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December 17[edit]

textbook equation doesn't work.[edit]

The equation: $2cos{\frac {\Pi }{6}}(5)+6\doteq 4.3$ is in my textbook. But I can't get any calculator to give me this results, radian mode or otherwise. Wolfram doesn't even return it. Any ideas? — Preceding unsigned comment added by 173.33.106.153 (talk • contribs) 02:18, 17 December 2010

Are you sure it's

2\cos({\frac {\pi }{6}})(5)+6=4.3

and not

2\cos({\frac {5\pi }{6}})+6=4.3

? 24.92.70.160 (talk) 01:24, 17 December 2010 (UTC)[reply]

I think the OP made a mistake too. In fact

\cos \left({\frac {5\pi }{6}}\right)=-{\frac {\sqrt {3}}{2}}\ \Rightarrow \ 2\cos \left({\frac {5\pi }{6}}\right)+6=6-{\sqrt {3}}=4.3\ (1\ {\mbox{d.p.}})

— Fly by Night (talk) 02:18, 17 December 2010 (UTC)[reply]

Are you sure you have your calculator set on radians rather than degrees? Michael Hardy (talk) 23:59, 18 December 2010 (UTC)[reply]

Figured it out- texbook's formula was $2cos{\frac {\pi }{6}}t+6$ , then they substituted in 5 to give $2cos{\frac {\pi }{6}}(5)+6\doteq 4.3$ . Some unclear notation on their part. Thanks! 173.33.106.153 (talk) 00:08, 19 December 2010 (UTC)[reply]

I don't see anything unclear about it. Michael Hardy (talk) 05:12, 20 December 2010 (UTC)[reply]

Derivative question[edit]

Is df(x)/dx in general equal to df(x²)/d(x²)? I would think so, but I'm having trouble proving it, and I used this for a math question and the answer I get seems wrong. 74.15.138.27 (talk) 04:33, 17 December 2010 (UTC)[reply]

No, it'll be the same derivative okay - but with x squared as the argument. Dmcq (talk) 08:19, 17 December 2010 (UTC)[reply]

Yeah, obviously, that was really stupid of me. Thanks. 74.15.138.27 (talk) 09:15, 17 December 2010 (UTC)[reply]

units in equations, cancelling[edit]

I realise this is a very basic question, but hopefully some of you will be willing to stoop to my level to explain it to me. I've always struggled with how units cancel in equations. Sometimes I get it right, other times I go horribly wrong and am never quite sure why. I'm studying solo and my textbook isn't explaining some of the basics (that I'm already supposed to have a clue about) and I have no teacher to ask. So, if I can get the formatting right, an example of my current confusion...

$v_{s}={\sqrt {\frac {\mu }{\rho }}}$

where μ is the rigidity modulus and

v s

is S wave speed and ρ is density.

Substituting in my figures, I end up with:

$v_{s}={\sqrt {\frac {3.0\times 10^{10}{\text{ kg m s}}^{-2}{\text{m}}^{-2}}{2.70\times 10^{3}{\text{ kg m}}^{-3}}}}$

I know that, for the final answer, I should get m s^–1. And my textbook tells me that, when I come to do the square root, I'll be working with the units m² s^–2, which is exactly what I want to get m s^–1. But how did they get to m² s^–2? They've cancelled out the kg top and bottom, which I'd have done, but I first tried to also use the m^–3 to cancel out the m and the m^–2. Why don't they cancel (apart from the fact that they can't because then I'd have no metres!)? But the bigger question is what has happened to make the answer at square root stage m²?! I have no idea how to get to m². I'm lost. What am I missing? I feel dreadfully stupid! I realise I should know this and one must wonder why I don't, but I guess I've always managed to blunder through and get by on knowing what the final units should be. Phantasten (talk) 13:28, 17 December 2010 (UTC)[reply]

In this case it's just about paying attention to the signs of the exponents.

m^{-2}={\frac {1}{m^{2}}}

and

{\frac {1}{m^{-3}}}=m^{3}

, so

{\frac {m\cdot m^{-2}}{m^{-3}}}=m\cdot m^{-2}\cdot {\frac {1}{m^{-3}}}=m\cdot {\frac {1}{m^{2}}}\cdot m^{3}={\frac {m\cdot m^{3}}{m^{2}}}={\frac {m^{4}}{m^{2}}}=m^{2}

.

Of course, this can be done with many less steps once you get the hang of it. -- Meni Rosenfeld (talk) 14:02, 17 December 2010 (UTC)[reply]

Another way to do it is to add all the (meter exponent) numerators together and subtract all the (meter exponent) denominators, being careful to maintain the signs:

(+1) + (-2) - (-3) = 1 - 2 + 3 = 2

So, this gives us m² (in the numerator). StuRat (talk) 17:59, 17 December 2010 (UTC)[reply]

You've both been really really helpful, thank you so much! Just what I needed. Phantasten (talk) 17:42, 18 December 2010 (UTC)[reply]

You're quite welcome. StuRat (talk) 05:01, 21 December 2010 (UTC)[reply]

Resolved

Rational points on a circle[edit]

I am solving a question which goes like this: Maximum number of points with rational co-ordinates on a circle whose center is $({\sqrt {3}},0)$ is (a) 1, (b) 2, (c) 4, (d) ∞. This question does not seem complete to me. Do we also need some information on the radius whether it is rational or irrational, or can we say about such points without knowing anything about radius. Thanks. - DSachan (talk) 23:02, 17 December 2010 (UTC)[reply]

That's why the word "maximum" is in the question. You're being asked for the maximum over all possible radii. Algebraist 23:07, 17 December 2010 (UTC)[reply]

Oh wow, very nice question. Thanks, I got the answer 2. - DSachan (talk) 00:45, 18 December 2010 (UTC)[reply]