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parametric equations - how do I equate coefficients

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? :frown:
(edited 11 years ago)
Reply 1
Original post by 2cool
I have the equation A(1+t^2) + (Bt+c)t

:frown:



This is not an equation?
Reply 2
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
Reply 3
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?
(edited 11 years ago)
Reply 4
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




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Reply 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


Hang on

Are you trying to do partial fractions?
Reply 6
In wich case

1=A(1+t2)+t(Bt+C)=(A+B)t2+Ct+A1 = A(1+t^2) + t(Bt+C) = (A+B)t^2 + Ct + A

Equating coefficients


(A+B) = 0
C = 0
A = 1
Reply 7
Original post by steve2005
Do you mean




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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

1=A(1+t2)+t(Bt+C)=(A+B)t2+Ct+A1 = A(1+t^2) + t(Bt+C) = (A+B)t^2 + Ct + A

Equating coefficients


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



why does A+B have to = 0 ?
Reply 8
Original post by 2cool

why does A+B have to = 0 ?


How many t^2 do you have on the LHS
Reply 9
Original post by TenOfThem
How many t^2 do you have on the LHS


one ....
Original post by 2cool
one ....


Where?

1 = (A+B)t^2 + Ct + A
(edited 11 years ago)
Reply 11
ahh there are 2, because multiplied out gets

At^2 + Bt^2 right?
(edited 11 years ago)
Original post by 2cool
ahh there are 2, because multiplied out gets

At^2 + Bt^2 right?


No

I will ask again

How many t^2 do you have on the Left Hand Side
Reply 13
Original post by TenOfThem
No

I will ask again

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


damn, talk about not reading th equestion,


none, you have just the constant 1
none

exactly

so you also have none on the RHS

that is why A+B = 0
Reply 15
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
(edited 11 years ago)
When equating coefficients you have an IDENTITY not an EQUATION that means that it holds true for all values of t
Reply 17
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 !
Original post by 2cool
ahhh ! ! thank you very much for the help ! can only give one +rep !


No Problem :smile:
Y u no learn to latex

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