Isolating variable in formula

[Twinbird24]

Like the description says, I need help isolating h from: mgh = k(h-l-L)²/2

I'm almost certain that the quadratic formula needs to be used to isolate h, but I'm not sure how to use it (haven't used it since grade 9).

Thanks!

[timo]

The starting point for the quadratic formula is "something * h² + otherthing * h + rest = 0", so rearrange your equation into this form. A rather obvious first step seems to be multiplying out the (h-l-L)² term, I think.

Edited January 3, 2011 by timo

[Twinbird24]

Thanks for the reply.

I'm not a very mathy person, could you explain this to me more?

Right now I have: k(-l-L)² * h² - mg * h = 0

I don't know if that is good for the quadratic formula or not (or if I rearrange the equation properly).

[swansont]

Since you have a quadratic term for h, that would be a good thing to try.

[timo]

Thanks for the reply.

I'm not a very mathy person, could you explain this to me more?

Right now I have: k(-l-L)² * h² - mg * h = 0

I don't know if that is good for the quadratic formula or not (or if I rearrange the equation properly).

It would be good for using the quadratic formula. Comparing with "something * h² + otherthing * h + rest = 0" which I wrote before you'd find that

something = k(-l-L)²,

otherthing = mg,

rest=0.

But I doubt that your term k(-l-L)² * h² - mg * h = 0 is correct. You should probably show the individual steps how you got from

mgh = k(h-l-L)²/2

to

k(-l-L)² * h² - mg * h = 0.

I am sure someone will be able to point out the mistake you made there. Hint: [math] (h-l-L)^2 \neq (-l-L)^2 \cdot h^2 [/math], for example for h=1, l=0, L=0.

[imatfaal]

Twinstar,

 

As Timo pointed out you have not multiplied out the bracket correctly; unfortunately you cannot just take h² out of the bracket. In the spirit of not giving an answer, perhaps this will help you realise/work out how to go about multiplying out brackets.

we can look at the simple example of all just two terms

(x-y)² this does NOT equal x² -y² this would be making the same as the mistake you made above. The way I would do it in longhand is as follows:

(x-y)² is also written as (x-y)(x-y)

now to get the correct answer we need to multiply each of the terms in the first bracket with each of the terms in last and add the results together. My maths teacher called this FOIL - you will see why

First terms x times x = x²

Outer terms x time -y = -xy

Inner terms -y times x = -yx

Last terms -y times -y = y²

so (x-y)(x-y) = x² -xy -yx +y²

^{obvious -xy is the same as -yz so we can rewrite as}

(x-y)² = (x-y)(x-y) = x² -2xy +y²

^{For your equation you still need to multiply each of the terms in the first bracket with each of the terms in the second bracket - This will give 9 terms, which will then simplify to 6 (because hL is the same as Lh etc.)}

^{Hope this has helped a bit. Come back when you have made a stab at it.}

[Twinbird24]

I can't use FOIL though because there are 3 terms inside the brackets and not just 2. When I expand I get: h² - hl - lh + l² - hL +lL - Lh + lL +L²

There doesn't seem to be anything that cancels out to reduce this to 6 terms.

How do I do this? Thanks!

[imatfaal]

No you cannot use FOIL - but I was illustrating the principle, and you seem to have got the idea. As these are just numbers represented by letters wouldn't you agree that Lh = hL, that lh=hl and that lL=Ll ; and I said simplify not cancel.

 

ie similar to

(x+y+z)² = = x² + xy + xz + yx +y² +yz + zx + zy +z² = x² +y² +z² + 2xy +2xz +2yz

you have to be a tiny bit more careful to get your signs right than the above example - but it seems you have that right so far.

[Twinbird24]

Okay, I rearranged and got this: h² + l² + L² -2lh - 2Lh + 2Ll

so then, the overall equation would be: mgh = k(h² + l² + L² -2lh - 2Lh + 2Ll) / 2

It just seems a lot more confusing now (h is now in 4 different places, with one of the h still being squared), how do I continue?

Thanks!

[timo]

My advice still stands:

The starting point for the quadratic formula is "something * h² + otherthing * h + rest = 0", so rearrange your equation into this form.

