Question

Invasive species often experience exponential population growth when introduced into a new environment. Zebra mussels are an invasive species that have recently made their way into midwestern lakes. In about six months, a zebra mussel population quadruples in size. Suppose 10 zebra mussels are newly introduced into a local lake. Find a function Z(t) that models the population of zebra mussels t years after they have been introduced. Use the function to determine how many zebra mussels there will be in 15 months. Use the function to determine how many months will pass before the lake has one million zebra mussels.

Answer 1

Answer:

If ‘a’ is the initial population of the Zebra mussels, then every six months, the population of Zebra mussels quadruples, i.e. it becomes 4a. In t years, i.e. in 2t intervals of 6 months each, the population of Zebra mussels can be computed with the help of a geometric series a, ar, ar2, ar3, …. arn-1 ( the series being finite with n terms) where a is the 1st term , r is the common ratio and n is the number of terms in the series..

Here, in 2t years, there will be 2t + 1 terms in the geometric series including the initial term. Thus in the above series, a = 10, r = 4 and n = 2t + 1. Then the population of the Zebra mussels after 2t years is the (2t +1)st term of the geometric series I.e. 10* ( 42t)                  

In 15 months, t = 15/12 = 1.25. Then the population of Zebra mussels after 15 months will be 10*(42.5 ) = 10 * 25 = 320

If after t years, the population of the Zebra mussels become 1 million , then we have

1000000 = 10 * (42t) or, 100000= 42t or,105 = 42t    Taking logarithms of both sides, we have 5 log 10 = 2t log 4 or, t = (5log10)/(2log4) = 5/1.20 years or (5/1.20) * 12 months = 50 months, i.e. 4 years and 2 months.

Step-by-step explanation:

Answer 2

Answer:

The model for the population of zebra mussels

$Z(t)=10e^{0.231t}$

At 15 months, the population is 315.

The population of zebra mussels will reach 1,000,000 in 50 months.

Step-by-step explanation:

We can model this as:

$dZ/dt=kZ\\\\dZ/Z=k\cdot dt \\\\ln(Z)+C=kt\\\\ Z=Ce^k^t\\\\\\Z(0)=10=C*e^0=C\\\\\\ Z(6)=10e^{k(6)}=40\\\\e^{6k}=4\\\\k=ln(4)/6=0.231$

Then, the model for the population of zebra mussels in time can be written as:

$Z(t)=10e^{0.231t}$

At 15 months, the population is:

$Z(15)=10e^{0.231\cdot 15}=10e^{3.465}=10*31.5=315$

The population at 15 months is Z=315.

Now, we have to calculate at which time the lake will have one million zebra mussels.

$Z(t)=10e^{0.231t}=10^6\\\\e^{0.231t}=10^5\\\\0.231t=5\cdot ln(10)\\\\t=5\cdot ln(10)/0.231=49.84\approx 50$

The population of zebra mussels will reach 1,000,000 in 50 months.