# Snail Invasion

Alignments to Content Standards: F-LE.A.2 F-LE.A.4

In 1966, a Miami boy smuggled three Giant African Land Snails into the country. His grandmother eventually released them into the garden, and in seven years there were approximately 18,000 of them. The snails are very destructive and had to be eradicated. According to the USDA, it took 10 years and cost \$1 million to eradicate them. 1. Assuming the snail population grows exponentially, write an expression for the population,$P$, in terms of the number,$t$, of years since their release. 2. How long does it take for the population to double? 3. Assuming the cost of eradicating the snails is proportional to the population, how much would it have cost to eradicate them if 1. they had started the eradication program a year earlier? 2. they had let the population grow unchecked for another year? ## IM Commentary The purpose of this task is to give students experience modeling a real-world example of exponential growth, in a context that provides a vivid illustration of the power of exponential growth, for example the cost of inaction for a year. There is an opportunity for further discussion based on part (c), since the ratio of costs from one year to the next is the same in each part. Although the task doesn't point this out explicitly, the teacher can bring it out in discussion, and it illustrates a key property of exponential growth. Students might choose to model exponential growth using either$P(t)=ae^{rt}$or$P(t)=ab^t$. Each has its advantages and disadvantages. The base$b$is the factor by which the population increases each year, and so is useful in part (c). And this interpretation of$b$yields directly the fact that$3b^7 = 18,\!000$, so that it can be calculated without logarithms. On the other hand, students might be more confortable working with base$e$and fitting$P(t) = a e^{rt}$to two data points, without engaging initially in the reasoning above. Depending on the goal that a teacher has in mind when using this task, he or she might suggest using one form or the other, or might use the opportunity to compare results when different students use different methods. ## Solutions Solution: Solution using base$e$1. Since it is assumed that the growth is exponential, we write$P(t)=ae^{rt}$for some constants$a$and$r$. We are given two data points, namely that$P(0)=3$and$P(7) =18,000$. (This last value is only an approximation.) From the first data point we get$3 = P(0) =ae^{0t}=a$, so we must have$a=3$. Then$P(7)=18,000$gives $$18,000=3e^{r(7)}.$$ Dividing both sides by 3 and rewriting the equation based on the definition of natural logarithm gives$7r=\ln(6000)$, so$r=\frac{\ln(6000)}{7}\approx 1.24$. So$P(t)=3e^{1.24t}$for$0 \le t \le 7$. We use this function to model the population, bearing in mind that it is based on approximate data. 2. The doubling time is the same no matter what the starting value. It takes the same time for the population to double from 3 to 6, then from 6 to 12, and so on. We calculate the time for the population to double from 3 to 6. So we want to find$t$such that $$6=3e^{1.24t}.$$ Dividing both sides by 3 and using the definition of natural logarithm to rewrite the equation gives$\ln(2)=1.24 t$, so$t\approx 0.56$. It takes just over half a year for the population to double. 3. 1. After 6 years the population is $$P(6) = 3e^{1.24(6)}\approx 5100.$$ If it costs \$1 million to clean up 18,000 snails, then it would cost $\frac{5100}{18000}\times \$1\text{ million}\approx \$280,000$ to clean up 5100 snails. This makes sense since the number of snails doubles almost two times per year, so the population (and the cost of dealing with it) should be a little more than $\frac14$ what it is in the 7th year.
2. After 8 years the population is $$P(8) = 3e^{1.24(8)}\approx 61000.$$ So, since it costs \$1 million to clean up 18,000 snails, it would cost$\frac{61000}{18000}\times \$1\text{ million}\approx \$ 3,400,000$to clean up 61,000 snails. This makes sense since the number of snails doubles almost two times per year, so the number of snails (and the cost of dealing with them) should be a little less than 4 times what it is in the 7th year. Solution: Solution that does not use logarithms 1. Since it is assumed that the growth is exponential, we write$P(t)=ab^t$for some constants$a$and$b$. We are given two data points, namely that$P(0)=3$and$P(7) =18,000$. (This last value is only an approximation.) From the first data point we get$3 = P(0) =ab^0=a$, so we must have$a=3$. Then$P(7)=18,000$gives $$18,000=3b^7.$$ Dividing both sides by 3 and raising both sides to the power of$\frac{1}{7}$gives$b=6000^{\frac{1}{7}}$, so$b\approx 3.47$. So$P(t)=3(3.47)^t$for$0 \le t \le 7$. We use this function to model the population, bearing in mind that it is based on approximate data. 2. The doubling time is the same no matter what the starting value. It takes the same time for the population to double from 3 to 6, then from 6 to 12, and so on. We calculate the time for the population to double from 3 to 6. So we want to find$t$such that $$6=3(3.47)^t.$$ Dividing both sides by 3 gives$2=3.47^t$. Without logarithms, we can solve by graphing$y = 2$and$y = 3.47^t$on the same$xy$-plane and finding$t$where the graphs intersect. We find$t\approx 0.56$. It takes just over half a year for the population to double. 3. 1. After 6 years the population is $$P(6) = 3(3.47)^6\approx 5100.$$ If it costs \$1 million to clean up 18,000 snails, then it would cost $\frac{5100}{18000}\times \$1\text{ million}\approx \$280,000$ to clean up 5100 snails. This makes sense since the number of snails doubles almost two times per year, so the population (and the cost of dealing with it) should be a little more than $\frac14$ what it is in the 7th year.
2. After 8 years the population is $$P(8) = 3(3.47)^8\approx 61000.$$ So, since it costs \$1 million to clean up 18,000 snails, it would cost$\frac{61000}{18000}\times \$1\text{ million}\approx \$ 3,400,000\$ to clean up 61,000 snails. This makes sense since the number of snails doubles almost two times per year, so the number of snails (and the cost of dealing with them) should be a little less than 4 times what it is in the 7th year.