answersmay

05-05-01
Our first question for the month of May is from Robin, the person responsible for initiating the Grimace discussion. He writes "My next question is about Hollywood,how every year there are two films that are very similar. for example,if we go a couple of years back we had the year of Deep Impact and Armagedon(which was a horrible film i might add),both about gigantic meteors crashing into earth.Then we had A Bug's Life and Antz ,both computer animated movies about insects with human like charecteristics!!And lets not forget Dantes peak and Volcanoe,both about(you guessed it)VOLCANOES!!!!and this year is no different,we got two movies coming out this summer about race car driving,Driven and The fast and the furious.So tell me,is it mere coincidence or is part of a big hollywood plot
that the public is unaware of????"
According to a "Hollywood" source with Miramax and Fox affiliation, it's all a coincedence, "it must be, in a town where everyone sues everyone for everything". I personally think that it's a coincedence too, after all, inspiration is derived from observation, and in an industry where almost everyone is located within a 10 mile radius of everyone else, there have to be some common observations...common inspiration...two very similar (and often lame) movies. Of course, we can never rule out a big hollywood conspiracy, because some movies, like Terminator and Terminator 2, Rocky3 and Rocky4, Flinstones and Flinstones Viva Rock Vegas, are just too similar to be coincedences.

Today's last question comes from Gary (note: this Gary and Gary LittleBear are two different people). Gary asks "[what happens to a particle if it's sucked into a black-hole, does it emerge in a different universe? how do black-holes work?]"
Some scientists do believe that "wormholes" form inside black-holes that make it possible for particles to travel to a different part of the universe, a different universe all-together, or back in time. However, the inside of a black-hole is a singularity (like the Big-Bang), meaning the laws of physics/science do not apply there, so we can't predict what happens. As far as the wormhomes go, they apply to black-holes that rotate or have electrical charge, and "connect" the black-hole to a white-hole (this is a time-reversed black-hole, a product of the symmetry of Einstein's general relativity, meaning that if you reverse the flow of time in the equation you will still get valid answer. Doing this to black-holes gives white-holes, which spit everything out rather than sucking everything in) so anything that falls through a wormhole in a black-hole will emerge in it's corresponding white-hole. Of course, just becasue something is mathematically possible, doesn't mean it exists; black-holes formed from the collapse of ordinary matter do not form wormholes, so most-liekly, no wormholes exist in our universe, even if they did exist, they would be very unstable, and would collapse if anything tried to go through them. So, to answer your first question, the particle will almost certainly not do any travelling, it will just hit a singularity, and we don't know what happens there. On to the second part, which come to think of it, I should've answered first. Black-holes are generally thought of as specific types of "dead" stars, though centers of galaxies can also be black-holes. Anyway, we'll only focus on the star thing, since that's the most common case. Stars are formed when a whole lot of gas starts to collapse in on itself due to gravitational atraction. As the gas particles start colliding, they get hot, so hot in fact, that soon instead of bouncing off after collisions, they get together to form new gases (mostly, it's hydrogen turning into helium) the heat from these fusion reactions increases the pressure of the gas until it becomes "pressury" enough to balance the gravitational forces and stop the shrinking of the gases...a star is born. (quick note: the fact that pressure plays a role in gravity leads to some other really interesting things, like quintessence and dark energy, read...umm...Febraury or March's Scientific American for more on dark energy). Eventually, after a long long time, the star will run our of nuclear fuel (the hydrogen gas) and won't be able to hold-off the effects of gravity, and will start to contract again. At this point, a few different things can happen. Depending on the stars mass/size, it would either become a white-dwarf or neutron star, explode, or turn into a black-hole ( a term, by the way, coined by John Wheeler in 1969). Here's how it works: as the star shrinks and becomes very small, it's particles get extremely close to each other, and, according to the Pauli exclusion principle (you might remember this one from your chemistry class, if not, look it up), the particles have to have different velocities. This means that they have to move away from each other, causing the star to expand. Eventually, Pauli's repulsion (which reminds me, I saw Bio-Dome like an hour ago) and gravity balance out, and the star maistains a constant radius; this is a white-dwarf, a neutron star is basicly formed the same way, except the protons and electrons will get so crammed that their quarks will recombine to form neutrons. If the mass of the star is larger than a certain limit, it will not become a white-dwarf or neutron star, but will instead either explode and lose enough mass to get below the limit and become a white-dwarf/neutron star, or would collapse in on itself, forming a black hole. A black-hole's gravity is so strong that nothing can escape it. The speed required to escape a body's gravitational pull is called an escape velocity, the escape velocity for black holes is greater than the speed of light, and since nothing moves faster than the speed of light, nothing can reach the escape velocity for a black hole; nothing can escape a black hole. The point at which light just barely fails to escape a black hole is called the black-hole's event-horizon; the event-horizon is the point of no return. The core of the black-hole is a singularity, which, as I said earlier, does not abide by any of the laws of physics. I'm kind fo tired, so that's about it for now. Sorry for the poor organization and lack of details, I'll improve this answer when I next update the site. Thanks for the awesome question, Gary.

