I'm reading Richard Hamming's "The Art of Doing Science and Engineering" and I'm trying to figure out whether there's an error in the book or I'm misapplying some change of variable rules.
In Chapter 2 he derives the basic logistic growth curve in terms of an exponential growth model with a natural upper limit. He states:
"Let $L$ be the upper limit. Then the next simplest growth equation seems to be $$ \begin{equation} \frac{dy}{dt} = ky(L -y) \end{equation}. $$ At this point we of course reduce it to a standard form that eliminates the constants. Set $y = Lz$ and $x = t/(kL^2)$, then we have $$ \frac{dz}{dx} = z(1 - z) $$ as the reduced form for the growth problem, where the saturation level is now 1."
The change $y = Lz$ makes sense to me, but it seems like the change of variable for $t$ should actually be $x = t\times(kL)$. This gives us $\frac{dz}{dy} = \frac{1}{L}$ and $\frac{dt}{dx} = \frac{1}{kL}$ so that $$ \begin{align*} \frac{dz}{dx} &= \frac{dz}{dy} \cdot \frac{dy}{dt} \cdot \frac{dt}{dx} \\ &= \frac{1}{L} \cdot kL^2z(1 - z) \cdot \frac{1}{kL}\\ &= z(1 - z). \end{align*} $$
I don't see how the constants would cancel out using the substitution $x = t/(kL^2)$. So is this a typo in the book or am I misapplying the chain rule?
$$\begin{equation} \frac{dy}{dt} = ky(L -y) \end{equation}.$$ Change the variable $y=zL$: $$\frac{dz}{dt} =kL z(1 -z)$$ We have that: $$x = tkL$$ $$dx =kL dt$$ Therefore: $$\frac{dz}{dt} \dfrac {dx}{dt}= kLz(1 -z)$$ $$\frac{dz}{dt} kL= kLz(1 -z)$$ $$\frac{dz}{dt}= z(1 -z)$$ You applied the chain rule correctly.