HELP -->A few questions. (1 Viewer)

[who_loves_maths]

Originally Posted by Slide_Rule
So taking into account the division by two earlier, and the limits:
4pi/sqrt(3) [arctan(tan(u/2)/sqrt(3))] from 5pi/2 to 3pi/2
= 4pi/sqrt(3)*[pi/6+pi/6]
=sqrt3.pi^2/27

I think that was too long. I think I made a mistake.

i hate to be the one looking for mistakes, but... how does 4pi/sqrt3*(pi/3) = sqrt3*(pi^2/27)? (in bold)

i think you mean 4pi/sqrt3*(pi/3) = (4sqrt3/9)*(pi^2) ? = [approx.] 7.60

 

[yrtherenonames]

I stand by my 7.595

 

[Slidey]

i hate to be the one looking for mistakes, but... how does 4pi/sqrt3*(pi/3) = sqrt3*(pi^2/27)? (in bold)

i think you mean 4pi/sqrt3*(pi/3) = (4sqrt3/9)*(pi^2) ? = [approx.] 7.60

For some reason I confused my binary operators (pi/6+pi/6 = pi^2/36). How anticlimactic.

 

[haboozin]

also, just adding to my last post, i think you have the term (in bold above) in that equation wrong: i think it should be 2@t^3, and not just 2t^3.

yea it is sorry....

 

[haboozin]

wtf is an arctan slide_rule?

is that inverse tan

tan^-1@

 

[Trev]

wtf is an arctan slide_rule?

is that inverse tan

tan^-1@

Yes, that is what he means by arctan.

 

[thunderdax]

I just realised, question 1 was in my last 4u test, except they lead you into it by first sketching y=x/(1-x²).

 

[haboozin]

yea because your teacher copied the catholic trial 2003...