At a friend's beach house for a bachelor party. Everyone is still passed out but me because the idiot neighbors started chit chatting outside my bedroom window at 8:30am


Not sure where they would be dumping charcoal here. The canal?

12/11/2011 10:22:49 AM

dig a hole in sand perhaps

12/11/2011 10:30:28 AM

Why are we dumping charcoal?

12/11/2011 10:51:29 AM

Do they have trash cans on the east coast yet?


[Edited on December 11, 2011 at 10:54 AM. Reason : l]

12/11/2011 10:54:13 AM

um, on the ground?

12/11/2011 11:04:13 AM

I would dump it in the grill then light it on fire and throw some meat on the grill but that's just me.

12/11/2011 11:21:27 AM

I find every response in this thread suitable

12/11/2011 11:23:11 AM

Quote :

"Why are we dumping charcoal?"

12/11/2011 11:27:27 AM

burn it to heat the beach house

12/11/2011 11:32:23 AM

Quote :

"I find every response in this thread suitable"

Also, reasonable.

12/11/2011 11:41:52 AM

You can douse it with water to ensure that its extinguished, bag it, and trash it (be very careful and attentive here, for obvious reasons).

Or just dig a hole and bury it. It'll decompose. If it was insta-light or had lighter fluid on it, I'd trash it.

12/11/2011 2:02:04 PM

Quote :

"dig a hole in sand perhaps"

As long as the charcoal isn't still red hot.

12/11/2011 3:10:00 PM

1) Dump it on top of the sand in a long straight line.
2) Setup a stand for a fire walking experience.
3) ...
4) Profit

12/11/2011 4:27:42 PM

12/11/2011 4:31:12 PM

Toilet? Will it flush?

12/11/2011 6:32:23 PM

Quote :

"As long as the charcoal isn't still red hot."

Don't wanna catch the sand on fire...sand is very flammable...

12/11/2011 6:33:51 PM

charcoal can stay hot a lot longer than you might think... i grilled one night and then the next afternoon i dumped my charcoal in the natural area by my apartment at the time... caught the damn pinestraw on fire... wasn't anything my fire extinguisher couldn't handle, but i was amazed it was still that hot almost 24 hours later.

12/11/2011 7:07:56 PM

Before my long spiel, I should mention that the part on your work after you started using u' and du' is wrong: The antiderivative of e^f(E), where f(E) is some arbitrary function of E, is not generally (1/f'(E))e^f(E), even though it is true that the antiderivative of f'(E)e^f(E) is e^f(E).



First consider the transformation E=x^2/2.29; then the limits remain the same, 1.036E=(518/1145)x^2, sinh(sqrt(2.29E))=sinh(x), and 0.453E dE=(453/1145)x^2 dx.

Then the integral becomes the integral from 0 to +infinity of (453/1145)x^2*e^(-(518/1145)x^2)*sinh(x) dx.

--------Begin Failed Solution Attempt--------
Then in integration by parts, let u=sinh(x) and dv=(453/1145)x^2*e^(-(518/1145)x^2) dx.

Then du=cosh(x), and to find v, do another integration by parts, letting U=x and dV=(453/1145)x*e^(-(518/1145)x^2) dx.

Then dU=dx and V=-(453/1036)e^(-(518/1145)x^2).

Therefore, v=-(453/1036)x*e^(-(518/1145)x^2)+(453/1036)Int(e^(-(518/1145)t^2) dt,0,x).

Then our original integral is lim(uv,x->+inf)-(uv)(0)-Int(v du,0,+inf), so keeping in mind that u=(1/2)e^x-(1/2)e^-x...
...hmm, I see a problem, it looks like one of the terms will go off to infinity!

Notice that one term will be sinh(x)*(453/1036)Int(e^(-(518/1145)t^2) dt,0,x), in the limit as x->+inf; by the relationship at the top, the right factor becomes a finite positive number, but the left factor becomes infinite.

