finding equation - model the growth of fish

[cherrybomb]

Find the equation for the length of the fish as a function of time if we know that the length of a newly hatched specimen in 0.1 mm and a mature specimen has length 670 mm. Any help or guide to getting the answer would be greatly appreciated, thank you in advanced

[KipIngram]

We have to know something about the growth profile. How long does it take the fish to mature from hatchling to adult? Is the growth profile exponential? Are there "growth spurts"? Etc.

With absolutely no input of that sort at all, I'd *guess* a 1-e^-kt sort of curve and do a fit. But I sure wouldn't bet money on it being right.

[cherrybomb]

We were also given this before the questions:

Ludwig von Bertalanfft (an Austrian biologist) introduced the most widely used method to model the growth of fish. His theory is that the length of a fish specimen L(t) at a time t after hatching satisfies the differential equation dL dt = K(L∞ − L) where K is a constant called the growth coefficient and L∞ is the length of a mature specimen.

[KipIngram]

Ok, good. I'm assuming that's dL/dt, and that you left out the slash. So now you can write this:

 

dL/dt = K(Li - L) (sorry, don't know how to do the infinity symbol)

 

dL/dt + KL = KLi

 

So you have a differential equation (a simple one) and you have data that you have to match at two points in time.

 

Have you studied differential equations?

[cherrybomb]

Sorry, no I haven't studied differential equations just yet but have been watching videos on youtube about the topic.

[Country Boy]

From [math]dL/dt= K(L_\infty- L)[/math] I would rather write it as [math]\frac{dL}{L_\infty- L)= Kdt[/math]. To integrate the left side, let [math]u= L_\infty- L[/math] so that [math]du= -dL[/math] and you have [math]-\frac{du}{u}= Kdt[/math]. That should be easy to integrate.

Edited May 22, 2017 by Country Boy

[KipIngram]

Ok, so let me give you a brief overview of the pertinent bits. Say you have an equation as follows:

 

df/dt + Af = B.

 

First you solve the "homogenous" equation, which just replaces B with 0:

 

df/dt + Af = 0

 

df/dt = -Af

 

With somewhat less than total mathematical rigor (meaning a lot more can be said about it), you can do this:

 

df/f = -A dt

 

and now you can integrate both sides of that using regular calculus:

 

ln(f) = -At + C

 

where C is an arbitrary constant.

 

Now exponentiate both sides:

 

f = e^(-At+C) = e^(-At)*e^C = C' e^(-At)

 

C' is still arbitrary; we just recognized that C' = e^C is still a constant.

 

So the homogenous solution is

 

f(t) = C e^(-At)

 

where I've dropped the prime from C for convenience. Now you have to add a "particular solution that takes into account your B != 0 in the original problem. If you set df/dt to 0, then you just have f = B/A. So you can write this:

 

f(t) = C e^(-At) + B/A

 

If you differentiate that the B/A goes away and you verify the homogeneous solution, and it also gives you the right answer at t=infinity when the fish is fully grown and df/dt = 0.

 

Now all that's left is to find C so that you get the hatchling length at t=0 and choose A to get the right period of growth.

 

Does that help?

Please note that A in my lesson corresponds to K in your problem, and B corresponds to K Li, so B/A is just Li.

[cherrybomb]

It's a little confusing still but thank you heaps for your input

[KipIngram]

You're welcome. I don't really see how to take it any further without just supplying the answer, which we're not supposed to do. Good luck with it! I would imagine that with most teachers if you included some of what we've discussed in your answer you'd get most of the credit even if you don't arrive at exact numbers for the constants. It would show a conceptual understanding of the process and it's underlying mathematics.

Meanwhile, you educated me a bit (Ludwig von Bertalanfft and his model), so thank you!