Maths/Physics help

[astarmathsandphysics]

That is what I did to get two integrals.

[omarsubei]

Ya I looked at that. I don't understand their method, but I can see that in your method, you put the same equation twice, you need to put one of them as +square root... and the other as -square root..., if you see what I mean

[omarsubei]

I don' get how you went from step one to two, how'd you put the integrals together?

[astarmathsandphysics]

You are right. I corrected my mistake.

[astarmathsandphysics]

I don' get how you went from step one to two, how'd you put the integrals together?

Exapnad the brackets and simplify. You can just add the integrands. The squared terms cancel and for one term I get--=+

[radam]

Hey;

I'd appreciate some help with this question;

(from the previous subsection b is an imaginary root of the equation z^5-1=0 and b^4+b^3+b^2+b+1=0)
________________________________________
Q 13, PART B, MATH HL NOV 08

c) If u=b+b^4 and v= b^2+b^3, show that

i) u+v=uv=-1
ii) u-v= sqrt(5) , given that u-v>0
_______________________________________

for i) Proving that u+v=-1 is not difficult... but i have no idea how to prove that uv=-1, or for ii) how u-v is sqrt(5).

Help would be greatly appreciated

[astarmathsandphysics]

Hey;

I'd appreciate some help with this question;

(from the previous subsection b is an imaginary root of the equation z^5-1=0 and b^4+b^3+b^2+b+1=0)
________________________________________
Q 13, PART B, MATH HL NOV 08

c) If u=b+b^4 and v= b^2+b^3, show that

i) u+v=uv=-1
ii) u-v= sqrt(5) , given that u-v>0


_______________________________________

for i) Proving that u+v=-1 is not difficult... but i have no idea how to prove that uv=-1, or for ii) how u-v is sqrt(5).

Help would be greatly appreciated

i)uv=(b+b^4)(b^2+b^3)=b^3+b^4+b^6+b^7 now use b^5=1 ie B^6=b and B^7=b^2 so we have b^3+b^4+b +b^2=-1 too.

I will answer the second part as soon as I come back, but I have to go RIGHT NOW and teach some maths.

ii)(u-v)^2=u^2-2uv+v^2=(b+b^4)^2-2(b+b^4)(b^2+b^3)+(b^2+b^3)^2=b^2+2b^5+b^8-2b^3-2b^4-2b^6-2b^7_+b^4+2b^5+b^6
b^5=1 so we have b^2+2+b^3-2b^3-2b^4-2b-2b^2+b^4+b=(b^2+b^3-2b^3-2b^4-2b-2b^2+b^4+b)+4
=-(b^3+b^4+b +b^2)+4=5

[radam]

Ah thanks!

I forgot that b^5=1;

I was thinking about the question with that bit of insight; and i was wondering if my method below is correct:

(u-v) is the same as sqroot (u-v)^2 (since u-v>0)

(U-v)=u^2-2uv+v^2= b^8+2b^5+b^2+2+b^4+2b^5+b^6

since uv=-1, and b^5=1

This gives b^4+b^3+b^2+b-6= 5

Hence  (u-v)^2= 5

u-v= sqrt(5)

The answer seems correct, but I'm wary of squaring as it normally adds complications to the matter... is there an easier and less time consuming way to do it?

Again thanks a lot for your help! With my math exam 2 weeks away, its a life-saver!

[Clerk]

Hi. I'm a standard level physics student. Does anyone know if I have to know how to "Solve problems using the equation of state of an ideal gas". There are some past papers exercises, but I think that this part of the syllabus is now only for high level. Please help me!

[astarmathsandphysics]

Ah thanks!

I forgot that b^5=1;

I was thinking about the question with that bit of insight; and i was wondering if my method below is correct:

(u-v) is the same as sqroot (u-v)^2 (since u-v>0)

(U-v)=u^2-2uv+v^2= b^8+2b^5+b^2+2+b^4+2b^5+b^6

since uv=-1, and b^5=1

This gives b^4+b^3+b^2+b-6= 5

Hence  (u-v)^2= 5

u-v= sqrt(5)

The answer seems correct, but I'm wary of squaring as it normally adds complications to the matter... is there an easier and less time consuming way to do it?

Again thanks a lot for your help! With my math exam 2 weeks away, its a life-saver!

I can't think of a simpler way to do it.

[astarmathsandphysics]

Post your questions and I will answer asap

[IGSTUDENT]

Can u pls suggest a good book for IB Higher maths ?Aslo for Chemisty HL?

[omega007]

This is going to help you with the books.

http://en.wikibooks.org/wiki/IB_Textbook_Reviews

[IGSTUDENT]

Thanks a lot.Will have a look.
All the same some of you who are already doing it..any suggestions?

[IGSTUDENT]

Thanks a lot.Will have a look.
All the same some of you who are already doing it..any suggestions?