the wave equation u_tt = c^2 (u_xx + u_yy) on triangular drum D. The membrane vibration is described by a function u(t,x,y) of three variables and it is assumed that u(t,x,y)=0 at the boundary of D for all t. Furthermore, u(0,x,y) and u_t(0,x,y) are known. \n       +        I remember having seen\n       | &#92;      such problems in Math21b \n       |  &#92;     and that it had to to do \n       | D &#92;    with Fourier theory but\n       +----+   where do I start?
122982
By sin(x)^5, I assume you mean (sinx)^5.  \nYou mean which of the following?\nFind sinx and then raise it to the fifth power or\nraise x to the fifth power and then take the sin of that number.\nI will work this out the first way as I do not know offhand how to do the second one.\nint = integral\nint[(sinx)^5dx]\nRewrite (sinx)^5 as sinx (sinx)^4\n(sinx)^4 = ((sinx)^2)^2\n(sinx)^2 = 1 - (cosx)^2\nThus, \n(sinx)^4 = ((sinx)^2)^2\n= (1-(cosx)^2)^2\nNow the integral is\nint[sinx(1-(cosx)^2)dx]\nUse substitution.\nLet u = cosx, then du = -sinxdx\nNow the integral is\nint[-(1-u^2)^2du]  \nExpand (1-u^2)^2 and multiply through by the -1\n-(1-u^2)^2 = -(1 - 2u^2 + u^4) = -1 + 2u^2 - u^4 \nThe integral now equals\nint[(-1 + 2u^2 - u^4 )du]\n= -u +(2/3)u^3 - (1/5)u^5 + C, where C is a constant\nReplace u with cosx to get the final answer.\n-(cosx) +(2/3)(cosx)^3 - (1/5)(cosx)^5 + C\n\nPlease note that the following is NOT true.\nIntegral [ sin(x)^5 ] = -Cos(x)^5 + C\nIf you take the derivative of -Cos(x)^5 (you have to use the chain rule) you will see why.
366056
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
231013
Look up this URL. It is very interesting pertaining to  his concepts of vibration...\n\nhttp://www.phy.davidson.edu/StuHome/jimn/Java/modes.html
219354
I admire mathematician's concise answers, but in this case some additional explanation may be due:\n\nWhen a wave is partially transmitted and partially reflected the two outgoing waves are shifted in phase from the incident wave. If energy or probability is to be conserved, the difference between the two phase shifts is pi/2 (90°). In this diagram each of the two possible outgoing  beams consist of components from two paths (R=Reflection,T=Transmission):\n\nOutgoing Horizontal: TRR + RRT\nOutgoing Vertical: TRT + RRR\n\nIn the first case the relative phase shift between the two components is 0 and in the second case it's pi (180°) ==> the two components are negatives of each other and interfere destructively.
115057
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.
89596
http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html\nhttp://en.wikibooks.org/wiki/Engineering_Acoustics/Bessel_Functions_and_the_Kettledrum
278559
FORMULA: speed = frequency * wavelenght (sorry cant get symbols to work on yahoo)\n\n(1) 0.2= 1/0.5 * w   ...(as f=1/T)\n=> 0.2=2w\n=> w=0.1\n\n(2) s= 25*(10^8) * 0.12    ...(12cm= 0.12m)\n=> s= 3*10^8\n=> s= 300000000 m/s\n\n(3) 2= f * 0.5\n=> f= 2/0.5\n=> f= 8\n\n8 waves strike dock each second.\n\npretty easy questions.
235735
sin x = (1/2i)(e^ix-e^-ix)\ncos x = (1/2)(e^ix+e^-ix)\n\nsolutions of a simple harmonic oscilator are\ny = A sin x + B cos x or\ny = C e^ix + D e^-ix
552849
The basic idea of a fourier series is to express a function - in this case the sawtooth wave - as a sum of sine and cosine functions of different frequencies (i.e. n omega0).  The result gotten for a sawtooth wave is the thing you described (the sideways 8 means "infinity").  That is, it's an infinite sum of sin curves, each one weighted by 1/n.  See a picture here: http://mathworld.wolfram.com/FourierSeriesSawtoothWave.html
432393
yes, it's a sine wave, and velocity is integral of that, ie a cosine\n(this is considering only fist order effects)
432189
Can someone help me with my precalculus problem? Can you please read the following problem and figure out the?\n\nCan someone help me with my precalculus problem? Can you please read the following problem and figure out the equation for me along with how you figure this out?\n\n\nA tsunami (commonly called a “tidal wave” because of its effect is like a rapid change in tide) is a fast moving ocean wave caused by an underwater earthquake. The water first goes down from its normal level then rises at an equal distance about its normal level, and finally returns to its normal level, the period is about 15 minutes.\n\nSuppose that a tsunami with an amplitude of 10 meters approaches pier Honolulu, where the normal depth of water is 9 meters.
