Complex equation (for a 9th grader :>)

[Avathacis]

Warning: This is mixed with physics, though all the physics work is done. Only formatting the equation is needed.

Homework Statement

L is needed from this equation:
L-l= 1/2 a(T-τ)²
Given:
t
τ
l

Homework Equations

T=√(2L/a)
a=2l/t²

The Attempt at a Solution

L= 1/[STRIKE]2[/STRIKE] [STRIKE]2[/STRIKE]l/t² (([STRIKE]2[/STRIKE]Lt²)/[STRIKE]2[/STRIKE]l-2*√((2Lt²)/2l)*τ+τ² ) + l


(the 2's at 1/2 2l/t^2 and 2lt^2/2l are removed right?)
Completely lost here. I think it would be better to attach a picture no?

 

[tiny-tim]

Welcome to PF!

Hi Avathacis ! Welcome to PF!

Sorry, but that's really difficult to read … can you type it again, putting in spaces, and using the X² tag just above the Reply box (for the ²)?

 

[Avathacis]

How does it look? :> I can still attach a screenshot of MS Word.

 

[tiny-tim]

Hi Avathacis!

L= 1/[STRIKE]2[/STRIKE] [STRIKE]2[/STRIKE]l/t² (([STRIKE]2[/STRIKE]Lt²)/[STRIKE]2[/STRIKE]l-2*√((2Lt²)/2l)*τ*τ² )

ah, that's better!

(i assume that last * is supposed to be a + ? )

that's ok (and yes, those 2s do cancel), but

i] where's the "-l" gone?

ii] it's probably easier to leave the ()² unexpanded.

ok, fix that, and then I suggest you get rid of the fractions by multiplying both sides of the equation by powers of t and l.

(after that, I'm not sure … it looks as if it's going to turn into a nasty quartic equation, but let's wait and see if something easy comes to light)

 

[Avathacis]

Hi Avathacis!


ah, that's better!

(i assume that last * is supposed to be a + ? )

that's ok (and yes, those 2s do cancel), but

i] where's the "-l" gone?

ii] it's probably easier to leave the ()² unexpanded.

ok, fix that, and then I suggest you get rid of the fractions by multiplying both sides of the equation by powers of t and l.

(after that, I'm not sure … it looks as if it's going to turn into a nasty quartic equation, but let's wait and see if something easy comes to light)

Err, what -l? Nvm, got your point. Editing.
Btw, thanks a lot for the help. (I fixed the * :p)

 

[Mark44]

Warning: This is mixed with physics, though all the physics work is done. Only formatting the equation is needed.

Homework Statement

L is needed from this equation:
L-l= 1/2 a(T-τ)²
Given:
t
τ
l

Homework Equations

T=√(2L/a)
a=2l/t²

The Attempt at a Solution

L= 1/[STRIKE]2[/STRIKE] [STRIKE]2[/STRIKE]l/t² (([STRIKE]2[/STRIKE]Lt²)/[STRIKE]2[/STRIKE]l-2*√((2Lt²)/2l)*τ*τ² )


(the 2's at 1/2 2l/t^2 and 2lt^2/2l are removed right?)
Completely lost here. I think it would be better to attach a picture no?

This is somewhat confusing. Is l lower case L? Also, you have T, t, and τ (Greek letter tau). Are all three of them needed?

Also, I don't see the need to replace a with 2l/t².

The original equation can be rewritten as
[tex]L - l = \frac{a(T - \tau)^2}{2} = \frac{a(T^2 - 2\tau T + \tau^2)}{2} = \frac{a(2L/a - 2\tau \sqrt{2L/a} + \tau^2}{2}[/tex]

[tex]\Rightarrow L - l = \frac{a(2L/a - 2\tau \sqrt{2L/a} + \tau^2}{2}[/tex]

This equation is quadratic in form. If you let u^2 = L, it will be a quadratic.

 

[Avathacis]

This is somewhat confusing. Is l lower case L? Also, you have T, t, and τ (Greek letter tau). Are all three of them needed?

Also, I don't see the need to replace a with 2l/t².

The original equation can be rewritten as
[tex]L - l = \frac{a(T - \tau)^2}{2} = \frac{a(T^2 - 2\tau T + \tau^2)}{2} = \frac{a(2L/a - 2\tau \sqrt{2L/a} + \tau^2}{2}[/tex]

[tex]\Rightarrow L - l = \frac{a(2L/a - 2\tau \sqrt{2L/a} + \tau^2}{2}[/tex]

This equation is quadratic in form. If you let u^2 = L, it will be a quadratic.

I) Yes all 3 are needed. There are 3 times.
II) l is lower case L.
III) a is unknown, so it is swapped for 2 known ..err.. letters. It is the same with T.
Also what are we going to do with -l?

 

[Mark44]

But there is no need to replace a that I can see, since it doesn't involve L. If you need to replace it, you can do that later, after you have solved for L.

If you make the substitution I suggested, you will have a u^2 term, a u term and everything else. l (ell) will be in amongst the "everything else."

 

[Avathacis]

But there is no need to replace a that I can see, since it doesn't involve L. If you need to replace it, you can do that later, after you have solved for L.

If you make the substitution I suggested, you will have a u^2 term, a u term and everything else. l (ell) will be in amongst the "everything else."

I still can't find a way to solve it with your equation :(. Sorry.

 

[Avathacis]

Also, my attempt at the solution might be wrong. Try redoing it yourselves :p.

 

[Mark44]

Here's the equation at the end of post 6.
[tex]L - l = \frac{a(2L/a - 2\tau \sqrt{2L/a} + \tau^2)}{2}[/tex]
[tex]\Rightarrow 2L - 2l = 2L - 2a\tau \sqrt{2L/a} + a\tau^2[/tex]

I didn't realize it before but the 2L terms drop out. Move the term with L in it to one side, and move everything else to the other side. Square both sides to get L out of the radical.

 

[Avathacis]

Here's the equation at the end of post 6.
[tex]L - l = \frac{a(2L/a - 2\tau \sqrt{2L/a} + \tau^2)}{2}[/tex]
[tex]\Rightarrow 2L - 2l = 2L - 2a\tau \sqrt{2L/a} + a\tau^2[/tex]

I didn't realize it before but the 2L terms drop out. Move the term with L in it to one side, and move everything else to the other side. Square both sides to get L out of the radical.

Oh, right! This is the "answer". It's pretty much done but i will need to reformat it again after i make a=2l/t².