Recent content by techninja

Many thanks to all of you. The answer wasn't necessary supposed to be in arctan, but I found the solution that is more similar to coomast's through some type of teacher-prodded guess-and-check (magic and really lucky). Of course, reversing the differentiation was very difficult, so I...

When I differentiate the solution I get, \frac{1}{2}arc\sec \left( {x^2 - 1} \right) , it only applies well for x>0, where defined.. Am I doing something funny?

My head just exploded. But I end up with an arcsec, which becomes decidedly unhappy in terms of range below the x=0 point... Is that just a limitation of the method? Thanks for your input, though. :)

Homework Statement \int {\frac{{dx}}{{\sqrt {x^2 - 2} \left( {x^2 - 1} \right)}}} Homework Equations \frac{d}{{dx}}\arctan (x)The Attempt at a Solution \frac{d}{{dx}}\arctan (x) seems to be part of it, I can't quite get much farther...

That's a trig identity, I believe. And, if sine is positive, it would be arccos(cos(omega)), and if not, it would be... 2*pi-arccos(cos(omega))? Would that be right? Thanks. (:

Nope; not at all. :smile:

Oh, sorry about that. Well, if we're going to go thataway, sine is positive in I and II, cosine is positive in I and III. Arcsine is defined in I and IV, arccosine is defined in I and II. Thanks!

Hi all, I'm working on a program, and it seems that I can't get the trigonometry in my head right. I have two equations, sin(omega) = something and cos(omega) = something, and I need to find what omega is. Given that there are two trig functions, I should be able to disambiguate what...

Thank you verily much for your input. I completely understand now. (:

If you're willing to wait a bit and plonk down a bit more cash for an on-the-edge calculator, may I suggest the TI-Nspire CAS? Seems like it comes out Q2 this year or so, but the screenshots and press coverage looks pretty decent.

Homework Statement A nozzle has a radius of 25mm. Water emerges at the rate of 750L/min. Find the force with which the nozzle must be held.Homework Equations F = m \times a \Delta m / \Delta t = pAv A_1 \times v_1 = A_2 \times v_2 R = vA The Attempt at a Solution I've so far been able to...

Thank you verily much! It suddenly makes sense! (I was wondering why the spring constant was perfectly related to part d) I still don't understand the second approach, however. Is it possible to find the other force? (Which I assume is just... force imparted upwards?)

I'm not sure what you mean. If I try to find position, I end up with: KE = ME_i - mgh Where ME_i is a constant, as are m and g. Thus, I get a linear equation. For the latter comment, I'm not sure how the free-body diagram should then be drawn as. There's force of gravity going...

http://img178.imageshack.us/img178/75/43802nv4.jpg A child's pogo stick (see figure) stores energy in a spring of spring constant k = 40000 N/m. At position A, x1 = -0.150 m, the spring compression is a maximum and the child is momentarily at rest. At position B, the spring is relaxed and the...