Wikipedia:Reference desk/Archives/Mathematics/2011 June 12

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June 12

Normal Deviation

The Menstrul Cycle of female is on an average, of 28 days and has standard deviation of 2 days A)in what % of women will the cycle differ by more than 3 days from mean ? B)mark out symmetrically around the mean the range in which 80% women will lie — Preceding unsigned comment added by 49.200.54.220 (talk) 16:40, 12 June 2011 (UTC)

This is obviously a homework question. We don't do people's homework for them here. Could you show us what you have done so far and then we will help you to understand what you need to do next. As it says at the beginning of this page: "If your question is homework, show that you have attempted an answer first, and we will try to help you past the stuck point. If you don't show an effort, you probably won't get help. The reference desk will not do your homework for you. — Fly by Night (talk) 17:49, 12 June 2011 (UTC)

I'll just point out that from a fully technical position, the question is underspecified. The answer depends *highly* on which probability distribution you use. Average and standard deviation are really only enough to fully specify a distribution if you're assuming a normal distribution (and some other, simple distributions). While it's possible that the length of the menstrual cycle is normally distributed (Central Limit Theorem and all that), as the problem is stated we can't be sure if the assumption is completely valid. -- 174.31.219.218 (talk) 19:14, 12 June 2011 (UTC)

Well, the fact that the OP entitled this section Normal Deviation is a bit of a clue, perhaps. Looie496 (talk) 23:18, 12 June 2011 (UTC)

The distribution can't be normal because that would imply the possibility of a negative time. The question also seems to assume (wrongly, as far as I know) that the cycle time is constant for any particular woman. AndrewWTaylor (talk) 19:36, 13 June 2011 (UTC)

A negative value would be 14 standard deviations away from the mean. I think we can safely ignore that problem. Few things are actually normally distributed, but a great many things are well approximately by the normal distribution within about 3 sd of the mean, which is what makes it such a useful distribution. --Tango (talk) 23:23, 13 June 2011 (UTC)

Newton's Second Law

"The second law states that the net force on a particle is equal to the time rate of change of its linear momentum p in an inertial reference frame:

$\mathbf {F} ={\frac {\mathrm {d} \mathbf {p} }{\mathrm {d} t}}={\frac {\mathrm {d} (m\mathbf {v} )}{\mathrm {d} t}},$

where, since the law is valid only for constant-mass systems the mass can be taken outside the differentiation operator by the constant factor rule in differentiation. Thus,

$\mathbf {F} =m\,{\frac {\mathrm {d} \mathbf {v} }{\mathrm {d} t}}=m\mathbf {a}$"

This is an extra from your article on Newton's laws of motion, specifically the section on the second law. I don't understand why it says that the law is valid only for constant mass systems, not least of all because on Isaac Newton, in the section entitled 'Laws of Motion', the second law is written as ${\vec {F}}={\frac {\mathrm {d} {\vec {p}}}{\mathrm {\mathrm {d} } t}}\,=\,{\frac {\mathrm {d} }{\mathrm {d} t}}(m{\vec {v}})\,=\,{\vec {v}}\,{\frac {\mathrm {d} m}{\mathrm {d} t}}+m\,{\frac {\mathrm {d} {\vec {v}}}{\mathrm {d} t}}\,.$, the form that I am familiar with.

I'm partly asking what the subtlety is that demands we treat the mass as constant in the first case, assuming it's not a mistake, and partly suggesting that someone who knows what they're doing clarifies the issue. I have studied mathematics and physics at university and if I am left confused by this then the layman will have no idea what's going on. Thanks. meromorphic [talk to me] 16:44, 12 June 2011 (UTC)

Can I just say, as a the layman of your words, I can follow the first version with mass constant, but would have more trouble with the rule if the rule is equally true in changing-mass systems. (I do understand the product rule, but still.) I couldn't say which is more familiar to people with physics or mathematics degrees, but I'm certainly more aware of f=ma than one adapted for use on changing-mass systems. I don't know which Newton used, but if the second is more "true", then it may still be preferable to keep the first and say "if mass is constant" rather than "since mass has to be constant", at least introductory-wise. Grandiose (me, talk, contribs) 16:52, 12 June 2011 (UTC)

The problem with ${\vec {F}}=\,{\vec {v}}\,{\frac {\mathrm {d} m}{\mathrm {d} t}}+m\,{\frac {\mathrm {d} {\vec {v}}}{\mathrm {d} t}}$ is that if the mass of an object is changing it must be picking up or expelling mass (or energy) from outside. The mass that's entering or leaving has momentum of its own, and the formula doesn't tell us how to accurately deal with that. For example if I have a moving object and a piece breaks off, but the original object and the piece both keep moving off at the same velocity, then I end up with a non-zero ${\vec {v}}\,{\frac {\mathrm {d} m}{\mathrm {d} t}}$, but can we really say there's a force acting on the object? Rckrone (talk) 18:24, 12 June 2011 (UTC)

I think the term for non-constant mass comes in when you take relativity into account. You can apply all the force you want but the velocity will always be less than c. So when you get close to c what the force does mostly is increase mass, which is where the dm/dt comes in. But this is the math helpdesk, we don't normally deal with soft science like physics.--RDBury (talk) 22:53, 12 June 2011 (UTC)

No response yet from physicists who might object to calling their science "soft"? Dbfirs 07:51, 17 June 2011 (UTC)

See here:

Hilbert said "Physics is too hard for physicists", implying that the necessary mathematics was generally beyond them; the Courant-Hilbert book made it easier for them.

Count Iblis (talk) 02:10, 18 June 2011 (UTC)

While you can do relativity using relativistic mass (and I often do when trying to give people an intuitive feels for how it works), it isn't the normal way of formulating it mathematically. You can see a derivation of the more common relativistic equivalent of F=ma here: Special_relativity#Force. --Tango (talk) 23:30, 13 June 2011 (UTC)