- obviously, a term h² is not going to kill you (now). It's what you were expecting, anyways.

- for the rearranging, note that a*h + b*h = (a+b)*h. In other words: bring everything on the left-hand side, then sort the terms by their powers of h (i.e. terms with an h², then those with h, then those without h), then you pretty much have the form I proposed and can start with the messy part of the problem.

[Twinbird24]

Thanks for the reply.

I moved everything to the left, now I have: 2mgh - k(h² + l² + L² -2lh - 2Lh + 2Ll) / 2 = 0

I understand that part about factoring out h from (a*h + b*h) but how do I sort out the terms?

Sorry If I seem really difficult, I understand what your telling me somewhat (really appreciated), but just not sure how to apply it to this specific equation.

Thanks!

Edited January 7, 2011 by Twinbird24

[imatfaal]

OK Tbird

 

You need to start again from your original formula mgh = k(h-l-L)²/2 - cos you have made a couple of errors (post#3 had no factors of two and post#11 has two )

 

1. Are you sure this is the correct starting point - I haven't checked its logic - I presume you are happy with it

2. Multiply through by two

3. multiply out the three way bracket (you have done this already in #11)

4. multiply the contents of your bracket by the k outside

5. group together the h² terms (should be only one), the h terms (should be three), and the no h terms (should be three)

6. Is there a common factor to the entire equation - if there is factor it out. (good form but doesnt apply here)

7. take the h² term outside a bracket and put its coefficients within the bracket (good form but doesnt apply here)

8. take the h term outside a bracket and put its coefficients within the bracket

 

You should now have an equation that look like this

0 = something.h2 - 2h(three somethings) + (three other somethings)

which is basically your standard quadratic

0 = ax² +bx +c

[Twinbird24]

Thanks for the reply. I asked a student in my class about this question, and the l and L can be combined into one variable (which I will call A), so the original equations looks like this: mgh = k(h-A)²/2 (I can use FOIL now). Here is what I did so far, I'm still stuck though (can't get the 0 = ax2 +bx +c format).

1) mgh = k(h-A)²/2

2) 2mgh = k(h-A)²

3) 2mgh = k(h² - hA - hA + A²)

4) 2mgh = k(h² - 2hA + A²)

5) 2mgh = kh² - k2Ah + kA²

 

If I ignore the 2mgh on the left side then I would have the 0 = ax2 +bx +c format, but now what do I do with the 2mgh on the left side? Thanks!

Edited January 13, 2011 by Twinbird24

[Fuzzwood]

Divide every term by h.

[imatfaal]

If you want to solve for h - which I seem to remember you do - I wouldn't divide by h You want a zero at one side of a quadratic in order to solve it - and the obvious way of doing that is to subtract 2mgh from both sides. You will end up with a very complicated h term - but then that was always gonna happen.

[Twinbird24]

Thanks for your help!

so the final equation would look like this, correct?

0 = kh² - 4mgkAh + kA²

[imatfaal]

Nope.

 

-2akh -2mgh DOES NOT equal -4mgkAh

 

You need to take out the common factors and put the rest in a bracket

Edited January 14, 2011 by imatfaal

[Twinbird24]

Thanks.

I have now: 0 = kh² + h(-2kA-2mg) + kA²

Is this good?

[Fuzzwood]

You can also take the 2 outside, and then you can make out h:

 

h = {2kAmg+/-[(2kAmg)²-4k²A²]}/2k

[imatfaal]

Twinbird

 

I would personally take write h term as ... - h(2kA+2mg)... Otherwise I think you have it. I havent had a look at your other thread on the derivation of the equation - I hope that bit is right after the amount of hard grind you have had on this side. Good Luck.

 

Fuzzwood

 

That looks like a the quadratic equation solution - however the coefficient of the h term is -2mg-2kA (ie two negative portions). and more importantly where is the square root term - the equation as I know it would be

 

[-b +/- (b^2 - 4ac)^1/2]/2a

 

in latex this looks nicer

 

[math] \frac{-b\pm\sqrt{b^2-4ac}}{2a}[/math]

[Twinbird24]

Thanks. The derivation is correct, just needed to get the equation into the quadratic form.