Express h in terms of D

[whatsoever]

1. The problem statement, all variables and given/known data

http://img88.imageshack.us/img88/1679/problemhup.png

2. Relevant equations



3. The attempt at a solution
http://img535.imageshack.us/img535/8075/solutionz.png
That is how far i have got, using that sinh(x)=1/2(e^(x)-e^(-x))
I have to express h in terms of D

[Mentallic]

Firstly let

\sqrt{\frac{2.09\cdot 10^6}{D}}=x

just to make things clearer and easier to write out. Now we have

0.5=\frac{e^{xh-10x}-e^{10x-xh}}{e^{xh}-e^{-xh}}

So we need to find h in terms of x (which is in terms of D, we can substitute back at the end)
Multiplying through by the denominator of the fraction, and then by 2e^{xh}[/tex] remember that e^x\cdot e^{-x}=1 and e^{x}\cdot e^{x}=e^{2x}
So we now have

[tex]e^{2xh}-1=2e^{2xh-10x}-2e^{10x}

and from here just rearrange, factorize out the exponents with h present, and solve from there using logs and such.

[whatsoever]

Seems like i have made a mistake sinh(x+y)=sinh(x).cosh(y)+sinh(y).cosh(x)
considering that and your help i've got this
http://img571.imageshack.us/img571/6678/probp.png
but i have no idea what to do next

[Mentallic]

Seems like i have made a mistake sinh(x+y)=sinh(x).cosh(y)+sinh(y).cosh(x)
I haven't studied sinh myself, so I took your word for it. I like that it's similar to the sin(a+b) expansion :wink:

considering that and your help i've got this
http://img571.imageshack.us/img571/6678/probp.png
but i have no idea what to do next

Again multiply through by exh and you can easily simplify things, such as the e^{2xh-10x}e^{-2xh-10x} term

[whatsoever]

I haven't studied sinh myself, so I took your word for it. I like that it's similar to the sin(a+b) expansion :wink:



Again multiply through by exh and you can easily simplify things, such as the e^{2xh-10x}e^{-2xh-10x} term

i've made a mistake when writing it its not e^{2xh-10x}e^{-2xh-10x}, its e^{2xh-10x}+e^{-2xh-10x}

[Mentallic]

Then multiply through by e2xh. You'll get a quadratic in e2xh, and if you can't see it, let u=e2xh and treat other terms such as e-10x as constants, then solve the quadratic in u, then substitute back.