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# parametric equations - how do I equate coefficients Watch

1. I have the equation A(1+t^2) + (Bt+c)t =1

How do I work out the value of A , B and C?
Below is how I try working it out, what am i doing wrong?

firstly, i multiply out the brackets so I get:
A + At^2 + Bt^2 +ct

I then collect constants and terms:
A = 1
At^2 + Bt^2 + tc = 0

now what do i do / where did I go wrong?
2. (Original post by 2cool)
I have the equation A(1+t^2) + (Bt+c)t

This is not an equation?
3. (Original post by steve2005)
This is not an equation?
I got it from the original parametric equation:

1/ t(1+t^2) = A/t + (Bt+c)/1+t^2
4. (Original post by 2cool)
I have the equation A(1+t^2) + (Bt+c)t =1

Sorry but is this an equation or an identity

If it is an equation then you cannot equate coefficients

Can you post the actual question?
5. (Original post by 2cool)
I got it from the original parametric equation:

1/ t(1+t^2) = A/t + (Bt+c)/1+t^2

Do you mean

6. (Original post by 2cool)
I got it from the original parametric equation:

1/ t(1+t^2) = A/t + (Bt+c)/1+t^2
Hang on

Are you trying to do partial fractions?
7. In wich case

Equating coefficients

(A+B) = 0
C = 0
A = 1
8. (Original post by steve2005)
Do you mean

i meant the first 1 , A sorry about my poor equation writing !

(Original post by TenOfThem)
Hang on

Are you trying to do partial fractions?
yes

(Original post by TenOfThem)
In wich case

Equating coefficients

(A+B) = 0
C = 0
A = 1

why does A+B have to = 0 ?
9. (Original post by 2cool)
why does A+B have to = 0 ?
How many t^2 do you have on the LHS
10. (Original post by TenOfThem)
How many t^2 do you have on the LHS
one ....
11. (Original post by 2cool)
one ....
Where?

1 = (A+B)t^2 + Ct + A
12. ahh there are 2, because multiplied out gets

At^2 + Bt^2 right?
13. (Original post by 2cool)
ahh there are 2, because multiplied out gets

At^2 + Bt^2 right?
No

How many t^2 do you have on the Left Hand Side
14. (Original post by TenOfThem)
No

How many t^2 do you have on the Left Hand Side

none, you have just the constant 1
15. none

exactly

so you also have none on the RHS

that is why A+B = 0
16. ahhh yes thanks... however something im a bit confused about in my head, maybe this is a wrong explanation

but

x = x^2 if x = 1

so wouldnt this mean that you dont have to have t^2 on both sides?

edit: I think because you have a t^2 + Ct + A , this would be like x^2 = x + 1 which isnt solvable which answers my own question
17. When equating coefficients you have an IDENTITY not an EQUATION that means that it holds true for all values of t
18. (Original post by TenOfThem)
When equating coefficients you have an IDENTITY not an EQUATION that means that it holds true for all values of t
ahhh ! ! thank you very much for the help ! can only give one +rep !
19. (Original post by 2cool)
ahhh ! ! thank you very much for the help ! can only give one +rep !
No Problem
20. Y u no learn to latex

Updated: April 6, 2012
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