613014
(dx/dy) = (xy^3) / (2y^4 + x^4)\n\nLet x = y*u.\nThen (dx/dy) = u + y*(du/dy).\n\nSubstituting these in the given equation we have,\n\nu + y*(du/dy) = (u*y^4) / {2y^4 + (u*y)^4} = u/(2 + u^4)\n\nor y*(du/dy) = u/(2 + u^4) - u = -u*(1 + u^4)/(2 + u^4)\n\nor ∫ [-(2 + u^4)/{u*(1 + u^4)}] du = ∫ [1/y] dy _________(1)\n\n\n\nNow, let us first express [-(2 + u^4)/{u*(1 + u^4)}] in easily integrable form.\n\n-(2 + u^4)/{u*(1 + u^4)} \n\n= -{2*(1 + u^4) - u^4}/{u*(1 + u^4)}\n\n= -{2*(1 + u^4)/{u*(1 + u^4)} + u^4}/{u*(1 + u^4)}\n\n= -2/u + u^3/(1 + u^4)\n\nTherefore, ∫ [-(2 + u^4)/{u*(1 + u^4)}] du\n\n= ∫ [ -2/u + u^3/(1 + u^4)] du\n\n= -2∫ [1/u] du + ∫ [ u^3/(1 + u^4)] du \n\n= -2*ln(u) + (1/4)*ln(1 + u^4) + k\n\nwhere k = arbitrary constant of integration\n\nHence, from (1) we have \n\n-2*ln(u) + (1/4)*ln(1 + u^4) + k = ln(y)\n\nor  (1/4)*ln(1 + u^4) + k = ln(y) + 2*ln(u) = ln(y) + ln(u^2) = ln(y*u^2)\n\nor k = ln(y*u^2) - (1/4)*ln(1 + u^4) \n\n= ln(y*u^2) - ln[(1 + u^4)^(1/4)]\n\n= ln[(y*u^2)/{(1 + u^4)^(1/4)}]\n\n\nNow, substituting u = x/y we have\n\nk = ln[(x^2/y)/{(1 + (x/y)^4)^(1/4)}]\n\nor k = ln[(x^2)/{(x^4 + y^4)^(1/4)}]\n\nLet k = ln(c).\n\nThen c = (x^2)/{(x^4 + y^4)^(1/4)}\n\nor x^2 = c*(x^4 + y^4)^(1/4)
266747
First, since 1/x is not defined for x = 0, before you can show  anything about your functions, you have to define values for g(0) and f(0). I'm going to assume that either the original statement of the problem had that or that you are supposed to find the value that makes the problem solvable.\n\nTo be true you have to define the problems with f(x) = 0 and g(x) = 0\n\n\nFor the continutiy of f(x) given epsilon > 0 find delta > 0 such that |f(x) - f(0)| < epsilon for all |x-0| < delta.\n\n|f(x) - f(0)| = |xsin(1/x) - 0| < |x| = |x - 0|\n\nGiven epsilon, pick delta = epsilon.\nIf |x-0| < delta, then |f(x) - f(0)| < epsilon.\n\n\nAs for the differentiability of g(x),\n\nf(0+h) - f(0)) = (0+h)^2sin(1/(0+h)) - 0\n\nlim(x->0)(f(0+h) - f(0))/h = \nlim(x->0)((0+h)^2sin(1/(0+h)) - 0)/h = \nlim(x->0) h^2(sin(1/h)/h = lim hsin(1/h) = 0\n\nSo the limit exists and is 0.
152466
if f(t) = [delta](0) (unit impulse at t = 0), its Laplace transform is F(s) = 1, therefore, the inverse transform of F(s) = 1 is f(t) = [delta](0).\n\n And given that {L(a*f(t)) = a*L(f(t))}\n\nL(2*[delta](0)) = 2 * (1).\n\nThe answer is:\n  2*[delta](0)
425520
I don't quite understand your notation, but here is how I would do this:\nThe general equation is y(x,t) = A sin 2 pi (t/T - x/L), where\nA = amplitude = 0.12 m\nL = wavelength = 0.3 m\nT = period\nThe latter is not given, but you can get it from the wavespeed v, via the formula v=L/T. Hence T = L/V = 0.3/6.4 = 0.0469 s.\nPlug A, L and T in the general equation and compare the result to the expression given in the problem.\nLooking at your expression, I would guess \na = A = 0.12 m \nb = 1/T = 21.3 s^-1\nc = 1/L = 3.33  m^-1
340097
OK, you already know (a), which is the total energy. At the peak displacement (which is the amplitude), all the energy is potential energy PE, in the form of spring deformation. PE = .5*k*x^2 = .5*35*.04^2 = .028 j. At other times, the energy is divided between PE and kinetic energy KE, and PE+KE=.028. So for (b), KE = .028-.5*35*.01^2 = .02625 j. Then the velocity = sqrt(2*KE/m) = 1.025 m/s. For (c), KE = .028-.5*35*.03^2, and for (d) PE = 5*35*.03^2.
84678
yeah, the guy was right, it's position and momentum, not energy and momentum. \n\nSee, if everything is a wave, it has a frequency spectrum associated with it (look up Fourier analysis). If something has one specific frequency only, then you can measure the frequency perfectly; the frequency gets you the wavelength, the wavelength gets you the momentum. But by definition, a wave at a single pure frequency extends to infinity in both directions, so you have no information about position. \n\nOn the other hand, if the wave looks like a single peak, you can calculate the position very nicely, but the frequency makeup (Fourier analysis) of a single peak contains an infinite number of frequencies, therefore an infinite number of momenta. \n\nFrom frequency /Fourier analysis, it's sort of obvious that the tighter you get the frequency/momentum, the wider the position spreads, and vice versa. No way you can specify a single frequency at a single point. (you can kind of see this intuitively; imagine a little chunk of a waveform, like just the tip of one peak; you can see the position precisely, but it's hard to estimate the frequency).