Testwiki:Reference desk/Archives/Mathematics/2008 June 16

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June 16

trigonometry

How can we calculate sin 18^o —Preceding unsigned comment added by Rohit max (talk • contribs) 05:36, 16 June 2008 (UTC)

See Exact trigonometric constants#How can the trigonometric values for sine and cosine be calculated?. --Prestidigitator (talk) 07:47, 16 June 2008 (UTC)

Mensuration

How can we calculate the volume of prism? And of Pyramid? —Preceding unsigned comment added by Rohit max (talk • contribs) 05:45, 16 June 2008 (UTC)

Try checking our articles on prisms and pyramids. Maelin (Talk | Contribs) 05:54, 16 June 2008 (UTC)

Image of a set

I feel myself very dumb, but... I have a R^2 set S whose image set Z in R^2 should be illustrated. The image set Z consists of vectors z1 and z2 that are represented by functions f1(x1,x2) and f2(x1,x2), respectively. I know how to illustrate S, but how to illustrate Z? How do I 'convert' the vectors in x1,x2 coordinates into vectors in z1,z2 coordinates? —Preceding unsigned comment added by 82.130.19.65 (talk) 09:26, 16 June 2008 (UTC)

You have a mapping $\theta :S\to Z:(x_1,x_2)\to (f_1(x_1,x_2),f_2(x_1,x_2))=(z_1,z_2)$. To find the z-vectors, you just need to know the x-vectors and the functions f_1 and f_2. -mattbuck (Talk) 10:04, 16 June 2008 (UTC)

Thanks mattbuck, that helps. But still, say I have a set S constrained by $a_1{x}_1^2+b_1{x}_2+c_1\le 0$ and $a_2{x}_1+b_2{x}_2+c_2\le 0$, and the functions are determined by $f_1=d_1{x}_1+h_1{x}_2$ and $f_2=d_2{x}_1+h_2{x}_2^2$, where the parameters can assume values from $R$. I just don't "get" how to relate Z to S. 82.130.19.65 (talk) 06:09, 17 June 2008 (UTC)

antiderivative.. help!

Anyone feel like having a crack at:

ʃe^kx/x² dx

or more specifically the function I started out with ʃe^kx(1-a²/x²) dx

(More specifically the integral between z and infinity where z is positive real, and z=a, k will be negative..).. Also the second (double) integral as well if you get the first..

Hints also appreciated. (87.102.86.73 (talk) 16:44, 16 June 2008 (UTC)

Integration by parts will reduce it to the exponential integral. Oded (talk) 17:01, 16 June 2008 (UTC)

is Exponential_integral#Convergent_series (these two covergent series) the best I can expect in terms of ease of calculation..? If so it looks like I would do 'just as well' as expressing the original function as a power series, integrating, and then evaluating (in terms of 'computer arithmetic time'). Thanks. Are there any other options?87.102.86.73 (talk) 18:58, 16 June 2008 (UTC)

I have to admit I was hoping for a clever substitution I hadn't thought of that would reduce the integral to a sumless form.. such is optimism.87.102.86.73 (talk) 19:00, 16 June 2008 (UTC)

The advantage in converting it to Ei is that Ei is probably better studied, in terms of its relationships to other functions. Perhaps there is also knowledge out there about efficient ways to calculate it and even ready-made routines. Oded (talk) 19:22, 16 June 2008 (UTC)

Percentages help

70% of my math mark is 68%. What if the other 30% (my final exam) is a big fat 0%? How would I put these together to find out my final mark?--Richard (Talk - Contribs) 22:46, 16 June 2008 (UTC)

Well, I assume you mean that you have scored 68% average over the first 70% of the course, rather than having 68% from 70% possible so far. It's pretty simple, multiply 0.7 by 0.68 to get your current percentage mark, then add on 0.3 times your exam mark. -mattbuck (Talk) 23:20, 16 June 2008 (UTC)

(ec) What you need to do is take a weighted average of the two scores, 68% and 0%, using respective weights of 70% and 30%. So you have 68%*70% + 0%*30% = 68% * 0.7 = 47.6%. Suppose the pass mark were 50%, then you could find out what you need to get in the final to pass - you need an extra 2.4%, and dividing that by 0.3 gives a total of 8 marks needed. (If you're interested, see what your maximum mark is, assuming you ace the final.) Confusing Manifestation(Say hi!) 23:22, 16 June 2008 (UTC)

That works, ConMan. Thanks. If I get a 0 on the exam, I'll have a 47.6%, though if I ace it, I'll get a 77.6%. Hopefully I'll get over 50 to pass... Thanks for your help guys.--Richard (Talk - Contribs) 00:21, 17 June 2008 (UTC)