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. (:
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...
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...