Recent content by Haths

Given; \frac{d^{2}u}{dx^{2}} = \frac{1}{c^{2}} \frac{d^{2}u}{dt^{2}} and; u(x,0) = \phi (x) \frac{d^{2}u(x,0)}{dt^{2}} = \theta(x) Show that the Fourier Transform of the u(x,t) w.r.t. to x is; \tilde{u}(k,t) = \tilde{\phi} (k) cos(ckt) +...

Now that would work if I was only interested in the area, but I'm interested in the curl of the vector field. The path around the triangle A-B-C etc. is in the same direction or opposite direction to the 'rotation' of the field. The scalar quanta from before points in the positive r direction...

Basically. How would I even begin to go about doing the surface integral of a triangle bound by the co-ordinates; (1,0,0) (0,1,0) (0,0,1) This is part of a larger question to compute; \int \int \vec{F} \cdot d \vec{r} Around the surface using stokes theorum, the scalar multiplier comes out...

Cheers that statement has cleared up my doubts. Also yes a simple dr integral would achieve the same result which would save a lot of the complicated answer. Haths

A particle is described by the normalised wavefunction; $ \psi (x,y,z) = Ae^{- \alpha ( x^{2} + y^{2} + z^{2} ) }$ Find the probability that a particle is in a dr shell of space. For what value of r is the probability of finding this particle greatest, and is this the same r value...

Thanks on the first section, yeah the jacobian is wrong :p. __________________________________________________________________ I've not seen the RHS of that equation before, but the first one appears to look just like the normal equation for calculating the curl of a field. So I'm not...

Two problems one that I have some idea about solving, the other I have no idea at all about where to start. 1. Find the surface integral of E . dS where E is a vector field given; E = yi - xj + 1/3 z3 and S is the surface x2 + z2 < r2 and 0 < y < b Well Gauss' theorum would be the place...

That is certianly one way to go about doing the problem. However... Because 75m isn't very big in magnitude, and it's close to the Earths surface, then the force on a 'cube of water' due to gravity. Can be considered; F=mg Because our relationship for g is linear. Then we can suitably...

Hmm... I = mr^{2} I = (3M)0^{2} + 2M(\frac{L}{2})^{2}) + ML^{2} Therefore; I = \frac{1}{2}ML^{2} + ML^{2} I = \frac{3}{2} ML^{2} I get the same as you. The definition of the moment of inertia is; SUM( miri2 ) So I'm preaty certian that we are following the correct method. So...

Is that a yes or a no to my question in the last post? ___________________________________________________________ Why do I want to solve; x''+3x=0 ? It's not the oridginal example, there is a -4 missing from it if it meant to be the oridginal Homogenious DE we are trying to solve, and I...

Well I can quickly realize that; 1/2(n)(n+1) = SUM( a , ... , an] ) Therefore; 1000 = 1/2 (n)(n+1) = SUM( a , ... , an ) Expanding out; 1/2 n2 + 1/2 n = 1000 Therefore; 2000 = n2 + n But that's not going to help you much trying to find the products of two consequtive...

See this is why for the first time I've been disappointed with the help I've received here, because simply saying 'solve' isn't a help when I'm asking what do you mean by 'solve'. If you ask me to solve a quadratic equation for its roots I know what your on about. If you ask me to find the...

Fair point Mark, you can assume it's a dx on the differential, but that's the only typo. I'm sorry, all of this is gobbledigook to me. It might be explaining it very well in your mind. But I don't know what "variation of parameters" is and a quick google didn't explain. Dick just says the same...

No no. I have no idea where you've thought the equation had a -4x term in it. It shouldn't matter what I write down as working, because I'm asking you what I should be doing at each step. If I've written down somthing wrong, just say; "No that's wrong, this is what you should do next." I was...

Sorry that doesn't help me at all. $ x_{p} = De^{itn}$ from what you've said just sounds like the; $ x_{p} = Ce^{itn}$ I already have. C is just some random constant, which is why I think what you've just said is a restatement of what I've identified already. [hr] However do...