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 save their own seeds, the fractional portion of their crop that remains free of hybrid plants t years later can be approximated by p(t)=(0.98)^t. a) using the model, predict the fractional portion of the crop that will be free of hybrid plants 15 yr after a neighboring farm begins to use purchased seeds. b) find p(15) and explain its meaning. c) when will half of the crop be hybrid plants?

Explanation:

Step1: Identify the function

The function for the fractional - portion of the crop free of hybrid plants is $P(t)=(0.98)^t$.

Step2: Substitute $t = 15$ into the function

We want to find the value of $P(t)$ when $t = 15$. So we calculate $P(15)=(0.98)^{15}$.
Using a calculator, $(0.98)^{15}\approx0.739$. To convert this to a percentage, we multiply by 100: $0.739\times100 = 73.9\%\approx74\%$.

Step3: Find when the crop is full of hybrid plants

The crop is full of hybrid plants when $P(t)=0$. We set up the equation $(0.98)^t = 0$.
Take the natural logarithm of both sides: $\ln((0.98)^t)=\ln(0)$. But $\ln(0)$ is undefined. In reality, we can find when $P(t)$ is very close to 0.
We know that for an exponential - decay function $y = a^x$ ($0\lt a\lt1$), as $x
ightarrow+\infty$, $y
ightarrow0$. We can also solve $(0.98)^t = 0.01$ (since when $P(t)=0.01$, 99% of the crop is hybrid).
Taking the natural logarithm of both sides: $\ln((0.98)^t)=\ln(0.01)$.
Using the property of logarithms $\ln(a^x)=x\ln(a)$, we get $t\ln(0.98)=\ln(0.01)$.
Then $t=\frac{\ln(0.01)}{\ln(0.98)}=\frac{- 4.60517}{-0.020203}\approx228$.

Answer:

a) 74
b) $P(15)=(0.98)^{15}\approx0.739$, which means 15 years after a neighboring farm begins to use purchased seeds, about 74% of the crop is free of hybrid plants.
c) Approximately 228 years.