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# Is it possible to make t the subject of x=t^3-8t? watch

1. in the title

Is it possible to make t the subject of x=t^3-8t ?
2. (Original post by 10614LW)
in the title

Is it possible to make t the subject of x=t^3-8t ?
No
3. (Original post by TenOfThem)
No
do you mind explaining why
4. (Original post by 10614LW)
do you mind explaining why
The same vlue of x can give multiple values of t, so you cannot invert it.
5. (Original post by 10614LW)
do you mind explaining why
It is not a one to one function

As James said
6. (Original post by 10614LW)
in the title

Is it possible to make t the subject of x=t^3-8t ?
Why do you want to do this?
Maybe we can help?
7. (Original post by 10614LW)
in the title

Is it possible to make t the subject of x=t^3-8t ?
No, just because t is a parameter of the equation where x is defined by this parameter. In other words, the value of x can be determined by certain values of parameter t.
8. (Original post by Kallisto)
No, just because t is a parameter of the equation where x is defined by this parameter. In other words, the value of x can be determined by certain values of parameter t.
That is not the reason

If x=t+3 then we could rearrange to get t=x-3
9. (Original post by MathMeister)
Why do you want to do this?
Maybe we can help?
part c of http://imgur.com/D06EjZb I was planning to rearrange the parametric equations and with the equation of the line (given in part b),
then solve the simultaneous equations to find the co-ord of B
10. (Original post by 10614LW)
part c of http://imgur.com/D06EjZb I was planning to rearrange the parametric equations and with the equation of the line (given in part b),
then solve the simultaneous equations to find the co-ord of B
There is no need for that

You can find dx/dy and then you have the gradient at A .... Since you already have the co-ordinates you then use the gradient and point to find the line

Then to find B you simply substitute for both x and y in your line .... Then solve for t
11. (Original post by TenOfThem)
There is no need for that

You can find dx/dy and then you have the gradient at A .... Since you already have the co-ordinates you then use the gradient and point to find the line

Then to find B you simply substitute for both x and y in your line .... Then solve for t
are you talking about part a and b? if so I done them a I got (7,1) it just c I don't get.

how to I get (x,y) from a cubic equation 2t^3-16t-10t^2-9=0
12. (Original post by 10614LW)
are you talking about part a and b? if so I done them a I got (7,1) it just c I don't get.

how to I get (x,y) from a cubic equation 2t^3-16t-10t^2-9=0
You solve that for t and use to to find x and y

You already know one value of t from A so that should make it easier to factorise

Edit ... You have 10t^2 and it should be 5t^2
13. You could rearrange it to,

T^3-8t-x=0

And then apply the horrible cubic formula with coefficients:

a=1
b=0
c=-8
d=-x

But that seems a bit excessive.

Though I dont feel comfortable with the notion of taking a function as a coefficient. It feels a bit iffy.
14. (Original post by TenOfThem)
You solve that for t and use to to find x and y

You already know one value of t from A so that should make it easier to factorise

I got t as 4.5 and -1

so do I sub it in?

y=(4.5)^3 - 8(4.5) = 55.125
and
x= (4.5)^2 = 20.25

it looks wrong tho?
15. (Original post by 10614LW)
I got t as 4.5 and -1

so do I sub it in?

y=(4.5)^3 - 8(4.5) = 55.125
and
x= (4.5)^2 = 20.25

it looks wrong tho?
Assuming 4.5 is correct then what looks wrong ?

Edit 4.5 is correct
16. (Original post by TenOfThem)
Assuming 4.5 is correct then what looks wrong ?
for co-ords it's just looks very large compared to A(7,1).

So the method is right then? I'll go back and check my workings.

Thank for the help
17. (Original post by james22)
Solve it for x=1 then, I bet that you cannot.
You are being trolled James. This poster has now been banned for this behaviour.
18. I have not quite read all the comments but I am surprised by some of them...

ANY CUBIC can be rearranged to give at least one rearrangement.

The OP in not talking about inverting a function; merely rearranging.

There is a standard method that can be applied by hand using the cosine triple angle formula or the cosh triple angle formula, depending on the discriminant of the cubic.

PS any decent computer algebra system should be able to rearrange this in fractions of a second.
19. (Original post by Mr M)
You are being trolled James. This poster has now been banned for this behaviour.
That seems a bit much.

20. (Original post by Xin Xang)
That seems a bit much.

He's been at it all day.

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