What is the amplitude, horizontal and vertical shift, and period of y=2cos(3x-(pi/2))+4?
956961
For an equation of the form y = a cos (bx + c) + d:\n\na is the amplitude\n2 pi / b is the period\nc is the phase shift\nd is the vertical shift.\n\nTherefore, the amplitude is 2, the period is 2 pi / 3, and the vertical shift is +4.
835859
First figure out whether to use sin or cos, I decided on cos because we look for the max and min.\n\nNext find the amplitude which is 6. a = amplitude/2 so a*cos(x).\n\nThird find the midpoint of the curve which is (max+min)/2 = m. a*cos(x) + m.\n\nThe fourth step is to get the max or min at the x cord. that you want. With cos you subtract max from x. a*cos(x-max) + m.\n\nFinally you find your wave length, which is simply (max-min)/pi = wl for cos. a*cos((x-max)/wl) + m\n\nso you go from (-2,1) to (10,7)\n\na = 3\nm = 4\nmax = 10\nwl = 12/pi to give us\n\n3*cos ((x-10)/(12/pi)) + 4
879813
you need to identify that there're two parts of the force, one acting horizontally, the other acting vertically.\nSo:\n1) the vertical part is:\n70sin30\n2) tje horizontal part is:\n70cos30
804155
a) T = 0.5 sec\n\nb) freq = 1/T = 2 Hz\n\nc) w = 2pi*freq = 4 pi\n\nd) w^2 = k/m\nk = m*w^2 = 0.5*16 pi^2 = 8*pi^2\n\ne)  V_max = A*2*pi*f = 0.35*2*pi*2 = 1.4 pi m/sec\n\nf) A_max = A*(2*pi*f)^2 = 0.35*(2*pi*2)^2 = 5.6*pi^2 m/sec^2
792458
I'm assuming the amplitude is the horizontal distance from vertical.  In that case, the amplitude is the base and the pendulum length is the hypotenuse of a right triangle.  Figure that vertical side, subtract it from the length of the pendulum and that's the vertical distance it falls through.  Energy is the weight multiplied by this distance.
986259
(a) Displacement is given by the equation:\n\nx = 6 cos[3π(t+(1/9))], so at t = 2 sec\n\nx = 6 cos[3π(2+(1/9))] = 3 m\n\n(b) Velocity is v = dx/dt = -18πsin[3π(t+(1/9))], so at t = 2 sec\n\nv = -18πsin[3π(2+(1/9))] = -48.97 m/s (in the negative x direction)\n\n(c) Acceleration is a = dv/dt = -54π^2cos[3π(t+(1/9))]\n\nso at t = 2 sec, a =  -266.48 m/s^2\n\n(d) Phase of the motion is θ = 3π(t+(1/9)), for t = 2\n\nθ = 19.89 radians\n\n(e) The frequency = ω/2π = 3π/2π = 1.5 revolutions per sec\n\n(f) Period = 2π/ω = 2/3 sec
775460
period (s) = 1/frequency (hz) \n\nso:\n\n(a) = 1/10 s = 0.1 s\n(b) = 1/0.2 s = 5 s\n(c) = 1/60 s = 0.01666... s
732315
This one's a little tricky.\n\nAllow me to use "t" instead of "delta"\n\n\nint ((cos(2t))^3)dt\n\nSince (cos(2t))^2 = 1 - (sin(2t))^2\n\nSubstitute.  this becomes:\n\nint ((1 - (sin(2t))^2) * cos(2t))dt\n\nSubstitute u = sin(2t).  Then du = cos(2t)dt.\nSo dt = du / cos(2t)\nThe key here is when you plug this in, the cos(2t) values cancel.\nSo we get:\n\nint (1 - u^2)du\n= u - (1/3)u^3 + C\n= sin(2t) - ((sin(2t))^3)/3 + C
748368
first get the time from the horizontal motion  t = .51 / 1.2 s (uniform motion, no acceleration here)\n\nthen use h = 1/2gt^2 for the vertical motion (with initial velocity zero)
107615
t=theta\n^=to the power of...\n\ncos (2t) + 2 sin ^2 (t) = cos (2t) + 2 - 2 cos ^2 (t)
132715
Formula for Sine, Cosine and Tangent:\nTangent= opp./adj.\nSine= opp./hyp.\nCosine= adj./hyp
624769
You can use sin and cosine.
277745
Answering...\n\nBasic Equation:\n\nSine:.. y = a sin bx\nCosine: y = a cos bx\nwhere:\namplitude = absolute value of a (always positive)\nPeriod = 360 degrees/absolute value of b or 2pi/abs b\n\nEX: y = sin 4x\namplitude = abs a = 1\nPeriod = 360 deg./abs. b = 360/4 = 90 deg.\n\nEXPLANATION:\n\nSine Function\n1. The graph looks like a regular sea wave.\n2. For positive function, along the y-axis, the graph starts at the origin and then it goes up like a wave until it reaches the peak (called crest).\n3. Then it goes down until it reaches its lowest part (called trough).\n4. Then the cycle is repeated.\n5. For a negative sine function, from the origin, the graph goes down to its trough and then up again.\n\nCosine Function:\n1. The graph also looks like a regular sea wave.\n2. For a positive function, the graph (along the y-axis) starts at the its crest.\n3. Then it goes down until it reaches its trough.\n4. Then it goes up again until it reaches its crest again.\n5. The cycle is repeated.\n6. If the function is negative, along the y-axis, the graph starts at its trough first. Then it goes up.
971390
{3XY^2M/1}-{2M/[X^-1Y^-1]^2\n+{[4X^2Y^2]/M^-1}-\n(3X^(3Y)^(3M)^-1/(XYM)^-2
85418
dx/dt+ax=b :P\n\n\nmathematica had some online find-the-derivitive program, last I remember
878725
antiderivative of:\n20-12cos2pi(x-0.1)\nis:\n\n20x - 12 sin( 2pi(x-0.1)) / 2pi +C\n=20x - 6 sin( 2pi(x-0.1)) / pi +C\nto double check just take the derivative of the function and see that you get:  20-12cos2pi(x-0.1) '
820986
1.\nPhases:\nf1 = (t/T1)*2*pi, where T1 = 1 sec\nf2 = (t/T2)*2*pi, T2 = 1 sec\nIn phase: f1 = f2 + 2*pi*n, where n is integer (0, +-1, +-2, ...)\n(t/T1)*2*pi = (t/T2)*2*pi + 2*pi*n\nt(1/T1 + 1/T2) = n\nT2 = 0.98 = 98/100 = 49/50, 1/T2 = 50/49\nt(49/49 - 50/49) = n\n-t/49 = n\nt = 49 seconds, n = -1\nNumber of oscillations if second, shorter pendulum (shorter pendulum has shorter period): N = t/T2 = 49/(49/50) = 50\n\n2.\nCoordinate: y(t) = 3*sin(20*pi*t)\nSpeed: v(t) = y'(t) = 20*pi*3*cos(20*pi*t)\nMean position: y(t) = 0. Take t0 = 0.\nMaximum displacement: y(t1) = amplitude = 3, speed = v(t1) = 0\nv(t1) = 0\ncos(20*pi*t1) = 0\n20*pi*t1 = pi/2\nt1 = 1/(2*20) = 1/40