Recent content by compliant

But that only makes use of the regular Fourier Transform, and not the Fourier Cosine Transform.

[PLAIN]http://img815.imageshack.us/img815/9894/fourier5313.png there was a minus sign. oops. In any case, relevant equations: F_{c}(f(x,t)) = \sqrt{\frac{2}{\pi}} \int^{\infty}_{0} cos (\lambda x) f(x,t) dx F_{c}(f''(x,t)) = -\sqrt{\frac{2}{\pi}} f'(0,t) - \lambda^{2} F_{c}(f(x,t)) which...

Homework Statement Find the solution u, via the Fourier sine/cosine transform, given: u_{tt}-c^{2}u_{xx}=0 IC: u(x,0) = u_{t}(x,0)=0 BC: u(x,t) bounded as x\rightarrow \infty , u_{x}(0,t) = g(t) 2. The attempt at a solution Taking the Fourier transform of the PDE, IC and BC...

Well, then I would have m_{2}\ddot{x_{2}}-m_{1}\ddot{x_{1}} = -kq ...which, from what I see, doesn't do a whole lot because I still can't factor the left side to do anything.

Homework Statement Two masses, m1 and m2, are connected to each other by a spring with a spring constant k. The system moves freely on a horizontal frictionless plane. Find the natural frequency of oscillation.Homework Equations F = -kx F = maThe Attempt at a Solution Let m1 be the mass on the...

bump.

Homework Statement u_t = -{{u_{x}}_{x}} u(x,0) = e^{-x^2} Homework Equations The Attempt at a Solution The initial state is a bell curve centred at x=0. The second partial derivative of u at t=0 is {4x^2}{e^{-x^2}}, which is a Gaussian function, which means nothing to me other than its...

sorry to do this, but bump.

tiny-tim, thanks for those. desperately needed. hallsofivy, I did draw the diagram, and found that it was rather inconveniently symmetrical, which was why I got stumped. going by your suggestion, from θ = 0 to θ = π/2, I would be integrating along the right side of the curve, where the...

Homework Statement Find \int{\int_{D}x dA} where D is the region in Q1 between the circles x2+y2=4 and x2+y2=2x using only polar coordinates. The Attempt at a Solution Well, the two circles give me r=2 and r=2 cos \theta, and the integrand is going to be r2cos \theta, but I have no...

*backtracks* ooooooooooookay. so. when x=0, f_x = -2, but f_x = 0, so x=0 is not a solution. which brings me back to the points (-1,0) and (1,2). looking at my calculations for (-1,0), turns out that it's horrendously wrong.At (-1,0), f_xx = -2(6-2) = -8, D = (-8)(-2) - (0-3+2)^2 = 16 - 1 > 0...

well, from f_y, I can deduce that x=0 or yx^2 - x - 1 = 0. If x=0, from f_x, y= -2. If yx^2 - x - 1 = 0, y = (x+1)/(x^2) Subbing into f_x, -2(x^2 - 1)(2x) - 2((x+1)-x-1)(2x(x+1)/(x^2) - 1) = 0 -2(x^2 - 1)(2x) = 0 x(x^2 - 1) = 0 x = 0 (as above), x = 1 or x = -1 If x = 1, y = 2 If x = -1...

It's on the right track, but to finish it off, you should also apply the second derivative test for partial deriviatives.

Homework Statement Show that f(x,y) = -(x^2 - 1)^2 - (yx^2-x-1)^2 has only two critical points, and both are maxima. The Attempt at a Solution Set partial derivatives (wrt x and y) to zero to find critical pts. f_x = -2(x^2 - 1)(2x) - 2(yx^2 - x - 1)(2xy - 1) = 0 f_y = -2(yx^2 - x -...

ahh. of course. r^2 = x^2 + y^2, and sin theta = y/r, where r is the square root of r^2. I indeed was making it too complicated for myself. thanks everyone.