The Underachiever asks "why is canada such a quiet country?? you never hear anything about canada in the news... we are never taught about canadas history in school... no one ever worries about them invading. do they even have any enemys?? whats so secret about canada?? didnt you send them a letter once?? anyways... blame canada!"
Though Canada is rather quiet, it's not extremely quiet, it just seems that way to us since we're far away and can't hear them very well. I heard something about Canada in the news a couple of weeks ago, during those stupid riots. You're right about us not being taught a lot of Canadian history in school, all I remember is the whole French thing, the War of 1812, the Warhawks (or whatever they were called, my US History teacher failed me (well, she gave me a B, but she failed me in the sense that she didn't teach me anything, my Lit. teacher on the other hand, failed me both literaly and figuratively), and something about Sword, which I think was the codename of the section of Normandy's coast the Canadians were in charge of securing during Operation Overloard. I don't think people worry about them invading much anymore, but they some folks did in the early 1800s (or late 1700s) and lobbied to have us invade them first. I don't think they have any major enemies, but I hear they and Finland aren't as close as we'd like them to be. I don't think there's anything secret about them, in fact they're very open, they're just very boring, so boring that we can't believe nothing happens to them and start to think that they keep their affairs private. Yes, I did send them a letter once (letter, decleration of war, what's the big difference?), though I think it got lost in the mail since I didn't know the address of their parlaiment and simply made it out to "Canada", mind you, this was about a year before that South Park movie came out, so I include the creators of South Park in the list of people who ripped off my ideas (along with Warren Beatty, Scandinavia [though they ripped me off before I was born], and whoever came up with the power rule in calculus, among many others).

The next one is from Polaris Dali "Hi, I was at the local Rite-Aid today when I was reminded that their slogan is: "It's not just a store, it's a solution" If their slogan holds any truth, then that means that there is no need for your site. If I need help with my calculus homework, how to perform open heart surgery, or learn to change the oil in my car I can go to them and they will help me out. What if they are lying though and they can't find a solution to my problem because let's say I need to get a divorce or create a monopoly without the government finding out. Then it IS just a store. Is this false advertising? and are you willing on going down to Rite Aid and testing it out to see if it is just a store? That is all."
     They sai "It's not just a store, it's a solution", they don't say "It's not just a store, it's the solution to everything". So, you shouldn't expect them to help you with calc homework, open heart surgery, or car maintenance. True, they must be the solution to at least one problem, or else it's false advertising, but they don't need ot be the solution to all problems. Like (1/a)arctan(u/a) is a solution to the integral of du/(u^2+a^2), but not the solution to problems associated with creating monopolies. I did go down to Rite-Aid to do a little investigation. Below is a list of some questions I asked, followed by whether or not they answered it (I didn't write down the answers they gave, only if they gave one).
Question                                                                                                     Answered?
How much is this cream soda?                                                                        Yes
What is Ampere's Law?                                                                                  No
How are you?                                                                                                Yes
Can you give me the change in dimes and pennies only?                                     Yes
How do I perform open heart surgery?                                                             No (weird look)
Where are the potato chips?                                                                                Yes
How much wood would a woodchuck chuck if a woodchuck could chuck wood?    No
What time is it?                                                                                               Yes
How do I integrade cos(x^3)/(x^5+6x)?                   Yes, wait that was the time question, so No
Is this register open?                                                                                        Yes

   So you see, it's not false advertising, since they did provide me with quite a few solutions, eventhough they only promised one. I really didn't enjoy doing this though, since I wouldn't like to be a cashier out there and be asked a bunch of annoying questions by some annoying guy, but hey, I asked some of the more annoying questions to teenage cashiers, so it's ok.

Juan asks "who came up with the name jokes such as "jack mehoff". was it someones actual name or what?"
I'm sure that there have been people with names like Jack Mehoff. But the actual idea to make prank calls using those phony names was invented almost right after the invention of the telephone. See, Alexander Graham Bell had an assistant names Jack Shapher. So, when he was testing out the telephone, Bell asked Mr. Shapher and a few of his other assistants to set up a station a mile away and wait for his call. When Bell called them, and they answered, he said "Jack Shapher, it works, it works!". The person who picked up the phone wasn't Jack though, it was Mr. William Praksley. Now, it's important to remember that Bell tried to sound very "dignified" and pronounced dance as dunce, and Jack as Juck (as if it were French). Also, the telephone still had some problems, and static blocked the very first part of his message. So, what Mr. Praksley heard was "Uck Shapher, it works, it works!". In those days, cock shaffer was quite an insult, so Mr. Praksley naturally thought "uck shapher" was pretty funny and started making up other names (Seymore Butts et al) and making crank calls, and that's how it all got started.
Note: none of the above has ever been documented, in fact, I made it all up, but it has never been documented to not be true either.

