Interpreting parameters of a differential equation to predict sea urchin growth I need to use this growth model fairly urgently. It should be simple but it has me (and everyone in my research institute) very confused. I have not worked with differentials for years so I assume I am misinterpreting something.
Basically, I what to have the model tell me how much the urchin has grown from a certain starting size and over a certain amount of time.
The formula given is the differential of the johnson growth equation:
$$
dS = k·S_t·dT\,·\,(\ln(S_∞) -\ln(S_t))^2
$$
And the other parameters are explained as:
where $S_t$, is the size on day $t$; $t =$number of days since the first new recruited urchins were spotted each year; $b$ is a scaling factor, equal to $(S - S_∞)/S_∞$; $k$ is the instantaneous growth rate coefficient; $S_∞$ is the asymptotic size; and $t_0$, in extrapolation, the time when $S_t= 0$ (Ebert, 1975)
Where:

*

*$k = 2.92$ (growth constant)

*$S_∞ = 90$ (asymptotic size)

The way I understand this (which  must be incorrect) is that $S_t$ is the starting size and $dT$ is the number of days the urchin grows for. But if this is inputted the answer does not make sense.
According to this paper, if a $10mm$ urchin growths for $60$ days, it should increase by about $20mm$.
We have tried all different things to work this out, but nothing makes sense. Does anyone understand how to use this equation? I have attached a screenshot of relevant parts of the paper and you can find the paper (DAFNI 1992) here: https://www.tandfonline.com/doi/abs/10.1080/00212210.1992.10688663?journalCode=tiee19
Thank you in advance
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 A: Too long for a comment :
All I can say is that, given Johnson's growth equation (transcribed from your image)
$$\tag{1}
S_t=S_\infty\,b\exp\Big(-\frac{1}{kt}\Big)
$$
we clearly have
$$\tag{2}
\frac{dS_t}{dt}=S_\infty\,b\,\exp\Big(-\frac{1}{kt}\Big)\,\frac{1}{kt^2}\,.
$$
By (1),
$$
-\frac{1}{t}=k\,\ln\Big(\frac{S_t}{S_\infty\,b}\Big)=k\,\Big(\ln S_t-\ln S_\infty-\ln b\Big)
$$
so that we can write (2) as
$$
\frac{dS_t}{dt}=S_t\,k\,\Big(\ln S_t-\ln S_\infty-\ln b\Big)^2\,.
$$
For $b=1$ and $dT=dt$ this is what you called the differential of the Johnson growth equation.
Please consider this and rewrite you question to meet MSE standards:

*

*no images to "display" formulas


*instead MathJax


*no links to inaccessible papers
A: The abstract says that $k$ is the yearly growth coefficient. So if $t$ is in days, one needs to change the formula to $S=S_\infty\exp(-365/(kt))$. Then size 10 occurs at day 57, and at day 117 the size via formula is about 31, indeed a growth by about 20.

To compare, the other model with the associated parameters has a faster start, but still raises from 10 at time 23 to 30 at time 83 in time difference 60.

