\n\nnow cos t = 1/sqrt(2) and sint = 1/sqrt(2)\n\nso t = pi/4 (as cos pi/4 = 1/sqrt(2) and sin is also positive so 1st quadrant)\n\nso t = pi/4 and r = sqrt(2)
247689
Use analytical trig (hint: sin function).  That should give you the answer you are looking for.
952205
You are missing a constraint in your problem.  If your only constraints are \n- theta must be less than or equal to pi/2 \n- you want to maximize theta\nthen theta = pi/2.\n\nSo how far should you be?  0 meters.
697093
Having fun with that homework? Mgh=Mglsin(theta).
698103
(degree/180)*pi=radians
49799
In polar the velocity vector is v = [ dr/dt, r * d(theta)/dt ]\n\nSo r(theta) = .1 * theta and theta(t) = 2t so:\n\nr(t) = r(theta(t)) = .2*t\n\nSo now:\n\ndr/dt = d/dt(.2 * t) = .2\nd(theta)/dt = d/dt(2 * t) = 2\n\nSo then the velocity vector becomes:\n\nv = [ .2 , 2 * .2 * t ]\nv = [ .2 , .4 * t ]\n\nAnd at time t = 1 we have\n\nv( t = 1 ) = [.2 , .4]\n\nSo finding the magnitude of the velocity vector:\n\n|v| = sqrt( .2^2 + .4^2 )\n\n----------------------------\nNow for acceleration:\n\na = [d^2(a)/dt^2 - r * ( d(theta)/dt )^2 )\n, 1/r * d( r^2 * d(theta)/dt )/dt ]\n\nAnyway, we note that\n\ndr^2/dt^2 = 0\nd(theta)/dt = 2\n\na = [ - .2 * t * 2^2, 1/(.2*t), .8 ]\na = [ -.8, .8 ]\n\nSo then |a| = sqrt( (-.8)^2 + (.8)^2 )
939411
Let center point be [0 0] of a rectangular coordinates system.\nAssume the object’s face be initial position on the plane; that is X-axis goes from [0 0] to object’s face, let Y-axis be X-axis rotated through [0 0] by +90 degrees. This way we have defined a rectangular system of coordinates on the plane. \nLet x & y be coordinates of a random point p in our rectangular coordinates system. \nNow lets consider polar coordinates system. Trace vector V from [0 0] to p, then\nx=r*cos(a) & y=r*sin(a), a degrees being angle from X-axis to V, r=sqrt(x^2+y^2). \nAngular position a of p=[x y] should be found as following.  \nAs tan(a)=y/x, (or cotan(a)=x/y), then a=Arctan(y/x) is general solution. Angular position a could be found for any pair of [x y] but [0 0] or when r=0.\nWe do not need general solution in our problem. Our solution must be:\nLet b=arctan(|y/x|) degrees, then a=b for x>0 & y>0, being 0<=a<=90; a=-b for x>0 & y<0, being -90<=a<0; a=180-b for x<0 & y>0, being 90<a<=180; a=-180+b for x<0 & y<0, being –180<a<-90.\nThus the object must be rotated by an angle of abs(a) degrees determined above in the direction of sign(a). \nRotating in opposite direction means to rotate the object by an angle of (360-a) or (–360+a). abs(360-a)=abs(-360+a) is always greater than abs(a).
546529
1. Make sure the angle of interest is in radians.  Radians are the "natural" (non)-units of angles, because they relate the angle measurement to the length of arc the angle cuts off on a unit circle.  So 2*Pi radians are equivalent to 360 degrees.  Basically, if you have that sucker in degrees, multiply it by 2 * Pi / 360.  Unit conversion is done. \n \n2. Make sure the angle is between -Pi and Pi.  Because sine is a periodic function, with period 2*Pi, just subtract or add 2*Pi enough times to get the angle between those two points.\n\nnow for the meat of the problem:\n\nsin(x) = x - x^3/3! + x^5/5! - x^7/7! + x^9/9! - ... + \n+ (-1)^(n-1)*x^(2n-1)/(2n-1)! + ...\n\nExample: sin(15)\n\nStep 1: 15 degrees = 15*(2*Pi/360) radians = Pi/12 radians \n \nStep 2: I need Pi/12 in decimal (let's see, Pi=3.14159 (memorize that much, it's pretty easy... "Pi when calculating sine, 3 point 1 4 1 5 9!")) \nPi/12 approx = .261799 (we only had 6 sig figs, so we can't add any more than that) \n \nNow to the series: \n         first term:  x=.261799 \n         1st estimate: .261799 \n         second term: -x^3/3! = -.0029905  \n         2nd estimate: .258809 \n         third term: x^5/5! = .0000102     \n         3rd estimate: .258819 \n         fourth term: -x^7/7! = -.167*10^-7  \n         stop here - past precision\n\nThe answer is positive \nsin(15) = 0.258819\nsin(18) = 0.309017
477554
If you assume that at the bottom left corner the coordinates are i = 0, j= 0. and at the other end (18,0).\nT has to be:\nT = 18/7.5\n\nFor Y coordinate, it must be zero:\n12t - 5t^2 = 0\nt = 0, or 12/5\n\nso 12/5 = 18/7.5 which is at the bottom right (18,0) and proves the question.
1126935
I'm gonna use t for theta\n\nr=3 cos(t)\n\nby definition x=rcos(t)\nre-arrange the definition\nsubstitute to get: \nr=3(x/r)\nre arrange this\nr^2 = 3x\nby definition:\nr^2 = x^2 + y^2\nso,\nx^2 + y^2 = 3x\nAnd then you can re-arrange it from there
784184
θ =0.3t² rad\nω = dθ/dt = 0.6t rad/s\nα = d²θ/dt² = 0.6 rad/s²\n\nWhen t = 5\na. ω = 0.6t\n= 0.6*5 = 3 rad/s\n\nb. v = rω\n= 10*3 = 30m/s\n\nc. α = 0.6 rad/s²\n\nd a = v²/r (= rω² = 10*3² = 90 m/s²)\n= 30²/10\n= 90 m/s²
435632
and the formula is:\n\ndB = k I dl sin theta / r^2\n\nwhere k = mu0/4pi
1081956
Multiply both sides by r thus:\nr^2=2*r[sin(theta)]\nso    X^2+y^2=2y\nNow rearrange.
691885
To map [32, 212] onto [0, 100], use Theta(x) = [(100-0)/(212-32)][x-32], or Theta(x)=5/9(x-32).  As you may know, this is the formual to convert Fahrenheit to Celsius.