So this was a multi-part group work problem. I have gotten through most of it and am down to the very last part.

Here it is, including some info found along the way through the problem:

Marina del Rey is connected to the ocean by a channel. About twice a day the ocean level rises and falls creating tides. This problem will relate the water level of a marina, $x(t)$, with that of the water level of the ocean, $y(t)$.

The differential equation that models the system is $x'+kx=ky$ that is, the rate of change of the water level in the marina is proportional to the difference between the water level in the ocean and the water level in the bay (I think this is a typo and my professor meant marina)

We are told that $y(t)=cos(\omega t)$, as well that we can assume tide is high every $4\pi$ hours. We are asked to find $\omega$, I assume then that $\omega$ is $1/2$ if $cos$ is to have a max value every multiple of $t=4\pi$ hours as $cos$ maxes at a value of $cos(Integer*2\pi)=1$ if $t=integer*4\pi$ then $\omega=1/2$ for this to be true. Or is my intuition flawed here? Perhaps there is some formulaic method to finding this $\omega$?

We are also given the complex equation $z'+kz=ke^{it/2}$, from this we find the steady-state part of the complex-valued solution to the equation, which I found to be $z_{complex}=(ke^{it/2})/((i/2)+k)$ we are told the real part of this solution, or $(k^{2}\cos(t/2)+(k/2)\sin(t/2))/(k^{2}+(1/4))$ is then the steady-state part of the solution to the original differential equation, $x'+kx=ky$. That is, $x(t)=(k^{2}\cos(t/2)+(k/2)\sin(t/2))/(k^{2}+(1/4))$. I believe that I did everything here correct but would appreciate some double checking.

From here, we are told to assume that the water level in the marina, $x(t)$, has a max height of $1/\sqrt{2}$.

Now here is where I am really stuck.

We are asked to use this info to solve for $k$ and $\phi$ to write the solution in both forms, $Acos(\omega t)+Bsin(\omega t)$ and $Rcos(\omega t-\phi)$ with $A^2+B^2=R^2$ and $\phi=arctan(B/A)$ I am really lost here and have tried manipulating the equation in various ways to solve for $k$ and nothing is clicking. Any help out there?


Rewrite as $$ x(t)=\frac{k}{\sqrt{k^2+\frac14}}\left(\frac{k}{\sqrt{k^2+\frac14}}\cos(t/2)+\frac{\frac12}{\sqrt{k^2+\frac14}}\sin(t/2)\right) $$ where now the inner coefficient pair is a point on the unit circle.

  • $\begingroup$ This then gets equated to the max value of $x(t)=1/\sqrt{2}$? Do I assume this occurs at multiples of $t=4\pi$, as was the case with $y(t)$? Leaving me with $x(4\pi)=k/(k^2+1/4)=1/\sqrt{2}$? Or am I getting ahead of myself. I think I am missing the logic behind using the unit circle point to help me find $k$, I'm sorry. $\endgroup$ – Joeseph Espinosa Sep 25 '16 at 22:04

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