Recent content by loveequation

Good luck. Here is another think that will help. Suppose you find set of roots for one (x, y). Then if you go to a neighboring (x, y) pair then obviously use the previous roots as a starting guess. This will go faster this way.

You want to explore a two-dimensional parameter space. That's easy. But you're going to have to write your own solver with zbrent at its core. The writing of a solver should take about two hours of human work. Here is a hybrid way to go, i.e., partially symbolic and partially numerical---I...

Hi: Give me an idea. What is the value of "n". I assume you have 6*n p's and q's to vary. What are the ranges of the p's and q's and what is the range of "t". For certain cases you can be certain there is no root. For example for t > 0 and all the p's and q's strictly positive there is...

Hi: You might want to post your question at the top level so it gets noticed. My guess is no, not for arbitrary coefficients p and q. The only special case which is analytically solvable is where the numerator of each term in the sum has the same root. If you decide you're going to bite the...

I think you have to write for each ith constituent of the gas \frac{\partial p_i}{\partial y} = -\rho_i(y)g and use the equation of state p_i = \rho_i R T/m_i where m_i is the molecular mass. Note that the molar density N_i = \rho_i/m_i. I assume you can do the rest. If not let me...

Let me think about it. In the meanwhile you might want to re-post it to see if others can be of more immediate assistance.

Consider a permanent magnet (fixed in the lab frame) and a charged particle moving in its vicinity. The force on the particle is \vec F = v \times \vec B + \vec E q The magnet does no work on the particle (\vec F \cdot \vec v = 0), the particle is deflected and its kinetic energy remains...

Correction. Instead of N_{total} for the total molar mass I meant to write M_{total} and I was unable to edit it for some reason.

Try the following. First write the total density as a function of height. You get this solving \partial p_{total} / \partial y = \rho_{total}(y) g [/itex] (using the equation of state p_{total}=\rho_{total} R T / M_{total}) where M_{total} is the total molar mass. Then write the total pressure...

Average speed = total distance/total time total time = time of first leg of journey + time of second leg of journey time of first leg of journey = distance of first leg/speed of first leg time of second leg = distance of second leg/speed of second leg Got it?

I maintain that the y components add. Think of the magnetic field lines of the Earth at the equator.

OK, don't introduce the separation of variables form u(x, t) = T(t) U(x). At this stage we are merely trying to derive the PDE, not solve it. The force on the left face of the differential element is F(x) = EA(x)\frac{\partial u}{\partial x} pointing to the left. The force on the...

Yes. The way I am thinking about it, the only role of friction is to enforce a non-slip pivot.

If the denominators of the different A_j(t) have common factors then that helps. Otherwise just use ZBRENT (a root finder). I assume you want all the solutions. If so, you will have to find all the intervals containing roots yourself before calling ZBRENT.

You're right. Due to repulsion, the charges will end up on the far ends of the spheres. Hence the distance between the charges will be greater than the center to center distance between the spheres.