Question

The population of a region is growing exponentially. There were 30 million people in 1980 (when t = 0) and 70 million people in 1990. Find an exponential model for the population (in millions of people) at any time t, in years after 1980A) P(t) = _____B) Whar population do you predict for the year 2000?C) Predicted population in the year 2000 = ____ million peopleD) What is the doubling time?Doubling time =____ years

Answer 1

`\begin{gathered} A)P(t)=30(1+0.088)^t\text{( in millions)} \\ B)P(20)=163.33\text{ Million people} \\ C)163.33\text{ Million people} \\ D)8.21years \end{gathered}`

Step-by-step explanation

you can model growth by a constant percent increase with the following formula:

`\begin{gathered} P(t)=A(1+r)^t \\ \text{where} \\ P(t)\text{ is the final amount} \\ A\text{ is }the\text{ initial amount} \\ r\text{ is the rate ( in decimal)} \\ t\text{ is the time ( years)} \end{gathered}`

so

Step 1

make the model.

A)Let

`\begin{gathered} P=70\text{ million} \\ \text{time}=\text{ 1990-180=10 years} \\ A=30\text{millones } \\ so \\ 70=30(1+r)^(10)\rightarrow equation \end{gathered}`

now, we need to solve for r

`\begin{gathered} 70=30(1+r)^(10)\rightarrow equation \\ (70)/(30)=(1+r)^(10) \\ ((70)/(30))^{(1)/(10)}=((1+r)^(10))^{(1)/(10)} \\ 1.088=1+r \\ \text{subtract 1 in both sides} \\ 1.088-1=1+r-1=0.0884229198 \\ r=0.0884229198 \\ or \\ r=8.8\text{ \%} \end{gathered}`

c) now, we can complete the model

`P(t)=30(1+0.088)^t`

Step 2

Whar population do you predict for the year 2000?

Let

`\begin{gathered} \text{time}=\text{ 2000-1980=20 years} \\ t=20 \\ P=30 \\ r=\text{0}.088 \end{gathered}`

replace

`\begin{gathered} P(20)=30(1+0.088)^(20) \\ P(20)=30(1+0.088)^(20) \\ P(20)=30(1+0.088)^(20) \\ P(20)=163.33\text{ Million people} \end{gathered}`

Step 3

D) What is the doubling time?

Doubling time =____ years

to solve this , we need to find the time , when population is double than currently, so

`\begin{gathered} \text{Double population= }2\cdot300\text{ milliones} \\ \text{Double population=}600\text{ Million} \end{gathered}`

then, let

`\begin{gathered} \text{time}=\text{ unknown= t} \\ A=30 \\ P=60 \\ r=0.088 \end{gathered}`

replace

`\begin{gathered} P(t)=30(1+0.088)^t \\ 60=30(1+0.088)^t \\ 60=30(1.088)^t \\ (60)/(30)=(1.088)^t \\ 2=(1.088)^t \\ \ln (2)=\ln (1.088)^t \\ \ln (2)=t\ln (1.088) \\ t=\frac{\ln 2}{\ln \text{ 1.088}} \\ t=8.2183 \end{gathered}`

therefore, the time is

D)8.21 years

I hope this helps you