Today's final question is from David from Arizona. He writes:
"Obviously you don't have all the answers, but do you think there is some source out there that does? The ancient Greeks believed that the oracle of Delphi was able to tap into some universal source of knowledge that had all the answers, and Edgar Casey claimed to have done the same. What do you think?"
Obviously. I don't care about this question becasue it is absolutely useless to care about this question, and here's why: there are two choices: 1. the source does exist; 2. that source doesn't exist. If the second choice is true, then it's useless to try to find such a source or to even think about it. If choice 1 is true, then we get to more possibilites: There exists a source that contains knowledge about everything and has all the answers; since it has all the answers, it will have the answer to this question, "Will humans ever find the 'source'?"; it can answer it in 2 ways, 1. Yes, or 2. No. If 1 is the answer, then we shouldn't worry about trying to find the source or think about it since we're bound to find it no matter what, if the answer is No, then we shouldn't worry about the source since we can never find it. You see, whether or not such a source exists, it's of absolutely no use for us to think or care about it.
That's it for today.

Juan asks "is it true that floresent lights give you cancer? if so why? how do street lights work? do their bulb ever go out?"
Okay, I'll answer your questions in reverse order.
Yes. For your third question, they work the same way as regular lights. If they're incandescent then they have a thin tungsten fillament in a glass bulb, the fillament offers a lot of resistance to the electricity running through it, and turns the electrical energy into heat, this heat causes the fillament to glow, and voila, we have light. If the street lights are fluorescent, then you've got a glass tube coated with phosphor on the inside, with an electrode at each end, there's gas containing argon and mercury inside the tube. A stream of electrones travels between the electrodes, the electrons bump the mercury atoms and excite them. The atoms later fall from their excited state back to an unexcited one, releasing ultraviolet light photons in the proccess. These photons collide with the phosphor coating and visible light is created. The reason street lightscome on by themselves whenever it gets dark is becasue they have a cadmium sulfide photo-resistor (CdS cell) inside them. When the cell gets hit by a lot of light, it has almost no resistance, when it gets very little light, it has a high resistance. So, when it gets dark and very little light reaches the CdS cell, it has high resistance anc conducts very little electricity, it basicly grounds the circiut and sends the light to the bulb. On to your first two questions, combined into a multi-questioning single question. Nothing has ever been conclusively proven to conclude that fluorescent lights cause cancer. There was one experiment that showed that pink fluorescent lights cause cancer more quickly than white oens, but that doesn't really mean anything, since it doesn't say that fluorescent lights cause a significant amount of cancer to begin with (I should've stated this better). Actually, there was an article in Science News magazine, in...1998, I think...that said that bright lights of any sort at night may "cause" cancer. When there's light, our brain produces seretonin (happy hormone), and when it's dark it produces melatonin (makes us drowsy and depressed, which is why so manny people in Finland and Norway and all those dark places kill themselves). Melatonin has a property which guards against cancer by prohibitinh the growth of a certain enzyme or something like that, I don't remember the exact details. Anyway, our pineal glands produce melatonin when it's dark, but if we have all these bright lights on when it's dark, then melatonin will not be produced and after a while of this decrease in melatonin production we will become more vulnerable to cancer. Again, I don't think this happens in significant degrees, so you shouldn't worry too much. Speaking of seretonin, that's how the drug extacy works, it produces high levels of seretoning, which is why you might find yourself feeling depressed after using it, since your brain used up it's supply of seretonin the night before and there's very little left for a while. So remember kids, don't do extasy (well, how's that for community service?). As for the cancer thing, no worries mate, it's like when they fed a rat a bathtub of nutrasweet a day and said it caused cancer, there's no need to be concerned unless you surround yourself in fluorescent lights and eat a tub of nutrasweet for a whole year.

