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The equations I have are

$\phi=(-ag/\omega)\cos(kx)\sin(\omega t)e^{kz}$


$\eta=a\cos(kx)\cos(\omega t)$

I know that $d\phi/dz=d\eta/dt$

but when I partially differentiate and rearrange I get

$gk e^{kz}= \omega^2$ and I don't know how to get rid of the exponential function.

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How does $h$ enter your equations? Do the parameters depend on $h$? – Johan Nov 26 '12 at 13:23
The original equation is $\phi=-ag/\omega cos(kx)sin(\omega t) (cosh(k[z+h])/ cosh(kH))$ but as h tends to inifity $\phi$ simplifies to the equation in the question. – Adam Nov 26 '12 at 13:37
Do you know why the original equation simplifies as h tends to infinity? How do I show that the original equations simplifies as h tends to infinity? – user52290 Dec 8 '12 at 23:41

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