Deriving and applying a non-linear function associated with a plausible real world scenario.

Question

Podunk’s population is experiencing uninhibited exponential growth. In 1990, it’s population was 23; in 1998, it was 31.

A) find the function that models Podunk’s population as a function of time t (i.e the function which gives Podunk’s population for a given year t)

B) According to your model in A), what will the population be in 2010?

C) According to your model in A), when will Podunk’s population reach 100?

Answer 1

The first thing we need to look at is the term exponential growth. The general formula for exponential growth will always be y = a(1+ x)^t.

Part A : Let’s use the general formula for our situation.

y will be the total population after t years. a will represent the initial population, which is 23 people. t is time in years. Lastly, r is rate, which is calculated by first data point divided by second data point (for us 23/31=0.7419)

After filling in what we know, we have y= 23(1 + 0.7419)^t.

Part B : We can figure out the population using the equation in part a and calculating the time passed in years by doing t = 2010-1990 = 20 years. Now we can solve.

y = 23 (1+ 0.7419)²⁰

y = 23 (1.7419)²⁰

y = 23 x 66,140.029

y = 1,521,220.6745 or 1,521,221 people

Part C : We will use the same equation from part A, but instead of being told the time, we are told the population.

100 = 23(1 + 0.7419)^t

100/23= 23(1 + 0.7419)^t/23

Now, we need to take the logarithm of both sides to take care of the exponent t.

log(100/23)= log((1 + 0.7419)^t)

using the exponent rule of logs, we can move the t outside of the logarithm

log(100/23)= t log(1 + 0.7419)

t = log(100/23)/log(1 + 0.7419) = 2.648 years

Answer 2

Exponential growth equation form is P_t = P₀(1 + r)^t P_t is population after t years, for a starting population of P₀ and a growth rate of r.

For this problem t = 1998 - 1990 = 8 years. 31 = 23(1 + r)⁸.

31/23 = 1.3478 = (1 + r)⁸

(1+ r) = ⁸√1.3478 = 1.038, so r = 0.038.

A) Exponential growth function is P_t = P₀(1.038)^t

Check: 23(1 + 0.038)⁸ = 23(1.038)⁸ = 23(1.3477) ≈ 31

B) For year 2010, t = 2010 - 1990 = 20, so equation is P₂₀₁₀ = 23(1.038)²⁰ You can calculate that.

C) How long to reach 100 pop'n? 100 = 23(1.038)^t

100/23 = 4.3478 = 1.038^t Take logs of both sides: t(log1.038) = log4.3478

t = log 4.3478/log 1.038 = 0.6383/0.0162 = 39.4 years

Check: 23(1.038)^39.4 = 23(4.3469) ≈ 100

Answer 3

The general exponential function is Y = a x b^X, There are two unknowns a and b.

To find a, solve the general equation for a to get a = Y / b^X, and b^X = Y / a.

Y represents the population and X the year. To make the problem manageable, let X = (X -1990). The equation for a becomes a = 23 / b^(X-1990) or a = 23 as b = 1 in 1990.

We now need to find b. b in 1998 is b^{(1998 - 1990)} = b⁸ = 31 / 23, so b = the eighth root of 31/23 which my calculator gives as 1.038 ( using the y^x function key).

The final equation is Y = 23 x 1.038^{(X - 1990)}. You should now be able to do parts B and C using the appropriate substitutions.