Wikipedia:Reference desk/Archives/Mathematics/2008 September 13

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September 13

Calculus Question

Integral of 1 over radical (6x-x^2) —Preceding unsigned comment added by 169.229.75.140 (talk) 01:10, 13 September 2008 (UTC)

Welcome to the Wikipedia Reference Desk. Your question appears to be a homework question. I apologize if this is a misevaluation, but it is our policy here to not do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn how to solve such problems. Please attempt to solve the problem yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. Thank you. Algebraist 01:13, 13 September 2008 (UTC)

I agree with Algebraist that this looks like a homework question, but I'm feeling generous so I'll give you a hint: My first step would be to use substitution to get rid of the 6x term, it should be easy from there. Good luck! --Tango (talk) 08:19, 13 September 2008 (UTC)

In my schooldays, many years ago, the integral of 1/(2ax-x^2)^0.5 was a standard form. Times change.—81.132.236.163 (talk) 14:25, 13 September 2008 (UTC)

You say that as if memorising a standard form over understanding the method is a good thing. 91.143.188.103 (talk) 18:25, 15 September 2008 (UTC)

It may well be in the back of a standard calculus text, I don't honestly remember - I'm pretty sure it is once you do the substitution, though (that's the easy bit I referred to!). It may depend on the textbook - if I were writing it, I'd keep the list short and expect readers to do the simple conversions to turn their problem into a standard one! --Tango (talk) 14:37, 13 September 2008 (UTC)

I was curious, so I dug out my calculus textbook from the bottom of the box of notes I brought it home for the summer in (Salas, Hille and Etgen if you're interested) and what you get after the substitution is on the back cover, the OP's question isn't. It seems Salas and I are of like mind! --Tango (talk) 14:42, 13 September 2008 (UTC)

Now the OP has had time to think. First substituting x=3+3t and later substituting t=sinv gives

(6x−x²)^−1/2dx = (6(3+3t)−(3+3t)²)^−1/2d(3+3t) = (18+18t−(9+18t+9t²))^−1/23dt = (1−t²)^−1/2dt

= (1−sin²v)^−1/2d(sinv) = (cos²v)^−1/2cosvdv = dv

So

(6x−x²)^−1/2dx = d(arcsin(x/3−1))

Bo Jacoby (talk) 01:24, 14 September 2008 (UTC).

OK, at least one nonsensical answer appears above and at least one that is needlessly complicated. Let's try being straightforward: if you're not instantly thinking completing the square then there's a gap in your understanding that you badly need to fill. So:

${\begin{aligned}6x-x^{2}&=-(x^{2}-6x)\\&=-(x^{2}-6x+9)+9{\text{ (add }}-9{\text{ to complete the square and}}\\&{}\qquad \qquad \qquad \qquad \qquad \qquad {\text{compensate by adding }}9)\\&=-(x-3)^{2}+9\\&=9-(x-3)^{2}\\&=9\left(1-{\frac {(x-3)^{2}}{9}}\right)\\&=9\left(1-\left({\frac {x-3}{3}}\right)^{2}\right)\\&=9(1-u^{2}).\end{aligned}}$

If

$u={\frac {x-3}{3}}$

then

$du={\frac {dx}{3}},\quad 3\,du=dx$

so

$\int {\frac {dx}{\sqrt {6x-x^{2}}}}=\int {\frac {3\,du}{3{\sqrt {1-u^{2}}}}}=\arcsin u+C=\arcsin \left({\frac {x-3}{3}}\right)+C.$

Pretty routine once you've practiced a bit. But if you don't recognize this immediately as a completing-the-square problem, you need to learn this: the purpose of completing the square is always to reduce a quadratic polynomial with a first-degree term to a quadratic polynomial with no first-degree term. Michael Hardy (talk) 03:27, 14 September 2008 (UTC)

