Question

farmers wishing to avoid the use of purchased seeds are increasingly concerned about inadvertently growing hybrid plants as a result of pollen drifting from nearby farms. assuming that these farmers raise their own seeds, the fractional portion of their crop that remains free of hybrid plants ( t ) years later can be approximated by ( p(t)=(0.93)^t ).

a) using the model, predict the fractional portion of the crop that will be free of hybrid plants 5 yr after a neighboring farm begins to use purchased seeds.

b) find ( p(5) ) and explain its meaning.

c) when will half of the crop be hybrid plants?

a) after 5 yr, (%) of the crop will be free of hybrid plants. (round to the nearest integer as needed.)

b) ( p(5)=%) of the crop per year. (round to one - decimal place as needed.)

Explanation:

Step1: Calculate the fraction of non - hybrid plants after 5 years

The formula for the fractional portion of the crop that remains free of hybrid plants $t$ years later is $P(t)=(0.93)^t$. Substitute $t = 5$ into the formula:
$P(5)=(0.93)^5$
$P(5)=0.93\times0.93\times0.93\times0.93\times0.93\approx0.70$ or $70\%$

Step2: Find the derivative of $P(t)$

First, recall that if $y = a^t$ where $a$ is a constant, then $y^\prime=a^t\ln(a)$. For $P(t)=(0.93)^t$, $P^\prime(t)=(0.93)^t\ln(0.93)$.
Substitute $t = 5$ into $P^\prime(t)$:
$P^\prime(5)=(0.93)^5\ln(0.93)$
$(0.93)^5\approx0.70$ and $\ln(0.93)\approx - 0.0725$.
$P^\prime(5)\approx0.70\times(- 0.0725)\approx - 5.1\%$ per year. The negative sign means the percentage of non - hybrid plants is decreasing.

Step3: Find when half of the crop is hybrid

If half of the crop is hybrid, then the fraction of non - hybrid plants $P(t)=0.5$.
Set $(0.93)^t = 0.5$.
Take the natural logarithm of both sides: $\ln((0.93)^t)=\ln(0.5)$.
Using the property of logarithms $\ln(a^b)=b\ln(a)$, we get $t\ln(0.93)=\ln(0.5)$.
Solve for $t$: $t=\frac{\ln(0.5)}{\ln(0.93)}\approx\frac{-0.6931}{-0.0725}\approx9.6$ years.

Answer:

a) $70\%$
b) $P^\prime(5)\approx - 5.1\%$ of the crop per year
c) Approximately $9.6$ years