Recent content by inutard

Yup! Thanks for all the help! This has cleared things up for me.

When I say the second integral is symmetric, I guess what i mean is that t is approaching infinity on both sides. But for the first integral, you can interpret it as an integral from a to 0 plus an integral from 0 to b, with 'a' going to negative infinity while 'b' goes to infinity. In that...

Hmm. Can you explain to me this then? http://www.wolframalpha.com/input/?i=integral+of+arctan%28x%29+from+negative+infinity+to+infinity The limit (2nd integral) converges to zero when I test it on mathematica tho. Also, what is the Cauchy Principal Value? That seems to be zero.

Thanks for the replies guys. I thought about the question a bit more. I believe that the two integrals are not the same. The first one can be split into the integral from -Infty to 0 + the integral from 0 to +Infty. These two integrals are both divergent and so the sum is divergent (there is...

Homework Statement Is $\intop_{-\infty}^{\infty}\arctan(x)\, dx$ convergent? What about $\lim_{t\rightarrow\infty}\intop_{-t}^{t}\arctan(x)\, dx$?Homework Equations The Attempt at a Solution I think the first integral may actually be divergent the way its written and the second one...

Mathematica would solve that sort of equation just fine. So would Maple and Matlab for that matter. www.wolframalpha.com would also suffice. As for solving it by hand, I don't see any clear way to do it without a whole lot of work.

No. You are not missing anything. When carrying out sig figs, it really should be two but many systems use three automatically.

yes. So youll notice that the mass does not actually matter in the question since it cancels out.

Perhaps you need a more precise answer like 1.75. If the homework system you are using is something like UTexas, 3 decimals are needed as they only tolerate 2% error.

Ok. Let us begin with this question. You are given the frequency of the LP (and thus the period as well), and you are given the mass of the penny. As you have written above, Frictional Force = co-eff * Normal and centripetal acceleration = v^2/r = 4Pi^2r*frequency^2. For the penny to stay on the...

Ahh. I see my mistake. My mistake was assuming that the Path Difference/ Wavelength = 0.5. A value of 0.5 for the P.D./wavelength is not the smallest path difference. From a bit of trial and error, you will see that 2.5 (i.e. n=2, n+0.5 = 2,5) creates the smallest path difference with an answer...

The work you did seems to be on the right track, but perhaps use of different notation will clear things up. Since one speaker is located at the origin (0, 0) and the other speaker is located at (0, 4), we do not need the co-ordinates of the Point P1-P2 but rather its distance: Since the Path...

You will need to know about standing waves and possible wavelengths to produce standing waves. Have you been taught the derivation of the formula Wavelength = 4L/(2n-1) where n is the nth harmonic? (Usually this is Taught as 4L/n where n = 1, 3, 5.. etc.

As pgardn mentions above, a magnetic field interacts with a charged particle in quite a peculiar way. Similar to a ball being spun on a string, the magnetic field exerts a centripetal force to the component of the charge's velocity that is perpendicular to the field (whew. that sounded like a...

Ah but think about it. Say we're looking at one interval of a cosine curve (0 to 2pi). How many max points are there? How many points cross y = 0? The points that cross y = 0 are like the nodes (when you multiply the cosine curve by -1 they remain invariant) while the antinodes are the points...