This must mean that I have a problem with my method, because as you can imagine, sinh(k*sqrt(E)) for constant k grows asymptotically as quickly as e^sqrt(E), which is more slowly than e^(-1.036E) reaches 0; then their product reaches 0 as quickly as e^-E, which is faster than E grows, in fact so quickly that the whole integrand also reaches 0 in exponential time, so the integral should be finite.
--------End Failed Solution Attempt--------

Let's go back to the step where u=sinh(x) and dv=(453/1145)x^2*e^(-(518/1145)x^2) dx; instead let's try u=x*sinh(x) and dv=(453/1145)x*e^(-(518/1145)x^2) dx.
Then du=sinh(x)+x*cosh(x) dx and v=-(453/1036)e^(-(518/1145)x^2).
Then the antiderivative is -(453/1036)x*sinh(x)e^(-(518/1145)x^2)+(453/1036)int(sinh(t)e^(-(518/1145)t^2)+t*cosh(t)e^(-(518/1145)t^2),0,x).

Now for the first term in that "int"-egral...letting U=e^(-(518/1145)t^2) and dV=sinh(t) dt, dU=-(1036/1145)t*e^(-(518/1145)t^2) dt and V=cosh(t), so the antiderivative becomes cosh(x)e^(-(518/1145)x^2)+(1036/1145)int(t*cosh(t)e^(-(518/1145)t^2) dt,0,x).
Now for the second term in that same "int"-egral...letting w=cosh(t) and dz=t*e^(-(518/1145)t^2) dt, dw=sinh(t) dt and z=-(1145/1036)e^(-(518/1145)t^2), so the antiderivative becomes -(1145/1036)cosh(x)e^(-(518/1145)x^2)+(1145/1036)int(sinh(t)e^(-(518/1145)t^2) dt,0,x).

If we denote 1036/1145 by k, then the second sentence means "so the antiderivative becomes -(1/k)cosh(x)e^(-(k/2)t^2)+(1/k)int(sinh(t)e^(-(k/2)t^2) dt,0,x)" and the first sentence means "so the antiderivative becomes cosh(x)e^(-(k/2)x^2)+k*int(t*cosh(t)e^(-(k/2)t^2) dt,0,x)" (to make it all cleaner).
Now there is one more integration by parts to do; if W=cosh(t) and dZ=t*e^(-(k/2)t^2) dt, then dW=sinh(t) dt and Z=-(1/k)e^(-(k/2)t^2), so that integral becomes -(1/k)cosh(x)e^(-(k/2)x^2)+(1/k)int(sinh(t)e^(-(k/2)t^2) dt,0,x).
Substituting back, some terms cancel out in "the first sentence" to yield just the innermost integral, int(sinh(t)e^(-(k/2)t^2) dt,0,x); then adding that to "the second sentence" yields the entire expression in that "int"-egral that I referred to earlier: -(1/k)cosh(x)e^(-(k/2)t^2)+(1+1/k)int(sinh(t)e^(-(k/2)t^2) dt,0,x).

Now wrapping up all the integrations by parts and back-substitutions so far, the antiderivative is
-(453/1036)x*sinh(x)e^(-(k/2)x^2)-(1/k)cosh(x)e^(-(k/2)t^2)+(1+1/k)int(sinh(t)e^(-(k/2)t^2) dt,0,x); this last unresolved integral does not appear amenable to integration by parts, and I don't quite know how to attack it, but when you evaluate the limit as x->+inf, that becomes finite. As for the two terms outside the integral, a similar asymptotic analysis shows that they go to 0 as x approaches 0 or +inf, so really what you're left with is (1+1/k)int(sinh(x)e^(-(k/2)x^2) dx,0,+inf), whatever that is.

12/11/2011 7:48:11 PM

[Edited on December 11, 2011 at 8:04 PM. Reason : ]

12/11/2011 8:00:19 PM

Bring it bitch.

[Edited on December 11, 2011 at 8:23 PM. Reason : .]

12/11/2011 8:23:41 PM

Quote :

"Don't wanna catch the sand on fire...sand is very flammable..."

I've actually seen someone get burnt because some jackass dumped their grill and kicked a little sand on top of the charcoals. It's all good until you realize your foot just turned into a giant blister.

12/12/2011 1:22:32 AM

.....at the charcoal dumping area?


/thread

12/12/2011 1:25:45 AM