World Population Growth: When Will 11 Billion Be Reached?

[MrNeWBiE]

Homework Statement

The population of the world was 5.7 billion people in 1995 and 6.2 billion in 2008. Assuming exponential growth by what year will the population reach 11 billion?

The Attempt at a Solution

i started by trying to find R

but am not sure the my way is right ,,, i started by 6.2/57 = 62/57

then i divided it by 13 ,, it's the difference between 2008 and 1995
,,, so (62/57)/13 = 62/741 ,,,, then i multiply it by 100 to get the percentage = 8.367

are my steps right until now ?

 

[Mark44]

Homework Statement

The population of the world was 5.7 billion people in 1995 and 6.2 billion in 2008. Assuming exponential growth by what year will the population reach 11 billion?

Where's the section for relevant equations? Since the assumption is that population growth is exponential, a very relevant equation would be the one that describes exponential growth.

The Attempt at a Solution

i started by trying to find R

What does R represent? You haven't shown the model you are using, so we have no idea what R is supposed to be.

but am not sure the my way is right ,,, i started by 6.2/57 = 62/57

You mean 6.2/5.7, which is equal to 62/57.

then i divided it by 13 ,, it's the difference between 2008 and 1995
,,, so (62/57)/13 = 62/741 ,,,, then i multiply it by 100 to get the percentage = 8.367

are my steps right until now ?

Not at all. Your work assumes that population growth will be linear, which is not the assumption in this problem. The ratio 62/57 is about 1.088, which says that the population increased by about 8.8%. If you divide by 13, the annual growth rate would be 8.8/13 or about 0.7%.

What's the equation you need to use for exponential growth (not linear growth)?

 

[MrNeWBiE]

and what is exponential growth ??

i know to answer only when i have % of the growth ,,, =.=

 

[MrNeWBiE]

and what is exponential growth ??

i know to answer only when i have % of the growth ,,, =.=

 

[Mark44]

Your textbook should have the general formula for exponential growth. Why don't you see if you can find it?

 

[MrNeWBiE]

if it's my textbook why i would ask you what is exponential growth ,,?

our beloved dc. give us question from other course ,,,, and we must solve it for the Assignment >.<!

 

[Dickfore]

By the year 2120.

 

[MrNeWBiE]

yo ,,, this forum for learning not for giving the answer ,,,,


if it is about the answer mark was going to answer ,,,

i want to learn here not to have answers only ,,,

at least show the steps =p

 

[Dickfore]

yo ,,, this forum for learning not for giving the answer ,,,,


if it is about the answer mark was going to answer ,,,

i want to learn here not to have answers only ,,,

at least show the steps =p

This forum is not about providing detailed solutions to your problems.

 

[Mark44]

if it's my textbook why i would ask you what is exponential growth ,,?

Are you saying that it's not in your textbook?

our beloved dc. give us question from other course ,,,, and we must solve it for the Assignment >.<!

dc? What's that?

Exponential growth: P(t) = P₀e^kt

Where P(t) means the population at a time t
P(0) - the population at some starting point
k - a constant that describes how quickly or slowly the population grows

In your problem - The population of the world was 5.7 billion people in 1995 and 6.2 billion in 2008. Assuming exponential growth by what year will the population reach 11 billion? - you can take 1995 to be year 0, so P(0) = 5.7 billion, and P(13) = 6.2 billion. These two values can be used to find the growth constant k. When you have that you can solve for t in the equation P(t) = 11 billion.

 

[Dickfore]

Maybe you have seen it under the name 'geometric sequence'.

 

[Susanne217]

Homework Statement

The population of the world was 5.7 billion people in 1995 and 6.2 billion in 2008. Assuming exponential growth by what year will the population reach 11 billion?

The Attempt at a Solution

i started by trying to find R

but am not sure the my way is right ,,, i started by 6.2/57 = 62/57

then i divided it by 13 ,, it's the difference between 2008 and 1995
,,, so (62/57)/13 = 62/741 ,,,, then i multiply it by 100 to get the percentage = 8.367

are my steps right until now ?

As mentioned use the very simple formula

[tex]y = b \cdot a^x[/tex]

Where you find a by

[tex]a = (\frac{y_2}{y_1})^{\frac{1}{x_2-x_1}}[/tex]

where [tex](x_1,y_1)[/tex] and where [tex](x_2,y_2)[/tex] are your points...

 

[MrNeWBiE]

so to find " k " ,,,,

6.2 = 5.7 e^k13

Ln6.2 / 5.7 = 13k lne

13k= 0.084 ... k = 0.65/100

then to find " t "

P(t)=5.7 e^0.65/100 t

11=5.7 e^0.65/100 t

Ln 11/5.7 = (0.65/100) t lne

0.6574=( 0.65/100)t

t= 101.133 ...

in 2096 not 2120 as dickfore said

right ?

 

[Mark44]

You're pretty close, but I got this for k: 0.00646793209311856953400952132232. I stored this value in memory in my calculator for later use.

I then solved 11/5.7 = e^kt, or ln(11/5.7) = kt, so t = (1/k)*ln(11/57). From this I got t = 101.64440326411666721034020332917.

The reason our answers are not quite the same is that you rounded off your value for k, and then rounded off your answer for ln(11/5.7).

 

[MrNeWBiE]

long numbers ,,, i only need the first 3 after the " . "

because i can't write a lot of numbers in my homework page ,,,,

 

[Mark44]

So mine rounded to 3 dec. places was 101.644 years, and yours was 101.133. My point is that if you round off before the end, you lose precision.