Okay, the next two questions were not specificly asked to the McMigran, but were posted on a forum. The McMigran answered them, and is putting them on his site becasue he needs to waste some room.
1. (int) tan^-1(x^.5) dx (from Lopnur)
there are probably easier ways of doing this, probably much easier, but I can't think of any. Here we go. note that { is supposed to be the integral sign and that i=-1^.5.
{arctan(x^.5)dx; u=x^.5, du=dx/(2x^.5), dx=2x^.5du
                                       dx=2udu
{arctan(x^.5)dx = 2{uarctan(u)du
now, we use integration by parts
dz=arctan(u)du; z={arctan(u)du; v=u; dv=du,
let's leave everything else alone for now and find z
{arctan(u)du, int. by parts where:
a=arctan(u) db=du
da=1/(u^2+1)   b=u
z = (u)arctan(u) - {[u/(u^2+1)]du
let u^2+1=c, dc=2u du, du=dc/2u
z = (u)arctan(u) - .5{dc/c
z = (u)arctan(u) - .5ln(c) = (u)arctan(u) - .5ln(u^2+1)
okay, now we have z, dz, v, and dv, we can continue with the integration by parts.
zv - {zdv
{uarctan(u)du = u^2arctan(u)-.5uln(u^2+1) - {[uarctan(u)-.5ln(u^2+1)]du
{uarctan(u)du = u^2arctan(u)-.5ln(u^2+1)-{uarctanudu + .5{ln(u^2+1)du
2{(u)arctan(u)du=u^2arctan(u)-.5ln(u^2+1)+.5{ln(u^2+1)
okay, now we're almost done, all we need is to integrade {ln(u^2+1)du, so lets leave the rest aside and do that.
u^2+1 = (1+i)(1-i) where i=-1^.5 (sqr.rt.-1)
{ln(u^2+1)du = {ln[(u+i)(u-i)]du
= {ln(u+i)du + {ln(u-i)du
using {ln(p)dp = pln(p)-p, and u(+/-)i=p, we get:
= (u+i)ln(u+i)-(u+i)+(u-i)ln(u-i)-(u-i)
= uln(u+i)+iln(u+i)+uln(u-i)-iln(u-i)-2u 8
= uln[(u+i0(u-i)]+iln[(u+i)/(u-i)]-2u
{ln(u^2+1) = uln(u^2+1)+iln[(u+i)/(u-i)]-2u
, now we have everything, so let's combine it all
2{(u)arctan(u)du = u^2arctan(u)-.5ln(u^2+1)+.5[uln(u^2+1)+iln[(u+i)/(u-i)]-2u]
now, substitute x^.5 for u and you're done. oh ya, don't forget to simplify, you might want to use some trig identity, possibly tan, for ln[(u+i)/(u-i)] if you don't want to have imaginary numbers in your answer, if not, then just leave it alone. Whew, that was a whole lot of steps.
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oh ya, the {ln(p)dp = pln(p)-p thing comes from integration by parts too.
let m=ln(P), dm=dp/p, dn=dp, n=p
we get pln(P)-{(p/p)dp
= pln(P)-{dp = pln(p)-p,

The integration of the ln(x^2+i) integral is courtesy of some guy whose name I don't know.

2. i need help for a calc BC final fast!!!!!!! any help will be greatly appreciated!!
S=integral sign. how do you calculate this?? S(x^5)dx/[(x^2)+ 4)] what method??
The above question is from MathWhiz (apparently not whizzy enough).
okay, here's the answer to the first question. { is the integral sign. Also, I'm going to be using a lot of substitution, and will leave the final answer in temrs of those letters, so you'll have to simplify it yourself.
{x^5/(x^2+4)dx
int. by parts.
u=x^5          dv=dx/(x^2+1)
du=5x^4          v=(1/2)arctan(x/2)
= vu - 2.5{x^4arctan(x/2)dx;
       2.5{x^4arctan(x/2)dx = K
{x^4arctan(x/2)dx; a=x^4       db=arctan(x/2)dx
                   da=4x^3 dx    b={arctan(x/2)dx
b. c=arctan(x/2)     de=dx
dc=.5/[(x/2)^2+1]   e=x
b=ec-.5{x/[(x/2)^2+1)
         z=x^2/4+1; dz=2xdx, dx=dz/2x
b = ec-.25{dz/z

b = ec-.25ln(z) = (x)arctan(x/2)-.25ln(x^2/4+1)
{x^4arctan(x)dx = ba-{bda
{x^4arctan(x)dx=ba-4{x^4arctan(x)dx + {x^3ln(x^2/4+1)
5{x^4arctan(x)dx=ba+{x^3ln(x^2/4+1)dx;
{x^3ln(x^2/4+1)
   p=(x^2/4)+1, dp=.5xdx, dx=dp/(.5x)
= {x^3ln(p)dp/(.5x) = 2{x^2ln(p)dp
= 2{4(p-1)ln(p) = 8{(p-1)ln(p)dp
= 8[{pln(p)dp - {ln(p)dp]
= 8[p^2(ln(p)-1) -(pln(p)-p)]
ok
5{x^4arctan(x) = ba+8[p^2(ln(p)-1)-(pln(p)-p)]
devide both sides by 2 to get:
K = .5ba+4[p^2(ln(p)-1)-(pln(p)-p)]

so, we're done.

{[x^5/(x^2+4)]dx = uv-K
u, v, and K are all defined somewhere above, just plug them in and simplify. Oh, and as always, double check everything to make sure there are

The McMigran now realizes that simple long division would have been a hell of a lot easier on both these problems, oops.