Question 7
The population, $P$, of a species satisfies the logistic differential equation
$$\frac{dP}{dt}=0.42P\left(1-\frac{P}{3600}\right)$$
where $P=240$ when $t=0$.
The solution of the differential equation can be expressed as
$$P=\frac{K}{1+Ae^{-0.42t}}$$
where $K$ and $A$ are constants.
The values of $K$ and $A$ are
That’s it. Step through the hints below to see why.
Not quite. You took $A=\dfrac{K}{P_0}=15$, forgetting to subtract $1$.
Not quite. You inverted the ratio, $A=\dfrac{P_0}{K-P_0}$ $=\dfrac{240}{3360}=\dfrac{1}{14}$.
Not quite. You used the point of fastest growth, $\dfrac{K}{2}=1800$, as $K$, then got $1+A=7.5$.
Read off the carrying capacity.
$$K=3600$$
Use $P(0)=\dfrac{K}{1+A}$.
$$240=\frac{3600}{1+A}\Rightarrow1+A=15\Rightarrow A=14$$
$$240=\frac{3600}{1+A}\Rightarrow1+A=15\Rightarrow A=14$$
$K=3600$ and $A=14$, option (A).
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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