Question

Tacoma's population in 2000 was about 200 thousand, and had been growing by about 9% each year. a. Write a recursive formula for the population of Tacoma b. Write an explicit formula for the population of Tacoma c. If this trend continues, what will Tacoma's population be in 2016? d. When does this model predict Tacoma’s population to exceed 400 thousand?

Answer 1

Answer:

a) The recurrence formula is $P_n = \frac{109}{100}P_{n-1}$.

b) The general formula for the population of Tacoma is

$P_n = \left(\frac{109}{100}\right)^nP_{0}$.

c) In 2016 the approximate population of Tacoma will be 794062 people.

d) The population of Tacoma should exceed the 400000 people by the year 2009.

Step-by-step explanation:

a) We have the population in the year 2000, which is 200 000 people. Let us write $P_0 = 200 000$. For the population in 2001 we will use $P_1$, for the population in 2002 we will use $P_2$, and so on.

In the following year, 2001, the population grow 9% with respect to the previous year. This means that $P_0$ is equal to $P_1$ plus 9% of the population of 2000. Notice that this can be written as

$P_1 = P_0 + (9/100)*P_0 = \left(1-\frac{9}{100}\right)P_0 = \frac{109}{100}P_0$.

In 2002, we will have the population of 2001, $P_1$, plus the 9% of $P_1$. This is

$P_2 = P_1 + (9/100)*P_1 = \left(1-\frac{9}{100}\right)P_1 = \frac{109}{100}P_1$.

So, it is not difficult to notice that the general recurrence is

$P_n = \frac{109}{100}P_{n-1}$.

b) In the previous formula we only need to substitute the expression for $P_{n-1}$:

$P_{n-1} = \frac{109}{100}P_{n-2}$.

Then,

$P_n = \left(\frac{109}{100}\right)^2P_{n-2}$.

Repeating the procedure for $P_{n-3}$ we get

$P_n = \left(\frac{109}{100}\right)^3P_{n-3}$.

But we can do the same operation n times, so

$P_n = \left(\frac{109}{100}\right)^nP_{0}$.

c) Recall the notation we have used:

$P_{0}$ for 2000, $P_{1}$ for 2001, $P_{2}$ for 2002, and so on. Then, 2016 is $P_{16}$. So, in order to obtain the approximate population of Tacoma in 2016 is

$P_{16} = \left(\frac{109}{100}\right)^{16}P_{0} = (1.09)^{16}P_0 = 3.97\cdot 200000 \approx 794062$

d) In this case we want to know when $P_n>400000$, which is equivalent to

$(1.09)^{n}P_0>400000$.

Substituting the value of $P_0$, we get

$(1.09)^{n}200000>400000$.

Simplifying the expression:

$(1.09)^{n}>2$.

So, we need to find the value of $n$ such that the above inequality holds.

The easiest way to do this is take logarithm in both hands. Then,

$n\ln(1.09)>\ln 2$.

So, $n>\frac{\ln 2}{\ln(1.09)} = 8.04323172693$.

So, the population of Tacoma should exceed the 400 000 by the year 2009.

Answer 2

Answer:

a) P(n)=1.09^n*P(0)

b) P(n)=200k*(1.09*1.09*...*1.09) (n-times)

c) 764k

d) By the year of 2009

Step-by-step explanation:

a) We can write down the formula of Tacoma population as below:

$P(n)=1.09^n*P(0)$

Where:

n is the last two digits of the year.

P(0)=200k

b) This formula can be written as below to be more explicit:

$P(n)=200k*(1.09*1.09*...*1.09) (n-times)$

c) Tacoma Population in 2016 is:

$P(16)=1.09^{16}*200k=794k$

d) When the value of (1.09)^n exceeds 2

$(1.09)^n\geq 2\\n\geq 9$