Question

Please help. The question is attached

Answer 1

That \(\frac{Y}{dX}\) doesn't make any sense. I think they might have meant \(\frac{Y}{X}\).

Sorry, that's all I can help. Is this a Calc 3 topic?

Answer 2

@KevinOrr actually that \[\frac{ Y }{ dX }\] is correct because \[K_w=dX\]
Its a differential equation type. Know anyone who might be able to help?

Answer 3

@dan815 Hey will u help me please with this question?

Answer 4

But just having one differential doesn't mean anything without another. \(dx = 5t \, dt\) sure, but \(dx = 5t\)?

Answer 5

My prof said that u can substitute the dX for K_w.
But am just stuck on how to nondimensionalize it @KevinOrr

Answer 6

So from reading http://en.wikipedia.org/wiki/Nondimensionalization#Nondimensionalization_steps, it doesn't seem that bad really. Let's do the krill growth model, and then see if you can do the whale growth model

Replace \(X\), \(Y\), and \(t\) with their defined scaled quantities:

\[
\begin{align}
\frac{dX}{dt} &= aX\left(1-\frac{X}{K}\right) - bXY \\
\frac{X^*}{t^*} \frac{dx}{d\tau} &= axX^{*}\left(1-\frac{xX^*}{K}\right) - b xX^* yY^* \\
\frac{dx}{d\tau} &= axt^{*}\left(1-\frac{xX^*}{K}\right) - bt^*x yY^*
\end{align}
\]

Now, you know that

\[
\frac{xX^*}{K} = x \Rightarrow X^* = K \\
\therefore \frac{dx}{d\tau} = axt^{*}(1-x) - bt^*xyY^*
\]

You also know that

\[
bt^*xyY^* = xy \Rightarrow t^* Y^* = \frac{1}{b}
\]

Now you have the krill growth model partially worked out. Do the same for the whale growth model, which will give you \(Y^∗\). However, it will give it to you in terms of \(K_w\). What you're supposed to do with that, idk.