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

10614LW

3

in the title

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

Reply 1

TenOfThem

18

Original post by 10614LW

in the title

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

No

Reply 2

10614LW

OP

3

Original post by TenOfThem

No

do you mind explaining why

Reply 3

james22

16

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.

Reply 4

TenOfThem

18

Original post by 10614LW

do you mind explaining why

It is not a one to one function

As James said

(edited 9 years ago)

Reply 5

MathMeister

10

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?

Reply 6

Kallisto

Entertainment Forum Helper

22

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.

Reply 7

TenOfThem

18

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

Reply 8

10614LW

OP

3

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

(edited 9 years ago)

Reply 9

TenOfThem

18

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

(edited 9 years ago)

Reply 10

10614LW

OP

3

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

(edited 9 years ago)

Reply 11

TenOfThem

18

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

(edited 9 years ago)

Reply 12

Xin Xang

14

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.

(edited 9 years ago)

Reply 13

10614LW

OP

3

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?

Reply 14

TenOfThem

18

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

(edited 9 years ago)

Reply 15

10614LW

OP

3

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

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.

Reply 17

TeeEm

19

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.