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An insect pest population doubles every 18 days. If an insecticide kills 90% of the insects, how often should it be applied to keep in insect population in check?

I tried using this exponential growth/decay model: $$A = P\left(\frac{A_1}{P_1}\right)^{t/t_1}$$ where $P_1$ (initial value), $A_1$ (new value), $t$ (time period).

However, would this be the proper way to do it? I'm not quite sure how to interpret (and thus solve) this problem.

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  • $\begingroup$ You need to fix the equation. I presume $A_1,\ P_1$ are initial values and $A$ the final value. $P$ and $t_1$ are not defined, while $t$ is the time period of what? $\endgroup$ – herb steinberg Aug 12 '18 at 3:32
  • $\begingroup$ I think this is poorly worded question, to answer you need to decide exactly what "keep in insect population in check" means $\endgroup$ – WW1 Aug 12 '18 at 3:39
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So you have an insect population that doubles every 18 days. Let's call it $m_0$. After 18 days, you will have $2m_0$, after another 18 days, you will have $4m_0$. The idea is that you model this growth with a function:

$f(t) = \text{something}$,

but we know some values of this function:

$f(0\ \text{days}) = m_0,$

$f(18\ \text{days}) = 2m_0,$

$f(36\ \text{days}) = 4m_0,$

$f(54\ \text{days}) = 8m_0,$ and so on.

We know the form of the function:

$f(t\ \text{days}) = m_0 \cdot n^t.$

To find $n$, we do:

$f(18\ \text{days}) = 2m_0 = m_0 \cdot n^{18\ \text{days}},$

$n=2^{1/18\ \text{days}}.$

Our function is:

$f(t\ \text{days}) = m_0 \cdot 2^{t/18\ \text{days}}.$

Remember that the insect population is always growing, and it will hit the desired number of insects (or a critical number of insects) at $t_c$ (critical time). Let's call that critical number of insects $m_c$. Then you apply the insecticide and now the insect population is:

$$f(t_c)=m_0 \cdot 2^{t_c/18} = 0.1m_c$$

Then you have a new function of insect population:

$f(t\ \text{days}) = 0.1m_c \cdot 2^{t/18\ \text{days}}.$

Since the question is "how often should it be applied to keep the insect population in check?", the insect population will grow and it will hit the desired number of insects, $m_c$ after some time $t$, we want to know that time (in days):

$0.1m_c \cdot 2^{t/18} = m_c,$

and the result is (solve for $t$!):

$t=59.79\ \text{days}.$

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