What are the chances

My wife went to work the other day and my son joked and said "We may never see her again." We laughed a bit but I got to wondering, what are the chances that we may never see here again? How does one go about establishing the probability that any one particular event like that is canoccur? I think that you would start let's say with the number of highway fatalities in the area in any one week or year, and then perhaps calculate the number or vehicals that travel on the roads that make it home safe every day and so on. I never studied statistics or probability - so I am curious how the though process in this sort of event is calcutate. Just so we are clear - I do wish my wife to come every day.142.68.147.80 (talk) 17:43, 13 September 2008 (UTC)

Yes, you would need to looks at how many fatalities there are doing a particular activity and divide it by how much that activity is done. You could improve the accuracy by restricting it to people like your wife (does she have a good no claims bonus? If so, you should look at how many people with good no claims bonuses die on the roads divided by how much they drive). You should probably base it on miles driven, rather than number of cars, although the risk may well not be proportional to the length of the journey. You would need to gather lots of data to get accurate statistics. Either that, or you could try googling - someone has probably done such a study already for most common activities (well, they certainly will have studied it in order to work out insurance premiums, but they may not have published the results). --Tango (talk) 18:05, 13 September 2008 (UTC)

And then double this number, because you yourself could die too. Sentriclecub (talk) 09:39, 14 September 2008 (UTC)

You can't necessarily double it, since you'll be doing different things and are a different person, so have different risks. Also, you can't just add the probabilities, you have to minus off the chance of your both dying otherwise it will be counted twice (this becomes obvious if the change of each of you dying is more than 50% since adding them together would get a probability of over 100%). --Tango (talk) 12:55, 15 September 2008 (UTC)

I originally thought about just writing as my first response and then double this number and leaving it at that. Allowing either of you two to figure out the subtle oversight you both made. Then I thought about the response that I eventually went with. But I feared that it may be misinterpreted that I literally mean to do all the suggested research, calculate a number, then multiply it by 2.000 and assume uncertainty of .001 but I gambled that my response would not dissected as literal advice of how to exactly measure the chance. I only wished to elucidate that the question implicitly assumes that the husband is immortal during the time period. For the elaborateness of the first response, nowhere is mentioned that there are two ways to meet "the condition". To be 100% accurate--here it goes-- the chance that "we may never see her again" is approximately double the chance of "she may die". Okay, but this statement can be refuted a million different ways, for starters you could say that the radius of the probability of the husband's death is .3% different than the probability of the wife's death. However, in the end, questioner gets a better and more accurate answer if he reads our first responce each, only. I guess its just that I am obsessed with probability puzzles. My favorite probability enigma is one that says 13% of people with 160 IQ couldn't pass a mensa entrance exam, if they are already a member of Prometheus socity (having truthfully passed an IQ test which they scored 160+). I merely believe that doubling the answer, is more accurate than to 1x it. If you thought I meant 2.000 and forbade 1.7 or 2.3, then its just misunderstanding on your part caused by failure of communication on my part. I only elaborate on this in great detail because I think Zain made a good point about your answer of the APR question, where your answer carried a broad generalization and you converted it into a special case. Nominal is only opposite of real, in a very limited number of cases. Inflation, sure, but APR no. I'm only sporadically involved ref desk contributor, so hope you can tolerate me until the end of the day. I'm obsessive, and its counterproductive for me (and for other contributors) to stray off from the original questioner's topic. So I apologize for excessiveness. Sentriclecub (talk) 21:50, 15 September 2008 (UTC) Forgot placeholder, oops. Made my point, from an hour ago, thanks.

You also need to take into account that both father and son could lose their eye sight. "Thus never seeing her again." —Preceding unsigned comment added by 203.202.144.223 (talk) 04:27, 16 September 2008 (UTC)

To make it really complex and assuming that god exists, the events that you said 'you may never see her again' and that 'you really don't see her again', may be dependent and in that case just saying that 'you won't see her again' may increase the probability of the event!

Topology Expert (talk) 13:28, 17 September 2008 (UTC)